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2020 AMC 10B Problem 21

2020 AMC 10B Problem 21AMC 10 Step by Step

Name: Date: September 13, 2026

Quadrilaterals & Polygon Areas · difficulty 4 of 5 · about 8 min · position P21-25

In square $ABCD$ , points $E$ and $H$ lie on $\overline{AB}$ and $\overline{DA}$ , respectively, so that $AE=AH.$ Points $F$ and $G$ lie on $\overline{BC}$ and $\overline{CD}$ , respectively, and points $I$ and $J$ lie on $\overline{EH}$ so that $\overline{FI} \perp \overline{EH}$ and $\overline{GJ} \perp \overline{EH}$ . See the figure below. Triangle $AEH$ , quadrilateral $BFIE$ , quadrilateral $DHJG$ , and pentagon $FCGJI$ each has area $1.$ What is $FI^2$ ?

A)
$\frac{7}{3}$
B)
$8-4\sqrt2$
C)
$1+\sqrt2$
D)
$\frac{7}{4}\sqrt2$
E)
$2\sqrt2$

Answer

Key insight

Extend CB, CD to line EH: the pentagon is a right isosceles triangle of altitude 2 sqrt(2) - 1 minus two 45-degree corners of leg FI.

Solution

The four regions tile the square, so its area is $4$ and its side is $2$ . Triangle $AEH$ is right isosceles with area $1$ , so $AE = AH = \sqrt2$ and $EH = 2$ ; its altitude from $A$ to $EH$ has length $1$ . Since $EH \perp AC$ , the distance from $C$ to line $EH$ is
$h = AC - 1 = 2\sqrt2 - 1.$

Extend $CB$ and $CD$ until they meet line $EH$ at $P$ and $Q$ . Triangle $CPQ$ has a right angle at $C$ and is symmetric about $AC$ , so it is right isosceles with altitude $h$ from $C$ to $PQ$ ; its hypotenuse is $2h$ and its area is $\tfrac12 \cdot 2h \cdot h = h^2$ .

The pentagon $FCGJI$ is this triangle with two corners cut off: triangle $FIP$ (right angle at $I$ , $45^\circ$ at $P$ ) and triangle $GJQ$ . Let $x = FI$ . Then $IP = x$ and the area of $FIP$ is $\tfrac12 x^2$ ; the same holds for $GJQ$ . Therefore
$[FCGJI] = h^2 - x^2 = 1.$

So
$x^2 = h^2 - 1 = (2\sqrt2 - 1)^2 - 1 = 8 - 4\sqrt2 + 1 - 1 = 8 - 4\sqrt2.$

The answer is $\boxed{\textbf{(B)}\ 8-4\sqrt2}$ .

Why this works

The whole figure is symmetric about diagonal $AC$ , and $EH$ is perpendicular to that diagonal, so every relevant line is at $45^\circ$ to the sides. That makes every triangle in sight a $45$ - $45$ - $90$ triangle, whose area is determined by a single length. Rather than locating $F$ and $G$ individually, enlarge the pentagon to a clean triangle and subtract the two congruent corners; the unknown $FI$ then appears alone.

Alternative approach

Coordinates with $A = (0,0)$ , $C = (2,2)$ , line $EH$ : $x + y = \sqrt2$ , and $F = (2, f)$ , $G = (f, 2)$ . Writing the pentagon as rectangle $FIJG$ plus triangle $FCG$ and setting the area to $1$ leads to a quadratic in $f$ ; after simplifying, $FI = \dfrac{2 + f - \sqrt2}{\sqrt2}$ satisfies $FI^2 = 8 - 4\sqrt2$ . It works but is far slower.

The trap

Setting up coordinates for F and G and solving a messy quadratic instead of using the symmetric right-isosceles structure along diagonal AC.

Common mistakes

Setting up coordinates for F and G and solving a messy quadratic instead of using the symmetric right-isosceles structure along diagonal AC.
Using the distance from $C$ to $EH$ as $2\sqrt2 - \sqrt2 = \sqrt2$ (confusing $AE$ with the altitude from $A$ ); the altitude of triangle $AEH$ is $1$ , not $\sqrt2$ .

Techniques

Add construction lines/points (drop altitudes, extend segments, connect centers) · Cut the figure into known shapes (triangles, rectangles, sectors)