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

 

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Solution

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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.

Adding the three identities gives

(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

    Write to:

    Charles Babbage Research Center
    P.O. Box 272, St. Norbert Postal Station
    Winnipeg, MB
    Canada R3V 1L6

Sangaku

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