The Academy Corner

193

THE ACADEMY CORNER

No. 33

Bruce Shawyer

All communications about this column should be sent to Bruce

Shawyer, Department of Mathematics and Statistics, Memorial University

of Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7

We present morereaders' solutions to some of the questions of the 1999

Atlantic Provinces Council on the Sciences Annual Mathematics Competition,

which was held last year at Memorial University, St. John's, Newfoundland

[1999 : 452].

1. Find the volume of the solid formed by one completerevolution about

the x{axis of the area in common to the circles with equations

2 2 2 2

x + y 4y +3 = 0 and x + y =3.

Solution by RichardTod, The Royal Forest of Dean, Gloucestershire,

England (adapted by the editor).

Here we have the two circles:

2 2 2 2

C : x +(y 2) =1 and C : x + y =3.

1 2

These intersect when 6 4y =0; that is, when y = . At this value,

we have x = .

The volume is therefore

2 p

V =4 2 1 x 1 dx = 3 .

6. Inside a squareofsider , four quarter circles aredrawn, with radius r

and centres at the corners of the square.

Find the area of the shaded region.

I. Solution by RichardTod, The Royal Forest of Dean, Gloucestershire,

England.

The area of interest is the `curved squareregion' in the centre of the

given square, and is denoted by c. The other regions of interest are

denoted by a and b.

194

D C

b b

c a a

b b

A B

Since the area of the squareisr , we have

4a +4b + c = r . (1)

. Hence we have Consider quarter circle AB C . This has an area of

2a +3b + c = . (2)

Consider equilateral triangle AB P . This has area .

. Consider the sixth of a circle AB P . This has area

The di erenceis r . We need two of these, plus the equi-

6 4

lateral triangle toget

r . (3) a +2b + c =

3 4

Solving (1), (2) and (3) simultaneously yields

3 +3 3

c = r .

II. Solution by Catherine Shevlin, Wallsend, England.

Let the origin be the centre of a squareofside1. Then the quarter circle

1 4x 4x +3

DP B has equation y = + .

2 2

1 4x 4x +3

Thus, the general area is 4r + dx.

2 2

3 1

2 2

2x+1

sin

4x 4x +3 (2x 1) x

. This is 4r +

2 8 2

+3 3 3

This simpli es togive r . 3

195

8. An arbelos consists of three semicircular arcs as shown:

A circle is placed inside the arbelos so that it is tangent to all three

semicircles.

Suppose that the radii of the two smaller semicircles are a and b,and

that the radius of the circle is r .

Assuming that a >b>rand that a, b and r are in arithmetic progres-

sion, calculate a=b.

Solution by Catherine Shevlin, Wallsend, England.

The radius of the large semi-circle is a + b. Thus, the centre of the

large semi-circle is distances a and b from the centres of the smaller

semi-circles as shown.

Applying the Cosine Rule to two triangles, we therefore have

2 2 2

(a + r ) b (a + b r ) = 2b(a + b r )cos ,

2 2 2

(b + r ) a (a + b r ) = 2a(a + b r )cos .

Eliminating cos , and solving for r , leads to

ab(a + b)

r = .

2 2

a + ab + b

If a, b and r are in arithmetic progression, we have

2 2

ab(a b ) ab(a + b)

= , 2b a =

2 2 3 3

a + ab + b a b

4 3 3 4

or a a b 2ab +2b =0. Writing x for a=b, we have

4 3 3

0=x x 2x +2 = (x 2)(x 1) .

Thus, a=b = x = 2.

196

THE OLYMPIAD CORNER

No. 206

R.E. Woodrow

All communications about this column should be sent toProfessor R.E.

Woodrow, Department of Mathematics and Statistics, UniversityofCalgary,

Calgary, Alberta, Canada. T2N 1N4.

For your pleasureover the summer break for the Olympiad Corner we

go on another quick tour of four Olympiad Contests. The rst problem set

we giveconsists of the problems of the First Round of the XXXIII Spanish

Mathematical Olympiad 1996{97. My thanks for collecting these go toour

regular contributor Francisco Bellot Rosado, I.B. Emilio Ferrari, Valladolid,

Spain; and to Richard Nowakowski, Canadian Team Leader to the IMO at

Buenos Aires.

XXXIII SPANISH MATHEMATICAL OLYMPIAD

1996{97

First Round, First Day

November 29, 1996 | Time: 4 hours

1. Show that any complex number z 6=0can be expressed as a sum of

two complex numbers such that their di erence and their quotient are purely

imaginary (that is, with real part zero).

2. Consider a circle of centre O ,radius r , and let P be an external

point. We draw a chord AB parallel to OP .

2 2

(a) Show that PA + PB is constant.

(b) Find the length of the chord AB which maximizes the area of the triangle

AB P .

3. Six musicians participate in a music festival. At each concert, some

of them play music, and the otherslisten. What is the minimal number of

concerts so that each musician listens to all the others?

4. The sum of two of the roots of the equation

3 2

x 503x +(a +4)x a =0

is equal to 4. Determine the value of a.

197

First Round, Second Day

November 30, 1996 | Time: 4 hours

5.Ifa, b, c are positivereal numbers, prove the inequality

2 2 2

a + b + c ab bc ca 3(b c)(a b) .

When is the = sign valid?

6. Find, with reasons, all the natural numbers n such that n has only

odd digits.

7. The triangle AB C has A =90 , and AD is the altitude from A.

\ \

The bisectors of the angles AB D and AD B intersect at I ; the bisectorsof

\ \

the angles AC D and AD C intersect at I .

Find the acute angles of AB C , given that the sum of distances from I

and I to AD is BC=4.

8. For each real number x, we denoteby bxc the largest integer which

is less than or equal to x. We de ne

q (n)= , n =1, 2, 3, ::: .

b n c

(a) Forming a table with the values of q (n) for 1 n 25,makeaconjecture

about the numbers n for which q (n) >q(n +1).

(b) Determine, with reasons, all the positiveintegers n such that

q (n) >q(n +1).

Next we moveto the problems of the 20 Austrian-Polish Mathemati-

cal Competition (1997). Thanks go toregular contributorsMarcin E. Kuczma,

Warszawa, Poland; and Walther Janous, Ursulinengymnasium, Innsbruck,

Austria, as well as to RichardNowakowski.

20 AUSTRIAN-POLISH MATHEMATICAL

COMPETITION (1997)

1. P is the common point of straight lines l and l .Two circles S and

1 2 1

S areexternally tangent at P and l is their common tangent line. Similarly,

2 1

two circles T and T areexternally tangent at P and l is their common

1 2 2

tangent line. The circles S and T havecommon points P and A, the circles

1 1

S and T havecommon points P and B , the circles S and T havecommon

1 2 2 2

198

points P and C , and the circles S and T havecommon points P and D .

2 1

Prove that the points A, B , C , D lieonacircle if and only if the lines l and

l are perpendicular.

2. Let m, n, p, q be positive integers. We havearectangular chess-

board of dimensions m n divided into m n equal squares. The squares

arereferred toby their coordinates (x; y ), where 1 x m, 1 y n.

Thereisapiece on each square. A piececan be moved from the square (x; y )

0 0 0 0

to the square (x ;y ) if and only if jx x j = p and jy y j = q . We want to

move each piece simultaneously so as to get again one piece on each square.

Find the number of ways in which such a multiple movecan be made.

3. On the blackboard thereare written the numbers 48, 24, 16, :::,

48 48

; that is, rational numbers with k =1, 2, :: :, 97. In each step,

97 k

two arbitrarily chosen numbers a and b arecancelled and the number

2ab a b +1 is written on the blackboard. After 96 steps there is only one

number on the blackboard. Determine the set of numbers which are possible

outcomes of the procedure.

4. Inaconvex quadrilateral AB C D the sides AB and CD areparallel,

the diagonals AC and BD intersect at point E , and points F and G are the

orthocentre s of the triangles EBC and EAD,respectively. Prove that the

mid-point of the segment GF lies on the line k perpendicular to AB such

that E 2 k .

5.Letp , p , p , and p be four distinct prime numbers. Prove that

1 2 3 4

3 2

there does not exist a cubic polynomial Q(x)= ax + bx + cx + d with

integer coecients such that

jQ(p )j = jQ(p )j = jQ(p )j = jQ(p )j =3.

1 2 3 4

6.Prove that there does not exist a function f : Z ! Z such that

f (x + f (y )) = f (x) y for all integers x and y .

7. (a) Prove that for all real numbers p and q the inequality

2 2

p + q +1 >p(q +1) holds.

(b) Determine the greatest real number b such that for all real numbers

2 2

p and q the inequality p + q +1 >bp(q +1) holds.

2 2

q the inequality p + q +1 >cp(q +1) holds.

8.Letn be a natural number and let M be a set with n elements. Find

the biggest integer k with the property: thereexistsa k -element family K of

three-element subsetsofM such that any two setsfrom K are non-disjoint.

199

9.LetP be a parallelepiped, let V be itsvolume, S its surfacearea,

and L the sum of the lengths of the edges of P .Fort 0 let P be the solid

consisting of points having distancefrom P not greater than t.Prove that

the volume of P is equal to

2 3

V + St + Lt + t .

4 3

Next we give Selected Problems from the Israel Mathematical Olympiads.

My thanks go to Richard Nowakowski for obtaining them for the Corner.

SELECTED PROBLEMS FROM

ISRAEL MATHEMATICAL OLYMPIADS

1.Prove that thereareatmost3 primes between 10 and 10 all of

whose digits in base ten are 1 (for example, 11).

2. Is there a planar polygon whose vertices haveinteger coordinates,

whose area is , such that this polygon is

(a) a triangle with at least two sides longer than 1000?

(b) a triangle whose sides are all longer than 1000?

3. Find all real solutions of

p p

4 4

13 + x + 4 x =3.

4.Prove that if two altitudes of a tetrahedron intersect, then so do the

other two altitudes.

5. Consider a partition of an n n square(composed of n 1 1

squares) intorectangles where one side of each rectangle has an integer length

and the other side is of length 1. What is the largest number of such partitions

such that no two partitions have an identical rectangle at the same place?

6.Asetofn +1 points in the plane, such that no three lie on a line,

is given. Each line segment connecting a pair of these pointsiscoloured by

either red or blue. A path of length k is a sequenceofk segments, where

the end of each segment is the beginning of the next one (except for the rst

segment). A simple path is a path that does not intersect itself (that is, two

segments meet if and only if theyareconsecutive segments of the path).

Prove that thereexists a monochromatic simple path of length n.

200

And as the 4 set we give the problems of the 1997 Chinese Mathe-

matical Olympiad. Once again thanks go to Richard Nowakowski, Canadian

Team Leader to the IMO at Buenos Aires, for collecting this set.

THE 1997 CHINESE MATHEMATICAL OLYMPIAD

First Day | January13,1997

Time: 4.5 hours

1.Letx , x , :: :, x be real numbers satisfying the following two

1 2 1997

conditions:

x 3 (i =1, 2, ::: , 1997); (a)

3. (b) x + x + + x = 318

1 2 1997

12 12 12

Find the maximum value of x + x + + x , and giveyour reason.

1 2 1997

2.LetA B C D be an arbitraryconvex quadrilateral and P agiven

1 1 1 1

point inside the quadrilateral. Suppose for vertex A , angles PA B and

1 1 1

PA D areacute. Similarly, for vertices B , C and D , angles PB A ,

1 1 1 1 1 1 1

PB C , PC B , PC D , PD A and PD C are all acute. De ne induc-

1 1 1 1 1 1 1 1 1 1

tively A , B , C and D to be the axisymmetric pointsofP about lines

k k k k

A B , B C , C D and D A respectively (k =2, 3,

k1 k1 k1 k1 k1 k1 k1 k1

:::). Consider the quadrilateral series

A B C D (j =1, 2, :::)

j j j j

and answer the following questions:

(a) Among the rst 12 quadrilaterals, which ones must be similar to the

1997 quadrilateral, and which ones are not necessarily so?

(b) Assume that the 1997 quadrilateral is inscribed in a circle. Among the

rst 12 quadrilaterals, which ones are inscribed in a circle, and which ones

are not necessarily so?

Giveyour proofs for the determinate answers of these questions, and

giveyour examples for those indeterminate ones.

3. Show that thereexist in nitely many positiveintegers n such that

one can arrange

1, 2, :::, 3n

in the following table

a , a , ::: , a

1 2 n

b ,b , :: : ,b

1 2 n

c , c , ::: , c

1 2 n

which satis es the following two conditions:

201

(1) a + b + c = a + b + c = = a + b + c is a multiple of 6;

1 1 1 2 2 2 n n n

and

(2) a + + a = b + + b = c + + c is also a multiple of 6.

1 n 1 n 1 n

Second Day | January 14, 1997

Time: 4.5 hours

4. Quadrilateral AB C D is inscribed in a circle. Line AB meets DC at

point P .LineAD meets BC at point Q.Tangent lines QE and QF touch

the circle at points E and F respectively. Prove that points P , E and F are

collinear.

5. Let A = f1, 2, 3, :::, 17g and, for a function f : A ! A, denote

[1] [k+1] [k]

f (x)= f (x) and f (x)= f (f (x)) (k 2 N).

Suppose that f : A ! A is a one-to-one function such that there

exists a natural number M satisfying the following properties:

(a) if m

[m] [m]

f (i +1) f (i) 6 1 (mo d 17) ,

[m] [m]

f (1) f (17) 6 1 (mo d 17) ;

(b) for 1 i 16,

[M ] [M ]

f (i +1) f (i) 1 or 1 (mo d 17) ,

[M ] [M ]

f (1) f (17) 1 or 1 (mo d 17) .

For all functions f satisfying the aboveconditions, nd the maximum

value of M , and proveyour conclusion.

6. Let a , a , ::: be non-negative numbers which satisfy

1 2

a a + a (m; n 2 N) .

n+m n m

Prove that

1 a a ma +

m n 1

for all n m.

Now we return toreaders' comments and solutions toproblems from

1998 numbersofthe Corner, and particularly to solutions toproblems of the

31 Spanish Mathematical Olympiad, First Round [1998: 452{453] for which

we gave some solutions last number [2000: 141{146].

202

2 2

6. Consider the parabolas y = cx + d, x = ay + b, with c>0,

d<0, a>0, b< 0. These parabolas have four common points. Show that

these four pointsareconcyclic.

Solution by Michel Bataille, Rouen, France.

Solution I.Let M (x ;y )(k =1, 2, 3, 4) be the four common points.

k k k

We suppose that theyare distinct (otherwise there is nothing toprove). We

2 2

have: y = cx + d and x = ay + b. Multiplying the rst equalityby a

k k

and the second one by c, adding and rearranging, we obtain:

1 1 ad + bc

2 2

x + y x x + =0 (k =1, 2, 3, 4).

k k

a c ac

1 ad+bc 1

2 2

x y + =0de nes either ;,ora Now, the equation x + y

a c ac

singleton, or a circle. Since it is satis ed by (x ;y )(k =1, 2, 3, 4),itdoes

k k

de ne a circle and the points M areconcyclic.

