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AMC 10 · 2020 · #21

Grade 8 geometry-2d
Archetype Compound Area by Decomposition / Difference
area-trianglessimilar-trianglesisosceles-trianglepythagorean-theorem identify-subproblemsconvert-to-algebracasework ↑ Prerequisites: area-trianglessimilar-trianglespythagorean-theorem
📏 Long solution 💡 3 insights 📊 Diagram

Problem

In square ABCDABCD, 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 AEHAEH, quadrilateral BFIEBFIE, quadrilateral DHJGDHJG, and pentagon FCGJIFCGJI each has area 1.1. What is FI2FI^2?

(A) 73(B) 8−42(C) 1+2(D) 742(E) 22\textbf{(A) } \frac{7}{3} \qquad \textbf{(B) } 8-4\sqrt2 \qquad \textbf{(C) } 1+\sqrt2 \qquad \textbf{(D) } \frac{7}{4}\sqrt2 \qquad \textbf{(E) } 2\sqrt2

Pick an answer.

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

Toolkit + CCSS Solution

Understand

Restated: In square $ABCD$, points $E$ on $AB$ and $H$ on $AD$ satisfy $AE = AH$. Points $F$ on $BC$ and $G$ on $CD$ have feet $I, J$ on segment $EH$ with $FI \perp EH$ and $GJ \perp EH$. The four regions — triangle $AEH$, quadrilateral $BFIE$, quadrilateral $DHJG$, pentagon $FCGJI$ — each have area $1$. Find $FI^2$.

Givens: Square $ABCD$ with all four regions of area $1$; $AE = AH$, so $\triangle AEH$ is right isosceles at $A$; $\overline{FI} \perp \overline{EH}$ and $\overline{GJ} \perp \overline{EH}$, with $F \in BC, G \in CD$; Answer choices: (A) $\tfrac{7}{3}$, (B) $8 - 4\sqrt{2}$, (C) $1 + \sqrt{2}$, (D) $\tfrac{7}{4}\sqrt{2}$, (E) $2\sqrt{2}$

Unknowns: The value of $FI^2$

Understand

Restated: In square $ABCD$, points $E$ on $AB$ and $H$ on $AD$ satisfy $AE = AH$. Points $F$ on $BC$ and $G$ on $CD$ have feet $I, J$ on segment $EH$ with $FI \perp EH$ and $GJ \perp EH$. The four regions — triangle $AEH$, quadrilateral $BFIE$, quadrilateral $DHJG$, pentagon $FCGJI$ — each have area $1$. Find $FI^2$.

Givens: Square $ABCD$ with all four regions of area $1$; $AE = AH$, so $\triangle AEH$ is right isosceles at $A$; $\overline{FI} \perp \overline{EH}$ and $\overline{GJ} \perp \overline{EH}$, with $F \in BC, G \in CD$; Answer choices: (A) $\tfrac{7}{3}$, (B) $8 - 4\sqrt{2}$, (C) $1 + \sqrt{2}$, (D) $\tfrac{7}{4}\sqrt{2}$, (E) $2\sqrt{2}$

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems, #9 Solve an Easier Related Problem, #8 Analyze the Units, #13 Convert to Algebra

Tool #1 (Diagram): place $A$ at the origin and use diagonal $AC$ as the axis of symmetry forced by $AE = AH$. Tool #7 (Subproblems): the key spot is that $FI \parallel GJ$ (both perpendicular to $EH$), making $FGJI$ a rectangle — so the pentagon splits cleanly into a right triangle on top and a rectangle below. Tool #9 (Easier Problem): symmetry collapses six unknowns into two ($c = CF$ and $p = FI$); two equations close the system. Tool #8 (Units/distances) and #13 (Algebra) seal the answer via the diagonal-distance equation and a one-line subtraction.

Execute — Answer: B

#7 Identify Subproblems 8.G.B.7 Step 1
  • Total area of $ABCD$ is $1 + 1 + 1 + 1 = 4$, so the side is $2$.
  • Triangle $AEH$ is right isosceles at $A$ with area $1$, so $\tfrac{1}{2}(AE)^2 = 1$, giving $AE = AH = \sqrt{2}$ and hypotenuse $EH = 2$.
$$s = 2, \;\; AE = AH = \sqrt{2}, \;\; EH = 2$$

💡 A right isosceles triangle of area $1$ has legs $\sqrt{2}$; four equal pieces give a side-$2$ square.

#1 Draw a Diagram 4.G.A.3 Step 2
  • Because $AE = AH$, the whole figure is symmetric across diagonal $AC$.
  • So $CF = CG$, $FI = GJ$, and line $FG$ is also perpendicular to $AC$.
$$CF = CG, \;\; FI = GJ$$

💡 Mirror symmetry across $AC$ equates the two sides.

#1 Draw a Diagram 4.G.A.2 Step 3
  • Key observation.
  • Both $FI$ and $GJ$ are perpendicular to $EH$, so $FI \parallel GJ$.
  • By symmetry $FI = GJ$, so quadrilateral $FGJI$ has two opposite sides equal and parallel — it is a parallelogram.
  • Since $IJ \subset EH$ and $FI \perp IJ$, $FGJI$ is in fact a rectangle.
$$FGJI \text{ is a rectangle with sides } FI \text{ and } IJ$$

💡 Two parallel perpendiculars of equal length make a rectangle.

