Angle Bisectors Bisect Arcs Below Was the Find Round of the 36Th

Volume 11, Number 2 April 2006 – May 2006

Olympiad Corner Angle Bisectors Bisect Arcs Below was the Find Round of the 36th

Austrian Math Olympiad 2005. Kin Y. Li

Part 1 (May 30, 2005) In general, angle bisectors of a triangle Solution. Let I be the incenter of ∆ABC.

Problem 1. Show that an infinite do not bisect the sides opposite the By the theorem, we have 2IR = AR + BR number of multiples of 2005 exist, in angles. However, angle bisectors > AB and similarly 2IP > BC, 2IQ > CA. which each of the 10 digits 0,1,2,…,9 always bisect the arcs opposite the Also AI + BI > AB, BI + CI > BC and occurs the same number of times, not angles on the circumcircle of the CI + AI > CA. Adding all these counting leading zeros. triangle! In math competitions, this fact inequalities together, we get is very useful for problems concerning Problem 2. For how many integer 2(AP + BQ + CR) > 2(AB + BC + CA). angle bisectors or incenters of a triangle values of a with |a| ≤ 2005 does the system of equations x2 = y + a, y2 = x + a involving the circumcircle. Recall that Example 2. (1978 IMO) In ABC, AB = have integer solutions? the incenter of a triangle is the point AC. A circle is tangent internally to the where the three angle bisectors concur. circumcircle of ABC and also to the Problem 3. We are given real numbers sides AB, AC at P, Q, respectively. a, b and c and define s as the sum s = an n n Theorem. Suppose the angle bisector of Prove that the midpoint of segment PQ + bn + cn of their n-th powers for is the center of the incircle of ∆ABC. non-negative integers n. It is known that ∠ BAC intersect the circumcircle of ∆ABC at X ≠ A. Let I be a point on the A s1 = 2, s2 = 6 and s3 = 14 hold. Show that line segment AX. Then I is the incenter 2 | sn − sn−1 ⋅ sn+1 |= 8 of ∆ABC if and only if XI = XB = XC. A holds for all integers n > 1.

Problem 4. We are given two P Q equilateral triangles ABC and PQR with I parallel sides, “one pointing up” and I “one pointing down.” The common area B C of the triangles’ interior is a hexagon. B C X Show that the lines joining opposite

corners of this hexagon are concurrent. X Solution. Let I be the midpoint of line

segment PQ and X be the intersection of (continued on page 4) Proof. Note ∠BAX =∠CAX =∠CBX. the angle bisector of ∠BAC with the arc So XB = XC. Then BC not containing A. Editors: ஻ Ի ஶ (CHEUNG Pak-Hong), Munsang College, HK I is the incenter of ∆ABC ଽ υ ࣻ (KO Tsz-Mei) By symmetry, AX is a diameter of the ∠CBI =∠ABI గ ႀ ᄸ (LEUNG Tat-Wing) ⇔ circumcircle of ∆ABC and X is the ∠IBX −∠CBX =∠BIX −∠BAX ୊ ፱ (LI Kin-Yin), Dept. of Math., HKUST ⇔ midpoint of the arc PXQ on the inside ؃ ᜢ ݰ (NG Keng-Po Roger), ITC, HKPU ∠IBX = ∠BIX ֔ ⇔ circle, which implies PX bisects .Artist: ྆ ؾ ़ (YEUNG Sau-Ying Camille), MFA, CU ⇔ XI = XB = XC ∠QPB . Now ∠ABX = 90˚ = ∠PIX Acknowledgment: Thanks to Elina Chiu, Math. Dept., so that X, I, P, B are concyclic. Then HKUST for general assistance. Example 1. (1982 Australian Math Olympiad) Let ABC be a triangle, and On-line: ∠IBX =∠IPX =∠BPX =∠BIX. http://www.math.ust.hk/mathematical_excalibur/ let the internal bisector of the angle A