Solution II. We introduce the complexrepresentation of the points

M : z = x + iy . We haveto show that the cross-ratio

k k k k

z z z z

4 2 4 1

z z z z

3 1 3 2

is a real number.

2 2

Since y = cx + d, y = cx + d, we obtain z z =(x x )

1 3 3 1 3 1

1 3

(1 + ic(x + x )) and analogous results for z z , z z , z z .

3 1 4 1 4 2 3 2

It is easyto see that we actually haveto show that the product

P =(1+ic(x + x ))(1 + ic(x + x ))(1 ic(x + x ))(1 ic(x + x ))

1 4 2 3 1 3 2 4

is a real number.

2 2

But, by eliminating y between y = cx + d and x = ay + b,wesee

2 4 2 2

that x , x , x , x are the rootsofac x +2acdx x + ad + b, so that

1 2 3 4

x + x + x + x =0. By multiplying out the rst two factors and the last

1 2 3 4

two factorsofP , it follows that P is a real number, as required.

Comment by Murray S. Klamkin, University of Alberta, Edmonton,

Alberta.

The following similar problem appeared in Math Horizons, February

1997:

\Prove that if two ellipses whose axes are parallel intersect in

four points, then the four points lie on a circle."

The solution is like the rst above.

203

7. Show that thereexists a polynomial P (x), with integer coecients,

such that sin 1 is a root of P (x)= 0.

Solutions by Michel Bataille, Rouen, France; by Pierre Bornsztein, Cour-

dimanche, France; by Geo reyA.Kandall, Hamden, CT,USA;by Murray

S. Klamkin, University of Alberta, Edmonton, Alberta; and by EdwardT.H.

Wang, Wilfrid Laurier University, Waterloo, Ontario. We give the solution

by Bornsztein.

We have

i =180

e = cos( )+ i sin( )

180 180

= cos 1 + i sin 1

i =180 180 i

and (e ) = e = 1 .

Then

180

cos + i sin = 1 .

180 180

2 2

Let b = sin , a = cos . Then a =1 b and we have

180 180

180

180 k 180k

1= (a + ib) = a (ib) .

k=0

If we take the real parts

180

2k 90k 1802k

1= a (1) b ;

k=0

that is,

180

k 2 2k 1802k

1+ (1) (1 b ) b =0,

k=0

and sin 1 is a root of a polynomial with integer coecients.

Now we moveto solutions of problems of the 31 Spanish Mathemat-

ical Olympiad, Second Round [1998: 453{454].

1. Consider sets A of 100 distinct natural numbers, such that the fol-

lowing property holds: \If a, b, c are elementsofA (distinct or not), there

exists a non-obtuse triangle of sides a, b, c." Let S (A) be the sum of the

perimeters of the triangles considered in the de nition of A. Find the mini-

mal value of S (A).

204

Solution by Pierre Bornsztein, Courdimanche, France.

First we note that the problem is ambiguous: if we choose a, b, c,isit

considered the same as if we choose b, c, a? For the rest of the solution we

consider that the triangle (a; b; c) is \equal" to (b; c; a) and then it will only

be counted one time in S (A).

Let A = fa , :::, a g be a set of 100 distinct natural numbers, with

1 100

1 2 100

and using the Law of Cosines, A satis es the desired property if and only if

2 2 2

for 1 i j k 100, a + a a . Moreover a a , a , a a ,

1 i j k 100

i j k

2 2

so that it suces to have 2a a .

1 100

Since a a +99 we must have

2) with a 2 N . a 99(1 +

1 1

Then a 240.

Then, if A has the property, we have

a 239 + i . (1)

It is easytoverify that A = f240, 241, :::, 339g has the property,

from the above. Consider A = fa , :::, a g with a = 239 + i.

1 100 i

Moreover, if B has the property, then, from (1) with B = fb , :::,

b g, b

100 1 2 100 i i

It is clear that S (B ) S (A).

Thus, the minimal value is S (A) with A = f240, :::, 339g. Now

S (A)=S + S + S where

1 2 3

X X

S = 2a + b , 3a , S =

s = a = a = =28950.

For S ,notethatx 2 A appearsinS ,

2 2

(i) twice for every time, we can choose b 2 Anfxg; that is, 99 times,

(ii) onceforevery time, we can choose a 2 Anfxg.

Then each x 2 A appearsinS exactly 3 99 times. Thus, we have

S =3 99 a =297s.

a2A

For S , each x 2 A appearsinS as many times as we can choose

3 3

(a; b) 2 Anfxg with a 6= b; that is, times.

S (A) = 3s + 297s + 4851s =5151s = 149 121 450 .

The minimal value is then 149 121 450.

3. A line through the barycentre G of the triangle AB C intersects the

side AB at P and the side AC at Q. Show that

PB QC 1

 .

PA QA 4

Solutions by Pierre Bornsztein, Courdimanche, France; and by Geo rey

A. Kandall, Hamden, CT,USA.WegiveKandall's generalization.

Let u = , v = . We will show that u + v =1. It will then follow

PA QA

1 1 1

u . that uv = u(1 u)= 

4 2 4

That u + v =1is a special case of the theorem below.

Theorem. Suppose AB C is any triangle and P , Q, R are arbitrarypoints

on the sides AB , AC , BC .LetG be the intersection of PQ and AR.Let

BR : RC = : , where + =1. Then

Proof.Draw lines through B and C that are parallel to AR, and extend PQ

to meet these lines at M and N , as in the diagram.

Then GR = MB+ NC, so that = + . Therefore,

When G is the barycentre(centroid), we have = and = = .

1 2

GA 2 2

+ , as asserted at the outset. Consequently, 1=

PA QA

4. Find all the integer solutions of the equation

p(x + y )= xy

in which p is a prime number.

Solutions by Michel Bataille, Rouen, France; by Pierre Bornsztein, Cour-

dimanche, France; by Geo reyA.Kandall, Hamden, CT,USA;andby

EdwardT.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. We give

Bornsztein's solution.

p(x + y )= xy is equivalent to (x p)(y p)=p .

2 2 2

Since p is a prime, the integral divisorsof p are p , p, 1, 1, p, p .

It follows that (x; y ) is a solution if and only if (x; y ) 2f(p(1 p);p 1),

(0; 0), (p 1;p(1 p)), (p +1;p(p +1)), (2p; 2p), (p(p +1);p +1)g.

5. Show that, if the equations

3 3 2 2 3 2

x +mxn =0, nx 2m x 5mnx2m n =0 (m 6=0, n 6=0)

haveacommon root, then the rst equation would have two equal roots,

and determine in this case the roots of both equations in terms of n.

Solutions by Michel Bataille, Rouen, France; and byPierre Bornsztein,

Courdimanche, France. We give Bataille's solution.

Let a be a common root of the two equations. We have a = n ma.

Thus, we also have

2 2 3 2

n(n ma) 2m a 5mna 2m n = 0

2 2

and a is a root of mx +3nx + m = 0 . (1)

Substituting a for x in

 

n 3n 9n

3 2

x + mx n = x +3 x + m x + x +2n ,

m m m

3 2

, and returning to (1), we deduce that 4m +27n =0. we obtain a =

From this, we easily get the following factorization:

3n 3n

x + , which shows that the rst of the given x + mx n = x

2m m

3n 3n

equations has a simple root and a double root .

m 2m

3 2

We note in passing that a = (since 4m +27n =0).

207

3 2 2

The second equation may be rewritten as nx 2m x 5mnx +

2m 25

n =0, so that the sum of the rootsis and the product of the roots

2 n

25n

is . If we denoteby b and c the roots other than a, we thus have

2m 3n 15n 25n 2m 25m

b + c = = and bc = = ,from which we

n 2m m 2 3n 3

15n

easily obtain b = c = .

1=3 1=3

n 4 n

3 2 1=3

Using 4m +27n =0, we see that = (where x 7! x

m 3

denotes the inverse function of x 7! x from R to R) and we calculate:

roots of the rst equation: (4n) ; ;

roots of the second equation: ; 5 ; 5 .

2 2 2

6. AB is a xed segment and C avariable point, internal to AB .

0 0

Equilateral triangles AC B and CBA areconstructed, in the same half-

plane de ned by AB , and another equilateral triangle AB C is constructed

in the opposite half-plane. Show that:

0 0 0

(a) the lines AA , BB and CC areconcurrent;

(b) if P is the common point of the lines of (a), nd the locus of P when C

varies on AB ;

00 00 00

(c) the centres A , B , C of the three equilateral triangles also form an

equilateral triangle;

00 00 00

(d) the points A , B , C and P areconcyclic.

Solutions by Michel Bataille, Rouen, France; and byPierre Bornsztein,

Courdimanche, France. We give Bataille's solution.

 0

(a) The rotation with centre C and angle 60 transforms B into A and

0 0 0 

B into A . Hence the lines AA and BB make an angle of 60 and, if we

denoteby P their point of intersection, we have: \AP B =120,from

0 0

which we deduce that the points A, P , B , C areconcyclic (with P and C in

0 0 

di erent half-planes de ned by AB ). Therefore, \C PB = \C AB =60

0 0

and the line C P makes an angle of 60 with the line BB .

Now, by the same reasoning as at the beginning, the line CC also

 0 0 0

makes an angle of 60 with the line BB . Thus, the lines CC and C P

0 0 0

coincide and AA , BB and CC areconcurrent at P .

(b) From (a) P belongs to the arc AB subtending 120 on the segment

AB .Conversely, if M is any point of this arc, let us rst construct C such

0 0

that 4AB C is equilateral, and M and C are not in the same half-plane

de ned by AB , and then C is the point of intersection of MC and AB .

0 0 0 0

The points A and B being obtained, the lines AA and BB intersect at the

0 

point P of CC such that \AP B =120 ; that is at M . Thus, M = P and

M is a point of the locus.

208

In conclusion, we nd that the locus is the arc AB subtending 120 on

the segment AB .

0 0

(c) The circles (AB C ) and (A BC ) intersect at P and C and have

00 00 00 00 00 00

respectivecentres B and A . Thus, A B ? PC. Similarly, A C ? PB.

00 00 00 00 

The lines A B and A C make the same angle as PC and PB; that is 60 .

00 00 00 00 00 00

We have the same result for lines A B and B C and for lines B C and

00 00 00 00 00

A C so that 4A B C is equilateral.

00 00

(d) 4A PC is isosceles at A , so that

 

 00 00 00

(180 \PA C )=90 \CBP \A PC = \PCA =

 

= 90 (120 \BC P )= \BC P 30 .

00 

Similarly, \B PC = \AC P 30 . Hence

00 00 00 00 

\B PA = \B PC + \A PC = \BC P + \AC P 60

 

= 180 60 =120

00 00 00

and P is on the circle (A B C ).

Remark. For the `degenerate' triangle AB C , P is the Fermat point and (c) is

Napoleon's theorem.

Tocomplete this number of the Corner and solutions for problems given

in 1998 numbers of the Corner, we turn to solutions of problems of the 46

Polish Mathematical Olympiad, 1994{95 given [1998: 455].

1. Find the number of those subsetsof f1, 2, :: :, 2ng in which the

equation x + y =2n +1 has no solutions.

Solutions by Mohammed Aassila, Strasbourg, France; byPierre

Bornsztein, Courdimanche, France; and by EdwardT.H. Wang, Wilfrid Lau-

rier University, Waterloo, Ontario. We give Aassila's solution, though they

were all very similar.

That the equation x + y =2n +1 has no solution means that x and

2n +1 x are not both chosen from x =1, 2, ::: , n.Hence we havethree

possibilities: to choose one, the other, or neither. The answer is 3 subsets

(counting the empty set).

3. Let p 3 be a given prime number. De ne a sequence (a ) by

a = n for n =0, 1, 2, ::: , p 1,

a = a + a for n p:

n n1 np

Determine the remainder left by a on division by p. p

209

Solution by Pierre Bornsztein, Courdimanche, France.

Lemma. Let n, k be integers such that n kp p. Then

 

a = a . (1)

n nipk+i

i=0

Proof. We use induction on k .

For k =1,letn ( p) be an integer. Then, from the de nition of fa g

a = a + a

n n1 np

which is (1) for k =1.

Let k be a given positive integer. Suppose that (1) holds for each

n kp.

We show (1) holds for n (k +1)p. The minimum value of n ip k + i

for i =0, :::, k is obtained for i = k , whereitisn kp.

Thus, for n (k +1)p we have n kp p.

Thus, in the following sum, we can use the de nition of (a ) for each

term.

 

a = a

n nipk+i

i=0

 

= (a + a )

nipk+i1

n(i+1)pk+i

i=0

 

k k

= a + a

nk1 nipk+i1

0 i

i=1

 

k k

+ a + a

n(i+1)pk+i n(k+1)p

i k

i=0

 

k k

= a + a

n(k+1) n(i+1)pk+i

0 i +1

i=0

 

k k

+ a + a .

n(i+1)pk+i n(k+1)p

i k

i=0

210

k k+1 k k+1 k+1 k k

Of course =1 = , =1= and = + ,for

0 0 k k+1 i+1 i+1 i

i =0, 1, :::, k 1. Thus,

 

k +1 k +1

a = a + a

n(k+1) n(i+1)pk+i

0 i +1

i=0

 

k +1

+ a

n(k+1)p

k +1

 

k+1

k +1

= a .

nip(k+1)+i

i=0

Thus, (1) holds for k +1, when n (k +1)p.

We deducefrom the lemma, that for n p

 

a = a

n nipp+i

i=0

(using (1) with k = p).

But, it is well known that, since p is a prime,

 

 0(mod p) , for i =1, 2;:::, p 1 .

Then a a + a (mo d p). But a = a + a .

n np n n1 np

Then, for n p , a a (mo d p), or equivalently, for t 0

a a (mo d p) .

t+p 1

The sequence formed by the remainders left by division of a by p is therefore

periodic, with period p 1.

3 2

Moreover, p = p(p 1) + p.

Then a a (mo d p). But a = a + a = p 1, so that we have

p p p1 0

a p 1(mod p) .

4. For a xed integer n 1 compute the minimum value of the sum

2 3 n

x x x

2 3 n

x + + + + ,

2 3 n

given that x , x , :::, x are positive numbers satisfying the condition

1 2 n

1 1 1

+ + + = n .

x x x

1 2 n

211

Solutions by Mohammed Aassila, Strasbourg, France; byPierre

Bornsztein, Courdimanche, France; and by Murray S. Klamkin, University

of Alberta, Edmonton, Alberta. We give the solution of Bornsztein, but all

were similar.

