Problem:
Rhombus ABCD has β BAD<90β. 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:
Let O be the incenter of ABCD for which βO is tangent to DA,AB, and BC at X,Y, and Z, respectively. Moreover, suppose that R,S, and T are the feet of the perpendiculars from P to DA,AB, and BC, respectively, such that RT intersects βO at P and Q.
We obtain the following diagram:
Note that β RXZ=β TZX=90β by the properties of tangents, so RTZX is a rectangle. It follows that the diameter of βO is XZ=RT=25.
Let x=PQ and y=RX=TZ. We apply the Power of a Point Theorem to R and T:
y2=9(9+x)y2=16(16βx)β
We solve this system of equations to get x=7 and y=12. Alternatively, we can find these results by the symmetry on rectangle RTZX and semicircle XPZ.
We extend SP beyond P to intersect βO and CD at E and F, respectively, where Eξ =P. So, we have EF=SP=5 and PE=25βSPβEF=15. On the other hand, we have PX=15 by the Pythagorean Theorem on right β³PRX. Together, we conclude that E=X. Therefore, points S,P, and X must be collinear.
Let G be the foot of the perpendicular from D to AB. Note that DGβ₯XP, as shown below:
As β PRX=β AGD=90β and β PXR=β ADG by the AA Similarity, we conclude that β³PRXβΌβ³AGD. The ratio of similitude is
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