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2020 AMC 12B Problem 18

Problem 18 of 25IntermediateGeometry

In square ABCD,ABCD, points EE and HH lie on AB‾\overline{AB} and DA‾,\overline{DA}, respectively, so that AE=AH.AE = AH. Points FF and GG lie on BC‾\overline{BC} and CD‾,\overline{CD}, respectively, and points II and JJ lie on EH‾\overline{EH} so that FI‾⊥EH‾\overline{FI} \perp \overline{EH} and GJ‾⊥EH‾.\overline{GJ} \perp \overline{EH}. See the figure below. Triangle AEH,AEH, quadrilateral BFIE,BFIE, quadrilateral DHJG,DHJG, and pentagon FCGJIFCGJI each has area 1.1. What is FI2?FI^2?

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Solution

The four regions have total area 4,4, so the square has side 2.2. Put A=(0,0),A = (0, 0), B=(2,0),B = (2, 0), C=(2,2),C = (2, 2), D=(0,2).D = (0, 2). Since △AEH\triangle AEH is an isosceles right triangle with area 1,1, we get AE=AH=2,AE = AH = \sqrt2, so E=(2,0)E = (\sqrt2, 0) and H=(0,2).H = (0, \sqrt2). Line EHEH is x+y=2.x + y = \sqrt2. Let F=(2,t).F = (2, t). Its perpendicular distance to line EHEH is FI=2+t−22.FI = \tfrac{2 + t - \sqrt2}{\sqrt2}. Write s=FI2=2+t−22,s=\frac{FI}{\sqrt2}=\tfrac{2+t-\sqrt2}{2}, so the foot of the perpendicular is I=(2−s,t−s).I=(2-s,t-s). The shoelace formula on B=(2,0),B=(2,0), F,F, I,I, and E=(2,0)E=(\sqrt2,0) gives [BFIE]=s2−(3−22).[BFIE]=s^2-(3-2\sqrt2). Since this area is 1,1, we get s2=4−22.s^2=4-2\sqrt2. Then FI2=2s2=8−42.FI^2 = 2s^2 = 8 - 4\sqrt2. Thus, the correct answer is B.

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Concepts: coordinate geometry · area decomposition · special right triangle

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