Problem:
In right β³ABC with hypotenuse AB,AC=12,BC=35, and CD is the altitude to AB. Let Ο be the circle having CD as a diameter. Let I be a point outside β³ABC such that AI and BI are both tangent to circle Ο. The ratio of the perimeter of β³ABI to the length AB can be expressed in the form nmβ, 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 Ο,CD=h, and k be the area of triangle ABI. Then h=yzβ, and so r=21βyzβ. Let s be the semiperimeter of β³ABI, so that s=x+y+z. On one hand k=sr=21β(x+y+z)yzβ, and on the other hand, by Heron's Formula, k=(x+y+z)xyzβ. Equating these two expressions and simplifying yields 4x=x+y+z, or 4x=x+AB. Thus x=3ABβ and 2s=2β 3ABβ+2β AB=38ββ AB. Hence m+n=8+3=11β.