The editors welcome contributions from all teachers and meet the circumcircle again at P. So XI = XB. By the theorem, I is the students. With your submission, please include your name, Define Q and R similarly. Prove that AP incenter of ∆ABC. address, school, email, telephone and fax numbers (if available). Electronic submissions, especially in MS Word, + BQ + CR > AB + BC + CA. are encouraged. The deadline for receiving material for the A Example 3. (2002 IMO) Let BC be a next issue is August 16, 2006. R diameter of the circle Γ with center O. For individual subscription for the next five issues for the Let A be a point on Γ such that 0˚ < 05-06 academic year, send us five stamped self-addressed Q envelopes. Send all correspondence to: I ∠AOB < 120˚. Let D be the midpoint Dr. Kin-Yin LI of the arc AB not containing C. The line Department of Mathematics B through O parallel to DA meets the line The Hong Kong University of Science and Technology C Clear Water Bay, Kowloon, Hong Kong AC at J. The perpendicular bisector of Fax: (852) 2358 1643 P OA meets Γ at E and at F. Prove that J is Email: [email protected] the incenter of the triangle CEF. Mathematical Excalibur, Vol. 11, No. 2, Apr. 06 - May 06 Page 2

A H, I on the circumcircle, we see that This implies J is the midpoint of KQ. P=BG∩CH, K=GJ∩HI and BI∩CJ= Hence the circle with center J and D F BD∩CE are collinear. Hence, BD∩CE is diameter KQ is tangent to circle O at K E on line PK, which is the same as line AP. and tangent to BC at Q. Since J is on the bisector of ∠BCA, this circle is J C O B Example 5. (2006 APMO) Let A, B be also tangent to AC. So this circle is O2. two distinct points on a given circle O and let P be the midpoint of line segment AB. Example 6. (1989 IMO) In an

Let O1 be the circle tangent to the line AB acute-angled triangle ABC the internal

at P and tangent to the circle O. Let ℓ be bisector of angle A meets the Solution. The condition ∠AOB < the tangent line, different from the line AB, circumcircle of the triangle again at A1. 120˚ ensures I is inside ∆CEF (when to O1 passing through A. Let C be the Points B1 and C1 are defined similarly. ∠ AOB increases to 120˚, I will intersection point, different from A, of ℓ Let A0 be the point of intersection of coincide with C). Now radius OA and and O. Let Q be the midpoint of the line the line AA1 with the external bisectors chord EF are perpendicular and bisect segment BC and O2 be the circle tangent of angles B and C. Points B0 and C0 are each other. So EOFA is a rhombus. to the line BC at Q and tangent to the line defined similarly. Prove that: Hence A is the midpoint of arc EAF. segment AC. Prove that the circle O2 is (i) the area of the triangle A0B0C0 is Then CA bisects ∠ECF. Since OA = tangent to the circle O. OC, ∠AOD = 1/2∠AOB = ∠OAC. twice the area of the hexagon Then DO is parallel to AJ. Hence N AC1BA1CB1, C ODAJ is a parallelogram. Then AJ = Z (ii) the area of the triangle A0B0C0 is at DO = EO = AE. By the theorem, J is least four times the area of the triangle L the incenter of ∆CEF. ABC. Q C Example 4. (1996 IMO) Let P be a J 0 C B point inside triangle ABC such that A P B 1

∠APB −∠ACB = ∠APC −∠ABC. K A M I Let D, E be the incenters of triangles Solution. Let the perpendicular to AB B A APB, APC respectively. Show that AP, 1 1 through P intersect circle O at N and M B BD and CE meet at a point. 0 C A0 with N and C on the same side of line AB. A By symmetry, segment NP is a diameter Solution. (i) Let I be the incenter of of the circle of O1 and its midpoint L is the ∆ABC. Since internal angle bisector J I center of O1. Let line AL intersect circle O and external angle bisector are K again at Z. Let line ZQ intersect line CM perpendicular, we have ∠B0BA0 = 90˚. H G at J and circle O again at K. By the theorem, A1I = A1B. So A1 must D E be the midpoint of the hypotenuse A0I Since AB and AC are tangent to circle O1, P of right triangle IBA0. So the area of AL bisects ∠ CAB so that Z is the B C ∆BIA0 is twice the area of ∆BIA1. midpoint of arc BC. Since Q is the F midpoint of segment BC, ∠ZQB = 90˚ = Cutting the hexagon AC1BA1CB1 into

∠LPA and ∠JQC = 90˚ =∠MPB. Next six triangles with common vertex I and Solution. Let lines AP, BP, CP applying a similar area fact like the last ∠ZBQ =∠ZBC =∠ZAC =∠LAP. intersect the circumcircle of ∆ABC statement to each of the six triangles, again at F, G, H respectively. Now So ∆ZQB, ∆LPA are similar. Since M is we get the conclusion of (i).

the midpoint of arc AMB, ∠APB −∠ACB =∠FPG −∠AGB (ii) Using (i), we only need to show the =∠FA G. ∠JCQ =∠MCB =∠MCA =∠MBP. area of hexagon AC1BA1CB1 is at least Similarly, ∠APC − ∠ABC = ∠FA H. So ∆JQC, ∆MPB are similar. twice the area of ∆ABC.

