Geometry for Middle School Teachers

Paul Yiu

Department of Mathematics Florida Atlantic University

Fall 2005

Part 1 of 3

Contents

1 The Pythagorean theorem 1 1.1 The hypotenuse of a right triangle ...... 1 1.2 The Pythagorean theorem ...... 3 1.3 Integer right triangles ...... 3

2 The equilateral triangle 5 2.1 Construction of equilateral triangle ...... 5 2.2 The medians of an equilateral triangle ...... 6 2.3 The center of an equilateral triangle ...... 7 2.4 The circumcircle and incircle of an equilateral triangle ...... 7 2.5 Trigonometry ...... 8 2.6 Cabri construction of a circle with given center and radius . . . . 10 2.7 Some interesting examples on equilateral triangles ...... 10

3 The square 13 3.1 Construction of a square on a segment ...... 13 3.2 The diagonals of a square ...... 14 3.3 The center of the square ...... 15 3.4 Some interesting examples on squares ...... 15

4 Some basic principles 17 4.1 Angle properties ...... 17 4.1.1 Parallel lines ...... 17 4.1.2 Angle sum of a triangle ...... 18 4.1.3 Angle properties of a circle ...... 18 4.2 Tests of congruence of triangles ...... 19 4.2.1 Construction of a triangle with three given elements . . . 19 4.2.2 Congruence tests ...... 19 iv CONTENTS

5 Circumcircle and incircle 23 5.1 Circumcircle ...... 23 5.1.1 The perpendicular bisector locus theorem ...... 23 5.1.2 Construction of circumcircle ...... 24 5.1.3 Circumcircle of a right triangle ...... 24 5.2 The incircle ...... 25 5.2.1 The angle bisector locus theorem ...... 25 5.2.2 Construction of incircle ...... 25 5.2.3 The incircle of a right triangle ...... 26 5.3 Tangents of a circle ...... 26 5.3.1 Tangent at a point on the circle ...... 26 5.3.2 The tangents from a point to a circle ...... 26 5.4 Some interesting examples on the circumcircle and incircle of a triangle ...... 27

6 Uses of congruence tests 29 6.1 Isosceles triangles ...... 29 6.2 Chords of a circle ...... 30 6.3 Parallelograms ...... 31 6.3.1 Properties of a parallelogram ...... 31 6.3.2 Quadrilaterals which are parallelograms ...... 32 6.3.3 Special parallelograms ...... 32 6.4 The midpoint theorem and its converse ...... 33 6.4.1 The midpoint theorem ...... 33 6.4.2 The converse of the midpoint theorem ...... 34 6.4.3 Why are the three medians concurrent? ...... 34 6.5 Some interesting examples ...... 35 Chapter 1

The Pythagorean theorem

1.1 The hypotenuse of a right triangle

This chapter is on the famous Pythagorean theorem. We certainly all know that this is an important relation on the sides of a right triangle. The side opposite to the right angle is called the hypotenuse and is the longest among all three sides. The other two are called the legs. If we denote the legs by a and b, and the hypotenuse by c, the Pythagorean theorem says that a2 + b2 = c2. B

c a

A b C

Why is this relation true? Suppose you have a cardboard whose length and breadth are 3 inches and 4 inches. How long is its diagonal? Your young students have not known the Pythagorean theorem yet. Help them calculate this length without throwing the Pythagorean theorem to them. Take two identical copies of the cardboard and cut each one along a diagonal. In this way there are 4 congruent right triangles. Arrange these 4 triangles in this way: 2 The Pythagorean theorem

3 2

In this ﬁgure, the 4 right triangles bound a large square outside and their hy- potenuses bound another smaller square inside. We calculate areas. (1) The outer side square has side length 3+4=7units. Its area is 72 =49 square units. (2) The 4 right triangles make up 2 rectangular cardboards each with area 3 × 4=12square units. The 4 triangles have total area 2 × 12 = 24 square units. (3) The inside square therefore has area 49 − 24 = 25 square units. Each of its sides has length 5 units. This method indeed applies to any right triangle with given legs.

Square on hypotenuse = square on sum of legs − 2 × rectangle with given legs as sides. b a

a 3 2 c c b

b cc a 41

a b Why is the inside region a square? 1.2 The Pythagorean theorem 3

1.2 The Pythagorean theorem

Rearrange the “puzzle” in §1.1 in another form:

2 b 4 b a a 3 1

Here, we see the same 4 right triangles inside the same outer square. Instead of the inside square, we now have two smaller squares, each built on a leg of the right triangle. This gives the famous Pythagorean theorem:

a2 + b2 = c2.

1.3 Integer right triangles

If we take any two numbers p and q and form

a =2pq, b = p2 − q2,c= p2 + q2, (1.1) then we have a2 + b2 = c2.

In particular, if p and q are integers, then so are a, b, and c. Sometimes, integer right triangles constructed from formula (1.1) can be re- duced. For example, with p =3and q =1, we obtain (a, b, c)=(6, 8, 10), which clearly is simply the right triangle (3, 4, 5) magniﬁed by factor 2. We say that an integer right triangle (a, b, c) is primitive if a, b, c do not have common divisor (other than 1). Every integer right triangle is a primitive one magniﬁed by an integer factor. A right triangle constructed from (1.1) is primitive if we choose the integers p, q of different parity 1 and without common divisors.

1This means that one of them is even and the other is odd. 4 The Pythagorean theorem

Here are all primitive integer right triangles with p, q < 10: p q a b c 2 1 4 3 5 3 2 12 5 13 4 1 8 15 17 4 3 24 7 25 5 2 20 21 29 5 4 6 1 6 5 7 2 7 4 7 6 8 1 8 3 8 5 8 7 9 2 9 4 9 8 How many distinct integer right triangles are there whose sides are not more than 100? Chapter 2

The equilateral triangle

Notations (O) circle with center O O(A) circle with center O, passing through A O(r) circle with center O, radius r

2.1 Construction of equilateral triangle

An equilateral triangle is one whose three sides are equal in length. The very ﬁrst proposition of Euclid’s Elements teaches how to construct an equilateral triangle on a given segment. C

A B

If A and B are the endpoints of the segments, construct the circles A(B) and B(A) to intersect at a point C. Then, triangle ABC is equilateral. An equilateral triangle is also equiangular. Its three angles are each equal to 60◦. 6 The equilateral triangle

2.2 The medians of an equilateral triangle

Let ABC be an equilateral triangle, and M the midpoint of BC. The segment AM is a median of the triangle. Since ABM ≡ACM by the test, this line AM is also (i) the bisector of angle A, (ii) the altitude on the side BC (since AM ⊥ BC), and therefore (iii) the perpendicular bisector of the segment BC.

B M C

This also means that one half of an equilateral triangle is a right triangle with angles 30◦ and 60◦. We call this an 30 − 60 − 90 right triangle. In a 30 − 60 − 90 right triangle,√ the hypotenuse is twice as long as the shorter leg. The two legs are in the ratio 3:1.

2 √ 3

1 2.3 The center of an equilateral triangle 7

2.3 The center of an equilateral triangle

Consider two medians of the equilateral triangle intersecting at O.

N O

B M C

Triangles AON and BOM are both 30 − 60 − 90 right triangles. Therefore, OA =2× ON and OB =2× OM. However, OA + OM and OB + ON have the same length. This means OM = ON, 1 OM = 1 × AM ON = 1 × BN and 3 , 3 . We have shown that any two medians of an equilateral triangle intersect at a point which divides each of them in the ratio 2:1. This is the same point O for all three medians, and is called the center of the equilateral triangle.

2.4 The circumcircle and incircle of an equilateral triangle

Since the three medians of an equilateral triangle have the same length, the center O is equidistant from the vertices. It is the center of the circumcircle of the equilateral triangle. The same center O is also the center of the incircle of the equilateral triangle.

1AM = BN ⇒ OA+OM = OB+ON ⇒ 2×ON+OM =2×OM+ON ⇒ OM = ON. 8 The equilateral triangle

A A

O O

B C B C

2.5 Trigonometry

For an acute angle θ, the trigonometric ratios sin θ, cos θ and tan θ are deﬁned by

a b a sin θ = , cos θ = , tan θ = , c c b

where (a, b, c) are the sides of a right triangle containing an acute angle θ with opposite leg a. If any one of the three ratios is known, then the other two can be found. Why? (1) sin2 θ +cos2 θ =1. tan θ = sin θ (2) cos θ . (3) sin(90◦ − θ)=cosθ. The precise determination of the trigonometric ratios of an angle is in general beyond the scope of elementary geometry. There are, however, a few special angles whose trigonometric ratios can be determined precisely. 2.5 Trigonometry 9

c a

θ A b C

θ 30◦ 60◦

sin θ

cos θ

tan θ 10 The equilateral triangle

2.6 Cabri construction of a circle with given center and radius

Given a point O and a segment AB, to contruct the circle O(AB), (1) construct an equilateral triangle OAC, (2) an equilateral triangle BCP such that P and O are on the same side of BC. Then, the circle O(P ) has radius equal to AB.

