Maths Olympiad Prep

Library / /121 of 348

Geometry Difficulty 4.8 AIME Find the answer

Let ABCDA B C D be a convex trapezoid such that BAD=ADC=90,AB=20,AD=21\angle B A D=\angle A D C=90^{\circ}, A B=20, A D=21, and CD=28C D=28. Point PAP \neq A is chosen on segment ACA C such that BPD=90\angle B P D=90^{\circ}. Compute APA P.

A number or a short expression. Spacing and $ signs are ignored.

Solution

Construct the rectangle ABXDA B X D. Note that BAD=BPD=BXD=90\angle B A D=\angle B P D=\angle B X D=90^{\circ} so ABXPDA B X P D is cyclic with diameter BDB D. By Power of a Point, we have CXCD=CPCAC X \cdot C D=C P \cdot C A. Note that CX=CDXD=CDAB=8C X=C D-X D=C D-A B=8 and CA=AD2+DC2=35C A=\sqrt{A D^{2}+D C^{2}}=35. Therefore, CP=CXCDCA=82835=325C P=\frac{C X \cdot C D}{C A}=\frac{8 \cdot 28}{35}=\frac{32}{5} and so AP=ACCP=35325=1435A P=A C-C P=35-\frac{32}{5}=\frac{143}{5}

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.