Denote

n 2

x x

n 2

+ + , S = x +

2 n

1 1

H = 1+ + + ,

2 n

w = , for i =1, 2, :::, n .

w =1. Thus, Then w > 0 for i =1, :::, n and

i i

i=1

n n

X Y

i i w

= w x (x )

i i

i=1 i=1

, from the weighted AM-GM. (Note that we haveusedn! copies of

2 3

n! x n! x

2 3

copies of , copies of ,etc.).

2 n! 3 n!

Thus,

1=H

x . 

i=1

Now after using the AM-GM inequality, we have

x =1,

i=1

with equality if and only if x = x = = x (=1). Then 1.

1 2 n

Thus S H with equality if and only if x = = x =1. Then the

1 n

minimal value of

2 3 n

x x x

2 3 n

x + + + +

2 3 n

1 1

1+ + + ,

2 n

and it is obtained if and only if x = x = = x =1.

1 2 n

That completes the Corner for this issue. Send me your nice solutions

and Olympiad materials.

212

BOOK REVIEWS

ALAN LAW

Conjecture and Proof by Miklos Laczkovich,

published byTypotex, Budapest, 1998.

ISBN # 963-7546-88X, softcover, 100+ pages, $24.00 (US).

Reviewed by Andy Liu, UniversityofAlberta, Edmonton, Alberta.

This is one of a few books Typotex publishes in English. It is essentially

the textbook of a course of the same name in the famed Budapest Mathe-

matics Semesters, a program in advanced mathematics for North American

undergraduate students, taught in English bytop-notch Hungarian mathe-

maticians.

The book deals with the existence or non-existenceofvarious math-

ematical objects, constructions or procedures. It is divided into two parts.

The rst part, which consistsofseven chapters, deals with non-existence.

2; that is, the Chapter 1 contains ve di erent proofs of the irrationalityof

non-existence of positiveintegers p and q such that = 2. Then come the

harder proofs of the irrationalityofe and .

Chapter2makes a switch from number theoryto geometry, and consid-

ers the problem of what lengths areconstructible by ruler and compass given

2 and cos arenotconstructible. a segment of length 1. It is shown that

This proves that two of the three classical problems, namely, the duplication

of the cube and the trisection of an arbitrary angle, cannot be solved.

The second part of the book consistsoften chapters and deals with

existence. A proof of existencecan be graded according to how directly it leads

to the existence ofagiven object. It may consist of an explicit construction

of the object or, alternatively, it may establish the existence of the object

without giving the slightest hint of how to nd it. The purest form of proof

of existenceusesreductio ad absurdum by assuming the non-existence of the

object and arriving at a contradiction.

Below is the table of contents of the second part:

8 The Pigeonhole Principle

9 Liouville numbers

10 Countable and uncountable sets

11 Isometries of R

12 The problem of invariant measures

13 The Banach-Tarski paradox

14 Open and closed setsinR | The Cantor set

15 The Peano curve

16 Borel sets

17 The diagonal method.

213

This is one of the most fascinating books I haveever read. Many gems

from various elds in mathematicsarebrought together in this compact pack-

age. The exposition is succinct and smooth, and leads the reader from the

basicsto the frontiers of mathematical research. It is recommended without

any reservation.

Note: The email address of Typotexis, for prices and

ordering information. My understanding is that one copycosts US$24,

two copies in the same order cost US$18 each, three in an order cost

US$16 each, and four or moreare US$15 each.

Winning Solutions by Edward Lozansky and Cecil Rousseau,

published by Springer-Verlag (Problem Books in Mathematics Series), 1996.

ISBN # 0-387-94743-4, softcover, 244+ pages, $34.95 (US).

Reviewed by Andy Liu, UniversityofAlberta, Edmonton, Alberta.

This book is in the Springer-Verlag Problem Books in Mathematics

series and is similar in format to another title in the series, Problem

Solving through Problems byLoren Larson. As the authors acknowledge

in the Preface, however, the present volume is more limited in scope. For

example, Euclidean geometryisomitted entirely. The sparseness in mater-

ial on graph theory is particularly surprising, since one of the authorsisan

expert in the eld.

The book is organized into three chapters titled Numbers, Algebraand

Combinatorics, with 8, 4 and 8 sections respectively. The basic concepts

arecovered in concise manner, supported byexamples of two types. The

rst type may be described as standard bookwork, constructed primarily to

illustrate the concepts. The second type consistsofproblems loosely based

on the textual material, and theycome primarily from various national and

international contests, on which many problem books of this kind draw. The

exercises, of which thereareplenty, are similarly divided intotwotypes along

these lines.

The following problem is taken from page 194 of the book. \Suppose

that the squares of an n n chessboardare labeled arbitrarily with the num-

bers 1 through n .Prove that thereare two adjacent squares (not diagonally)

whose labels di er in absolutevalue by at least n." After Erd}os mentioned it

in a talk in Texas, Ben Walter, then a student, came up with a solution which

is essentially the same as the one given in this book, and far from easy. There

is a related but simpler problem, A2, in the 1981 Putnam Competition.

This book is a welcome addition to the shelves of anyone associated

with mathematicsproblem solving in general and mathematicscompetitions in particular.

214

APanorama of Harmonic Analysis bySteven G. Krantz,

published by The Mathematical Association of America, 1999,

The Carus Mathematical Monographs, Number 27.

ISBN # 0-88385-031-3, hardcover, 357+xi pages, $41.50 (US).

Reviewed by Alan Law, UniversityofWaterloo, Waterloo, Ontario.

This is the most recent volume in the MAA Carus monograph series.

These monographs are intended toprovide expositions of mathematical sub-

jectsthatarecomprehensible not only toteachers and students specializing

in mathematics, but also to scienti c workers in other elds. Steven Krantz's

contribution is a rst-class addition to this outstanding series.

The book assumes that the reader has some acquaintance with elemen-

taryreal analysis. A short overview of rudimentsfrom measure theoryand

elementary functional analysis precedes the chapter on basics of Fourier se-

ries: interweaving of historical developments along with the author's easy

style bring the reader, simply, through summability, kernels, pointwise and

norm convergence, and the Hilbert transform.

Chapter 2 provides a readable introduction to the Fourier transform

and properties, including the Uncertainty Principle, in E . Chapter 3 ad-

dresses multiple Fourier series and some of the accompanying complexities

|problems are illustrated through discussions in an E setting. The next two

chapterstouch on a substantial amount of mathematical structure, including

spherical harmonics, fractional and singular integrals, and Hardy spaces (a

real-variable approach is outlined, too). Chapter 6 gives a clear introduction

of substanceto modern theories of integral operators, including motivation

for and discussion about the celebrated T (1) Theorem.

This reviewer particularly liked Chapter 7: \Wavelets". It illustrates

some restrictions in applications of classical Fourier analysis and how wave-

letsareproviding a powerful extension of harmonic analysis. The chapter

supplies a nice introduction to the one-dimensional Haar wavelet basis, gives

a mathematical description for a multi-resolution analysis, and shows the

construction of a continuous wavelet.

APanorama of Harmonic Analysis is well-titled and well-placed in the

Carus monograph series. The author provides a most-readable tourofhar-

monic analysis from its beginnings in the 1800's toits current stage of evo-

lution. I recommend the book toprofessionals, graduate students and inter-

ested senior undergraduates.

215

The Devil's Dartboard

Trevor Lipscombe and Arturo Sangalli

Abstract

Toproduce the most-dicult dartboard possible, we devise a

polynomial-time algorithm. The problem, and the algorithm, are shown

to be closely related totheTravelling Salesman Problem.

AMS Subject Classi cation: 68Q25 (Analysis of algorithms and

problem complexity), 05A99 (Classical combinatorial problems, and

00A08 (Recreational mathematics).

Introduction

Students of mathematicscan often be found playing dartsinbars. The

reason, no doubt, is their love of the subject. For example, recent analyses

such as [1] and [2] have shown that the mathematically adept, yet physically

inept, darts players should aim at the bull's eyeinorder to maximize their

potential score. This raises the question: what could dartboard manufac-

turersdotoovercom e this strategy? The answer could be torenumber the

boardcompletely. One mathematician toconsider the bene ts and strategies

of possible renumbering was IvarsPeterson, in a popular article [3].

The numbers in a dartboardare (sort of) arranged so that thereisa

penalty for error. For instance, the number 20 at the top of the boardis

sandwiched between the far smaller numbers 1 and 5. Aim at the 20 with

three darts and, if you are not very good, you risk a scoreof20 + 5 + 1 = 26.

Likewise, 19 is anked by 3 and 7 (for a possible scoreof29), and the 17 has

a 2 and 3 on either side (=22).

Therearemore than 1:2 10 circular arrangements of the numbers

1, 2, :::, 20 | the exact number being 19! (that is, factorial 19). The chal-

lenge is to ndtheonearrangement among them that gives the least re-

ward for throwing error. More mathematically, what is the arrangement for

which the standarddeviation of the three-sums (that is, the sums of three

consecutive numbers) is the smallest? We call this arrangement \The Devil's

Dartboard."

We shall consider n{number dartboards (n> 3) and ask: Is therea

polynomial-time[4] algorithm that will generate the Devil's Dartboardfor

each n?ADevil's DartboardProblem (DDP) consists of nding such an al-

gorithm or proving that it does not exist. For the record, the actual dart-

board, found in drinking establishments across the globe, is the permutation:

20, 1, 18, 4, 13, 6, 10, 15, 2, 17, 3, 19, 7, 16, 8, 11, 14, 9, 12, 5.

216

A Decent Algorithm

Here is a polynomial-time algorithm that, if it does not always produce

aDevil's Dartboard, seems tocome pretty close, especially for large n.

The permutation p is de ned by induction. Choose p(k +1) such that

the 3{sum p(k 1) + p(k )+p(k +1) is as close as possible tothemean

m =3(n +1)=2 of the 3{sums. In the case of a tie, select p(k +1) so that

consecutive 3{sums (di erent from m) fall on opposite sides of m, beginning

with a 3{sum greater than m.To get the induction started, set p(1) = n and

p(2) = 1.

For example, if n =6, we have m = 3(6 + 1)=2=10:5. The algorithm

then produces the Devil's Dartboard 6, 1, 4, 5, 2, 3. The sequenceof3{sums

is 11, 10, 11, 10, 11, 10,andthestandarddeviation =0:5, clearly the

smallest one possible. It is not, however, the only Devil's Dartboard for 6

numbers. EveryDevil's Dartboard has its `opposite'; that is, the permuta-

tion obtained byreading the numbers right to left, which is another Devil's

Dartboard. These two permutations correspond to the two ways of writing

the numbersona circular board: clockwise or counterclockwise.

For n =20, the algorithm generates the permutation 20, 1, 11, 19,

2, 10, 18, 4, 9, 17, 6, 8, 16, 7, 12, 13, 5, 14, 15, 3, having =2:54.Itis

notaDevil's Dartboard, though, because the permutation that ends :::, 3,

15, 14, 5 | and coincides with the previous permutation elsewhere | has

 =2:52.

Devil's Dartboards and Travelling Salesmen

The Devil's DartboardProblem, which concerns permutations and min-

imization, is reminiscent of a famous optimization puzzle, the so-called Trav-

elling Salesman Problem (TSP) [5]: Given a set f1, 2, :: :, ng of n \cities"

and the set of distances d(i; j ) between cities i and j , nd an ordering of

cities | that is, a tour | of minimum length (the length of a tour is the sum

of the distances between consecutive cities.) The \cities" may be arbitrary

objects, and the \distances" d(i; j ) any non-negativereal-valued function.

The so-called NP problems (for Nondeterministic Polynomial Time

Veri able) [6] are, roughly speaking, those problems whose solutions might

be dicult to nd but are easyto check. For example, it is straightforward

toverify whether a given tour of the cities in a TSP has length less than, say,

1; 000. The TSP is therefore anNPproblem, but it is actually more than that:

it has the property that any other NP problem may be transformed intoan

instance of the TSP in polynomial time. In other words, the TSP is \as hard

as" any other NP problem. This implies that any polynomial-time algorithm

for the TSP could be used tosolveevery other NP problem also in polynomial

time. Problems that have this propertyare known as NP complete. In prac-

tical terms, it is considered highly unlikely that NP-completeproblems could

217

be \eciently" solved | that is, within polynomial time | although strictly

speaking the question is still open. Strategies to nd near-optimal tours

rather than exact solutions are known as heuristics. The nearest-neighbour

heuristic for the TSP is the common-sense rule of travelling to the nearest

city not yet called upon, with the initial city being chosen at random. In other

words, a tour p is constructed step bystep such that each new city p(k +1)

adds the minimal length to the partial tour p(1), p(2), :::, p(k ).

Consider a generalized travelling salesman problem (GTSP): Thereisa

(reasonably simple) function L which assigns toevery permutation p of the n

cities and to each k (= 1, 2, :::, n), a non-negativereal number L (k ). Find

the permutation p of the cities for which the sum L (1) + L (2) + :::+ L (n)

p p p

is a minimum.

The standard TSP is obtained bytaking L (k )= d(p(k );p(k + 1)); the

Devil's DartboardProblem resultsfrom taking L (k )= (p(k )+ p(k +1)+

p(k +2) 3(n +1)=2) . Moreover, in this general setting, our algorithm is

the equivalent of the nearest-neighbour heuristic.

Mathematically sophisticated salespersons, therefore, will be able to

spend more time in the bar, and less on the road, and be far better at darts

than their colleagues.

References

1. David Percy, MathematicsToday 35,p.54.

2. Robert Matthews, `Go For It' New Scientist 17, April 1999, p. 16.

3. IvarsPeterson, IvarsPeterson's MathLand, May 19, 1997.

4. Arturo Sangalli, The Importance of Being Fuzzy,Princeton University

Press, Princeton, 1998.

5. E.L. Lawler et al. (eds), The Travelling Salesman Problem, John Wiley&

Sons, New York, 1985.

6. M.R. Garey and D.S. Johnson, ComputersandIntractability: A Guide to

the Theory of NP Completeness,W.H.Freeman, San Francisco, 1979.

Trevor Lipscombe Arturo Sangalli

Princeton UniversityPress Department of Mathematics

41 William Street Champlain Regional College

Princeton Lennoxville Campus

NJ 08540, USA. PO Box 5003 Lennoxville QC, Canada J1M 2A1.

218

THE SKOLIAD CORNER

No. 46

R.E. Woodrow

For your summer problem pleasure, we are giving the rst rounds of

two contests. I hope you enjoy the problems. My thanks go to Richard

Nowakowski, Canadian Team Leader to the IMO in Buenos Aires for collect-

ing them.

BUNDESWETTBEWERB MATHEMATIK 1997

Federal Contest in Mathematics (Germany)

First Round

1. Can you always choose 15 from 100 arbitraryintegers so that the

di erence of any two of the chosen integers is divisible by 7?

What is the answer if 15 is replaced by 16?(Proof!)

2. Determine all primes p for which the system

p +1 = 2x

2 2

p +1 = 2y

has a solution in integers x, y .