C A

2.7 Some interesting examples on equilateral triangles

(1) For an arbitrary point P on the minor arc BC of the circumcircle of an equilateral triangle ABC, AP = BP + CP. A

Q B C

P 2.7 Some interesting examples on equilateral triangles 11

Proof. If Q is the point on AP such that PQ = PC, then triangle CPQ is equilateral since . Note that ∠ACQ = ∠BCP. Thus, ACQ ≡BCP by the test. From this, AQ = BP, and AP = AQ + QP = BP + CP. (2) Consider a triangle ABC whose angles are all less than 120◦. Construct equilateral triangles BCX, CAY , and ABZ outside the triangle. Let F be the intersection of the circumcircles of the equilateral triangles. Note that ∠CFA = ∠AF B = 120◦. It follows that ∠AF B = 120◦, and F lies on the circumcircle of the equilateral triangle BCX as well. Therefore, ∠XFC = ∠XBC =60◦, and ∠XFC + ∠CFA = 180◦. The three points A, F , X are collinear. Thus, AX = AF + FX = AF + BF + CF by (1) above. Similarly, B, F , Y are collinear, so do C, F , Z, and BY , CZ are each equal to AF + BF + CF.

B C

Theorem 2.1 (Fermat point). If equilateral triangles BCX, CAY , and ABZ are erected on the sides of triangle ABC, then the lines AX, BY , CZ concur at a point F called the Fermat point of triangle ABC. If the angles of triangle ABC are all less than 120◦ (so that F is an interior point), then the segments AX, BY , CZ have the same length AF + BF + CF. (3∗) Napoleon’ theorem. If equilateral triangles BCX, CAY , and ABZ are erected on the sides of triangle ABC, the centers of these equilateral triangles form another equilateral triangle. 12 The equilateral triangle

B C

X Chapter 3

The square

3.1 Construction of a square on a segment

Given a segment AB, construct (1) an equilateral triangle ABP , (2) an equilateral triangle AP Q (different from ABP ), (3) the segment PQand its midpoint M, (4) the ray AM, (5) the circle A(B) to intersect this ray at D, (6) the circles B(A) and D(A) to intersect at C. Then ABCD is a square.

D C M Q P

A B

This construction gives an oriented square. If you apply it to AB and BA, you will get two different squares on the two sides of AB. 14 The square

3.2 The diagonals of a square

A diagonal of a square divides it into two congruent right triangles. Each is an isosceles right triangle or a 45 −√45 − 90 triangle. If each side of a square has unit length, a diagonal has length 2.

√ √ 2 2 1 1

1 1 θ 45◦

sin θ

cos θ

tan θ

The diagonals of a square are perpendicular to each other. They divide the square into 4 isosceles right triangles. 3.3 The center of the square 15

3.3 The center of the square

The intersection of the diagonals is the center of the square. It is the common center of the circumcircle and the incircle of the square.

3.4 Some interesting examples on squares

(1) If ABCbCa and ACBcBa are squares erected externally on the sides AB and AC of a triangle ABC, the triangle ACaBa has the same area as triangle ABC.

Ca A Bc

B C

∗ (2 ) The perpendicular from A to BaCA passes through the midpoint of BC. Likewise, those from B to CbAb and from C to AcBc pass through the midpoints of CA and AB respectively. These three perpendiculars meet at the centroid G of triangle ABC. 16 The square

C a A Bc

G C b B C

Ab Ac

∗ (3 ) The midpoint of BcCb does not depend on the position of the vertex A.It is also the same as the center of the square erected on the BC, on the same side as A.

M Cb

B C Chapter 4

Some basic principles

4.1 Angle properties

4.1.1 Parallel lines

Consider two parallel lines 1, 2, and a transversal L.

α 1 γ δ

β 2

(1) The corresponding angles α and β are equal. (2) The alternate angles β and γ are equal. (3) The interior angles β and δ are supplementary, i.e., β + δ = 180◦. The converses of these statements are also true. In other words, if any one of (1), (2), (3) holds, then the lines 1 and 2 are parallel. 18 Some basic principles

4.1.2 Angle sum of a triangle The angle sum of a triangle is always 180◦. This is because when a side of the triangle is extended, the external angle formed is equal to the sum of the two remote internal angles. A

α β B C

4.1.3 Angle properties of a circle (1) Let P be a point on the major arc AB of a circle (O). ∠AOB =2∠AP B.

O O O

Q Q

A B A B A B P

(2) Let P be a point on the minor arc AB of a circle (O). ∠AOB +2∠AP B = 360◦.

(3) The opposite angles of a cyclic quadrilateral are supplementary. (4) The angle contained in a semicircle is a right angle. (5) Equal chords subtend equal angles at the center. 4.2 Tests of congruence of triangles 19

P A

A O B O

B D

4.2 Tests of congruence of triangles

4.2.1 Construction of a triangle with three given elements

A triangle has six elements: three sides and three angles. Consider the construction of a triangle given three of its six elements. The triangle is unique (up to size and shape) if the given data are in one of the following patterns. (1) SSS Given three lengths a, b, c, we construct a segment BC with length a, and the two circles B(c) and C(b). These two circles intersect at two points if (and only if) b + c>a[triangle inequality]. There are two possible positions of A. The resulting two triangles are congruent. (2) SAS Given b, c, and angle A<180◦, the existence and uniqueness of triangle ABC is clear. (3) ASA or AAS These two patterns are equivalent since knowing two of the angles of a triangle, we easily determine the third (their sum being 180◦). Given B, C and a, there is clearly a unique triangle provided B + C<180◦. (4) RHS Given a, c and C =90◦.

4.2.2 Congruence tests

These data patterns also provide the valid tests of congruence of triangles. Two triangles ABC and XY Z are congruent if their corresponding elements are equal. Two triangles are congruent if they have three pairs of equal elements in one the ﬁve patterns above. The ﬁve valid tests of congruence of triangles are as follows. 20 Some basic principles

(1) SSS: ABC ≡XYZ if AB = XY, BC = YZ, CA= ZX.

X A

B C Y

(2) SAS: ABC ≡XY Z if AB = XY, ∠ABC = ∠XY Z, BC = YZ.

X A

B C Y

(3) ASA: ABC ≡XYZ if ∠BAC = ∠YXZ, AB= XY, ∠ABC = ∠XY Z.

X A

B C Y

(4) AAS. We have noted that this is the same as ASA: ABC ≡XY Z if ∠BAC = ∠YXZ, ∠ABC = ∠XY Z, BC = YZ, . 4.2 Tests of congruence of triangles 21 X A

B C Y

(5) RHS. The ASS is not a valid test of congruence. Here is an example. The two triangles ABC and XY Z are not congruent even though ∠BAC = ∠YXZ, AB= XY, BC = YZ.

B Y

A C X Z However, if the equal angles are right angles, then the third pair of sides are equal: AC2 = BC2 − AB2 = YZ2 − XY 2 = XZ2, and AC = XZ. The two triangles are congruent by the SSS test. Without repeat- ing these details, we shall simply refer to this as the RHS test. ABC ≡XY Z if ∠BAC = ∠YXZ =90◦,BC= YZ, AB= XY.

B Y

A C Z X The congruence tests for triangles form a paradigm for proofs in euclidean geometry. We shall illustrate this with numerous examples in Chapter 6. Here, we give only one example, the converse of the Pythagorean theorem. 22 Some basic principles

Theorem 4.1 (Converse of Pythagoras’ theorem). If the lengths of the sides of ABC satisfy a2 + b2 = c2, then the triangle has a right angle at C.

B Y

c a a

C b A Z b X

Proof. Consider a right triangle XY Z with ∠Z =90◦, YZ = a, and XZ = b. By the Pythagorean theorem, XY 2 = YZ2 + XZ2 = a2 + b2 = c2 = AB2.It follows that XY = AB, and ABC ≡XY Z by the test, and ∠C = ∠Z =90◦. Chapter 5

Circumcircle and incircle

5.1 Circumcircle

The perpendicular bisector of a segment is the line perpendicular to it through its midpoint.

5.1.1 The perpendicular bisector locus theorem Theorem 5.1. A point P is equidistant from B and C if and only if P lies on the perpendicular bisector of BC.

B M C

Proof. Let M be the midpoint of BC. (⇒) If PB = PC, then PBM ≡PCM by the test. This means ∠PMB = ∠PMC and PM ⊥ BC. The point P is on the perpendicular bisector of BC. (⇐) If P is on the perpendicular bisector of BC, then PMB ≡ PMC by the test. It follows that PB = PC. 24 Circumcircle and incircle

5.1.2 Construction of circumcircle

Given a triangle ABC, there is a circle (O) through the three vertices. The center O of the circle, being equidistant from B and C, must lie on the perpendicular bisector of BC. For the same reason, it also lies on the perpendicular bisectors of CA and AB. This is called the circumcenter of triangle ABC.

