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How To Solve An Old Japanese Geometry Problem

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Tangent circles in the ratio 2 : 1

Hiroshi Okumura and Masayuki Watanabe

Inthis article we consider the followingold Japanese geometry problem (see Figure 1), whose statement in [1, p. 39] is missingthe condition that two of the vertices are the opposite ends of a diameter. (The authors implicitly correct the omission in the proof they provide on page 118.) We denote by

O (r )the circle with centreO, radiusr.

Problem[1, Example 3.2]. The squaresACB0D0 andABC0Dhave a com- mon vertex A, and the vertices C andB0, C0 andB lie on the circle O (R) whose diameter isB0C0,Alying within the circle. The circleO1(r1)touches

AB andAC and also internally touchesO (R), andO2(r2)is the incircle of triangleABC. Show that

r

1

= 2r

2. (1)

B 0

C 0

D B

C

D 0

A

Figure 1. Figure 2.

We shall see that B0C0, being a diameter, is merely a sucient condition.

Figure 2 shows that some additional property involvingB0andC0is required for deducing (1). In Theorem 1 we give a simple condition that implies (1).

Theorem 1. For any triangle ABC, ifA0 is the re ection inBC of a point

A, then the radius of one of the circles internally touching the circumcircle ofA0BCand also touchingABandAC is twice the size of the radius of the incircle ofABC.

Copyright c 2001 Canadian Mathematical Society

(2)

Proof. Let AA0 intersect BC at P and the circumcircle (say) of A0BC again at D, and let Qbe the foot of the perpendicular from C to BA0 (see Figure 3). Then\BA0D =\QCBsince the right trianglesBA0P andBCQ share a common angle \A0BC. Moreover \BA0D = \BCD. Hence,

\QCB = \BCD. This implies thatH andDare symmetric inBC, where

H is the orthocentre of A0BC. Thus, D is the orthocentre of ABC, and therefore, D BC and ABC share a common nine-point circle (say), and

touches the incircle (say) ofABC internally by Feuerbach's Theorem.

Therefore, the dilatation of magni cation2with centre Acarries into the circumcircle ofD BC, which is , and into one of the circles touchingAB,

AC and internally. This implies that the last circle is twice the size of , and the proof is complete.

We mentioned \one of the circles" in the theorem, but if we introduce orientations of lines and circles, the circle is determined uniquely. Let us assumethat andABChave counter-clockwise orientations. Thenthe circle of twice the size of (illustrated by a dotted line in Figure 3) is the one which touches ,;AB!and;CA!so that the orientations at the points of tangency are the same (see the arrows in Figure 3), and the point of tangency of the circle with is the re ection ofAin the point of tangency of and .

A 0

B

A C

Q

H

D

P

Figure 3.

The following two properties follow immediately (see Figures 4 and 5).

They can be found without a proof in [1, p. 29].

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Corollary 1. IfABC is a right triangle with right angle atA, then the circle touching the circumcircle ofABC internally and also touchingABandAC is twice the size of the incircle ofABC.

Corollary 2. IfCB0Ais an isosceles triangle with CB0 = CA, andB is a point lying on the lineB0A, then one of the circles touching the circumcircle of BCB0 internally and also touchingAB and AC is twice the size of the incircle ofABC.

Figure 4.

C B

A

Figure 5.

B

B 0

A C

The last corollary holdssince the re ection inBCofAlieson the circumcircle of BCB0. Though Figure 5 illustrates only the case where A lies on the segmentBB0, and this is the case stated in [1], the corollary does not need this condition (see Figure 6b).

Conversely, for a triangleABC, letA0be the re ection inBCofA, and letB0be the intersection of the lineABand the circumcircle ofA0BC. Then

CA=CB

0holds. Hence, with the two sidesABandACand their intersec- tions with the circumcircle, we can construct two similar isosceles triangles (see Figures 6a and 6b, where the ratio of the two smaller circles is2 : 1).

Alies within the circumcircle if and only if\CAB >90, and Figure 1 can be obtained by letting\CAB=135.

Figure 6a.

B A

0

C

B 0

C 0 A

A B

0

C

A 0 C

0

B

Figure 6b.

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The following property with a proof using trigonometric functions can be found in [2, p. 75] (see Figure 7).

Corollary 3. IfACB0D EandABC0D0E are two regular pentagons sharing the side AE, then the circle internally touching the circumscribed circle of

CB 0

C 0

Band the sidesABandACis twice the size of the incircle ofD ED0.

C 0

B 0

B C

D 0

D A

E

Figure 7.

B

n+1 A

n+1

B

n A

n

B

2n A

2n B

2n+1 A

2n+1 B

1 A

1 B

2 A

2

"

A

2n+1;[n=2]

Figure 8.

The corollary can be generalized yet further (see Figure 8).

Theorem 2. If A1, A2, , A2n+1 are vertices of a regular (2n+ 1){gon lying in this order, and Bi is the re ection ofAi in the lineA1A2n+1, and

is the circle passing through An, An+1, Bn+1, Bn, then one of the cir- cles touching the lines A1A2, B1B2 and internally is twice the size of the incircle of the triangle made by the lines A2nA2n+1, B2nB2n+1 and

A

2n+1;[n=2]

B

2n+1;[n=2], where[x]is the largest integer which does not ex- ceedx.

Proof. A2n+1is the centre of , and the re ection ofA2n+1;[n=2]in the line through the centres of the two regular polygons isA1+[n=2]. Let us produce

A

2n+1 A

1toP, whereP lies on andA1lies on the segmentA2n+1P, and letQbe the intersection ofA1PandA1+[n=2]B1+[n=2]. By simplecalculation we have

A

2n+1

P = 2rcos



2(2n+1)

, A2n+1A1 = 2rsin 

2n+1

, where r is the circumradius ofA1A2A2n+1. Now let us suppose thatn is odd; then[n=2]=(n;1)=2and we have

A

1

Q = rsin



2

2n+1

 n;1

2 +



2n+1



; A

2n+1 A

1

2

= r



sin n

2n+1

;sin 1

2n+1



,

A

2n+1

Q = A

2n+1 A

1 +A

1 Q

= r



sin n

2n+1

+sin 1

2n+1



,

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A

1 Q+A

2n+1

Q = 2rsin n

2n+1

 = 2rsin



1

2

; 1

2(2n+1)





= 2rcos 1

2(2n+1)

.

Therefore, we get A1Q+A2n+1Q = A2n+1P, and this implies that Q is the mid-point of A1P. Similarly, we can prove the same fact in the case of

nbeing even. Thus, the ratio of the two similar isosceles triangles made by the linesA1A2,B1B2,A1+[n=2]B1+[n=2], and A1A2,B1B2, the tangent of

atP, is1:2and the theorem is proved.

A

2n+1 A1

B2 Q P

B1+(n;1)=2 A1+(n;1)=2

Figure 9 (whennis odd).

The authors would like to thank the referee for suggesting helpful comments.

References

1. H. Fukagawa and D. Pedoe, Japanese Temple Geometry Problems, Charles Bab- bage Research Centre Canada, 1989.

2. H. Fukagawa and D. Sokolowsky, Japanese mathematics - how many problems can you solve? Volume 2, Morikita Shuppan, Tokyo, 1994 (in Japanese).

Hiroshi Okumura Masayuki Watanabe

Maebashi Institute of Technology

460-1 Kamisadori Maebashi Gunma 371-0816, Japan

okumura@maebashi-it.ac.jp watanabe@maebashi-it.ac.jp

References

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