3. A square S is inscribed in an acute-angled triangle AB C by placing

two corners on the side BC and one corner on AC and AB ,respectively. In

a similar way, squares S and S are inscribed in AB C .

b c

For which kind of triangle AB C do the sides of S , S and S have

a b c

equal length?

4. In a park thereare 10 000 trees, placed in a square latticeof100

rows and 100 columns. Determine the maximum number of trees that can

be cut down satisfying the condition: sitting on a stump, you cannot see any

other stump.

MACEDONIAN MATHEMATICAL

COMPETITIONS 1997

Round 1 | Part 1

1. Find a number with three di erent digits, ve times smaller than the

sum of all the other numbers with the same digits. Determine all solutions!

219

p p p

2. Calculate 7 48 + 5 24 + 3 8.

3.Prove that, if a> 0, b> 0, c>0, then

a b c 3

+ + .

b + c c + a a + b 2

Prove that the equalityisvalid if and only if a = b = c.

4.Determine all the ordered pairs (x; y ), x 2 N, y 2 N, for which

three of the stated properties are true, and only one is false:

(i) y j (x +1),

(ii) x =2y +5,

(iii) 3 j (x + y ),

(iv) x +7y is a prime number.

5. For which natural numbers n is the sum

2 2 2 2

n +(n +1) +(n +2) +(n +3)

divisible by 10?

6. Three tired travellersarrived at an inn and asked for a meal. The

innkeeper did not have anything else to o er them but potatoes. While the

potatoes were being baked the travellers fell asleep. After a while, when the

of the potatoes and potatoes were done, the rst traveller woke up, ate

went back to sleep without waking up the others. Then the second traveller

woke up, ate of the rest of the potatoes and went back to sleep. At last,

the thirdtraveller woke up, ate of the rest of the potatoes and went back to

sleep. The innkeeper watched all of these carefully and, in the end, when he

counted the rest of the potatoes, there were 8. How many did the innkeeper

bake?

Last issue we gavetheproblems of the Kangarou des Mathematiques, 

 

EpreuveEUROPEENNE Cadets, 1997. Hereare the solutions.

1. c 2. c 3. c 4. b 5. b

6. c 7. b 8. e 9. e 10. a

11. d 12. b 13. c 14. c 15. c

16. c 17. a 18. d 19. c 20. b

21. b 22. b 23. d 24. c 25. d

26. e 27. b 28. a 29. e 30. e

That completes the Skoliad Corner for this number. Send me your com-

ments and suggestions as well as suitable contest materials.

220

MATHEMATICAL MAYHEM

Mathematical Mayhem began in 1988 as a Mathematical Journal for and by

High School and University Students.Itcontinues, with the same emphasis,

as an integral part of Crux Mathematicorum with Mathematical Mayhem.

All material intended for inclusion in this section should be sent to

Mathematical Mayhem, Department of Mathematics, University of Toronto,

100 St. George St., Toronto, Ontario, Canada. M5S 3G3. The electronic

address is

[email protected]

The Assistant Mayhem Editor is Cyrus Hsia (University of Western On-

tario). The rest of the sta consists of Adrian Chan (Harvard University),

Jimmy Chui (UniversityofToronto), and David Savitt (Harvard University)

Mayhem Problems

The Mayhem Problems editorsare:

Adrian Chan Mayhem High School Problems Editor,

Donny Cheung Mayhem Advanced Problems Editor,

David Savitt Mayhem Challenge BoardProblems Editor.

Note that all correspondence should be sent to the appropriate editor |

see the relevant section. In this issue, you will nd only solutions | the

next issue will feature only problems.

We warmly welcome proposals for problems and solutions. With the

schedule of eight issues per year, we request that solutions from the previous

issue be submitted in time for issue 4 of 2001.

High School Solutions

Editor: Adrian Chan, 1195 HarvardYard Mail Center, Cambridge, MA,

USA 02138-7501

H253. Find all real solutions to the equation

p p p

2 2 2

3x 18x +52+ 2x 12x + 162 = x +6x +280.

Solution by EdwardT.H. Wang, Wilfrid Laurier University, Waterloo,

Ontario; and Lino Demasi, student, St. Ignatius High School, Thunder Bay,

Ontario.

221

Notice that bycompleting the square,

2 2

3x 18x +52 = 3(x 3) +25,

2 2

2x 12x +162 = 2(x 3) +144,

2 2

x +6x +280 = (x 3) +289.

p p

25 + 144=5+12=17and the RHS is Therefore, the LHS is at least

289=17, with equality for both when x =3. Therefore, x =3is at most

the only solution.

Also solved by Murray S. Klamkin, UniversityofAlberta, Edmonton,

Alberta.

H254. Proposed byAlexandreTrichtchenko, 1 year, Carleton Uni-

versity, Ottawa, Ontario.

Let p and q be relatively prime positiveintegers, and n a multiple of pq .

Find all ordered pairs (a; b) of non-negative integers that satisfy the diophan-

tine equation n = ap + bq .

Solution by Masoud Kamgarpour, student, Carson Graham Secondary

School, North Vancouver, BC; and EdwardT.H. Wang, Wilfrid Laurier Uni-

versity, Waterloo, Ontario.

If n is negative, then thereare no non-negative solutions in a and b.

So, assume that n is non-negative. Since n is a multiple of pq ,let n = kpq

for some non-negativeinteger k . Then kpq = ap + bq ==) pjbq ==) pjb,

since gcd(p; q )= 1. Therefore, b = sp for some non-negative integer s.

Similarly, q jap ==) q ja, since gcd(p; q )= 1. Therefore, a = tq for some

non-negative integer t. So, we have kpq = tpq + spq ==) k = t + s,and

(a; b)=(sp; tq ) where k = t + s for non-negative integers t, s.

H255.Wehave a set of tiles which contains an in nite number of

regular n{gons, for each n =3, 4, :::. Which subsets of tiles can be chosen,

so that they taround a common vertex? For example, we can choose four

squares, or four triangles and a hexagon.

Solution. Let n n ::: n be the number of sides of the

1 2 t

n{gons. Recall that the interior angle of a regular n{gon measures

180(n 2)=n =180 360=n degrees. Then

 

t t

X X

360 1

180 = 180t 360 =360 (1)

n n

i i

i=1 i=1

180t 360 t 2 1

= = . (2) ==)

n 360 2

i=1

Since 1=n > 0, t 3. Also, n 3 for all i, yielding

i i

i=1

180 360=n 60 for all i,so

 

360

180 60t , 360 =

i=1

222

which implies that t 6.

Case 1: t =3.

Equation (2) becomes

1 1 1 1

+ + = .

n n n 2

1 2 3

Since n n n , 1=n (1=2)=3= 1=6,son 6.

1 2 3 1 1

If n =3, then 1=n +1=n =1=6, and the solutions are

1 2 3

(n ;n ;n )=(3; 7; 42), (3; 8; 24), (3; 9; 18), (3; 10; 15), (3; 12; 12).

1 2 3

If n =4, then 1=n +1=n =1=4, and the solutions are

1 2 3

(n ;n ;n )=(4; 5; 20), (4; 6; 12), (4; 8; 8).

1 2 3

If n =5, then 1=n +1=n =3=10, and the only solution is

1 2 3

(n ;n ;n )=(5; 5; 10).

1 2 3

If n =6, then 1=n +1=n =1=3, and the only solution is

1 2 3

(n ;n ;n )=(6; 6; 6).

1 2 3

Case 2: t =4.

Equation (2) becomes

1 1 1 1

+ + + =1.

n n n n

1 2 3 4

Since n n n n , 1=n 1=4,son 4.

1 2 3 4 1 1

If n = n =3, then 1=n +1=n =1=3, and the solutions are

1 2 3 4

(n ;n ;n ;n )= (3; 3; 4; 12), (3; 3; 6; 6).

1 2 3 4

If n =3and n 4, then 1=n +1=n +1=n =2=3, so that

1 2 2 3 4

1=n (2=3)=3= 2=9,orn 9=2,son =4.

2 2 2

Hence, 1=n +1=n =5=12, and the only solution is

3 4

(n ;n ;n ;n )= (3; 4; 4; 6).

1 2 3 4

If n 4, then n 4 for all i, and the only solution is

1 i

(n ;n ;n ;n )= (4; 4; 4; 4).

1 2 3 4

Case 3: t =5.

Equation (2) becomes

1 1 1 1 1 3

+ + + + = .

n n n n n 2

1 2 3 4 5

Since n n n n n , 1=n (3=2)=5= 3=10,son 10=3,

1 2 3 4 5 1 1

or n =3.

Then

1 1 1 1 3 1 7

+ + + = = ,

n n n n 2 3 6

2 3 4 5

223

so 1=n (7=6)=4= 7=24,or n 24=7,son =3.

2 2 2

Then again,

1 5 1 1 1 7

= , + + =

n n n 6 3 6

3 4 5

so 1=n (5=6)=3= 5=18,or n 18=5,son =3.

3 3 3

Finally,

1 1 1 1 5

= , + =

n n 6 3 2

4 5

so the solutions are (n ;n ;n ;n ;n )= (3; 3; 3; 3; 6), (3; 3; 3; 4; 4).

1 2 3 4 5

Case 4: t =6.

Equation (2) becomes

1 1 1 1 1 1

+ + + + + =2.

n n n n n n

1 2 3 4 5 6

Since 1=n 1=3 for all i, we must have equality for all i, so the only

solution is (n ;n ;n ;n ;n ;n )= (3; 3; 3; 3; 3; 3).

1 2 3 4 5 6

a b c

H256. Let A =2 p p , where p and p are primes, possibly equal

1 2

to each other and to 2, and a, b, and c are positive integers. It is known that

a b c

p p (mo d 4), b c (mo d 2), and that 2 , p , p arethree consecu-

1 2

tiveterms of an arithmetic sequence, not necessarily in that order. Find all

possible values for A.

a b c

Solution. Since 2 , p , and p form an arithmetic sequence, p is even

1 2

if and only p is even.

Case 1: p , p arebotheven.

1 2

a b c

Since p , p are prime, p = p =2,and2 , 2 , 2 form an arithmetic

1 2 1 2

sequence in some order. Without loss of generality, assume that

a c b+1 ca b+1a

1 a b c. Then 2 +2 =2 ==) 1+ 2 =2 . The

only solution is c a =0and b +1 a =1, which implies that a = b = c,

a b c 3a a

so A =(2 )(2 )(2 )=2 =8 .

Case 2: p p 1 (mo d 4).

1 2

b c

Without loss of generality, assume that p

1 2

b a c

order of the arithmetic sequence must be p , 2 , p , which implies that

1 2

b c a+1 b c b c

p + p =2 . But p p 1 (mo d 4). Thus, p + p 2 (mo d 4).

1 2 1 2 1 2

2 a+1 a+1

But, 2 j2 ==) 2 0(mod4). Thereare no solutions for this case.

Case 3: p p 1 (mo d 4).

1 2

b c

Without loss of generality, assume that p

1 2

b a c

order of the arithmetic sequence must be p , 2 , p , which implies that

1 2

b c a+1 b c

p + p =2 . But p + p 2(mod4) if b c 0 (mo d 2) and

1 2 1 2

a+1

2 (mo d 4) if b c 1 (mo d 2). Again, 2 0(mod4). Thereareno

solutions for this case either.

The only solution is A =8 for some positiveinteger A.

224

Advanced Solutions

Editor: Donny Cheung, c/o Conrad Grebel College, UniversityofWa-

terloo, Waterloo, Ontario, Canada. N2L 3G6

A229.Intetrahedron SABC , the medians of the faces SAB, SBC ,

and SC A,taken from the vertex S , make equal angles with the edges that

they lead to. Prove that jSAj = jSBj = jSC j.

Solution by Murray S. Klamkin, University of Alberta, Edmonton,

Alberta.

Let the vectors SA, SB, SC be denoted by 2A, 2B, 2C,respectively.

Then the medians to the faces SAB, SBC ,andSC A are given by A+B,

B+C,andC+A,respectively. The cosines of the equal angles made are

given by

(B C) (B + C) (C A) (C + A) (A B) (A + B)

= = .

jB CjjB+Cj jC AjjC+Aj jA BjjA+Bj

These three fractions are equal toafraction whose numerator is the sum of

the three numerators and whose denominator is the sum of the three denom-

inators. Since this latter fraction equals 0, each of the other three numerators

must also be 0; that is, jBj = jCj = jAj.

A230. Proposed by Naoki Sato.

For non-negativeintegers n and k ,letP (x) denote the rational func-

n;k

tion

n n n k1

(x 1)(x x) (x x )

k k k k1

(x 1)(x x) (x x )

Show that P (x) is actually a polynomial for all n, k .

n;k

Solution by Murray S. Klamkin, University of Alberta, Edmonton,

Alberta.

Factoring out all the x's, we get

n n1 nk+1

(x 1)(x 1) (x 1)

. (1)

k k1

(x 1)(x 1) (x 1)

If k n k +1, then, we can cancel out like factors in the numerator and

denominator. So we can assume that k

(n)(n1)(nk+1)

Since is always an integer, all the factors of the

(k)(k1)(1)

(k i)'s, i =0, 1, :::, k 1, divide intothefactorsofthe(n i)'s,

i =0, 1, :: :, k 1. Hence all the factors of the denominator of (1) which

are of the form (x ! ) where ! is a root of unityarecancelled bylikefactors

in the numerator. Hence, P (x) is actually a polynomial for all n, k .

n;k

Also solved by Vishaal Kapoor, Simon Fraser University, Burnaby, BC.

225

A231. Proposed by Mohammed Aassila, CRM, Universite de Montreal, 

Montreal, Quebec. 

For the sides of a triangle a, b, and c,prove that

2 2 2

13 (a + b + c)(a + b + c )+ 4abc 1

 .

27 (a + b + c) 2

Solution by Murray S. Klamkin, University of Alberta, Edmonton,

Alberta.

We can eliminate the triangle constraintsby letting, as usual, a = y + z ,

b = z + x, and c = x + y where x, y , z 0. Then a + b + c =2T ,

2 2 2 2

a + b + c =2(T T ), and abc = T T T , where T = x + y + z ,

2 1 2 3 1

T = yz + zx + xy ,and T = xy z . The left hand inequality now becomes,

2 3

after simpli cation, T 27T , which follows by the AM-GM inequality.

The right hand inequalitybecomes 4T 0. There is equality in the left

hand inequality if and only if the triangle is equilateral. There is equalityin

the right hand inequality if and only if the triangle is degenerate with sides

p, p, 0.

Also solved byVedula N. Murty, Dover, PA, USA; Vishaal Kapoor,

Simon Fraser University, Burnaby, BC; and Masoud Kamgarpour, student,

Carson Graham Secondary School, North Vancouver, BC.