B C

5.1.3 Circumcircle of a right triangle

To construct a circle through the vertices of a right triangle ABC, mark the midpoint M of the hypotenuse AB. The circle M(C) passes through A and B, and is the circumcircle of the right triangle.

A M B 5.2 The incircle 25

5.2 The incircle

5.2.1 The angle bisector locus theorem The distance from a point P to a line is the distance between P and its pedal (orthogonal projection) on .

K P

A H

Theorem 5.2. A point P is equidistant from two lines if and only if P lies on the bisector of an angle between the two lines. Proof. Let A be the intersection of the two lines, and H, K the pedals of a point P on these lines. (⇒)IfPH = PK then AP H ≡AP K by the test. It follows that ∠PAH = ∠PAK, and P lies on the bisector of angle HAK. (⇐)If∠PAH = ∠PAK, then AP H ≡AP K by the test. It follows that PH = PK.

5.2.2 Construction of incircle The incircle of a triangle is the one which is tangent to each of the three sides of the triangle. Its center is the common point of the angle bisectors, and can be located by constructing two of them. This incenter I is equidistant from the three sides. A

B C 26 Circumcircle and incircle

5.2.3 The incircle of a right triangle If d is the diameter of the incircle of a right triangle, then a + b = c + d.

ac d

C b A

5.3 Tangents of a circle

5.3.1 Tangent at a point on the circle A tangent to a circle is a line which intersects the circle at only one point. Given a circle O(A), the tangent to a circle at A is the perpendicular to the radius OA at A.

A A O O M P

5.3.2 The tangents from a point to a circle If P is a point outside a circle (O), there are two lines through P tangent to the circle. Construct the circle with OP as diameter to intersect (O) at two points. These are the points of tangency. The two tangents have equal lengths since OAP ≡OBP by the test. 5.4 Some interesting examples on the circumcircle and incircle of a triangle27

5.4 Some interesting examples on the circumcircle and incircle of a triangle s = 1 (a + b + c) ABC (1) Let 2 be the semiperimeter of triangle . The lengths of the tangents from the vertices to the incircle are as follows.

tangent from A =s − a, tangent from B =s − b, tangent from C =s − c.

ac r r

C b A (2) The following rearrangement shows that for an arbitrary triangle, the radius of its incircle is given by

r = area of triangle . semi-perimeter of triangle

r 28 Circumcircle and incircle

(3∗) Construct the circumcircle (O) of triangle ABC and select an arbitrary point P on it. Construct the reflections of P in the sidelines of the given triangle. These three reflection points always lie on a straight line . Furthermore, as P varies on (O), the line passes through a fixed point.

B C

(4∗) Construct the circumcircle (O) and incircle (I) of a triangle ABC. Select an arbitrary point P on (I) and construct the perpendicular to IP at P . This is a tangent to (I). Let it intersect the circle (O) at Y and Z. Construct the tangents from Y and Z to (I). These two tangents always intersect on the circumcircle (O).

A Y

Z O I

B C

X Chapter 6

Uses of congruence tests

6.1 Isosceles triangles

An isosceles triangle is one with two equal sides.

Proposition 6.1. A triangle is isosceles if and only if it has two equal angles.

A A

B M C B D C

Proof. (⇒) Let ABC be a triangle in which AB = AC. Let M be the midpoint of BC. Then ABM ≡ACM by the test. From this conclude that ∠BAM = ∠CAM. (⇐) Let ABC be a triangle in which ∠B = ∠C. Construct the perpendicular from A to BC, and let it intersect BC at D. Then ABD ≡ACD by the test. From this AB = AC and the triangle is isosceles. 30 Uses of congruence tests

6.2 Chords of a circle

(1) Let M be a point on a chord BC of a circle (O). M is the midpoint of BC if and only if OM ⊥ BC.

B M C

Proof. (⇒)IfM is the midpoint of BC, then OMB ≡OMC by the test. Therefore, ∠OMB = ∠OMC and OM ⊥ BC. (⇐)IfOM ⊥ BC, then OMB ≡OMC by the test. It follows that BM = CM, and M is the midpoint of BC. (2) Equal chords of a circle are equidistant from the center. Conversely, chords equidistant from the center are equal in length. Proof. Suppose BC and PQ are equal chords of a circle (O), with midpoints M and N respectively. Then BM = PN, and OBM ≡OPN by the test. Therefore, OM = ON, and the chords are equidistant from the center O. Conversely, if OM = ON, then OBM ≡OPN by the test. It follows that BM = PN, and BC =2· BM =2· PN = PQ. 6.3 Parallelograms 31

6.3 Parallelograms

A parallelogram is a quadrilateral with two pairs of parallel sides.

6.3.1 Properties of a parallelogram (1) The opposite sides of a parallelogram are equal in length. The opposite angles of a parallelogram are equal.

D C

A B

Proof. Consider a parallelogram ABCD. Construct the diagonal BD. Note that ABD ≡CDB by the test. It follows that AB = CD and AD = CB, and ∠BAD = ∠DCB. Similarly, ∠ABC = ∠CDA. (2) The diagonals of a parallelogram bisect each other.

D C

A B

Proof. Consider a parallelogram ABCD whose diagonals AC and BD intersect at M. Note that AD = CB and AMD ≡CMB by the test. It follows that AM = CM and DM = BM. The diagonals bisect each other. 32 Uses of congruence tests

6.3.2 Quadrilaterals which are parallelograms (1) If a quadrilateral has two pairs of equal opposite sides, then it is a parallelogram.

D C

A B

Proof. Let ABCD be a quadrilateral in which AB = CD and AD = BC. Con- struct the diagonal BD. Then ABD ≡CDB by the test. It follows that and AD//BC. Also, and AB//DC. Therefore, ABCD is a parallelogram.

(2) If a quadrilateral has one pair of equal and parallel sides, then it is a parallelogram.

D C

A B

Proof. Suppose ABCD is a quadrilateral in which AB//DC and AB = CD. Then ABD ≡CDB by the test. It follows that AD = CB and ABCD is a parallelogram since .

6.3.3 Special parallelograms A quadrilateral is (1) a rhombus if its sides are equal, (2) a rectangle if its angles are equal (and each 90◦), (3) a square if its sides are equal and its angles are equal. 6.4 The midpoint theorem and its converse 33

These are all parallelograms since (1) a rhombus has two pairs of sides, and (2) a rectangle has two pairs of sides. Clearly, then a square is a parallelogram.

(1) A parallelogram is a rhombus if and only if its diagonals are perpendicular to each other. (2) A parallelogram is a rhombus if and only if it has an angle bisected by a diagonal. (3) A parallelogram is a rectangle if and only if its diagonals are equal in length.

6.4 The midpoint theorem and its converse

6.4.1 The midpoint theorem Given triangle ABC, let E and F be the midpoints of AC and AB respectively. The segment FE is parallel to BC and its length is one half of the length of BC. A

E F D

B C

Proof. Extend FE to D such that FE = ED. Note that CDE ≡AF E by the test. It follows that (i) CD = AF = BF, (ii) ∠CDE = ∠AF E, and CD//BA. Therefore, CDFB is a parallelogram, and F E//BC. Also, BC = FD =2FE. 34 Uses of congruence tests

6.4.2 The converse of the midpoint theorem Let F be the midpoint of the side AB of triangle ABC. The parallel through F to BC intersects AC at its midpoint. A

E F D

B C

Proof. Construct the parallel through C to AB, and extend FE to intersect this parallel at D. Then, CDFB is a parallelogram, and CD = BF = FA. It follows that AEF ≡ CED by the test. This means that AE = CE, and E is the midpoint of AC.

6.4.3 Why are the three medians concurrent? Let E and F be the midpoints of AC and AB respectively, and G the intersection of the medians BE and CF. Construct the parallel through C to BE, and extend AG to intersect BC at D, and this parallel at H..

E F G

B D C

H 6.5 Some interesting examples 35

By the converse of the midpoint theorem, G is the midpoint of AH, and HC = 2 · GE Join BH. By the midpoint theorem, BH//CF . It follows that BHCG is a parallelogram. Therefore, D is the midpoint of (the diagonal) BC, and AD is also a median of triangle ABC. We have shown that the three medians of triangle ABC intersect at G, which we call the centroid of the triangle. Furthermore,

AG =GH =2GD, BG =HC =2GE, CG =HB =2GF.

The centroid G divides each median in the ratio 2:1.

6.5 Some interesting examples

(1) Why are the three altitudes of a triangle concurrent? Let ABC be a given triangle. Through each vertex of the triangle we construct a line parallel to its opposite side. These three parallel lines bound a larger triangle ABC. Note that ABCB and ACBC are both parallelograms since each has two pairs of parallel sides. It follows that B A = BC = AC and A is the midpoint of BC.

