Problem:
Rhombus ABCD has ∠BAD<90∘. There is a point P on the incircle of the rhombus such that the distances from P to the lines DA,AB, and BC are 9,5, and 16, respectively. Find the perimeter of ABCD.
Solution:
Answer (125):
Let X,Y, and Z be the projections of P onto lines DA,AB, and BC, respectively, so that PX=9, PY=5, and PZ=16. Let R,S, and T be points where the incircle of the rhombus is tangent to these three lines, respectively, and let Q be the second intersection point of XZ with the incircle. Note that the given conditions imply that P lies on minor \operatorname{arc} \wideparen{S R}.
The height of the rhombus ABCD is PX+PZ=25. By symmetry, QX=PZ=16, and hence, by the Power of a Point Theorem, RX=PX⋅QX=12. Analogous reasoning applied to SY yields SY=5⋅(25−5)=10.
Let x=AX. By equal tangents, AR=AS, so AY=AS−10=AR−10=(12+x)−10=x+2, and applying the Pythagorean Theorem to triangles △AXP and △AYP yields
AP2=x2+92=(x+2)2+52,
from which x=13. Thus AR=AX+XR=13+12=25. Because the diagonals of the rhombus are perpendicular and meet at its incenter O,△AOD is right with height OR=225. Thus △ARO∼△ORD and RD:RO=RO:RA=1:2; that is, RD=21RO=425. The perimeter of ABCD is