So AF bisects ∠HAG. Let K be the B incenter of ∆HAG. Then K is on AF By the intersecting chord theorem, AP·BP and lines HK, GK pass through the = NP·MP = 2LP·MP. Using the similar D H A2 A midpoints I, J of minor arcs AG, AH triangles above, we have respectively. Note lines BD, CE also 1 LP ⋅ MP ZQ ⋅ JQ A = = . 1 pass through I, J as they bisect ∠ABP, 2 AP ⋅ BP BQ ⋅CQ ∠ACP respectively. By the intersecting chord theorem, KQ·ZQ C = BQ·CQ so that Applying Pascal’s theorem (see vol.10, (continued on page 4) no. 3 of Math Excalibur) to B, G, J, C, KQ = (BQ·CQ)/ZQ = 2JQ. Mathematical Excalibur, Vol. 11, No. 2, Apr. 06 - May 06 Page 3

Problem Corner (Source: 1965 Soviet Union Math not relatively prime, but every three of Olympiad) them are relatively prime.

We welcome readers to submit their Solution. Jeff CHEN (Virginia, USA), (b) Determine with proof if there solutions to the problems posed below Koyrtis G. CHRYSSOSTOMOS exists an infinite sequence of positive for publication consideration. The (Larissa, Greece, teacher), G.R.A. 20 Math integers satisfying the conditions in (a) solutions should be preceded by the Problem Group (Roma, Italy) and Alex O above. solver’s name, home (or email) address Kin-Chit (STFA Cheng Yu Tung Secondary School). (Source: 2003 Belarussian Math and school affiliation. Please send submissions to Dr. Kin Y. Li, Olympiad) L Department of Mathematics, The Hong Solution. G.R.A. 20 Math Problem Kong University of Science & P Group (Roma, Italy) and YUNG Fai. Technology, Clear Water Bay, Kowloon, Hong Kong. The deadline for O R (a) We shall prove by induction that the submitting solutions is August 16, conditions are true for every positive integer k ≥ 3. 2006. A Q

Problem 251. Determine with proof For k = 3, the numbers 6, 10, 15 satisfy Let the spy plane be at Q when the rocket the conditions. Assume it is true for the largest number x such that a cubical was fired. Let L be the point on the circle some k ≥ 3 with the numbers being a1, gift of side x can be wrapped obtained by rotating Q by 90˚ in the a2, …, ak. Let p1, p2, …, pk be distinct completely by folding a unit square of forward direction of motion with respect prime numbers such that each pi is wrapping paper (without cutting). to the center A. Consider the semicircle greater than a1a2…ak. For I = 1 to k, let with diameter AL on the same side of line Problem 252. Find all polynomials bi= aipi and let bk+1= p1p2…pk. Then f(x) with integer coefficients such that AL as Q. We will show the path from A to n L along the semicircle satisfies the gcd(bi, bj)=gcd(ai, aj) >1 for 1≤ i < j ≤k, for every positive integer n, 2 − 1 is conditions. divisible by f(n). gcd(bi, bk+1) = pi > 1 for 1 ≤ i ≤ k,