C A B

Z H

B X C

A 36 Uses of congruence tests

Consider the altitude AX of triangle ABC. Seen in triangle ABC, this line is the perpendicular bisector of BC since it is perpendicular to BC through its midpoint A. Similarly, the altitudes BY and CZ of triangle ABC are perpendicular bisectors of CA and AB. As such, the three lines AX, BY , CZ concur at a point H. This is called the orthocenter of triangle ABC.

(2∗) The butterﬂy theorem. Let M be the midpoint of a chord PQof a circle (O). AB and CD are two chords passing through M. Join BC to intersect PQat X and AD to intersect PQat Y . Then MX = MY.

P Y M X Q

B D Exercise A

Exercises marked with † are to be done with Cabri. Those marked with ∗ are more difﬁcult.

Problem A1†. (a) Inscribe an equilateral triangle in a given circle. (b) Circumscribe an equilateral triangle about the same circle. (c) How do the areas of the two equilateral triangles compare?

Problem A2†. (a) Inscribe a square in a given circle. (b) Circumscribe a square about the same circle. (c) How do the areas of the two squares compare?

Problem A3†. (a) Inscribe a regular hexagon in a given circle. (b) Circumscribe a regular hexagon about the same circle. (c) How do the areas of the two hexagons compare?

Problem A4†. (a) Inscribe a regular octagon in a given circle. (b) Circumscribe a regular octagon about the same circle. (c) How do the areas of the two octagons compare?

Problem A5†. (a) Inscribe a regular hexagon and an equilateral triangle in the same circle. (b) How do the areas of the two regular polygons compare? 38 Uses of congruence tests

Problem A6†. An equilateral triangle of side 2a is partitioned symmetrically into a quadrilateral, an isosceles triangle, and two other congruent triangles. If the inradii of the√ quadrilateral√ and the isosceles triangle are equal, the common inradius is ( 3 − 2)a. Construct the partition.

Problem A7†. If the equilateral triangle has area A, what is the area of the shaded triangle?

Problem A8†. Construct the square externally on the hypotenuse of a right triangle. Join the center of the square to the right angle vertex. Why does this line bisect the right angle? 6.5 Some interesting examples 39

Problem A9. If each side of the equilateral triangle has√ length 6, what is the radius of the circle? Leave your answer in the form a + b c for integers a, b, c.

Problem A10. The equilateral triangle has side length 3. What is the radius of the circle?

Problem A11. If each side of the square and the equilateral triangle has length 1, what is the radius of the circle? Why ? A

B C O

E D 40 Uses of congruence tests

Problem A12. (a) ABCD is a square and CMD is an equilateral triangle. Show that AMB is an isosceles triangle with base angle 15◦.

A B

D C

(b∗) Show that if AMB is an isosceles triangle with base angle 15◦, then CMD is an equilateral triangle.

Problem A13. In the diagram below, BCD is an isosceles right triangle, and AD = DC. Make use of the diagram to calculate tan 22.5◦. Leave your answer in a form with an integer denminator. C

A D B

Problem A14. In the diagram below, DCB is a 30 − 60 − 90 right triangle, and AD = DC. Make use of the diagram to calculate tan 15◦. Leave your answer in a form with an integer denminator. C

A D B 6.5 Some interesting examples 41

Problem A15∗. ABC is an isosceles triangle with angle A =20◦. E and F are points on AC and AB such that ∠EBC =60◦ and ∠FCB =50◦. Calculate ∠BEF.

20◦

60◦ 50◦ B C

Problem A16. A man has a square ﬁeld, 60 feet by 60 feet, with other property adjoining the highway. He put up a straight fence in the line of 3 trees, at A, P , Q. If the distance between P and Q is 91 feet, and that from P to C is an exact number of feet, what is this distance?

A 60 B

P 60 91 ? Q D C 42 Uses of congruence tests

Problem A17. Two chords AB and CD of a circle (O) have lengths 16 and 30 units, and are 7 units apart. Find the radius of the circle. [Hint: ﬁrst ﬁnd the distance from the center to the nearer chord].

16 A B 7 C D 30

Problem A18∗. The right triangles below have the shape of the 3 − 4 − 5 triangle. What is the least number of matches of equal lengths to make up the conﬁguration ? 6.5 Some interesting examples 43

Problem A19. Calculate the area of the 13 − 14 − 15 triangle by cutting it into two integer right triangles. What is the radius of its incircle?

13 15

C ? ? A

Problem A20∗. It is known that exactly 13 convex polygons can be formed from the Chinese tangram.There are 1 triangle, 6 quadrilaterals, 2 pentagons, and 4 hexagons. Show them. Geometry for Middle School Teachers

Paul Yiu

Department of Mathematics Florida Atlantic University

Fall 2005

Part 2 of 3

Contents

7 The regular hexagon, octagon and dodecagon 45 7.1 The regular hexagon ...... 45 7.2 The regular octagon ...... 46 7.3 The regular dodecagon ...... 47

8 Similar triangles 49 8.1 Tests for similar triangles ...... 49 8.2 The parallel intercepts theorem ...... 51 8.3 The angle bisector theorem ...... 52 8.4 Some interesting examples of similar triangles ...... 53

9 Tangency of circles 57 9.1 External and internal tangency of two circles ...... 57 9.2 Three mutually tangent circles ...... 58 9.3 Some interesting examples of tangent circles ...... 58

10 The regular pentagon 61 10.1 The golden ratio ...... 61 10.2 The diagonal-side ratio of a regular pentagon ...... 62 10.3 Construction of a regular pentagon with a given diagonal . . . . 63 10.4 Various constructions of the regular pentagon ...... 64 10.5 Some interesting constructions of the golden ratio ...... 66

11 The regular decagon 69 11.1 Constructions ...... 69 11.2 Some relations among sides of regular polygons inscribed in a circle ...... 70 iv CONTENTS

12 Construction of regular n-gons 73 12.1 Gauss’ theorem ...... 73 12.2 Construction of a regular 15-gon ...... 73 12.3 Construction of a regular 17-gon ...... 74 Chapter 7

The regular hexagon, octagon and dodecagon

A regular polygon of n sides can be constructed (with ruler and compass) if it is possible to divides the circle into n equal parts by dividing the 360◦ at the center into the same number of equal parts. This can be easily done for n =6(hexagon) and 8 (octagon). 1 We give some easy alternatives for the regular hexagon, octagon and dodecagon.

7.1 The regular hexagon

A regular hexagon can be easily constructed by successively cutting out chords of length equal to the radius of a given circle.

1Note that if a regular n-gon can be constructed by ruler and compass, then a regular 2n-gon can also be easily constructed. 46 The regular hexagon, octagon and dodecagon

7.2 The regular octagon

(1) Successive completion of rhombi beginning with three adjacent 45◦-rhombi.

(2) We construct a regular octagon by cutting from each corner of a given square (of side√ length 2) an isosceles√ right triangle√ of (shorter) side x. This means 2 − 2x : x = 2:1, and 2 − 2x = 2x; (2 + 2)x =2, √ 2 2(2 − 2) √ x = √ = √ √ =2− 2. 2+ 2 (2 + 2)(2 − 2)

D C D C

Q Q x

A x 2 − 2x P x B A P B

√ Therefore AP =2− x = 2, which is half of the diagonal of the square. The point P , and the other vertices, can be easily constructed by intersecting the sides of the square with quadrants of circles with centers at the vertices of the square and passing through the center O. 7.3 The regular dodecagon 47

7.3 The regular dodecagon

(1) Trisection of right angles.

(2) A regular dodecagon can be formed from 4 equilateral triangles inside a square. 8 of its vertices are the intersections of the sides of these equilateral triangles, while the remaining 4 are the midpoints of the sides of the squares formed by the vertices of the equilateral triangles inside the square.

3 An easy dissection shows that the area of the regular dodecagon is 4 of that of the (smaller) square containing it. 48 The regular hexagon, octagon and dodecagon

(3) Successive completion of rhombi beginning with ﬁve adjacent 30◦-rhombi.

This construction can be extended to other regular polygons. If an angle of θ := 360 ◦ measure 2n can be constructed with ruler and compass, beginning with n − 1 adjacent θ-rhombi, by succesively completing rhombi, we obtain a regular 2n-gons tesellated by 1 (n − 1) + (n − 2) + ···+2+1= n(n − 1) 2 rhombi. Chapter 8

Similar triangles

8.1 Tests for similar triangles

Two triangles are similar if their corresponding angles are equal and their corresponding sides proportional to each other, i.e., ABC ∼XYZ if and only if ∠A = ∠X, ∠B = ∠Y, ∠C = ∠Z, and BC CA AB = = . YZ ZX XY It is enough to conclude similarity of two triangles when they have three pairs of (1) AAA: if the triangles have two (and hence three) pairs of equal angles.

Z Y

B C 50 Similar triangles

CA AB = ∠A = ∠X (2) proportional SAS:ifZX XY and .

Z Y

B C

BC CA AB = = (3) proportional SSS:ifYZ ZX XY .