A232.Five distinct points, A, B , C , D , and E lie on a line (in this

order) and jAB j = jBC j = jCDj = jDE j. The point F lies outside the

line. Let G be the circumcentre of triangle AD F and H the circumcentreof

triangle BEF . Show that the lines GH and FC are perpendicular.

(1997 Baltic Way)

Comment. Attimeofpress, we do not have a solution for this problem.

Challenge Board Solutions

Editor: David Savitt, Department of Mathematics, Harvard University,

1 Oxford Street, Cambridge, MA, USA 02138

C85. Proposed by Christopher Long, graduate student, Rutgers Uni-

versity, NJ, USA. From a set of course notes on analytic number theory.

Let C =(c ) be an m n matrix with complex entries, and let D be

i;j

areal number. Show that the following statementsare equivalent:

(i) For any complex numbers a , j =1, :: :, n,

m n n

X X X

a c D ja j .

j i;j j

i=1 j =1 j =1

226

(ii) For any complex numbers b , i =1, :::, m,

n m m

X X X

b c D jb j .

i i;j i

j =1 i=1 i=1

Solution by Michael Lambrou, UniversityofCrete, Crete, Greece.

It suces to show (i) implies (ii). Put

x = b c .

j i ij

i=1

Then the left-hand side of (ii) is

1 0

n m m n n

X X X X X

A @

x c b = b c jx j = x

j ij i i ij j j

j =1 i=1 i=1 j =1 j =1

0 1

1=2

m m n

X X X

B C

 jb j x c

@ A

i j ij

i=1 i=1 j =1

0 1

1=2

m n

X X

2 2

1=2

@ A

 jb j D j x j ,

i j

i=1 j =1

by Cauchy's inequality and by (i). This easily becomes

n m

X X

2 2

jx j D jb j ,

j m

j =1 i=1

which is the statement (ii).

Torephrase this using highbrow terminology, de ne the norm of an

m n matrix A to be the maximum value of the length of the vector Av ,

with the maximum taken over all vectors v 2 C of length 1. Observe then

D , and (ii) says that the that (i) says simply that the norm of C is at most

norm of the adjoint C of C (the complexconjugate of the transpose of C )

is at most D . Therefore the equivalence of (i) and (ii) is just the statement

that the norm of C is equal to the norm of its adjoint, which is a standard

fact.

C86.LetK denote the completegraph on n vertices; that is, the

graph on n vertices with all possible edges present. Show that K can be

decomposed into n 1 disjoint paths of length 1, 2, :::, n 1.Forexample,

for n =4, the graph K ,withvertices A, B , C , and D , decomposes into the

paths fAC g, fBD; DAg,andfAB ; B C; C D g.

227

Can we require that the paths in the decomposition be simple? (We say

that a path is simple if it passes through each vertex at most once; that is, if

no vertex is the end-point of more than two edges along the path.)

I. Solution.Ifn is odd, then at any vertexofK the number of incident

edges is even. Hence, by a famous theorem due to Euler, it is possible totrace

an unbroken path which traverses each edge of K exactly once. Break such

a path into pieces of length 1, 2, :::, n 1 in any way we choose, and we

obtain a solution for n odd.

However, we must try something di erent when n is even, as Euler's

theorem does not apply. Instead, draw the vertices of K as if theywere

evenly spaced around a circle, and label the vertices A , ::: , A . Now,

1 n

draw the zigzag path

A ! A ! A ! A ! A !!A ! A ! A .

1 2 n 3 n1

n=2 n=2+2 n=2+1

One may easily check that drawing the same path rotated by 2j =n (in other

words, with the subscripts of the A's all augmented by j modulo n) for inte-

gers j

n 1. Leaving one such path intact, splitting the second such path intoa

path of length 1 and a path of length n 2, and so on, gives the desired

decomposition of K , and in fact this decomposition is simple.

It remains to answer whether or not our paths can be chosen tobe

simple in the case where n is odd. In this case, select any n 1 of the

vertices of our graph, and call the remaining vertex A. Using the method

of the preceding paragraph, construct the (n 1)=2 zigzags on the chosen

n 1 vertices. Join A to the endpoints of each zigzag, and we obtain a

decomposition of our K into (n 1)=2 disjoint loops of length n,each

passing through everyvertex of the graph. Finally, split these loops into

paths of length 1 and n 1, of length 2 and n 2, and so on, which shows

that indeed these paths can be taken to be simple when n is odd.

II. Solution byRoman Muchnik, Yale University, New Haven, CT,USA.

We prove the result using induction. The result is easily checked for

n =2and n =3. Assume the result is true for n = t for some positive

integer t.

Let A , A , ::: , A denote the vertices of K . By assumption, K

1 2 t t t

can be decomposed into paths of length 1, 2, :: :, t 1. We add two ver-

tices P and Q, and the appropriate edges toformK . Consider the path

t+2

PA QA PA :::, alternating between P and Q, passing through everyver-

1 2 3

texofK , and terminating on P or Q after A . This path has length 2n,and

t t

adding PQ, forms a path of length 2t +1. Divide this path into one of length

t and the other of length t +1. Thus, K is decomposed into paths of

t+2

length 1, 2, :::, t 1, t, t +1.

Therefore, by induction, the result is true for all positive integers n.

228

Problem of the Month

Jimmy Chui, student, UniversityofToronto

Problem.LetP be a point inside triangle AB C such that

\AP B \AC B = \AP C \AB C .

Let D , E be the incentres of triangles AP B , AP C ,respectively. Show that

AP , BD, and CE areconcurrent.

(1996 IMO, Problem 2.Proposed by Canada)

We present two solutions to this problem. One solution involves pedal

triangles, and the other solution uses inversion. The solutions start o iden-

tically.

B C

Let X , Y , and Z be the feet of the perpendicularsfrom P to the sides

BC , CA, and AB respectively. Let = \AP B \AC B = \AP C

\AB C .So,\AP B = C + and \AP C = B + .

Note that BD is the angle bisector of \AB P ,from properties of

incentres. Let the intersection of BD and AP be Q; then, from the angle

bisector theorem, we have

AQ AB

= .

BP QP

Similarly, if R is the intersection of CE and AP ,then

AR AC

= .

CP RP

Now, AP , BD, and CE concur if and only if Q and R coincide, or

equivalently

AQ AR AB AQ AR AC

= (=) = = = .

QP RP BP QP RP CP

229

I. Solution | Pedal Triangles.

Consider the pedal triangle XY Z.From properties of pedal triangles,

\YZX = \AP B \AC B = and \ZY X = \AP C \AB C = .So

triangle YZX is isosceles with XY = XZ.

Now, XY = CP sin C and XZ = BP sin B ,from pedal triangle prop-

erties. So BP sin B = CP sin C . Then,

sin C AB BP

= = ,

CP sin B AC

with the last equalitycoming from the Sine Law. Thus

AB AC

= ,

BP CP

and so AP , BD, and CE areconcurrent.

II. Solution | Inversion.

0 0

Invert about A with a radius of 1, and let B , C , and P map onto B , C ,

and P respectively. Now, from inversion properties of angles,

0 0 0 0

\AC B = \AB C = B , and AC P = \AP C = B + . Hence,

0 0 0 0 0 0 0 0 0 0

\B C P = \P C A \B C A = . Similarly, \C B P = . There-

0 0 0 0 0 0 0

fore, triangle B C P is isosceles with B P = C P .

From inversion properties of line segments,

AB 1 BP

0 0

(=) = , B P =

0 0

AB AP BP AP B P

and

AC 1 CP

0 0

(=) = . C P =

0 0

AC AP CP AP C P

Thus,

AC AB

= ,

BP CP

and so AP , BD, and CE areconcurrent.

230

J.I.R. McKnight Problems Contest 1996

1. (a) Prove log x + log x +log x +log x + :: : =2log x.

2 4 16 256 2

1ab

a b

2(1b)

. (b) If 60 =3and 60 =5, nd 12

2. (a) Given that f (x +1) f (x)= 4x +5 and f (0) = 6, nd f (x).

(b) Given g (x=2) 3g (2=x)= 16x, nd g (x).

3. A regular 18-sided polygon is inscribed in a circle, and triangles are

formed by joining any 3 of the 18 vertices.

(a) How many triangles are there?

(b) How many right angled triangles are there?

(c) How many obtuse triangles are there?

(d) How many acute triangles are there?

4. Let P be the centre of the square

constructed on the hypotenuse AC

of the right-angled triangle AB C .

Prove that BP bisects angle AB C .

5. If three vertices of a cube in 3-dimensional space havecoordinates

(4; 6; 3), (0; 9; 3) and (3; 5; 8), nd the coordinates of the other ve

vertices.

6. An oil drum is lying on its side. It

has a radius of 50 cm and a length of

150 cm, and oil is leaking from it at

arateof45 litres per minute. How

fast is the oil level changing, when

the oil is still 25 cm deep? (Note:

1 litre=1000 cubic cm.)

6 3 4

7. Prove that sin cos 3 =4 .

8. Prove that

(a) cos A sin B = [sin(A + B ) sin(A B )]=2.

(b)

cos(n +1) sin n n

cos r = + .

2sin 2

r =1

231

Constructive Geometry, Part I

Cyrus C. Hsia

student, UniversityofToronto

Introduction

Geometric constructions haveplayed an important role in the devel-

opment of classical mathematics. Euclid used various constructions in his

proofs in the Elements. In retrospect, a construction may seem straightfor-

ward and trivial, but coming up with the construction in the rst placecan be

rather dicult. In this part, we will give the rules for construction in Euclid-

ean geometry and begin the journey intoconstructing some of the most basic

gures and shapes. In the second and third parts of this article, we will delve

into some olympiad problems and some extensions.

Tools

Thereare two basic fundamental tools used by the Greeks in their con-

structions of geometric gures. The limitations of these two objectsgive rise

to many obstacles and creative solutions in many problems. So, for any given

problem in Euclidean geometry that we will discuss lateron,weare allowed

to use only the following:

1. A Straight Edge

This is simply any straight edge with no markings on it to determine

length. What can you do with this? For any two points in the plane,

you may draw a straight line through them.

2. A Collapsible Compass

This compass consists of two arms joined together at one of their ends.

When the compass is lifted from the paper, the two arms collapse to-

gether. What can you do with this? For any two points in the plane,

you may draw a circular arc(orcompletecircle) with one arm pivoted

at one of the points, allowing the compass torotate about this point.

Strategy

Here is some simple advice that the author suggests following in solving

a geometric construction problem:

1. Work the problem backwards.

If you are asked toconstruct a point, then assume the point existsand

see how itsproperties arerelated to the given information.

232

2. Give the construction.

Onceyou have the solution, give the construction step-by-step accord-

ing to the rules of Euclidean geometry.

3. Prove that your construction does satisfy the given problem.

Show that the construction is indeed the desired one, and that all the

conditions of the problem are satis ed.

Basic Constructions

The best way to learn how tosolve these types of problems is todo

a lot of them. So, let us get right intosomeconstruction problems. We

will give the solutions to a few of them and leave the remainder for the

reader tocomplete. Also, a later problem can be solved morereadily by

using previously known constructions.

Problem 1A. Construct the mid-point of a given line segment.

Solution. First, attempt to nd a solution on your own with only a

straight-edge and compass. Onceyou have the solution, compare it with the

construction given here.

r r

A B

(a) Label the end-points A and B as shown in the diagram.

(b) Draw a circular arccentred at A with radius AB as shown.

(c) Draw a circular arccentred at B with radius AB as shown.

(d) Let the two intersection pointsbeC and D .

(e) Draw a straight line through C and D .

(f) The intersection of the two lines, E , is the required mid-point.

233

Now we show that E is indeed the mid-point. To see this, note that

it is sucient to show that AB C D isaparallelogram; it is well known that

the diagonals of a parallelogram bisect each other. One way of showing that

AB C D is a parallelogram is to show that the opposite sides are equal in

length. Now, byconstruction, AC = AB = BD,since these lengths are the

radii of the two circles. Similarly, AD = BC . Thus, AB C D is a parallelo-

gram and the diagonals bisect each other at E .

Problem 1A { Generalization. Divide a given line segment into n equal

line segments, n 1.

Problem 1B. Construct the perpendicular bisector of a given line seg-

ment.

Solution. Leftasanexercise for the reader.

Problem 1C. Construct a right-angled triangle with hypotenuse on a

given line segment.

r r r

A O B

Solution.

(a) Take two points, A and B , on the line and construct the mid-point O .

(b) Construct the circle centred at O with radius OA.

(c) Pick any point, C , on the circle not at A or B .

(d) Triangle AB C is a right triangle with the right angle at vertex C .

The proof that this is the required triangle is left tothereader.

Problem 2. Construct a line through a point parallel toagiven line.

: s

234

Problem 3. Construct a line through a point perpendicular toagiven

line.

Problem 4. Construct the centre of a circle, given the circle.

Solution. Work this problem backwards to see what properties arecom-

mon to all chords of a circle and the centre of the circle. Onceyou have played

around with this a bit, we may come up with the following construction.

(a) Draw any two distinct chords of the circle which are not parallel.

(b) Construct the perpendicular bisectors of both chords.

(c) The intersection of these perpendicular bisectors is the centre of the

given circle.

The proof that this gives the centre of the circle is left to the reader.

Problem 5. Construct the tangent from a given point outside a given

circle to that circle.

235

Problem 6. Construct an equilateral triangle.

Problem 7. Construct the angle bisector of a given angle. That is, draw

a line that divides an angle into two equal halves as shown.

Angle bisector

Exercises

All of the following constructions should be performed using straight

edge and compass only.

1. Giveconstructions for the problems above whose solutions arenotpro-

vided.

2. Construct two circles tangent to each other.

3. Construct the common tangentsto these circles.

4. Construct the common tangents for two non-intersecting circles. A221,

[1998 : 411, 1999 : 489]

5. Construct a 15 angle. How many ways can you do this?

Cyrus C. Hsia

21 Van Allen Road

Scarborough

Ontario

M1G 1C3

[email protected]

236

PROBLEMS

Problem proposals and solutions should be sent toBruce Shawyer, Department

of MathematicsandStatistics, Memorial University of Newfoundland, St. John's,

Newfoundland, Canada. A1C 5S7. Proposals should be accompanied by a solution,

together with references and other insights which arelikely tobeofhelptotheedi-

tor. When a submission is submitted without a solution, the proposer must include

sucient information on why a solution is likely. An asterisk (?) after a number

indicates that a problem was submitted without a solution.

In particular, original problems are solicited. However, other interesting prob-

lems may also be acceptable provided that theyare not too well known, and refer-

ences are given as to their provenance. Ordinarily, if the originator of a problem can

be located, it should not be submitted without the originator's permission.

