Problem:
Isosceles △ABC has a right angle at C. Point P is inside △ABC, such that PA=11,PB=7, and PC=6. Legs AC and BC have length s=a+b2, where a and b are positive integers. What is a+b ?
Answer Choices:
A. 85
B. 91
C. 108
D. 121
E. 127 Solution:
Let D,E, and F be the reflections of P about AB,BC, and CA, respectively. Then ∠FAD=∠DBE=90∘, and ∠ECF=180∘. Thus the area of pentagon ADBEF is twice that of △ABC, so it is s2.
Observe that DE=72,EF=12, and FD=112. Furthermore, (72)2+122=98+144=242=(112)2, so △DEF is a right triangle. Thus the pentagon can be tiled with three right triangles, two of which are isosceles, as shown.
It follows that
s2=21⋅(72+112)+21⋅12⋅72=85+422
so a+b=127.
{OR}
Rotate △ABC90∘ counterclockwise about C, and let B′ and P′ be the images of B and P, respectively.
Then CP′=CP=6, and ∠PCP′=90∘, so △PCP′ is an isosceles right triangle. Thus PP′=62, and BP′=AP=11. Because (62)2+72=112, the converse of the Pythagorean Theorem implies that ∠BPP′=90∘. Hence ∠BPC=135∘. Applying the Law of Cosines in △BPC gives
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