Katayama-hiko Shrine (1873), Osafune-cho, Setouchi City, Okayama Prefecture (09)

Problem

There are two large circles and two small circles inside a rhombus as shown in the figure.

If the longer diagonal of the rhombus is $85 \ inches$ and the shorter is $42 \ inches$, find the diameters of the large and small circles.

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$\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$

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Solution

Let the diameters of the larger and smaller circles be $D$ and $d$, respectively.

If the longer diagonal of the rhombus is $a$ and the shorter is $b$, the length of one side $AB$ of the rhombus is as follows:

$$AB=\sqrt{\left(\dfrac{a}{2}\right)^2+ \left(\dfrac{b}{2}\right)^2}=\frac{\sqrt{a^2+b^2}}{2}.$$

Then,

$$AB∶AE=OB∶OF,$$

$$\frac{\sqrt{a^2+b^2}}{2}∶\frac{b}{2}=\left(\dfrac{a}{2}-\dfrac{D}{2}\right)∶\frac{D}{2},$$

$$∴ \ D=\frac{ab}{b+\sqrt{a^2+b^2}}. \qquad [1]$$

Next,

$$PE=\sqrt{\left(\dfrac{D+d}{2}\right)^2-\left(\dfrac{D}{2}\right)^2}=\frac{\sqrt{2Dd+d^2}}{2}. \qquad [2]$$

Also,

$$AB∶BE=AP∶PG,$$

$$\frac{\sqrt{a^2+b^2}}{2}∶\frac{a}{2}=AP∶\frac{d}{2},$$

$$∴ \ AP=\frac{d \sqrt{a^2+b^2}}{2a}. \qquad [3]$$

Since $[2]+[3]=AE$,

$$\frac{\sqrt{2Dd+d^2}}{2}+\frac{d \sqrt{a^2+b^2}}{2a}=\frac{b}{2},$$

$$a \sqrt{2Dd+d^2}=ab-d \sqrt{a^2+b^2}. \qquad [4]$$

Squaring both sides of $[4]$ and rearranging for $d$,

$$a^2 (2Dd+d^2 )=a^2 b^2-2abd \sqrt{a^2+b^2}+d^2 (a^2+b^2 ),$$

$$b^2 d^2-2(ab \sqrt{a^2+b^2}+a^2 D) d+a^2 b^2=0,$$

$$∴ \ d=\frac{ab \sqrt{a^2+b^2}+a^2 D \pm \sqrt{(ab \sqrt{a^2+b^2}+a^2 D)^2-a^2 b^4}}{b^2},$$

$$∴ d=\frac{ab \sqrt{a^2+b^2}+a^2 D \pm a \sqrt{a^2 b^2+2abD \sqrt{a^2+b^2}+a^2 D^2}}{b^2}. \qquad [5]$$

Substituting $a=85$ and $b=42$ into $[1]$ and $[5]$,

$$D≒26.0945, \qquad and \qquad d≒585.1653 \quad or \quad 12.3469.$$

However, $d≒585.1653$ is too large. Thus,

$$d≒12.3469.$$

$$\begin{eqnarray} (Answer) \quad large \ circle: \quad approx. 26.0945 \ inches \\ and \qquad small \ circle: \quad approx. 12.3469 \ inches. \end{eqnarray}$$

Commentary

The questioner’s answer to this problem is as follows:

$$D=21.559 \qquad and \qquad d=19.854.$$

These are different from the answers found above.

This problem is a quotation from Guow-Sanpow (1699) by Kenryu Miyake, and both the problem text and the numerical values are the same as in the original.

However, the diagram was drawn incorrectly when quoting, and the correct diagram in the original source is as follows:

Then, let us calculate the diameters of the two circles on the left and right and the two on the top and bottom.

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Solution

Let the diameters of the circles on the left and right and on the top and bottom be $D$ and $d$, respectively.

If the longer diagonal of the rhombus is $a$ and the shorter is $b$, then the length of one side $AB$ of the rhombus is as follows:

$$AB=\sqrt{\left(\dfrac{a}{2}\right)^2+ \left(\dfrac{b}{2}\right)^2}=\frac{\sqrt{a^2+b^2}}{2}.$$

Then,

$$AB∶BE=AP∶PG,$$

$$\frac{\sqrt{a^2+b^2}}{2}∶\frac{a}{2}=\frac{b-d}{2}∶\frac{d}{2},$$

$$\sqrt{a^2+b^2}:a=(b-d):d,$$

$$d \sqrt{a^2+b^2}=ab-ad,$$

$$d (a+\sqrt{a^2+b^2})=ab,$$

$$∴ \ d=\frac{ab}{a+\sqrt{a^2+b^2}}. \qquad [a]$$

Then,

$$EO=\sqrt{\left(\dfrac{D+d}{2}\right)^2-\left(\frac{d}{2}\right)^2}=\frac{\sqrt{D^2+2Dd}}{2}. \qquad [b]$$

Also,

$$AB∶AE=OB∶OF,$$

$$\frac{\sqrt{a^2+b^2}}{2}∶\frac{b}{2}=OB∶\frac{D}{2},$$

$$∴ \ OB=\frac{D \sqrt{a^2+b^2}}{2b}. \qquad [c]$$

Since $[b]+[c]=EB$,

$$\frac{\sqrt{D^2+2Dd}}{2}+\frac{D \sqrt{a^2+b^2}}{2b}=\frac{a}{2},$$

$$b \sqrt{D^2+2Dd}=ab-D \sqrt{a^2+b^2}. \qquad [d]$$

Squaring both sides of $[d]$ and rearranging for $D$,

$$b^2 (D^2+2Dd)=a^2 b^2-2abD \sqrt{a^2+b^2}+D^2 (a^2+b^2),$$

$$a^2 D^2-2(ab \sqrt{a^2+b^2}+b^2 d)D+a^2 b^2=0,$$

$$∴ \ D=\frac{ab \sqrt{a^2+b^2}+b^2 d \pm \sqrt{(ab \sqrt{a^2+b^2}+b^2 d)^2-a^4 b^2}}{a^2},$$

$$∴ \ D=\frac{ab \sqrt{a^2+b^2}+b^2 d \pm b \sqrt{a^2 b^2+2abd \sqrt{a^2+b^2}+b^2 d^2}}{a^2}. \qquad [e]$$

Substituting $a=85$ and $b=42$ into $[a]$ and $[e]$,

$$d≒19.8543, \qquad and \qquad D≒81.8340 \quad or \quad 21.5558.$$

However, $D≒81.8340$ is too large. Thus,

$$D≒21.5558.$$

For $d$, it matches the answer of the Sangaku, and for $D$, it matches the answer up to the second decimal place.

Reference

Hidetoshi, Fukagawa and Tony Rothman (2008) Sacred Mathematics: Japanese Temple Geometry, p.147; pp.164-165. Princeton University Press.

Yoshikazu Yamakawa, ed. (1997) Okayama ken no Sangaku (Sangaku in Okayama Prefecture), p.40; pp.350-352.