To facilitate their consideration, please send your proposals and solutions

on signed and separatestandard 8 "11"orA4 sheets of paper. These may be

typewritten or neatly hand-written, and should be mailed to the Editor-in-Chief,

to arrivenolater than 1 December 2000. They may also be sent by email to

[email protected]. (It would be appreciated if email proposals and solu-

tions were written in LT X).Graphics les should be in epic format, or encapsulated

postscript. Solutions received after the abovedatewillalsobeconsidered if there

is sucient time before the date of publication. Please note that we do not accept

submissions sent byFAX.

For the information of proposers who submitted problems in 1998, we

have either used them, or will not be using them.

2539. Proposed by Ho-joo Lee, student, Kwangwoon University,

Kangwon-Do, South Korea.

Let AB C D be a given convex quadrilateral. Let X and Y be pointson

the line segments BC and AD respectively, and P be an arbitrary point

on XY and inside AB C D . Suppose that \AP B < \PXB, and that

\DP C < \PXC.Prove that AD BC 4PX PY .

2540. Proposed byPaul Yiu, Florida Atlantic University, BocaRa-

ton, FL, USA.

Given an equilateral triangle AB C , let P and Z be points on the incircle

such that P is the mid-point of AZ and BZ < C Z . The segment CZ and

the extension of BZ meet the incircle again at X and Y respectively. Show

that:

1. triangle XY Z is equilateral;

2. the points A, X and Y arecollinear; and

3. each of the segments XA, YB and ZC is divided in the golden ratio

by the incircle.

237

2541. Proposed byEdwardT.H. Wang, Wilfrid Laurier University,

Waterloo, Ontario.

Show that, for all natural numbers n and k 2,

 

n2i 2

k k 4

is divisible by 2 .

2542. Proposed by Hassan A. Shah Ali, Tehran, Iran.

Suppose that k is a natural number and 0, i =1, ::: , n,and

= .Prove that

n+1 1

1 0

X X

kj j 1

A @

i i+1

1in

1j k

Determine the necessary and sucient conditions for equality.

2543. Proposed by Ho-joo Lee, student, Kwangwoon University,

Kangwon-Do, South Korea.

In quadrilateral AB C D ,wehave \AB D = \AD B = \BDC =40 ,

\DB C =10 and AC and BD meet at P . Show that BP = AP + PD.

2544. Proposed byAram Tangboondouangjit, Carnegie Mellon Uni-

versity, Pittsburgh, PA, USA.

For any triangle AB C , ndtheexact value of

cos A +cosB

1+cosA +cosB cos C

cyclic

2545. Proposed byAram Tangboondouangjit, Carnegie Mellon Uni-

versity, Pittsburgh, PA, USA.

For any triangle AB C ,prove that

sin A

=1.

2 2 2

sin A ( cos A + cos B + cos C )+sinB cos (A B ) + sin C cos(A C )

cyclic

2546. Proposed byAram Tangboondouangjit, Carnegie Mellon Uni-

versity, Pittsburgh, PA, USA.

Prove that triangle AB C is equilateral if and only if

4 4 4

a + b + c

a cos(B C )+ b cos(C A)+ c cos(A B )= . abc

238

2547. Proposed byAram Tangboondouangjit, Carnegie Mellon Uni-

versity, Pittsburgh, PA, USA.

 2 4

In triangle AB C with angles , and , and area ,prove that

7 7 7

2 2 2

a + b + c

7 . =4

2548 . Proposed by Clark Kimberling, UniversityofEvansville,

Evansville, Indiana, USA.

Let a(1)=1and, for n 2, de ne a(n)= ba(n1)=2c, if this is not in

f0, a(1), :: :, a(n 1)g,anda(n)= 3a(n 1) otherwise.

(a) Does any positive integer occur more than once in this sequence?

(b) Does every positive integer occur in this sequence?

2549 . Proposed by Mohammed Aassila, CRM, Universitede 

Montreal, Montreal, Quebec .

Is it possible to choose four points in the plane such that all the dis-

tances that theydetermine are odd integers?

2550. Proposed by Catherine Shevlin, Wallsend upon Tyne,

England.

Given are two semi-circles, C and C of di erent radii a and b, and a

a b

rectangle AB C D such that the diameters of the semi-circles lie contiguously

on the side AB as shown, and the common tangent to the semicircles passes

through the vertex D of the rectangle.

D C

A B

Find, in terms of a and b, the ratio in which the common tangent divides the

side BC .

Correction

2495 Proposed byG.Tsintsifas, Thessaloniki, Greece.

Let P be the interior isodynamic point of 4AB C ; that is,

AP BP CP

= = (a, b, c are the side lengths, BC , CA, AB ,of4AB C ).

bc ca ab

Prove that the pedal triangle of P has area F , where F is the area

2 2 2

a + b + c

of 4AB C and d = +2 3F . 2

239

SOLUTIONS

No problem is ever permanently closed. The editor is always pleased to

consider for publication new solutions or new insightsonpastproblems.

2431. [1999: 172] Proposed byJillTaylor, student, Mount Allison

University, Sackville, New Brunswick.

Let n 2 N.Prove that thereexist triangles with integer area, integer

side lengths, one side n and perimeter 4n, where n is not necessarily prime.

For a given n,are such triangles uniquely determined?

[Compareproblem 2331.]

Combination of solutions by GerryLeversha, St. Paul's School, London,

England, and Kenneth M. Wilke, Topeka, KS, USA.

Let n, b and c be the sides of the triangle. Since the perimeter n + b +

c =4n,ifb< nthen c>2n,contradicting the triangle inequality. Hence

we can take n

c arecoprime; for, if (n; b)= d>1 say, then djc since c =3n b,and

the triangle would be similar to a smaller triangle whose sides arecoprime

[and whose area is still an integer, sinceitisrational (A=d ); byHeron's

formula, any triangle with integer sides and semiperimeter and rational area

has integer area. | Ed.]

Let b = n + k for some integer k , so that c =2n k . Then byHeron's

formula, the area A of the triangle satis es

2 2

A = s(s n)(s b)(s c)=2n k (n k ) ,

where s =2n is the semiperimeter. Since (n; b)= (n; c)= (b; c)= 1there

2 2

are positive integers x and y , such that either n k = x and k =2y ,or

2 2 2 2

n k =2y and k = x , so that either way n = x +2y .

2 2

Conversely if n = x +2y for positive integers x and y ,wecan take

2 2 2

k = x toproduce the triangle n, n + x , n +2y where the perimeter is

2 2 2 2 2 2

3n + x +2y =4n and A =2n(n)(2y )(x )=(2nxy ) byHeron.

[Editorial comment.Therefore it has been proved that for a positive

integer n there is a triangle satisfying the conditions of the problem if and

2 2

only if n is of the form j (x +2y ) for positive integers j , x, y (wherewe

may as well assume that (x; y )= 1). The solution now goes on to identify

these numbersasprecisely those n having at least one prime factor of the

form 8t +1 or 8t +3.]

2 2

It is known (for example, [1], pp. 188{191) that p = x +2y for

positive integers x and y whenever p is a prime of the form 8t +1 or 8t +3.

2 2

Conversely, if n = x +2y for relatively prime x and y ,andp is a prime

2 2

factor of n,thenx +2y 0 (mo d p). So, multiplying by the inverse of y

in the group of residues [that is, the unique element z 2f1, 2, ::: , p 1g

such that yz 1 (mo d p)]weobtain (xz ) +2 0(mod p). Thus 2 is

240

a quadratic residue of p. Itisnowastandard number-theoretic result (for

example, [1], p. 139) that p 1 or 3(mod8).

If n is a product of more than one such prime, one can construct the

2 2

necessaryrepresentation of n in the form x +2y from the corresponding

representations for each prime factor by using the identity

2 2 2 2 2 2

(A +2B )(C +2D )= (AC 2BD) +2(AD BC ) ,

and the resulting triangles are not uniquely determined. For example, for

2 2 2 2

n =33we have 3= 1 +2 1 and 11 = 3 +2 1 , so that

2 2 2 2 2 2

33 = (3 2 1) +2(3 1) or 33 = 5 +2 2 =1 +2 4 ,

and the corresponding triangles are 33, 41, 58 with perimeter 132 and area

660, and 33, 34, 65 with perimeter 132 and area 264.

Reference:

[1] T. Nagell, Introduction to Number Theory, second edition, Chelsea Pub-

lishing Company, 1964.

Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; NIKOLAOS

DERGIADES, Thessaloniki, Greece; CHARLES R. DIMINNIE, Angelo StateUniversity, San An-

gelo, TX, USA; RICHARD I. HESS, Rancho Palos Verdes, CA, USA; WALTHER JANOUS, Ursuli-

nengymnasium, Innsbruck, Austria; MICHAEL LAMBROU, UniversityofCrete, Crete, Greece;

D.J. SMEENK, Zaltbommel, the Netherlands; and the proposer.

The characterization of those n for which the problem has a solution was essentially

found bymostsolvers, including the proposer, though nobody included every detail of the

argument. Some solvers seemed uncertain whether the proposer wished the problem solved

for every positiveinteger n. In fact she did not: the problem as published was the editor's

(misleading) restatement of the original proposal.

2438. [1999: 173] Proposed byPeter Hurthig, Columbia College,

Burnaby, BC.

Show how to tile an equilateral triangle with congruent pentagons.

Re ections are allowed. (Compareproblem 1988.)

Solution by Mark Lyon and Max Shkarayev, students, Universityof

Arizona, Tuscon, AZ, USA.

De ne O as the circumcentre of the triangle. If we connect the vertices

V , V , V of the triangle to O we get a partition of the triangle intocongruent

1 2 3

triangles. If we now \bend" each of the segments V O in the middle, we

will end up with a partitioning of the triangle bycongruent pentagons, as

required.

2 3

241

This solution could be extended to a partition of the equilateral triangle

into three congruent n{gons, for any integer n 3. In the case of odd n the

procedureisvery similar; we just haveto bend each V O (n 3)=2 times. In

the case of even n,we rst de ne M , M , M to be the mid-points of the

1 2 3

sides. Drawing the segments M O will partition the triangle intocongruent

quadrilaterals. We now bend the M O 's (similar totheoddcase) (n 4)=2

times to obtain a partition of the triangle intocongruent n{gons.

V V

1 1

M M

3 2

V V

2 2

3 3

n even n odd

Also solved by RICHARD I. HESS, Rancho Palos Verdes, CA, USA; MICHAEL LAMBROU,

UniversityofCrete, Crete, Greece; J. SUCK, Essen, Germany; and the proposer. All solutions

were the same, except for the proposer's, whose solution used 24 congruent pentagons:

Lambrou gave the same construction as aboveto divide the triangle into three congruent

n{gons, for any integer n 3. Suck went further to similarly divide the regular n{gon into

congruent k{gons, for any integers n, k 3.

The proposer would liketo see a solution (perhaps with more pentagons) in which the

congruent pentagons are convex. (Or a proof that it is impossible?)

2439. [1999: 238] Proposed byToshio Seimiya, Kawasaki, Japan.

Suppose that AB C D is a square with side a.LetP and Q be points

on sides BC and CD respectively, such that \PAQ =45. Let E and

F be the intersections of PQ with AB and AD respectively. Prove that

AE + AF 2 2a.

242

I Solution byNikolaos Dergiades, Thessaloniki, Greece.

A F

The circle centred at A with radius a is tangent at B , D to BC and CD

respectively. Let PQ be the tangent at H to the abovecircle. Then AP is

the bisector of \BAH = ! , AQ is the bisector of \HAD = , and hence

0 

we have \PAQ =45 because ! + =90 .From this we conclude that

Q Q. Thus

 

AH AH 1 1

AE + AF = + = a +

cos ! cos cos ! cos 

 a 2 =2 2a

! + 

cos

by Jensen's inequality since the function f (x)= 1= cos x is convexfor

1 + 2 tan x

0 0. The equality holds when

cos x

! = or \BAP =22:5 .

II Solution byEckard Specht, Otto-von-Guericke University, Magde-

burg, Germany.

Let BP = x, DQ = y , BE = u, DF = v and \PAB = . (See gure

on page 243.) Thus x = a tan . Then \QAD = =4 and

 

tan tan

1 tan a(a x) 

= a = a = . y = a tan

4 1+tan tan 1 + tan a + x

a(ax x )

Evidently 4PBE 4FDQ,souv = xy = f (x).

a + x

By the AM-GM inequality, we have

uv . (1) u + v 2

It is clear that the lower bound of the sum u + v can be found by maximizing

the product uv . Hence

2 2

a 2ax x

x = a( 2 1) , =0 for f (x)= a

(a + x)

243

A E

yielding

p p

a x

2 2

x)=ax = a ( 2 1) , 2 uv =2a( 2 1) . (2) uv = f (

a + x

The second derivativeis

2 2

(a + x ) +(a 2ax x ) 2a

f (x)= 2a < 0 , =

(a + x) a + x

which veri es the maximum. Thus we havefrom (1) and (2)

p p

AE + AF =2a +(u + v ) 2a +(2 2a 2a)= 2 2a .

III Solution by the proposer.

A F

B C

P X

We may assume that \BAP \DAQ. Let G be the point of re ec-

tion of B across AP . Then \PAG = \PAB, \AGP = \AB P =90

 

and AG = AB . As \PAQ =45 we get \BAP + \DAQ =45,so

that \GAQ = \DAQ. Since we have AG = AB = AD , we get

244

4AGQ 4AD Q. Thus \AGQ = \AD Q =90. Hence P , G, Q are

collinear, so that AG ? PQ.

Let X , Y be the intersections of AG with the lines BC , CD,respec-

tively. By the assumption that \BAP \DAQ we get

\PAG = \PAB \DAQ =45 \CAQ = \PAC ,

so that X is a point on the side BC . Since AX ? PQ,wehave

\AX P = \AE P . As \EAP = \BAP = \GAP = \XAP , we get

4AE P 4AX P , whence AE = AX . Similarly, we have

4AF Q 4AY Q, and AF = AY .

Suppose X does not coincide with the point C . Then X 6= Y .LetM

be the mid-point of XY .As\XCY =90 , we have XM = CM = MY .

Hence we have \MCX = \MXC > \AC X =45. Thus \AC M =

2a. Since M is the \AC X + \MCX > 90 . Therefore, AM > AC =

mid-point of XY ,weget

AX + AY =2AM > 2 2a .

Since AX = AE and AF = AY , we have

AE + AF 2 2a .

We include equality in the abovesince equality holds when X coincides with

C ; that is, when \BAP = \DAQ.