Z Y

B C 8.2 The parallel intercepts theorem 51

8.2 The parallel intercepts theorem

The intercepts on two transversals between a set of parallel lines are proportional to each other: if 0, 1, 2, ..., n−1, n are parallel lines and L1 and L2 are two transversals intersecting these lines at A0, A1, A2,...,An−1, An, and B0, B1, B2, ...,Bn−1, Bn, then A A A A A A 0 1 = 1 2 = ···= n−1 n . B0B1 B1B2 Bn−1Bn

L2 L1

A0 B0 C0 0

A1 B1 C1 1 A2 B2 2

An−1 Bn−1 Cn−1 n−1

An Bn n

Proof. Through B1, B2,...,Bn construct lines parallel to L1, intersecting 0, 1, ...,n−1 at C0, C1,...,Cn−1 respectively. Note that (i) B1C0 = A1A0, B2C1 = A2A1,...,BnCn−1 = AnAn−1, and (ii) the triangles B0B1C0, B1B2C1,...,Bn−1BnCn−1 are similar. It follows that B C B C B C 1 0 = 2 1 = ···= n n1 . B0B1 B1B2 Bn−1Bn Hence, A A A A A A 0 1 = 1 2 = ···= n−1 n . B0B1 B1B2 Bn−1Bn 52 Similar triangles

8.3 The angle bisector theorem

The bisector of an angle divides the opposite side in the ratio of the remaining two sides, i.e.,if AX bisects angle A, then BX AB = . XC AC D

B X C Proof. Construct the parallel through B to the bisector to intersect the extension of CA at D. Note that ∠ABD = ∠BAX = ∠CAX = ∠CBD = ∠ABD.

This means that triangle ABD is isosceles, and AB = AD. Now, from the parallelism of AX and DB,wehave AX DA AB = = . XB AC AC 8.4 Some interesting examples of similar triangles 53

8.4 Some interesting examples of similar triangles

(1) The altitude from the right angle vertex divides a right triangle into two triangles each similar to the given right triangle.

A D B

Proof. Let ABC be a triangle with a right angle at C.IfD is a point on AB such that DC ⊥ AB, then ACD ∼CBD ∼ABC. Here are some interesting consequences. AD CD ACD ∼CBD = (a) From ,wehaveCD BD. It follows that CD2 = AD · CD. BD BC CBD ∼ABC = (b) From ,wehave CB AB so that BC2 = BD · AB. AD AC ACD ∼ABC = Similarly, from ,wehave AC AB so that AC2 = AD · AB.

It follows that BC2 + AC2 = BD · AB + AD · AB =(BD + AD) · AB = AB · AB = AB2.

This gives an alternative proof of the Pythagorean theorem. (a) and (b) above give simple constructions of geometric means. 54 Similar triangles

Construction 8.1. Given a point P on a segment AB, construct (1) a semicircle with diameter AB, (2) the perpendicular to AB at P , intersecting the semicircle at Q. Then PQ2 = AP × PB.

Q Q

θ A O P B A O P B

(2) The radius of the semicircle is the geometric mean of AP amd BQ.

B Q

A P This follows from the similarity of the right triangles OAP and QBO.If r AP r = OA = OB = r2 = AP · BQ , then QB r . From this, . 8.4 Some interesting examples of similar triangles 55

(3∗) Heron’s formula for the area of a triangle.

r I

Y s − b C s − c Y s − a A

The A-excircle of triangle ABC is the circle which is tangent to the side BC and to the extensions of AB and AC.IfY and Z are the points of tangency with AC and AB, it is easy to see that AY = AZ = s, the semiperimeter. Hence, CY = s − b. If the incircle touches AC at Y , we have known from Example 1 of §5.4 that CY = s − c and AY = s − a. From these we can ﬁnd the radii r of the incircle, and r of the A-excircle quite easily. From the similarity of triangles AIY and AI Y ,wehave r s − a = . r s r s − c ICY ∼CIY = . Note that also , from which we have s − b r and r · r =(s − b)(s − c).

Multiplying these two equations together, we obtain (s − a)(s − b)(s − c) r2 = . s From Example 2 of §5.4, the area of triangle ABC is given by = rs. Hence, we have the famous Heron formula = s(s − a)(s − b)(s − c). 56 Similar triangles Chapter 9

Tangency of circles

9.1 External and internal tangency of two circles

Two circles (O) and (O) are tangent to each other if they are tangent to a line at the same line P , which is a common point of the circles. The tangency is internal or external according as the circles are on the same or different sides of the common tangent .

O P O O O P

The line joining their centers passes through the point of tangency. The distance between their centers is the sum or difference of their radii, according as the tangency is external or internal. 58 Tangency of circles

9.2 Three mutually tangent circles

Given three non-collinear points A, B, C, construct (i) triangle ABC and its incircle, tangent to BC, CA, AB respectively at X, Y , Z, (ii) the circles A(Y ), B(Z) and C(X). These are tangent to each other externally.

Y X

C Z A

9.3 Some interesting examples of tangent circles

(1) In each of the following diagrams, the shaded triangle is similar to the 3−4−5 triangle. 9.3 Some interesting examples of tangent circles 59 60 Tangency of circles

(2∗) Given triangle ABC, let D, E, F be the midpoints of the sides BC, CA, AB respectively. The circumcircle of triangle DEF is always tangent internally to the incircle of triangle ABC.

F E

B D C Chapter 10

The regular pentagon

10.1 The golden ratio

The construction of the regular pentagon is based on the division of a segment in the golden ratio. 1 Given a segment AB, to divide it in the golden ratio is to construct a point P on it so that the area of the square on AP is the same as that of the rectangle with sides PB and AB, i.e.,

AP 2 = AP · PB.

A construction of P is shown in the second diagram.

A P B A P B

1In Euclid’s Elements, this is called division into the extreme and mean ratio. 62 The regular pentagon

Suppose PB has unit length. The length ϕ of AP satisﬁes

ϕ2 = ϕ +1.

This equation can be rearranged as 1 2 5 ϕ − = . 2 4

Since ϕ>1,wehave 1 √ ϕ = 5+1 . 2 Note that √ AP ϕ 1 2 5 − 1 = = = √ = . AB ϕ +1 ϕ 5+1 2 This explains the construction above.

10.2 The diagonal-side ratio of a regular pentagon

Consider a regular pentagon ACBDE. It is clear that the ﬁve diagonals all have equal lengths. Note that (1) ∠ACB = 108◦, (2) triangle CAB is isosceles, and (3) ∠CAB = ∠CBA = (180◦ − 108◦) ÷ 2=36◦. In fact, each diagonal makes a 36◦ angle with one side, and a 72◦ angle with another. C

A B P

E D 10.3 Construction of a regular pentagon with a given diagonal 63

It follows that (4) triangle PBC is isosceles with ∠PBC = ∠PCB =36◦, (5) ∠BPC = 180◦ − 2 × 36◦ = 108◦, and (6) triangles CAB and PBC are similar. Note that triangle ACP is also isosceles since (7) ∠ACP = ∠AP C =72◦. This means that AP = AC. Now, from the similarity of CAB and PBC,wehaveAB : AC = BC : PB. In other words AB · AP = AP · PB,orAP 2 = AB · PB. This means that P divides AB in the golden ratio.

10.3 Construction of a regular pentagon with a given diagonal

Given a segment AB, we construct a regular pentagon ACBDE with AB as a diagonal. (1) Divide AB in the golden ratio at P . (2) Construct the circles A(P ) and P (B), and let C be an intersection of these two circles. (3) Construct the circles A(AB) and B(C) to intersect at a point D on the same side of BC as A. (4) Construct the circles A(P ) and D(P ) to intersect at E. Then ACBDE is a regular pentagon with AB as a diameter. C

A B P

E D 64 The regular pentagon

10.4 Various constructions of the regular pentagon

(1) To construct a regular pentagon with vertices on a given a circle O(A), (i) construct a radius OP perpendicular to OA, (ii) mark the midpoint M of OP and join it to A, (iii) bisect the angle OMA and let it intersect OA at Q, (iv) construct the perpendicular to OA at Q to intersect the circle at B and E, (v) mark C, D on the circle such that AB = BC and AE = ED. Then ABCDE is a regular pentagon.

Q E B

P O M

D C

(2) A compass-only costruction of the regular pentagon:2 construct (i) (O)=O(A), (ii) A(O) to mark B and F on (O), (iii) B(O) to mark C, (iv) C(O) to mark D, (v) D(O) to mark E, (vi) A(C) and D(AC) to intersect at X, (vii) Y = C(OX) ∩ E(OX) inside (O), (viii) P = A(OY ) ∩ (O), (ix) Q = P (A) ∩ (O), (x) D(AP ) to intersect (O) at R and S (so that R is between Q and D), (xi) S(R) to intersect (O) at T . Then AQRST is a regular pentagon inscribed in (O).

