2020 AMC 10B Problem 21

Attempt Problem 21 of the 2020 AMC 10B below, then check your answer against the video solution and professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2020 AMC 10B solutions, or check the answer key.

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21.

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?
73\dfrac73
8−428-4\sqrt2
1+21+\sqrt2
742\dfrac74\sqrt2
222\sqrt2
Answer: B
Concepts:square (geometry)area decompositionspecial right triangle
Small hint:
The four given regions fill the square, so the square’s side length is known
Big hint:
Extend the perpendicular through FF to meet AB,AB, and compare the two resulting right isosceles triangles
Video solution:
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Written solution:
The four named regions fill the square and each has area 1,1, so the square has area 44 and side length 2.2. Since triangle AEHAEH is right isosceles with area 1,1, we have AE=AH=2.AE=AH=\sqrt2.
Extend FIFI to meet ABAB at K,K, and set x=BFx=BF and t=BE=2−2.t=BE=2-\sqrt2. Because EHEH has slope −1,-1, line FKFK has slope 1,1, so BF=BK=x.BF=BK=x. If KK were on segment EB,EB, then region BFIEBFIE would lie inside triangle BFK,BFK, whose area would be at most t22<1,\frac{t^2}{2}<1, a contradiction. Thus KK lies to the left of E,E, and EK=x−t.EK=x-t.
Triangle BFKBFK is right isosceles with area x22.\frac{x^2}{2}. Triangle EIKEIK is right isosceles with hypotenuse EK=x−t,EK=x-t, so its area is (x−t)24.\frac{(x-t)^2}{4}. Since their difference is region BFIE,BFIE, 1=x22−(x−t)24.1=\frac{x^2}{2}-\frac{(x-t)^2}{4}. Therefore 4=2x2−(x−t)2=(x+t)2−2t2. \begin{aligned} 4&=2x^2-(x-t)^2\\ &=(x+t)^2-2t^2. \end{aligned}
Also, FK=x2FK=x\sqrt2 and KI=x−t2,KI=\frac{x-t}{\sqrt2}, so FI=FK−KI=x+t2.FI=FK-KI=\frac{x+t}{\sqrt2}. It follows that FI2=(x+t)22=2+t2=2+(2−2)2=8−42. \begin{aligned} FI^2&=\frac{(x+t)^2}{2}\\ &=2+t^2\\ &=2+(2-\sqrt2)^2\\ &=8-4\sqrt2. \end{aligned}
Thus, the correct answer is B .
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