 

Also solved by SEFKET ARSLANAGIC,UniversityofSarajevo, Sarajevo, Bosnia and

Herzegovina; MICHEL BATAILLE, Rouen, France; FRANCISCO BELLOT ROSADO, I.B. Emilio

Ferrari, Valladolid, Spain; PAUL BRACKEN, CRM, UniversitedeMontr eal, Montreal, Quebec; 

CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; IAN JUNE L. GARCES, and RICHARD

B. EDEN, student, Ateneo de Manila University, The Philippines; RICHARD I. HESS, Ran-

cho Palos Verdes, CA, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria;

MASOUD KAMGARPOUR, student, Carson Graham Secondary School, North Vancouver, BC;

 

MURRAY S. KLAMKIN, University of Alberta, Edmonton, Alberta; VACLAVKONECNY, Fer-

ris StateUniversity, Big Rapids, MI, USA; MICHAEL LAMBROU, UniversityofCrete, Crete,

Greece; KEE-WAI LAU, Hong Kong; R. LAUMEN, Antwerp, Belgium; HO-JOO LEE, stu-

dent, Kwangwoon University, Kangwon-Do, South Korea; GERRY LEVERSHA, St. Paul's

School, London, England; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; HEINZ-

J URGEN SEIFFERT, Berlin, Germany; D.J. SMEENK, Zaltbommel, the Netherlands; and

PARAGIOU THEOKLITOS, Limassol, Cyprus, Greece.

2440. [1999: 238] Proposed byToshio Seimiya, Kawasaki, Japan.

Given: triangle AB C with \BAC =90 . The incircle of triangle AB C

touches BC at D .LetE and F be the feet of the perpendicularsfrom D to

AB and AC respectively. Let H be the foot of the perpendicular from A to

BC .

Prove that the area of the rectangle AE D F is equal to . 2

245

I. Solution by Dimitar Mitkov Kunchev, student, Baba Tonka School of

Mathematics, Rousse, Bulgaria.

Let AB = c, AC = b, and BC = a. Then if s is the semiperimeter of

4AB C we have

a + b c a + c b

CD = s c = and DB = s b = .

2 2

Since 4EBD and 4AB C are similar, we have

BD a + c b b ED

= ==) ED = ,

AC BC 2 a

and using 4FDC 4AB C , we similarly obtain:

a + b c c

FD = .

2 a

If we denoteby S the area of the rectangle AE D F , then

2 2

(a + c b)(a + b c)bc a (c b)

S = ED FD = = bc

2 2

4a 4a

 

2 2 2 2 2 2

2b c 1 bc AH a c +2bc b

bc = = = . =

2 2

4a 4a 2 a 2

We used the Theorem of Pythagoras and the formula for the altitude in a

right triangle: AH =(bc=a).

r 2

45 B

A B

II. Solution byVaclav Konecn y, Ferris State University, Big Rapids, MI,

USA.

Let r be the inradius of triangle AB C .From the gure:

2 2 2 2

(AH ) (r + r 2 cos(45 B )) r (1 + cos B + sin B )

= =

2 2 2

(1 + cos B +sin B + 2 cos B +2sinB +2cosB sin B ) =

= r (1 + cos B + sin B +cosB sin B )

= (r + r cos B )(r + r sin B )=ED AE .

246

Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Spain; SEFKET

ARSLANAGIC,UniversityofSarajevo, Sarajevo, Bosnia and Herzegovina; ATENEO PROBLEM-

SOLVING GROUP,Ateneo de Manila University, The Philippines; SAM BAETHGE, Nordheim,

TX, USA; MICHEL BATAILLE, Rouen, France; CHRISTOPHER J. BRADLEY, Clifton College,

Bristol, UK; NIKOLAOS DERGIADES, Thessaloniki, Greece; RICHARD I. HESS, Rancho Pa-

los Verdes, CA, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MASOUD

KAMGARPOUR, student, Carson Graham Secondary School, North Vancouver, BC; MICHAEL

LAMBROU, UniversityofCrete, Crete, Greece; R. LAUMEN, Antwerp, Belgium; HO-JOO LEE,

student, Kwangwoon University, Kangwon-Do, South Korea; GERRY LEVERSHA, St. Paul's

School, London, England; HEINZ-J URGEN SEIFFERT, Berlin, Germany; ANDREI SIMION, stu-

dent, Brooklyn Technical High School, New York; D.J. SMEENK, Zaltbommel, the Netherlands;

PARAGIOU THEOKLITOS, Limassol, Cyprus, Greece; KENNETH M. WILKE, Topeka, KS, USA;

PETER Y. WOO, Biola University, La Mirada, CA, USA; APLAKIDES YIANNIS, Veria, Greece;

JEREMY YOUNG, student, Nottingham High School, Nottingham, UK; and the proposer.

2441. [1999: 239] Proposed byPaul Yiu, Florida Atlantic University,

BocaRaton, FL, USA.

Suppose that D , E , F are the mid-points of the sides BC , CA, AB

of 4AB C . The incircle of 4AE F touches EF at X , the incircle of 4BF D

touches FD at Y , and the incircle of 4CDE touches DE at Z .

Show that DX , EY , FZ areconcurrent. What is the intersection

point?

Solution byNikolaos Dergiades, Thessaloniki, Greece.

We shall prove that DX , EY , and FZ all pass through the incentre

of 4AB C . Because it is clear that DX passes through the incentre when

AB = AC , assume that AB > AC and let L denote the point where the

incircle of 4AB C touches BC . [Note that the dilatation with centre A and

ratio 2 takes the incircle of 4AF E and itscontact point X to the incircle of

4AB C and the point L.] Then X lies on AL and AX = XL.LetAH be

the bisector of \A with H on BC , and call P the point where AH meets

DX .

Claim: P is the incentre.

Toprove the claim, let s be the semiperimeter of 4AB C ; then

BL = s AC , LC = s AB ,

2DL = AB AC , and 2DH = BH HC .

HP AX LD

Menelaus's theorem for D , P , X on 4AH L gives =1,

PA XL DH

HP 2HD BH HC

= =

PA 2DL AB AC

BH BH

= , =

AB AB

1 AB

247

where the last equality holds because AH is the angle bisector (so that

HC : BH = AC : AB ). From this we conclude that in 4AB H , BP is the

bisector of \B , and therefore, P is the incentreof4AB C . In the same way

we prove that EY and FZ also pass through the incentre, as promised.

Also solved by MICHEL BATAILLE, Rouen, France; FRANCISCO BELLOT ROSADO,

 

I.B. Emilio Ferrari, and MAR IA ASCENSION LOPEZ CHAMORRO, I.B. Leopoldo Cano,

Valladolid, Spain (2 solutions); CHRISTOPHER J. BRADLEY, Clifton College, Bris-

tol, UK; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MASOUD

KAMGARPOUR, student, Carson Graham Secondary School, North Vancouver, BC

(only the concurrencepart); MITKO KUNCHEV, Baba Tonka School of Mathematics,

Rousse, Bulgaria; MICHAEL LAMBROU, UniversityofCrete, Crete, Greece (2 solutions);

GERRY LEVERSHA, St. Paul's School, London, England; HEINZ-J URGEN SEIFFERT,

Berlin, Germany; TOSHIO SEIMIYA, Kawasaki, Japan; ANDREI SIMION, New York,

NY, USA; ECKARD SPECHT, Magdeburg, Germany; D.J. SMEENK, Zaltbommel, the

Netherlands; PARAGIOU THEOKLITOS, Limassol, Cyprus, Greece; PETER Y. WOO,

Biola University, La Mirada, CA, USA; JEREMY YOUNG, student, Nottingham High School,

Nottingham, UK; and the proposer.

The Nagel point of a triangle is de ned to be the intersection point of the lines from

avertexto the point of the opposite side where that side touches itsexcircle. Many readers

provided accessible references | too many to list | and showed how the solution to 2441

follows easily from familiar properties of the Nagel point. In particular, the intersection point

of our problem is the Nagel point of 4DE F (because X , Y ,andZ areprecisely the points

where the excircles of 4DE F touch the sides).Astandardtheorem states that the incentreof

a triangle is the Nagel point of its mid-point triangle. Other approaches treat the Nagel point

as the isotomic conjugate of the Gergonne point, which is almost easier toprove than to say!

The storycan be found in the many references.

2442. [1999: 239] Proposed by Michael Lambrou, Universityof

Crete, Crete, Greece.

1 1 1 1

Let fa g , fx g , fy g , ::: , fz g ,bea nite number of given

n n n n

1 1 1 1

sequences of non-negative numbers, wherealla > 0. Suppose that a

n n

P P P

is divergent and all the other in nite series, x , y , :: :, z ,are

n n n

convergent. Let A = a .

n k

k=1

(a) Show that, for every > 0, thereisann 2 N such that, simultaneously,

A x A y A z

n n n n n n

0 <, 0 <, :: : , 0 <.

a a a

n n n

A x

n n

exists, it must havevalue zero. (b) From part (a), it is clear that if lim

n!1

Construct an example of sequences as above such that the stated limit does

not exist.

Solution by Achilleas Sinefakopoulos, student, University of Athens,

Greece.

A x

n n

(a) Let > 0 be given and let X = fn 2 N :0 <g, Y = fn 2 N :

A y

A z

n n

<g, ::: , Z = fn 2 N :0 <g. 0 

a a

n n

Suppose to the contrary that X \ Y \ ::: \ Z = ;. Now, if n 2 N,

then n 62 X or n 62 Y , :::,orn 62 Z . Since x 0, y 0, ::: , z 0,in

n n n

248

any case, we have

a x + y + + z

n n n n

. 0 < 

A 

x + y + + z

n n n

It is clear that the series converges. This, together

with use of the comparison test, contradicts a theorem of Abel, which implies

that the series (a =A ) diverges. Thus, X \ Y \ ::: \ Z 6= ;, and the

n n

proof of (a) is complete.

1 1

if n is a perfect square, and x = (b) For all positive n 2 N,setx =

n n

if n is not a perfect square.

P P

Also, let a =1for all n 2 N. Then a is divergent, x is

n n n

, but convergent because its partial sums are bounded aboveby 2

n=1

A x A x

n n n n

does not exist because lim inf = lim inf nx =0and lim

a a

n!1 n!1 n!1

n n

A x

n n

= lim sup nx =1.Weare done. lim sup

n!1 n!1

Also solved by RICHARD I. HESS, Rancho Palos Verdes, CA, USA (part (a) only); and the

proposer.

2443. [1999: 239] Proposed by Michael Lambrou, Universityof

Crete, Crete, Greece.

Without the use of any calculating device, nd an explicit example of an

integer, M , such that sin(M ) > sin(33)( 0:99991). (Of course, M and 33

areinradians.)

Solution byVaclav Konecn y, Ferris State University, Big Rapids, MI,

USA.

M = 11 will do. Recall that sin(3A)= 3sin(A) 4sin (A). Thus,

sin(33) sin(11) = 4 sin(11) 4 sin (11) = 4 sin(11) cos (11) < 0,since

< 11 < 4 .] Thus sin(11) < 0, the angle 11 is in the 4 quadrant. [Ed:

and hence, sin(11) > sin(33).

Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; NIKOLAOS

DERGIADES, Thessaloniki, Greece; CHARLES R. DIMINNIE and TREY SMITH (jointly), Angelo

StateUniversity, San Angelo, TX, USA; SHAWN GODIN, Cairine Wilson S.S., Orleans, Ontario;

RICHARD I. HESS, Rancho Palos Verdes, CA, USA; WALTHER JANOUS, Ursulinengymnasium,

Innsbruck, Austria; MICHAEL JOSEPHY, Universidad de Costa Rica, San Jose, CostaRica;

KEE-WAI LAU, Hong Kong, China; GERRY LEVERSHA, St. Paul's School, London, England;

DIGBY SMITH, Mount Royal College, Calgary, Alberta; ECKARD SPECHT, Otto-von-Guericke

University, Magdeburg, Germany; DAVID R. STONE, Georgia Southern University, Statesboro,

GA, USA; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer.

The answer M = 11 was also given by Diminnie and T. Smith, Josephy, and D. Smith.

Other answers included M = 322, 366, 699,and156522 (each given by two solvers);

M = 677, 1032 (each given byonesolver).Stone gavea36{digit \monster" for his M ,

while the proposer's answer is another \monster": M = b10 c +40.Konecn yc ommented

that thereare an in nite number of solutions and stated that for any given N , the number of

249

6621

N ,or0:004226N ;forexample, solutions M satisfying 0

for N =10; 000,weexpect 42 positive solutions, and he provided a computer print-out which

actually lists all 42 solutions in the range 0

answers M =322and 366 given by some solversare indeed the smallest positive solutions.

Using a computer, he also found that the next solution less than 11 is M = 344.

Both Godin and Josephy commented that, byexamining the continued fraction expansion

of ,onecan get an in nite sequence of solutions M such that sin(M ) tends to 1.

2444. [1999: 239] Proposed by Michael Lambrou, Universityof

Crete, Crete, Greece.

0 0 11

n n

X X

1 ln(n!) 1

@ @ AA

Determine lim ln(k ) .

n!1

n n j

k=1 j =k

Solution by Heinz-Jur gen Sei ert, Berlin, Germany.

Based on the identity ln(k ) = ln(n!), a simple induction argu-

k=1

ment yields

1 0

n n n

X X X

ln(k !) 1

A @

= . ln(k )

j k

k=1 k=1 j =k

It is easily veri ed that

0 1

 

n n n

X X X

1 ln(n!) 1 ln(k !) 1

@ A

= ln(k ) ln(k ) . (1)

n n j n k

k=1 k=1 j =k

Stirling's Formula gives

 

ln(n!)

lim ln(n) =1.

n!1

Hence, by a well-known result due to Cauchy, the right hand expression of

(1) tends to 1 as n tends to in nity. From (1), it now follows that the required

limit exists and equals 1.

 

Also solved by SEFKET ARSLANAGIC, UniversityofSarajevo, Sarajevo, Bosnia and

Herzegovina; MICHEL BATAILLE, Rouen, France; PAUL BRACKEN, CRM, Universitede 

Montreal, Montreal, Quebec; RICHARD I. HESS, Rancho Palos Verdes, CA, USA; WALTHER

JANOUS, Ursulinengymnasium, Innsbruck, Austria; KEE-WAI LAU, Hong Kong, China;

ACHILLEAS SINEFAKOPOULOS, student, University of Athens, Greece; and the proposer.

The Editor was surprised at how many di erentwaysthisproblem could be solved!

250

2445. [1999: 240] Proposed by Michael Lambrou, Universityof

Crete, Crete, Greece.

Let A, B be a partition of the set C = fq 2 Q :0

A, B are disjoint sets whose union is C ).

Show that thereexist sequences fa g, fb g of elementsofA and B

n n

respectively such that (a b ) ! 0 as n !1.

n n

Solution byJeremy Young, student, Nottingham High School, Notting-

ham, UK.

We assume that elements may recur in a sequence, since this is neces-

sary if one of A, B is of nitesize.

The question assumes A, B are nonempty. Thereforeleta , b be

1 1

any elementsofA, B respectively, and construct the sequences fa g, fb g

n n

recursively. Given a , b ,consider (a + b )=2:

i i i i

 if (a + b )=2 2 A, let a =(a + b )=2 and b = b ;

i i i+1 i i i+1 i

 if (a + b )=2 2 B ,let a = a and b =(a + b )=2.

i i i+1 i i+1 i i

Then ja b j = ja b j. By induction,

i+1 i+1 i i

1 1

ja b j = ja b j < ,

n+1 n+1 1 1

n n

2 2

which goes to 0 as n !1.