2Modiﬁcation of a construction given by E. P. Starke, 3 of a regular pentagon in 13 steps using compass only. 10.4 Various constructions of the regular pentagon 65

Q C B R P

D O Y A

S E F T

(3∗) Another construction of an inscribed regular pentagon. Let O(A) be a given circle. AXY 5 (i) Construct an isosceles triangle whose height is 4 of the radius of the circle. (ii) Construct A(O) intersecting O(A) at B and C. (iii) Mark P = AX ∩ OB and Q = AY ∩ OC. (iv) Join P and Q by a line intersecting the given circle at B and C. Then AB and AC are two sides of a regular pentagon inscribed in the circle.

B C P Q B C

O X Y 66 The regular pentagon

10.5 Some interesting constructions of the golden ratio

(1) Construction of 36◦, 54◦, and 72◦ angles. Each of the following constructions begins with the division of a segment AB in the golden ratio at P .

◦ ◦ 36◦ 36 B A P

C C 54◦ 72◦

36◦ B A P

◦ 54 ◦ 72 B A P D

(2∗) Odom’s construction. Let D and E be the midpoints of the sides AB and AC of an equilateral triangle ABC. If the line DE intersects the circumcircle of ABC at F , then E divides DF in the golden ratio. A

D E F

B C 10.5 Some interesting constructions of the golden ratio 67

(3) Given a segment AB, erect a square on it, and an adjacent one with base BC.IfD is the vertex above A, construct the bisector of angle ADC to intersect AB at P . Then P divides AB in the golden ratio.

A P B C

Proof. If AB =1, then √ AP : PC =1 : 5, √ AP : AC =1 : 5+1 √ AC :2AB =1 : 5+1 √ AB : AP = 5+1:2 =ϕ :1.

This means that P divides AB in the golden ratio. (4) Hofstetter’s parsimonious construction. 4 Given a segment AB, construct (i) C1 = A(B), (ii) C2 = B(A), intersecting C1 at C and D, (iii) the line AB to intersect C1 at E (apart from B), (iv) C3 = E(B) to intersect C2 at F (so that C and F are on opposite sides of AB), (v) the segment CF to intersect AB at G. Then the point G divides the segment AB in the golden section.

4K. Hofstetter, Another 5-step division of a segment in the golden section, Forum Geom.,4 (2004) 21–22. 68 The regular pentagon

D F C2

E G A B

(5) Hofstetter’s rusty compass construction. 5 (i) Construct A(B) and B(A) intersecting at C and D. (ii) Join CD to intersect AB at its midpoint M. (iii) Construct M(AB) to intersect B(A) at E on the same side of AB as C. (iv) Join DE to intersect AB at P . Then P divides the segment in the golden ratio. E C

A B M P

5K. Hofstetter, Division of a segment into golden ratio with ruler and rusty compass, Forum Geom., 5 (2005) 133–134. Chapter 11

The regular decagon

11.1 Constructions

(1) Successive completion of rhombi beginning with four adjacent 36◦-rhombi.

(2) Another construction. 70 The regular decagon

11.2 Some relations among sides of regular polygons inscribed in a circle

Given a circle (O), we denote by sn the length of a side of a regular n-gon inscribed in the circle. Note that the radius of the circle is equal to s6.

Theorem 11.1 (Euclid). s6 + s10 : s6 = ϕ :1

Proof. Consider a regular pentagon ABCDE inscribed in a circle (O). Let F be the midpoint of the arc AB (so that AF is a side of a regular decagon inscribed in (O)). Extend AF and OB to intersect at G.

A F

E B

C D

(1) Triangle OAF is isosceles, with angles (36, 72, 72). (2) Triangle GOA is also isosceles with base angles 72◦. Therefore, F divides GA in the golden ratio. Note also that GF = FO = s6.

∗ 2 2 2 Theorem 11.2 ( Euclid). s5 = s6 + s10.

Proof. Let ABCDE be a regular pentagon inscribed in a circle (O). Let F be the midpoint of the arc AB so that AF is a side of a regular decagon inscribed in the same circle. Let P be the midpoint of the arc AF (so that AP is a side of a regular 20-gon inscribed in the circle). Join OP to intersect AB at H. (1) Since F is the midpoint of the arc AB, triangle FAB is isosceles. The radius OP is the perpendicular bisector of AF . Since H lies on this radius, it is equidistant from A and F . The triangle HAF is also isosceles. The two triangles 11.2 Some relations among sides of regular polygons inscribed in a circle 71

A P

F H

B E

C D

FAB and HAF , sharing a common base angle at A, are similar. Therefore, AB : AF = AF : AH, and

2 2 AB × AH = AF = s10.

(2) Triangle OAB is clearly isosceles, with angles 72◦, 54◦, and 54◦. In trian- HBO ∠B =54◦ ∠BOH =54◦ 3 ∠BOA gle , clearly . Also, , being 4 of . There- fore, it is also isosceles with the same base angle as triangle OAB. From the similarity of OAB and HOB,wehaveAB : OB = OB : HB, and

2 2 AB × HB = OB = s6.

(3) The result follows from noting that

2 2 AB × AH + AB × HB = AB = s5. 72 The regular decagon Chapter 12

Construction of regular n-gons

12.1 Gauss’ theorem

The construction problem of regular n-gons was solved by Carl Friedrich Gauss (1777-1855), who ﬁrst discovered in 1796 a construction of the regular 17-gon, and later established the following theorem.

Theorem 12.1 (Gauss). A regular n-gon can be constructed with ruler and com- k pass if and only if n is an integer of the form 2 · p1 · p2 ···pk, where p1, p2,...pk are distinct prime numbers of the form 2m +1.

If a number 2m +1is prime, then m =2n for some integer n. We write 2n Fn =2 +1. The ﬁrst 5 values of Fn are all prime numbers.

n 0 1 2 3 4

Fn 3 5 17 257 65537

It is not known if there are other prime numbers of the form Fn. For example, Leonhard Euler (1707-1783) has shown that

25 32 F5 =2 +1=2 + 1 = 4294967297 = 641 × 6700417.

12.2 Construction of a regular 15-gon

Let A1A2A3A4A5 be a regular pentagon inscribed in a circle (O). Construct equilateral triangles A1A2B5, A2A3B5, A3A4B1, A4A5B2, A5A1B3 with the extra 74 Construction of regular n-gons vertices inside the circle. Extend the sides of these equilateral triangles to intersect the circle again. The 10 intersections, together the vertices of the regular pentagon, form the vertices of a regular 15-gon inscribed in the circle.

O B1 A1

12.3 Construction of a regular 17-gon

To construct two vertices of the regular 17-gon inscribed in a given circle O(A). OB OA J OJ = 1 OA (1) On the radius perpendicular to , mark a point such that 4 . E OA ∠OJE = 1 ∠OJA (2) Mark a point on the segment such that 4 . (3) Mark a point F on the diameter through A such that O is between E and F and ∠EJF =45◦. (4) With AF as diameter, construct a circle intersecting the radius OB at K. (5) Mark the intersections of the circle E(K) with the diameter of O(A) through A. Label the one between O and A points P4, and the other and P6. (6) Construct the perpendicular through P4 and P6 to intersect the circle O(A) at 1 A4 and A6.

1 Note that P4 is not the midpoint of AF . 12.3 Construction of a regular 17-gon 75

Then A4, A6 are two vertices of a regular 17-gon inscribed in O(A). The polygon can be completed by successively laying off arcs equal to A4A6, leading to A8, A10,...,A16, A1 = A, A3, A5,...,A15, A17, A2.

A6 A4

A P6 F O E P4 76 Construction of regular n-gons Exercise B

Exercises marked with † are to be done with Cabri. Those marked with ∗ are more difﬁcult.

Problem B1†. Construct the following diagram, beginning with the isosceles triangle.

Problem B2†. Construct the following diagram, beginning with the two segments which are perpendicular and have equal lengths. 78 Construction of regular n-gons

Problem B3†. Construct the following diagram, beginning with the equilateral triangle.

Problem B4. AB and BC are tangent to the circle (Q) at T and C respectively. If each circle has radius 1, ﬁnd the length of BC.

A P Q C

Problem B5∗. AT is tangent to the circle (Z). If each circle has radius 5, what is the length of BC?

T C B

A X Y Z 12.3 Construction of a regular 17-gon 79

Problem B6†. (a) Two circles centers A and B have equal radii a, each containing the center of (K) the other. The circle is tangent to both√ circles and to the line joining their√ (K) 3a T 3a centers. It is known that has radius 4 and the point of tangency is 2 from O, the midpoint of AB. Construct the diagram.

T A O B

(b∗) Justify the lengths given in (a).

Problem B7∗. From a disk of unit diameter, how large can a smaller disk be removed so that the remainder balances at a point on the boundary?

P A B

Problem B8†. Construct the following inscribed equilateral triangles in a regular pentagon. Which of the two squares has larger area? 80 Construction of regular n-gons

Problem B9†. Construct the following inscribed square in a regular pentagon. Which of the two squares has larger area?