Problem 253. Suppose the bisector of For any point P on the arc QL, let the gcd(bh, bi, bj) = gcd(ah, ai, aj) = 1 ∠BAC intersect the arc opposite the radius AP intersect the semicircle at R. for 1≤ h ≤ i < j ≤ k and angle on the circumcircle of ∆ABC at Let O be the midpoint of AL. Since gcd(bi, bj, bk+1) = 1 for 1 ≤ i < j ≤ k, A . Let B and C be defined similarly. 1 1 1 ∠QAP =∠RLA = 1/2∠ROA Prove that the area of ∆A1B1C1 is at completing the induction. least the area of ∆ABC. and AL = 2AO, the length of arc AR is the same as the length of arc QP. So the (b) Assume there are infinitely many Problem 254. Prove that if a, b, c > 0, conditions are satisfied. positive integers a1, a2, a3, … satisfying then the conditions in (a). Let a have Finally, the rocket will hit the spy plane at 1 2 exactly m prime divisors. For i = 2 to abc( a + b + c) + (a + b + c) L after 5π/1000 hour it was fired. m + 2, since each of the m + 1 numbers ≥ 4 3abc (a + b + c). gcd(a , a ) is divisible by one of these Comments: One solver guessed the path 1 i m primes, by the pigeonhole principle, Problem 255. Twelve drama groups should be a curve and decided to try a are to do a series of performances (with circular arc to start the problem. The other there are i, j with 2 ≤ i < j ≤ m + 2 such some groups possibly making repeated solvers derived the equation of the path by that gcd(a1, ai) and gcd(a1, aj) are performances) in seven days. Each a differential equation as follows: using divisible by the same prime. Then group is to see every other group’s polar coordinates, since the spy plane has gcd(a1, ai, aj) > 1, a contradiction. performance at least once in one of its a constant angular velocity of 1000/10 = day-offs. Commended solvers: CHAN Nga Yi 100 rad/sec, so at time t, the spy plane is at Find with proof the minimum total (10, 100t) and the rocket is at (r(t), θ(t)). (Carmel Divine Grace Foundation number of performances by these Since the rocket and the spy plane are on Secondary School, Form 6) and groups. the same radius, so θ(t) = 100t. Now they CHAN Yat Sing (Carmel Divine have the same speed, so Grace Foundation Secondary School, ***************** (r'(t))2 + (r(t)θ '(t))2 = 106. Form 6). Solutions Then Problem 248. Let ABCD be a convex **************** quadrilateral such that line CD is r'(t) = 100. tangent to the circle with side AB as Problem 246. A spy plane is flying at 100 − r(t)2 the speed of 1000 kilometers per hour diameter. Prove that line AB is tangent along a circle with center A and radius Integrating both sides from 0 to t, we get to the circle with side CD as diameter if 10 kilometers. A rocket is fired from A the equation r = 10 sin(100t) = 10 sin θ, and only if lines BC and AD are which describes the path above. parallel. at the same speed as the spy plane such that it is always on the radius from A to Problem 247. (a) Find all possible Solution. Jeff CHEN (Virginia, USA) the spy plane. Prove such a path for the and Koyrtis G. CHRYSSOSTOMOS rocket exists and find how long it takes positive integers k ≥ 3 such that there are k positive integers, every two of them are (Larissa, Greece, teacher). for the rocket to hit the spy plane. Mathematical Excalibur, Vol. 11, No. 2, Apr. 06 - May 06 Page 4

Problem 250. Prove that every region Problem 4. The function f is defined D F with a convex polygon boundary cannot for all integers {0, 1, 2, …, 2005}, C be dissected into finitely many regions assuming non-negative integer values with nonconvex quadrilateral boundaries. in each case. Furthermore, the following conditions are fulfilled for Solution. YUNG Fai. all values of x for which the function is A B Assume the contrary that there is a defined: E dissection of the region into nonconvex f(2x + 1) = f(2x), f(3x + 1) = f(3x) quadrilateral R1, R2, …, Rn. For a and f(5x + 1) = f(5x). nonconvex quadrilateral Ri, there is a vertex where the angle is θi > 180˚, which How many different values can the Let E be the midpoints of AB. Since we refer to as the large vertex of the function assume at most?

