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Inradius as the GCD of sides (Posted on 2024-02-14) Difficulty: 3 of 5
Positive integers a<b<c are lengths of sides of a right triangle whose inradius is equal to gcd(a+1,b)2. Find a,b,c.

No Solution Yet Submitted by Danish Ahmed Khan    
Rating: 4.0000 (1 votes)

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Solution Solution | Comment 1 of 2
One inradius formula is Area=radius*semiperimeter.
Plugging in the sides of a right triangle and simplifying yields radius = a*b/(a+b+c).

Now lets use a parameterization of the right triangle: u^2-v^2, 2uv, u^2+v^2. Plug this into the radius formula to get radius = v*(u-v).
This value is to be a perfect square.

Trying small values I quickly find v=1 and u=10 to give a solution.
a=2*1*10=20, b=10^2-1^2=99, c=10^2+1^2=101, and radius=1*(10-1)=9.
gcd(20+1,99)^2=3^2=9.  These match, so the solution is verified.  a,b,c are 20,99,101.

Edited on February 18, 2024, 10:53 am
  Posted by Brian Smith on 2024-02-17 19:22:28

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