Also solved by MICHEL BATAILLE, Rouen, France; NIKOLAOS DERGIADES, Thes-

saloniki, Greece; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MICHAEL

JOSEPHY, Universidad de Costa Rica, San Jose, CostaRica; MICHAEL PARMENTER, Memorial

University of Newfoundland, St. John's, Newfoundland; ACHILLEAS SINEFAKOPOULOS, stu-

dent, University of Athens, Greece; TREY SMITH and CHARLES R. DIMINNIE, Angelo State

University, San Angelo, TX, USA; PETER Y. WOO, Biola University, La Mirada, CA, USA; and

the proposer.

Josephy and Parmenter gave the same solution as Young.

2446. [1999: 240, 306] Proposed by Catherine Shevlin, Wallsend

upon Tyne, England.

A sequence of integers, fa g with a > 0, is de ned by

n 1

if a 0 (mo d 4),

3a +1 if a 1 (mo d 4),

n n

a =

n+1

2a 1 if a 2 (mo d 4),

n n

a +1

if a 3 (mo d 4).

Prove that thereisaninteger m such that a =1.

(Compare OQ.117 in OCTOGON,vol 5, No. 2, p. 108.)

251

Solution byNikolaos Dergiades, Thessaloniki, Greece, slightly modi ed

by the editor.

The sequence fa g is clearly bounded below, sincenoterm can be neg-

ativeorzero. Suppose that there is no natural number m for which a =1.

We will deriveacontradiction by showing that for everyterm a of the

sequence, thereisaterm a

m+j m

observations:

If a 0(mod4),thena =4k for some k 6=0and it follows that

m m

a =2k< a .

m+1 m

If a 3(mod4), then a =4k +3 for some k 0, so that we have

m m

a = k +1

m+1 m

If a 2(mod4), then a =4k +2 for some k 0, so that we have

m m

a =8k +3, giving a =2k +1

m+1 m+2 m

If a 1(mod4), then a =4k +1 for some k>0, so that we have

m m

a =12k +4 giving a =6k +2. Hence, either a =3k +1

m+1 m+2 m+3 m

if a 0 (mo d 4),or a =12k +3 if a 2 (mo d 4). In the

m+2 m+3 m+2

latter case, we have a =3k +1

m+4 m

 

Also solved by SEFKET ARSLANAGIC,UniversityofSarajevo, Sarajevo, Bosnia and

Herzegovina; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; CON AMORE PROB-

LEM GROUP,Royal Danish School of Educational Studies, Copenhagen, Denmark; RICHARD

I. HESS, Rancho Palos Verdes, CA, USA; MICHAEL LAMBROU, UniversityofCrete, Crete,

Greece; GERRY LEVERSHA, St. Paul's School, London, England; KATHLEEN E. LEWIS, SUNY

Oswego, Oswego, NY, USA; MICHAEL PARMENTER, Memorial University of Newfound-

land, St. John's, Newfoundland; HARRY SEDINGER, St. BonaventureUniversity, St. Bon-

aventure, NY, USA; SKIDMORE COLLEGE PROBLEM GROUP,Saratoga Springs, NY, USA:

DAVID R. STONE, Georgia Southern University, Statesboro, GA, USA; JEREMY YOUNG, stu-

dent, Nottingham High School, Nottingham, UK; ROGER ZARNOWSKI, Angelo State University,

San Angelo, TX, USA; and the proposer. There was one partially incorrect solution.

Most of the submitted solutions are similar totheonegiven. The Con AmoreProblem

Group also submitted a proof for the \incorrect" version [1999: 240].

2447. [1999: 240] Proposed by GerryLeversha, St. Paul's School,

London, England.

Two circles intersect at P and Q.Avariable line through P meets the

circles again at A and B . Find the locus of the orthocentre of triangle AB Q.

Solution by John G. Heuver, Grande Prairie Composite High School,

Grande Prairie, Alberta.

Let O and O be the centres of the two circles.

1 2

252

Join Q to A, B , O and O .Triangle QO O is similar to triangle QAB

1 2 1 2

1 1

\QO P = \QAB and \QO O = \QO P = (because \QO O =

1 2 1 2 1 2

2 2

\QB A). Consequently, triangle QAB is obtained from triangle QO O by

1 2

In triangle AB Q,draw the altitudes from A and B , which intersect

in H and meet QA and QB at E and D respectively. Consider triangle

0 0

EQH , and let triangle E QH be its equivalent in triangle QO O .Wehave

1 2

\EQH = \BAQ, QE = QB cos(\AQB ), QH =2R cos(\AQB ),

where R is the circumradius of triangle AB Q.

It follows that triangle EQH is obtained by spiral similarityfrom trian-

0 0

gle E QH with centre Q,arotation by angle and similaritycoecient k .

The locus of H is then a circle passing through Q with the exclusion of

the point Q itself.

Also solved by MICHEL BATAILLE, Rouen, France; NIELS BEJLEGAARD, Stavanger,

Norway; NIKOLAOS DERGIADES, Thessaloniki, Greece; WALTHER JANOUS, Ursulinengym-

 

nasium, Innsbruck, Austria; VACLAVKONECNY, Ferris StateUniversity, Big Rapids, MI,

USA; MITKO KUNCHEV, Baba Tonka School of Mathematics, Rousse, Bulgaria; MICHAEL

LAMBROU, UniversityofCrete, Crete, Greece; TOSHIO SEIMIYA, Kawasaki, Japan;

D.J. SMEENK, Zaltbommel, the Netherlands; PARAGIOU THEOKLITOS, Limassol, Cyprus,

Greece; PANOS E. TSAOUSSOGLOU, Athens, Greece; PETER Y. WOO, Biola University, La Mi-

rada, CA, USA; and the proposer. Some solversdidnotexplicity mention the exclusion of the

point Q, but the editor has been generous.

Janous and Konecn ybothc alculated the radius of the locus circle as

2 2 2

d(a + b d )

(a + b + d)(a + b d)(b a + d)(a b + d)

where a and b aretheradii of the given circles [containing A and B respectively],and

d = a cos A + b cos B .

253

2448. [1999: 240] Proposed by GerryLeversha, St. Paul's School,

London, England.

Suppose that S is a circle, centre O , and P is a point outside S . The

tangentsfrom P to S meet the circle at A and B . Through any point Q on S ,

the line perpendicular to PQ intersects OA at T and OB at U .Prove that

OT OU = OP .

Solution byPeter Y. Woo, Biola University, La Mirada, CA, USA.

First we note that points T and U exist only if Q 6= A and Q 6= B .Let

us consider the case when Q is on the major arc AB . The solution is similar

when Q is on the minor arc AB .

We prove that 4OP U is similar to 4OT P . Since PA and PB are

tangentsto the circle S , \POA = \POB.Also,\AO U = \BOT , so that

\UOP = \POT . (1)

Now, AP T Q is a cyclic quadrilateral, so

\OT P = \AQP = \AQB \BQP .

Since BP U Q isacyclic quadrilateral, \BQP = \BU P . Also,

\AQB = \AO B = \BOP .

Hence

\OT P = \BOP \BU P = \OP U .

Thus

\OT P = \OP U . (2)

From (1) and (2), 4OP U is similar to 4OT P . Therefore,

OP OT

= ,

OP OU

which completes the proof.

 

Also solved by SEFKET ARSLANAGIC,UniversityofSarajevo, Sarajevo, Bosnia and

Herzegovina; MICHEL BATAILLE, Rouen, France; FRANCISCO BELLOT ROSADO, I.B.

Emilio Ferrari, Valladolid, Spain; NIKOLAOS DERGIADES, Thessaloniki, Greece; RICHARD

I. HESS, Rancho Palos Verdes, CA, USA; WALTHER JANOUS, Ursulinengymnasium, Inns-

 

bruck, Austria; VACLAVKONECNY, Ferris StateUniversity, Big Rapids, MI, USA; MICHAEL

LAMBROU, UniversityofCrete, Crete, Greece (two solutions); MICHAEL PARMENTER,

Memorial University of Newfoundland, St. John's, Newfoundland; TOSHIO SEIMIYA, Kawa-

saki, Japan; D.J. SMEENK, Zaltbommel, the Netherlands; JEREMY YOUNG, student, Notting-

ham High School, Nottingham, UK; and the proposer.

254

2449. [1999: 240] Proposed by GerryLeversha, St. Paul's School,

London, England.

Two circles intersect at D and E . Theyaretangent to the sides AB and

AC of 4AB C at B and C respectively. If D is the mid-point of BC ,prove

that DA DE = DC .

Solution byNikolaos Dergiades, Thessaloniki, Greece.

Since \BED = \AB C and \DE C = \AC B , then

\BAC + \BEC =180 .

Therefore, the points A; B ; E and C lie on a circle. Let AD meet this circle

at the point A . We have

0 0

\A CB = \A EB = \AB C

and

0 0

\A BC = \A EC = \AC B ,

0 0

so that the triangles AB C and A BC are equal and then DA = DA. By the

0 2

Intersecting Chords Theorem, DA DE = DB DC ,or, DA DE = DC ,

as desired.

 

Also solved by SEFKET ARSLANAGIC,UniversityofSarajevo, Sarajevo, Bosnia and

Herzegovina; MICHEL BATAILLE, Rouen, France; RICHARD I. HESS, Rancho Palos Verdes,

 

CA, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; VACLAVKONECNY,

Ferris State University, Big Rapids, MI, USA; MICHAEL LAMBROU, UniversityofCrete, Crete,

Greece; JOHN G. HEUVER, Grande Prairie Composite High School, Grande Prairie, Alberta;

TOSHIO SEIMIYA, Kawasaki, Japan; D.J. SMEENK, Zaltbommel, the Netherlands; JEREMY

YOUNG, student, Nottingham High School, Nottingham, UK; PETER Y. WOO, Biola Univer-

sity, La Mirada, CA, USA; and the proposer.

Most of the solvers found the same or quite similar solution. Therewerealsoseveral

solutions by analytic geometryandonebyinversion. Woo proved that DB bisects the angle

AD E . This implies 4AD C and 4CDE are similar, and the claim follows immediately.

2450. [1999: 240] Proposed by GerryLeversha, St. Paul's School,

London, England.

(2k )!

(k !)

k=0

Find the exact value of .

(k !)

k=0

I. Solution by Michel Bataille, Rouen, France.

Clearly, both of the given series have positiveterms and areconvergent

(byratio test).

(2n)!

and A = u . Using the well-known result Let u =

n n

(n!)

n=0

 

n 2n

, =

k n

k=0

255

we have

 

n n

X X

1 1 1 2n n n!

= = u = ,

2 2

n! n n! k (k !) ((n k )!)

k=0 k=0

and so

 

1 1 1 1

X X X X

n! 1 n! 1

= A =

2 2 2 2

(k !) ((n k )!) (k !) ((n k )!)

n=0

k=0 n=k k=0

byrearrangement.

1 1

X X

n! 1

Let us compute a = . We have a = = e,

k 0

((n k )!) n!

n=0

n=k

n(n 1) (n k +1)

(k)

and for k 1, a = = f (1), where f

(n k )!

n=k

denotes the k derivative of the function f given by

1 1

n+k n

X X

x x

k x

f (x)= x e = = .

n! (n k )!

n=0

n=k

Now, by Leibniz's formula, we have

 

k k

X X

k k k !

k (p) x (kp) (k) kp x

(x ) (e ) = f (x) = x e

p p (k p)!

p=0 p=0

kp 2 x

x , = (k !) e

p!((k p)!)

p=0

k 1

X X

1 1

and so a = e(k !) . Substituting in A = a

k k

2 2

p!((k p)!) (k !)

p=0

k=0

 

1 1

X X

1 A

= . But this series is just the Cauchy prod- gives

e p!((k p)!)

p=0

k=0

1 1 1

X X X

1 1 1

uct of the series and ,soitsvalue is e .

2 2

(k !) k ! (k !)

k=0 k=0 k=0

[Ed: Convergence of the product series is not a problem, since both series

are absolutely convergent.]

2 2

and hence the required value is e . It follows that A = e

(k !)

k=0

256

II. Solution by the proposer.

 

x +

2 2

2 x 1=x

 e e e . The usual We make use of the identity e

series expansion yields

 

1 1 1

X X X

1 1 x 1

x + e .

k ! x k ! k !x

k=0 k=0 k=0

Equating the constant terms of both sides of the identityabove then

 

1 1

X X

1 1 2k

yields = e and hence the required value is e .

k ! k k !

k=0 k=0

[Ed: If any reader feels that this \proof" is not rigorous, the following com-

mentsby the proposer himself might help you toaccept a \compromise"

attitude:

\This argument shows a cavalier disregard for the ner points of rigorous

analysis but is surely in the spirit of Euler and Gauss, and that is good enough

for me!"]

Also solved by CON AMORE PROBLEM GROUP,Royal Danish School of Educational

Studies, Copenhagen, Denmark; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Aus-

tria; MICHAEL LAMBROU, UniversityofCrete, Crete, Greece; KEE-WAI LAU, Hong Kong,

China; HEINZ-J URGEN SEIFFERT, Berlin, Germany. Two other readersconjectured the correct

answer, but wereunableto supply a proof.

Both Janous and Lau obtained the answer as immediateconsequenceoftheknownresults

that

1 1

X X

(2k)! x

k 2x

= I (2x) and x = e I (2x),where I (x) denotes the modi-

0 0 0

2 3

(k!) (k!)

k=0 k=0

ed Bessel function of the rst kind. The answer e is obtained by setting x =1in the identities

above. Theygave the following references in which these identities can be found.

[1] A.P. Prudnikovetal.,Integrals and Series (Elementary Functions) [in Russian],Moscow

1981.

[2] Eldon R. Hansen, ATable of Series and Products,Prentice Hall Inc. Englewood Cli s, N.J.

1975.

Lambrou gaveadirect proof by using \term-by-term" di erentiation of some power

series. Sei ert's solution is based on Legendre's Doubling Formula for Beta functions and the

1 1

2cosu

e du. known identity =

(k!) 

k=0

Crux Mathematicorum

Founding Editors/Redacteurs-fondateurs: Leopold Sauve& Frederick G.B. Maskell

Editors emeriti / Redact eur-emeriti: G.W. Sands, R.E. Woodrow, Bruce L.R. Shawyer

Mathematical Mayhem

Founding Editors/ Redacteurs-fondateurs: Patrick Surry& Ravi Vakil

Editors emeriti / Redact eurs-emeriti: Philip Jong, Je Higham,

J.P.Grossman, Andre Chang, Naoki Sato, Cyrus Hsia