CD is tangent to the circle, the distance quadrilateral. The three other vertices, Problem 5. Determine all sextuples from E to line CD is h1 = AB/2. Let F where the angles are less than 180˚ will be (a,b,c,d,e,f) of real numbers, such that be the midpoint of CD and let h2 be the referred to as small vertices. the following system of equations is distance from F to line AB. Observe fulfilled: that the areas of ∆CEF and ∆DEF = Since the boundary of the region is a 4 4 CD·AB/8. Now convex polygon, all the large vertices are 4a=(b+c+d+e) , 4b=(c+d+e+f) , 4 4 in the interior of the region. At a large 4c=(d+e+f+a) , 4d=(e+f+a+b) , line AB is tangent to the circle 4 4 vertex, one angle is θ > 180˚, while the 4e=(f+a+b+c) , 4f=(a+b+c+d) . with side CD as diameter i remaining angles are angles of small h =CD/2 Problem 6. Let Q be a point in the ⇔ 2 vertices of some of the quadrilaterals and areas of ∆AEF, ∆BEF, ∆CEF and interior of a cube. Prove that an ⇔ add up to 360˚ − θ . Now ∆DEF are equal to AB·CD/8 i infinite number of lines passing AD EF, BC EF n through Q exists, such that Q is the ⇔ ∥ ∥ (360o −θ ) ∑ i mid-point of the line-segment joining ⇔ AD∥BC. i=1 the two points P and R in which the line accounts for all the angles associated with and the cube intersect. Problem 249. For a positive integer n, all the small vertices. This is a contradiction since this will leave no more if a1,⋯, an, b1, ⋯, bn are in [1,2] and angles from the quadrilaterals to form the 2 2 2 2 then prove a1 +L + an = b1 +L + bn , angles of the region. Angle Bisectors Bisect Arcs that (continued from page 2) 3 3 a1 an 17 2 2 +L+ ≤ (a1 +L+ an ). b b 10 Let H be the orthocenter of ∆ABC. Let 1 n Olympiad Corner line AH intersect BC at D and the (continued from page 1) Solution. Jeff CHEN (Virginia, USA). circumcircle of ∆ABC again at A2. Note

For x, y in [1,2], we have ∠ A BC = ∠A AC Part 2, Day 1 (June 8, 2005) 2 2 1/2 ≤ x/y ≤2 = ∠DAC ⇔ y/2 ≤ x ≤ 2y Problem 1. Determine all triples of = 90˚ −∠ACD ⇔ (y/2 − x)(2y − x) ≤ 0 positive integers (a,b,c), such that a + b +c = ∠HBC. 2 2 ⇔ x + y ≤ 5xy/2. is the least common multiple of a, b and c. Similarly, we have ∠A2CB = ∠HCB. Let x = a and y = b , then a 2 + b 2 ≤ i i i i Problem 2. Let a, b, c, d be positive real Then ∆BA2C ≅ ∆BHC. Since A1 is the 5aibi/2. Summing and manipulating, numbers. Prove midpoint of arc BA1C, it is at least as we get far from chord BC as A2. So the area of a + b + c + d 1 1 1 1 n 2 n 4 n ∆ BA1C is at least the area of ∆ BA2C. 2 2 2 ≤ 3 + 3 + 3 + 3 . − aibi ≤ − (ai + bi ) = − ai . abcd ∑ 5 ∑ 5 ∑ a b c d Then the area of quadrilateral BA1CH i=1 i=1 i=1 is at least twice the area of ∆BHC.

Problem 3. In an acute-angled triangle Let x = (a 3/b )1/2 and y = (a b )1/2. Then i i i i ABC, circle k with diameter AC and k Cutting hexagon AC BA CB into three x/y = a /b in [1,2]. So a 3/b + a b ≤ 1 2 1 1 1 i i i i i i with diameter BC are drawn. Let E be the quadrilaterals with common vertex H 5a 2/2. i foot of B on AC and F be the foot of A on and comparing with cutting ∆ABC into Summing, we get BC. Furthermore, let L and N be the three triangles with common vertex H n 3 n n points in which the line BE intersects with in terms of areas, we get the conclusion a 5 2 i + a b ≤ a . k (with L lying on the segment BE) and K of (ii). ∑ b ∑ i i 2 ∑ i 1 i=1 i i=1 i=1 and M be the points in which the line AF Remarks. In the solution of (ii), we Adding the two displayed inequalities, intersects with k2 (with K on the segment saw the orthocenter H of ∆ABC has the we get AF). Prove that KLMN is a cyclic quadrilateral. property that ∆BA2C ≅ ∆BHC (hence, 3 3 also HD = A2D). These are useful facts a1 an 17 2 2 + + ≤ (a1 + + an ). for problems related to the orthocenters b L b 10 L Part 2, Day 2 (June 9, 2005) 1 n involving the circumcircles.