Problem B10†. Given a circle (O), construct a symmetrically inscribed pentagon ABCDE such that AB = CD = AE = the radius of the circle. If P and Q are the midpoints of the sides BC and DE, what can you say about triangle AP Q?

B E

P Q

C D

Problem B11. ABC is a right triangle with a right angle at C, AC =2and BC =1. Show that the incircle of the triangle divides the side BC in the golden ratio. 12.3 Construction of a regular 17-gon 81

Problem B12†. Outside a given circle place a chain of 8 circles each tangent to its neighbors and to the given circle.

Problem B13∗. ABCD and ABEF are squares with C, D, F , F on a circle (O). The line AB is extended to intersect (O) at P . Show that AP is divided at B in the golden ratio.

D C

A P O B

F E

Problem B14. Resolve the paradox: the area increased after rearrangement.

D 8 C

3 3 8 5 Q P Y 5 3

5 5 5 5

A B 3 X 5 5 8 82 Construction of regular n-gons

Problem B15†. Construct the following diagram.

Problem B16†. Construct the following diagram given that the shaded triangle is a 3 − 4 − 5 triangle.

Problem B17†. Construct the following diagram. 12.3 Construction of a regular 17-gon 83

Problem B18†. Construct the following diagram.

Problem B19∗†. Construct the following diagram given that the tangent line passes through the midpoint of the top edge of the square.

Problem B20∗†. Construct the following diagram given that the shaded triangle is a 3 − 4 − 5 triangle. Geometry for Middle School Teachers

Paul Yiu

Department of Mathematics Florida Atlantic University

Fall 2005

Part 3 of 3

Contents

13 Regular solids 85 13.1 The faces of a regular solid ...... 85 13.2 Edge lengths of inscribed regular solids ...... 86

14 The regular tetrahedron 87 14.1 Construction ...... 87 14.2 Measurements ...... 87 14.2.1 The height of a regular tetrahedron ...... 87 14.2.2 The circumsphere of a regular tetrahedron ...... 88 14.3 The angle between a line and a plane ...... 89 14.4 The angle between an edge and a face of a regular tetrahedron . . 89 14.5 The angle between two planes ...... 90 14.6 The dihedral angle of a regular tetrahedron ...... 90 14.7 Inscribed regular tetrahedron ...... 91 14.8 The angles Ω and Ψ ...... 92

15 The cube 93 15.1 The face diagonals and the space diagonals of a cube ...... 93 15.2 The angle between a space diagonal and a face...... 93 15.3 Inscribed cube ...... 94 15.3.1 Inscribed cube with two faces parallel to the equator . . . 94 15.3.2 Inscribed cube with a vertex at the north pole ...... 94

16 The square pyramid and the regular octahedron 95 16.1 The square pyramid ...... 95 16.2 Dihedral angle between a face and the base ...... 95 16.3 Dihedral angle between two adjacent equilateral triangles . . . . 96 16.4 Inscribed regular octahedron ...... 96 iv CONTENTS

16.4.1 Inscribed regular octahedron with a vertex at the north pole 96 16.4.2 Inscribed regular octahedron with a face parallel to the equator ...... 96

17 The regular icosahedron 99 17.1 Construction ...... 99 17.2 Dihedral angle of the regular icosahedron ...... 100

18 The regular dodecahedron 101 18.1 Euclid’s construction of the regular dodecahedron ...... 101 18.1.1 Two golden points above a mid-line of a square ...... 101 18.1.2 The pentagonal faces ...... 101 18.2 Dihedral angle between two adjacent faces of a regular dodecahedron ...... 103 18.3 Inscribed regular dodecahedron with a vertex at the north pole . . 104 84 CONTENTS Chapter 13

Regular solids

13.1 The faces of a regular solid

A regular solid is a convex solid whose faces are congruent regular polygons. Euclid has shown in his Elements that there are only ﬁve regular solids, and he gave explicit constructions of these solids.

Theorem 13.1. The faces of a regular solid can only be equilateral triangles, squares, or regular pentagons.

Proof. Around each vertex of a regular solid there are at least three faces, and the sum of the angles must be smaller than 360◦. Therefore, each angle in the regular 360◦ = 120◦ polygonal faces is smaller than 3 . These faces can only be equilateral triangles, squares, or regular pentagons.

If the faces are equilateral triangles, there are either 3, 4, or 5 triangles around each vertex. If these are squares or regular pentagons, then there must be 3 faces around each vertex.

Faces around a vertex regular solid (1) 3 equilateral triangles regular tetrahedron (2) 4 equilateral triangles regular octahedron (3) 5 equilateral triangles regular icosahedron (4) 3 squares cube (5) 3 regular pentagons regular dodecahedron 86 Regular solids

13.2 Edge lengths of inscribed regular solids

Given a sphere, we inscribe a regular solid of each type above. The length of an edge of the solid is related to the radius of the sphere as follows.

solid length√ of an edge (1) s = 2 6 R regular tetrahedron t √3 (2) regular octahedron so = 2R √ 2( 5−1) (3) regular icosahedron si = √ R √ 5 2 3 (4) sc = R cube √3 √ √ (5) s = 2 3 R = 3( 5−1) R regular dodecahedron d 3ϕ 12 The diagram below is a vertical section of a sphere with the edge lengths of the regular solids inscribed in the sphere.

sd D si

sc = ϕ · sd I

C so

st O

S Chapter 14

The regular tetrahedron

14.1 Construction

The regular tetrahedron has 4 faces, which are congruent equilateral triangles. It has four vertices and six edges. An easy way to construct a regular tetrahedron is to fold an equilateral triangle along segments joining the midpoints of its sides.

A B

D C D

14.2 Measurements

14.2.1 The height of a regular tetrahedron Let ABCD be a regular tetrahedron. We shall assume each edge of length a. Consider the equilateral triangle ABC as the base of the pyramid, with center 88 The regular tetrahedron

O. The vertex D is vertically above O. In the right triangle ADO, AD = a, AO = √1 a 3 . Therefore, a2 2 6 DO2 = AD2 − AO2 = a2 − = a2 = a2. 3 3 9 √ h = DO = 1 6a The height of the regular tetrahedron is 3 .

K C

O Ω

A M B

14.2.2 The circumsphere of a regular tetrahedron

Let R be the radius of the sphere containing the four vertices of the regular tetrahe-√ a h = 1 6a dron. Suppose each side of the tetrahedron has length√ . Its height is 3 . ABC r = 1 3a The radius of the circle containing is 3 . If the circumsphere has center K and radius R, then AK = R, OK = h − R, AO = r, and R2 =(h − R)2 + r2 = h2 − 2hR + R2 + r2; 2hR =h2 + r2, h2 + r2 a2 3 h2 3 R = = = 2 = h. 2h 2h 2h 4 The center of the circumsphere of a regular tetrahedron divides an altitude in the ratio 3:1. 14.3 The angle between a line and a plane 89

14.3 The angle between a line and a plane

The angle between a line and a plane is the angle between the line and its orthogonal projection on the plane. If a line intersects a plane Π at a point A,we choose a point P on and construct from it the perpendicular to Π to intersect the plane at Q. Q is called the orthogonal projection of P on Π. The line AQ is the orthogonal projection of on Π, and ∠PAQis the angle between and Π.

Q A

14.4 The angle between an edge and a face of a regular tetrahedron

The angle Ω between an edge of a regular tetrahedron and a face not containing it is given by

r 1 cos Ω = = √ . a 3

This is approximately 54.74◦. 90 The regular tetrahedron

14.5 The angle between two planes

Consider two planes Π1 and Π2 intersecting in a line . Let A be a point on .On each plane draw a line through A perpendicular to . The angle between the two lines is the angle between the planes Π1 and Π2.

Π2

A Π1

14.6 The dihedral angle of a regular tetrahedron

The dihedral angle Ψ of a regular tetrahedron is the angle between two faces. This OMD 1 70.53◦ is equal to angle , whose cosine is clearly 3 . This is approximately . 14.7 Inscribed regular tetrahedron 91

K C

A M B 14.7 Inscribed regular tetrahedron

Note that if the height of a regular tetrahedron is h, then its circumradius is R = 3 h h = 2 · 2R 4 . This gives 3 . Therefore, in a given sphere, we can inscribe a regular tetrahedron by choosing an arbitrary point as a vertex. Regard this vertex as the north pole. The remaining three vertices are on the latitude circle which is two thirds of the diameter below the north pole. N

(1) If st is the length of a side of a regular tetrahedron inscribed in the sphere, √ 2 2 s = √ R. t 3 (2) The latitude containing the opposite face of the north pole is 90◦ −Ψ south, 92 The regular tetrahedron

cos Ψ = 1 where 3 .

14.8 The angles Ω and Ψ

√ 3 √ √ 2 3 2 2

Ω

Ω ≈ 54.74◦ Ψ ≈ 70.53◦ 1 1 Chapter 15

The cube

15.1 The face diagonals and the space diagonals of a cube

A cube has 6 faces, 8 vertices, and 12 edges. The diagonals of a cube are of two kinds: the face diagonals and the space diagonals. If each side of a cube has length a, then √ (i) a face diagonal has length 2√a, and (ii) a space diagonal has length 3a.

