MathDB
2010 PUMaC Geometry B4: BF+DH=FH

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August 21, 2011
geometryrectangletrigonometry

Problem Statement

Unit square ABCDABCD is divided into four rectangles by EFEF and GHGH, with BF=14BF = \frac14. EFEF is parallel to ABAB and GHGH parallel to BCBC. EFEF and GHGH meet at point PP. Suppose BF+DH=FHBF + DH = FH, calculate the nearest integer to the degree of FAH\angle FAH. [asy] size(100); defaultpen(linewidth(0.7)+fontsize(10)); pair D2(pair P) { dot(P,linewidth(3)); return P; } // NOTE: I've tampered with the angles to make the diagram not-to-scale. The correct numbers should be 72 instead of 76, and 45 instead of 55. pair A=(0,1), B=(0,0), C=(1,0), D=(1,1), F=intersectionpoints(A--A+2*dir(-76),B--C)[0], H=intersectionpoints(A--A+2*dir(-76+55),D--C)[0], E=F+(0,1), G=H-(1,0), P=intersectionpoints(E--F,G--H)[0]; draw(A--B--C--D--cycle); draw(F--A--H); draw(E--F); draw(G--H); label("AA",D2(A),NW); label("BB",D2(B),SW); label("CC",D2(C),SE); label("DD",D2(D),NE); label("EE",D2(E),plain.N); label("FF",D2(F),S); label("GG",D2(G),W); label("HH",D2(H),plain.E); label("PP",D2(P),SE); [/asy]