Problem:
In right △ABC with hypotenuse AB,AC=12,BC=35, and CD is the altitude to AB. Let ω be the circle having CD as a diameter. Let I be a point outside △ABC such that AI and BI are both tangent to circle ω. The ratio of the perimeter of △ABI to the length AB can be expressed in the form nm​, where m and n are relatively prime positive integers. Find m+n.
Solution:
Let E and F be the points of tangency on AI and BI, respectively. Let IE=IF=x,AE=AD=y,BD=BF=z,r= the radius of the circle ω,CD=h, and k be the area of triangle ABI. Then h=yz​, and so r=21​yz​. Let s be the semiperimeter of △ABI, so that s=x+y+z. On one hand k=sr=21​(x+y+z)yz​, and on the other hand, by Heron's Formula, k=(x+y+z)xyz​. Equating these two expressions and simplifying yields 4x=x+y+z, or 4x=x+AB. Thus x=3AB​ and 2s=2⋅3AB​+2⋅AB=38​⋅AB. Hence m+n=8+3=11​.