Circles Lined on the Legs of a Right Triangle: What Is This About?
A Mathematical Droodle
What if applet does not run? 
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Copyright © 19962018 Alexander Bogomolny
The applet purports to suggest an extension of the following sangaku [Temple Geometry, p. 36]:
ABC is a rightangled triangle at C, and 8 circles of the same radius r lie within the triangle and have various contacts. O_{1}(r) touches BC and AB. O_{2}(r) touches BC and O_{1}(r), O_{3}(r) touches BC and O_{2}(r), O_{4}(r) touches BC and O_{3}(r), and O_{5}(r) touches both BC and AC and also O_{4}(r). O_{6}(r) touches AC and O_{5}(r), O_{7}(r) touches AC and O_{6}(r), and finally O_{8}(r) touches AC and AB and O_{7}(r). Show that the radius of the incircle of triangle ABC is equal to 3r. 
(The problem appears on a surviving 1892 tablet in the Hyogo prefecture. )
Draw a line parallel to the b = AC side of the triangle through the upper most point of tangency of the circles. We get two similar right triangles with legs a and (a  6r) and inradius, say, R and r. Which gives a proportion
(1)  r/R = (a  6r) / a = 1  6r/a. 
For a triangle similarly obtained in the A corner we have
r/R = (b  8r) / b = 1  8r/b. 
Comparing the two expressions we see that
8a = 6b or 4a = 3b. 
If, for example, a = 3 then b = 4 and the hypotenuse equals 5 giving a 345 triangle. The inradius can be found from
R = (a + b  c)/2, 
or, in this case, R = (3 + 4  5)/2 = 1. On the other hand, from say (1),
r = aR / (a + 6R), 
so that r = 1/3 and, indeed, R = 3r, as required.
As the applet shows (and calculations are entirely generic), any Pythagorean triple
References
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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Activities Contact Front page Contents Geometry
Copyright © 19962018 Alexander Bogomolny