Problem

If the length of the hypotenuse of a right-angled triangle is $a$ and the diameter of the circle inscribed in the right-angled triangle is $d$, what are the lengths of the short and long sides of the right-angled triangle?　(In the figure below, $AB$ is the short side and $BC$ is the long side.)

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Solution

From the figure, it follows that

$$AB+BC=a+d, —–[1]$$

$$(AB+BC)^2=AB^2+BC^2+2AB\cdot BC=a^2+d^2+2ad,$$

$$AB^2+BC^2=a^2+d^2+2ad-2AB\cdot BC.$$

However, obviously $AB^2+BC^2=a^2$, so

$$d^2+2ad-2AB\cdot BC=0,$$

$$∴ 2AB\cdot BC=d^2+2ad.$$

Furthermore, it turns out that

$$(AB-BC)^2=AB^2+BC^2-2AB\cdot BC=a^2-(d^2+2ad),$$

$$∴ AB-BC=-\sqrt{a^2-2ad-d^2}, —–[2]$$

$(∵ AB<BC)$

From [1]+[2], we see that

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

$$∴ AB=\frac{a+d-\sqrt{a^2-2ad-d^2}}{2}.$$

From [1]-[2], we have

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

$$∴ BC=\frac{a+d+\sqrt{a^2-2ad-d^2}}{2}.$$

(Answer) $AB=\frac{a+d-\sqrt{a^2-2ad-d^2}}{2},$

$BC=\frac{a+d+\sqrt{a^2-2ad-d^2}}{2}.$

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Reference

Yoshikazu Yamakawa, ed. (1997) Okayama ken no Sangaku (Sangaku in Okayama Prefecture), p.50; pp.310-311.