Problem:
In โณABC with AB=12,BC=13, and AC=15, let M be a point on AC such that the incircles of โณABM and โณBCM have equal radii. Let p and q be positive relatively prime integers such that CMAMโ=qpโ. Find p+q.
Solution:
Let CMAMโ=k and the common altitude of โณAMB and โณCMB be h. Because the radius of the incircle of triangle equals twice its area divided by its perimeter, the ratio of the areas of two triangles with equal inradii is the same as the ratio of their perimeters. Thus 13+CM+BM12+AM+BMโ= 21โCMโ
h21โAMโ
hโ=k. Replacing AM by kโ
CM and solving for BM yields BM=1โk13kโ12โ. The fact that BM>0 implies that 1312โ<k. Because CMAMโ=k and AM+CM=15, it follows that CM=k+115โ and AM= k+115kโ. Applying the Law of Cosines to triangles ABM and BCM and to angles โ BMA=ฮฑ and โ CMB=ฯโฮฑ respectively yields
122=AM2+BM2โ2AMโ
BMcosฮฑ, and
132=BM2+CM2+2BMโ
CMcosฮฑ.
Using AM=kโ
CM, multiplying the second equation by k, and adding the two equations yields
132k+122=BM2(k+1)+AM2+CM2k.
Substituting into the above equation produces
169k+144=(1โk13kโ12โ)2(k+1)+(k+115kโ)2+(k+115โ)2k
Simplifying this equation yields 4k(69k2โ112k+44)=0. Its solutions are k=0, k=32โ, and k=2322โ. Because k>1312โ, only the last solution is valid, and so p+q=45โ.
The problems on this page are the property of the MAA's American Mathematics Competitions