Ω

√ R = 3 a The radius of the sphere containing the vertices of the cube is 2 .

15.2 The angle between a space diagonal and a face

The angle between a space diagonal and a face of a cube is 90◦ − Ω. 94 The cube

15.3 Inscribed cube

There are two natural ways to inscribe a cube in a given sphere.

15.3.1 Inscribed cube with two faces parallel to the equator If two opposite faces are parallel to the equator, here is a vertical section through two face diagonals on the latitude circles. The northern latitude circle can be determined by erecting an equilateral triangle on EW (two antipodes on the equator). The center of the triangle is the center of the latitude circle.

N N

s C Ω t Ω Ω Ψ O W E O O

S S

15.3.2 Inscribed cube with a vertex at the north pole If we put one vertex at the north pole, then the opposite vertex is at the south pole. The remaining 6 vertices form antipodal two equilateral triangles on two latitude circles. These two latitude circles divide the diameter NS into three equal parts. Chapter 16

The square pyramid and the regular octahedron

16.1 The square pyramid

Let ABCD be a square and E a point vertically above the center O of ABCD such that EAB, EBC, ECD, and EDA are equilateral triangles. The height of E above O is the same as OA. This means that each slant edge of the square pyramid makes and angle 45◦ with the square base.

D C

A B

16.2 Dihedral angle between a face and the base √ Note that in the right triangle EMO, OE : OM = 2:1. This means that ∠EMO =Ω. 96 The square pyramid and the regular octahedron

D C

O Ω

A M B 16.3 Dihedral angle between two adjacent equilateral triangles

If we take an identical copy of the square pyramind and identify the square faces, we obtain a solid with 8 congruent equilateral triangles as faces. This is a regular octahedron. The dihedral angle of each edge is clearly 2Ω. This is the same for the angle between two equilateral triangle faces of the square pyramid.

16.4 Inscribed regular octahedron

16.4.1 Inscribed regular octahedron with a vertex at the north pole If a regular octahedron has a vertex at the north pole, its opposite vertex is at the south pole. The other four vertices form a square on the equator. If the sphere√ has radius R, the length of a side of an inscribed regular octahedron is so = 2R.

16.4.2 Inscribed regular octahedron with a face parallel to the equator If a face of a regular octahedron is parallel to the equator, say an equilateral triangle ABC on the north θ-latitude circle, its opposite face ABC is an antipodal equilateral triangle at the south θ-latitude circle. The orthogonal projections X, Y , Z of A, B, C on the south latitude circle are such that AZBXCY is a regular hexagon. 16.4 Inscribed regular octahedron 97 N

S If the radius of the latitude circle is r, then (i) each side of the regular hexagon AZBXCY has length r, (ii) each side√ of the equilateral triangle (and hence each edge of the octahedron) has length 3r, √ ( 3r)2 − r2 = √and (iii) the separation of the two latitude circles is given by 2r. These are the same two latitudes to inscribe a cube with two faces parallel to the equator.

A r A A

√ 2 r 2 √ Ω Ω 3r W E O O

Ω X C r

S 98 The square pyramid and the regular octahedron Chapter 17

The regular icosahedron

17.1 Construction

In Book XIII of the Elements, Euclid shows how to inscribe a regular icosahedron whose faces are congruent equilateral triangles with vertices on a given sphere. N

s I i

S In a vertical great circle of the sphere, inscribe a rectangle of dimensions 2:1. Place a regular pentagon on the latitude circle at level I along the longer side of the rectangle. These vertices, their antipodes on the sphere, and the north and south poles, form the 12 vertices of a regular icosahedron inscribed in the sphere. There are 20 faces and 30 edges. If the sphere has radius R, each side of the inscribed regular icosahedron has length √ 2( 5 − 1) s = √ R. i 5 100 The regular icosahedron

17.2 Dihedral angle of the regular icosahedron

Let A1A2A3A4A5 be a regular pentagon on the latitude circle through I.IfM is the midpoint of the side NA1, then the angle A2MA5 is the dihedral angle of the two faces containing the edge NA1. If each side of the regular pentagon has length s, then A2A5 = ϕ · s. The dihedral angle is equal to angle A2PA5 in the diagram below. This is approximately 138.183◦.

A2 A5 Chapter 18

The regular dodecahedron

18.1 Euclid’s construction of the regular dodecahedron

18.1.1 Two golden points above a mid-line of a square Given a square with center O, let A and A be the midpoints of two opposite sides of the square. We call AA a mid-line of the square. Divide the segment OA (respectively OA) in the golden ratio at a point B (respectively B), and erect a square on the segment OB (respectively OB), in a plane perpendicular to the square. Denote the vertex of the square opposite to O by P (respectively P ). We choose P and P on the same side of the square, and call them the two golden points above the mid-line AA of the square.

P K P

A A B O B

18.1.2 The pentagonal faces Consider a cube with a mid-line on each of its six faces: 102 The regular dodecahedron

Faces Mid-line top and bottom left-right left and right front-rear front and rear top-bottom

Construct two golden points above the indicated mid-line of each face, all externally of the cube. Euclid showe that the 8 vertices of the cube, together with the 12 golden points constructed above the faces, form the vertices of a regular dodecahedron.

Theorem 18.1 (Euclid). The two vertices common to the top and the front faces, the two golden points above the left-right mid-line of the top face, and the “top” golden point above the top-bottom mid-line of the front face form a planar regular pentagon.

P K P

X A B O C Y

D Q

Proof. Assume OB =1so that OA = ϕ. The lengths of the various segments are very easily related to these two. (1) We ﬁrst show that the triangle XQY and the quadrilateral XY PP are coplanar. Note that the points Q, DC, OK, K are on the same vertical plane, and tan OCK = OK = 1 = ϕ − 1 (i) OC ϕ , tan DQC = DC = ϕ−1 = ϕ − 1 (ii) DQ 1 . This shows that Q, C, K are along the same line, and the two polygons are coplanar. 18.2 Dihedral angle between two adjacent faces of a regular dodecahedron103

(2) Note that

XP2 =XB2 + BP 2 = AB2 + OC2 + BP 2 =OA2 + AB2 + OB2 =ϕ2 +(ϕ − 1)2 +12 =2ϕ2 − 2ϕ +2 =2(ϕ +1)− 2ϕ +2=4.

This means XP =2. By symmetry, YP = XQ = YQ =2=PP. Therefore, the pentagon QXP P Y is equilateral. (3) A similar calculation shows that PY = XY = XP . Therefore, triangle XPP and YPP are isosceles triangles with top angles 108◦. (4) Now it is easy to see that ∠PXQand P YQare also 108◦. This completes the proof that QXP P Y is a regular pentagon.

Theorem 18.1 holds if we replace the front face replaced by the rear face. These two regular pentagons share a common edge, namely, the segment joining the two golden points above (the left-right mid-line of) the top face. Similarly, above the indicated mid-line of each face, there is a pair of regular pentagons. This gives a regular solid with 12 faces which are regular pentagons. It has 30 edges.

Corollary 18.2. The length of a side of a regular dodecahedron inscribed in a given sphere is 1 s = · s , d ϕ c where sc is the length of a side of an inscribed cube.

18.2 Dihedral angle between two adjacent faces of a regular dodecahedron

The dihedral angle between two adjacent faces of a regular dodecahedron is 2Φ ≈ 116.565◦. In the diagram below, CC is the front-rear mid-line of the top face, and K is the midpoint of the segment PP above the left-right mid-line of the same face. 104 The regular dodecahedron K

Φ Φ 1

O C 2ϕ C

18.3 Inscribed regular dodecahedron with a vertex at the north pole

Suppose each face of an inscribed dodecahedron is a regular pentagon of side length s and diagonal d. If one vertex of a regular dodecahedron is at the north pole, its opposite vertex is at the south pole, the remaining 18 vertices lie on 4 latitude circles, the 3 neighbors of the north pole on one latitude, and 6 on a lower latitude each at a distance d from the north pole. N

s D d

ϕ · sd

S (1) On the latitude circle at the level of C, which is one-third of the diameter below the north pole, inscribe a hexagon whose alterate sides have two lengths in the golden ratio. This hexagon can be constructed as follows. First inscribe an equilateral triangle in the circle. Divide the sides of the triangle in the golden ratio. Join each vertex to the point of division on its opposite side to meet the circle again. These six points deﬁne the hexagon on the latitude circle. D 2 R (2) On the latitude circle (level ) of radius 3 , construct an equilateral triangle whose vertices are on the meridians of the midpoints of the longer sides of the hexagon on the latitude at level C. 18.3 Inscribed regular dodecahedron with a vertex at the north pole 105

(3) The 9 points on these two latitudes circles, together with the antipodal points, and the north and south poles, form the vertices of a regular dodecahedron inscribed in the given sphere.