## Three Incircles In a Right Triangle: What Is This About? A Mathematical Droodle

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Solution The applet purports to suggest the following sangaku [Temple Geometry, #2.3.2, p. 29]:

ABC is right-angled at C, and CD is the perpendicular from C onto AB. If O1(r1), O2(r2), O3(r3) are the incircles of the respective triangles ABC, ADC, and BDC, show that

 r1 + r2 + r3 = CD.

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(This is an undated Sangaku from the Iwate prefecture.)

The three triangles are right-angled and, therefore, similar. Let the sides of ΔABC be a, b, c, in the usual manner. Then corresponding sides of ΔADC are b²/2, ab/c, b, and those of ΔBDC are ab/c, a²/c, and a. The inradii of the three triangles are easily found to be

 (1) r1 = (a + b - c)/2 = c/c·(a + b - c)/2, r2 = (b²/2 + ab/c - b)/2 = b/c·(a + b - c)/2, r3 = (ab/c + a²/c - a)/2 = a/c·(a + b - c)/2.

(2)
 r1 + r2 + r3 = (a + b - c)(a + b + c) / 2c = [(a + b)² - c²] / 2c = [(a² + b² - c²) + 2ab] / 2c = ab / c,

where we have used the Pythagorean theorem. For the area S of ΔABC we have 2S = ab, on the one hand, and, on the other, 2S = ch, where h is the altitude CD. Comparing the two shows that ab/c = h, as required.

Note that the derivation (1)-(2) may be seen as the converse of one of the proofs of the Pythagorean theorem.

### References

1. H. Fukagawa, D. Pedoe, Japanese Temple Geometry Problems, The Charles Babbage Research Center, Winnipeg, 1989

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