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. What is the smallest possible area of the triangle if a, b, c and r are distinct integers?
(In reply to
Some progress by Brian Smith)
Continuing from my previous post, r is an integer such that r = a*b*c^2 / ((a+b+c)*(a+b)^2) with (a,b,c) is a Pythagorean triple.
In order for r to be an integer (a+b) must divide the numerator. Assume (a,b,c) is a primitive Pythagorean triple. Then a,b,c are pairwise coprime and exactly one of a and b is even with c always being odd.
In general if a and b are coprime then any polynomial f(a,b) with integer coefficients then a is coprime to f if and only if one of the terms is a power of b. The converse is also true that b is coprime it f if and only if one of the terms is a power of a.
Then a is coprime to a+b, b is coprime to a+b, and c^2=a^2+b^2 is coprime a*b. c^2 also equals (a+b)^2-2*a*b, which then implies that c^2 is coprime to (a+b)^2.
Then for any solution, the base primitive Pythagorean triple must be scaled up by a factor of at least (a+b)^2 for r to become an integer. The smallest primitive Pythagorean triple is (3,4,5) which needs to be scaled up by a factor of 49 to yield (147, 196, 245). This makes r=25 and the triangle area of 14406.
re: Some progress
