ABC is a right triangle with legs a and b and hypotenuse c. Two circles of radius r are placed inside the triangle, the first tangent to a and c, the second tangent to b and c, and both circles externally tangent to each other. Draw a third circle of radius s tangent externally to the first two circles, and to the hypotenuse. What is the smallest possible radius of the third circle if a, b, c, r and s are distinct integers?
It's easy to show s=r/4.
From the half-angle tangent formula we can get a formula for c in terms of a,b,r:
c=rb/(c-a)+2r+ra/(c-b)
or
r=c/(c^2+2ab-ac-bc)
From here I just imported a list of Pythagorean triples into a spreadsheet and searched. For integer values of s. The smallest is
a=84, b=112, c=140, r=20, s=5. (This is just 28 times a 3-4-5)
In retrospect, r is just a fraction with c on top, so we should seek integer multiples of small triples and the numerator of s can't be smaller than c. (3-4-5 gives s=5/28 so this must be the minimum.)
The next smallest solution (besides double the previous) is
5-12-13 gives s=13/34.
Edited on February 12, 2019, 8:58 am
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Posted by Jer
on 2019-02-11 15:34:08 |
Solution