Problem

When the area of a right-angled triangle is $S$, and the diameter of a circle inscribed on the three sides is $d$, find the short side, long side, and hypotenuse of the right-angled triangle.　(In the figure below, $AB$ is the short side, $BC$ is the long side, and $CA$ is the hypotenuse.)

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Solution

From the figure we can see that

$$AB+BC=CA+d,$$

$$∴ AB+BC-CA=d. —–[1]$$

We also know that

$$S=\frac{1}{2}×AB×BC,$$

$$∴ AB\cdot BC=2S. —–[2]$$

Furthermore, from Heron’s formula, we have

$$S=\frac{1}{2}×\frac{d}{2}×(AB+BC+CA),$$

$$∴ AB+BC+CA=\frac{4S}{d}. —–[3]$$

From [3]-[1], it follows that

$$2CA=\frac{4S}{d}-d=\frac{4S-d^2}{d},$$

$$∴ CA=\frac{4S-d^2}{2d}.$$

Thus, it turns out that

$$AB+BC=CA+d=\frac{4S+d^2}{2d}, —–[4]$$

Moreover, from the Pythagorean theorem, we see that

$$AB^2+BC^2=CA^2,$$

$$(AB-BC)^2=AB^2+BC^2-2AB\cdot BC=CA^2-4S=\frac{16S^2-24Sd^2+d^4}{4d^2},$$

$$∴ AB-BC=-\sqrt{\frac{16S^2-24Sd^2+d^4}{2d}}. —–[5]$$

$(∵ AB<BC)$

From [4]+[5], we have

$$2AB=\frac{4S+d^2-\sqrt{16S^2-24Sd^2+d^4 }}{2d},$$

$$∴ AB=\frac{4S+d^2-\sqrt{16S^2-24Sd^2+d^4}}{4d}.$$

From [4]-[5], we see that

$$2BC=\frac{4S+d^2+\sqrt{16S^2-24Sd^2+d^4}}{2d},$$

$$∴ BC=\frac{4S+d^2+\sqrt{16S^2-24Sd^2+d^4}}{4d}.$$

(Answer) $AB=\frac{4S+d^2-\sqrt{16S^2-24Sd^2+d^4}}{4d},$

$BC=\frac{4S+d^2+\sqrt{16S^2-24Sd^2+d^4}}{4d},$

$CA=\frac{4S-d^2}{2d}.$

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Reference

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