Asuhayama Shrine is located 750 $m$ north of Ibara Station on the Ibara Line of the Ibara Railway.

 

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Problem

As shown in the figure, the side $BC$ is one side of a regular pentagon, which has a diagonal $DF$.

When we know the lengths of sides $AB$ and $AC$ and the angle $∠A$, find the length of the diagonal $DF$ of the regular pentagon.

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Solution

Let $AB=a, \ AC=b, \ ∠A=θ$, and $G$ be the foot of the perpendicular line from point $B$ to $AC$. Then,

$$cos⁡θ=\frac{AG}{a},$$

$$∴ \ AG=a×cos⁡θ.$$

Moreover,

$$tan⁡θ=\frac{BG}{AG}=\frac{BG}{a×cos⁡θ},$$

$$∴ \ BG=a×cos⁡θ×tan⁡θ=a×sin⁡θ. \qquad \left(∵ \ tan⁡θ=\frac{sin⁡θ}{cos⁡θ}. \right)$$

Since $CG=AC-AG=b-a×cos⁡θ$,

$$BG^2+CG^2=BC^2,$$

$$BC^2=a^2×sin^2⁡θ+(b-a×cos⁡θ )^2,$$

$$BC^2=a^2×(sin^2⁡θ+cos ^2⁡θ )-2ab×cos⁡θ+b^2,$$

$$BC^2=a^2-2ab×cosθ+b^2,$$

$$∴ \ BC=\sqrt{a^2-2ab×cosθ+b^2}.$$

Next, let $H$ be the foot of the perpendicular line drawn from the vertex $E$ of the regular pentagon to the diagonal $DF$.

Since $∠D \ (=∠F) \ =36°$ in the above diagram,

$$DH=cos36°×DE,$$

$$∴ \ DF=2 \ cos36°×DE.$$

Now, find the value of $cos⁡36°$.

Using the double angle formulae $sin⁡2θ=2 \ sin⁡θ×cos⁡θ$  and  $cos⁡2θ=1-2 \ sin^2⁡θ$,

$$sin72°=2 \ sin36°×cos36°=2×(2 \ sin18°×cos18°)(1-2 \ sin^2⁡ 18°)$$

$$=(4 \ sin18°×cos18°)(1-2 \ sin^2 18°).$$

Furthermore, since $sin⁡(\frac{π}{2}-θ)=cos⁡θ$,

$$sin72°=cos18°,$$

$$cos18°=(4 \ sin18°×cos18°)(1-2 \ sin^2 18°),$$

$$4 \ sin18°×cos18°-8 \ sin^3⁡ 18°×cos18°-cos18°=0,$$

$$∴ \ 8 \ sin^3 18°×cos18°-4 \ sin18°×cos18°+cos18°=0. \qquad (1)$$

By dividing both sides of equation $(1)$ by $cos⁡18° \ (≠0)$ and setting $sin18°=x$,

$$8x^3-4x+1=0. \qquad (2)$$

The cubic equation $(2)$ can be solved as follows:

$$(2x-1)(4x^2+2x-1)=0.$$

First, since $sin18°≠\frac{1}{2} \ (=sin30°)$, $\langle 2x-1=0 \rangle$ is inappropriate.

Therefore,

$$4x^2+2x-1=0,$$

$$∴ \ x=\frac{-1±\sqrt{1+4}}{4}=\frac{-1±\sqrt{5}}{4}.$$

Clearly

$$sin18°>0.$$

$$∴ \ x \ (=sin18°) \ =\frac{\sqrt{5}-1}{4}.$$

Using the half-angle formula $sin^2⁡θ=\frac{1-cos⁡2θ}{2}$,

$$sin^2 18°= \left( \frac{\sqrt{5}-1}{4} \right)^2=\frac{6-2 \sqrt{5}}{16}=\frac{1-cos⁡36°}{2},$$

$$1-cos36°=\frac{6-2 \sqrt{5}}{8},$$

$$∴ \ cos36°=\frac{1+\sqrt{5}}{4}.$$

Since $DE=BC$ and the value of $BC$ was calculated above,

$$DF=2 \ cos36°×DE=\frac{1+\sqrt{5}}{2}×BC,$$

$$∴ \ DF=\frac{1+\sqrt{5}}{2}× \sqrt{a^2-2ab×cosθ+b^2}.$$

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Reference

Hidetoshi, Fukagawa and Tony Rothman (2008) Sacred Mathematics: Japanese Temple Geometry, p.133. Princeton University Press.

Yoshikazu Yamakawa, ed. (1997) Okayama ken no Sangaku (Sangaku in Okayama Prefecture), p.44; pp.331-332.