#7 Identify Subproblems 6.G.A.1 Step 4
  • Decompose the pentagon $FCGJI$ (vertices in order $F, C, G, J, I$).
  • Drawing diagonal $FG$ splits it into the right-isosceles triangle $\triangle FCG$ (above) and the rectangle $FGJI$ (below).
  • With $c = CF = CG$, we have $[\triangle FCG] = c^2/2$ and $FG = c\sqrt{2} = IJ$.
  • So $$1 \;=\; [FCGJI] \;=\; \tfrac{c^2}{2} + c\sqrt{2} \cdot FI.$$
$$\tfrac{c^2}{2} + c\sqrt{2}\,FI = 1$$

💡 Pentagon = right triangle on top + rectangle on bottom.

#8 Analyze the Units 8.G.B.8 Step 5
  • Now the second equation.
  • Place coordinates $A = (0,0), B = (2,0), C = (2,2), D = (0,2)$, so $E = (\sqrt{2}, 0), H = (0, \sqrt{2})$ and line $EH: x + y = \sqrt{2}$.
  • By symmetry $F = (2, 2-c)$ and $G = (2-c, 2)$, so line $FG: x + y = 4 - c$.
  • The perpendicular distance from $C$ to line $EH$ is $\tfrac{|4 - \sqrt{2}|}{\sqrt{2}} = 2\sqrt{2} - 1$, and it splits as (distance $C \to FG$) $+$ (distance $FG \to EH$) $= \tfrac{c}{\sqrt{2}} + FI$.
  • So $$\tfrac{c}{\sqrt{2}} + FI = 2\sqrt{2} - 1 \;\Longleftrightarrow\; c + FI \sqrt{2} = 4 - \sqrt{2}.$$
$$c + FI \sqrt{2} = 4 - \sqrt{2}$$

💡 Along diagonal $AC$, add the two perpendicular widths to recover the total distance $C$-to-$EH$.

#13 Convert to Algebra 8.EE.C.7 Step 6
  • Square the distance equation: $(c + FI \sqrt{2})^2 = (4 - \sqrt{2})^2 = 18 - 8\sqrt{2}$, i.e., $$c^2 + 2 c \sqrt{2} \cdot FI + 2 FI^2 \;=\; 18 - 8\sqrt{2}.$$ Double the pentagon equation: $c^2 + 2 c \sqrt{2} \cdot FI = 2$.
  • Subtracting: $$2 FI^2 \;=\; 16 - 8\sqrt{2} \;\Longleftrightarrow\; FI^2 \;=\; 8 - 4\sqrt{2}.$$
$$FI^2 = 8 - 4\sqrt{2} \Rightarrow \textbf{(B)}$$

💡 Squaring the linear distance equation produces the $c^2 + 2c\sqrt{2}\cdot FI$ block; subtract the pentagon equation to isolate $2 FI^2$.

[1] #7 8.G.B.7 Total area of $ABCD$ is $1 + 1 + 1 + 1 = 4$, so the side is $2$. Triangle $AEH$
[2] #1 4.G.A.3 Because $AE = AH$, the whole figure is symmetric across diagonal $AC$. So $CF =
[3] #1 4.G.A.2 Key observation. Both $FI$ and $GJ$ are perpendicular to $EH$, so $FI \parallel
[4] #7 6.G.A.1 Decompose the pentagon $FCGJI$ (vertices in order $F, C, G, J, I$). Drawing diag
[5] #8 8.G.B.8 Now the second equation. Place coordinates $A = (0,0), B = (2,0), C = (2,2), D =
[6] #13 8.EE.C.7 Square the distance equation: $(c + FI \sqrt{2})^2 = (4 - \sqrt{2})^2 = 18 - 8\s

Review

Reasonableness: Numerical check: $FI^2 \approx 2.343$, so $FI \approx 1.531$. Constraints: $FI <$ distance $C$-to-$EH = 2\sqrt{2} - 1 \approx 1.828$ ✓. Recover $c = (4 - \sqrt{2}) - FI\sqrt{2} \approx 4 - 1.414 - 2.165 = 0.421$, inside $(0, 2)$ ✓. Verify pentagon: $c^2/2 \approx 0.089$, rectangle $c\sqrt{2}\,FI \approx 0.595 \cdot 1.531 \approx 0.911$, sum $\approx 1.000$ ✓. Choice $(B)$ confirmed.

Alternative: Tool #13 (Convert to Algebra): set $F = (2, t)$ via symmetry, write $[BFIE] = 1$ from a shoelace, solve for $t$, then square $FI = (2 + t - \sqrt{2})/\sqrt{2}$. Tool #10 (Physical): cut a paper square of side $2$, mark $E, H$ at distance $\sqrt{2}$ from $A$, and slide $F, G$ until the four regions match. The pentagon-rectangle decomposition above is fastest because the parallel-perpendicular insight turns one quadratic into a clean subtraction.

CCSS standards used (min grade 8)

  • 4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines (Recognizing that $FI \parallel GJ$ (both perpendicular to $EH$), so $FGJI$ is a rectangle.)
  • 4.G.A.3 Recognize a line of symmetry for a two-dimensional figure (Using diagonal $AC$ as the symmetry axis forced by $AE = AH$, giving $CF = CG$ and $FI = GJ$.)
  • 5.G.A.2 Represent real-world and mathematical problems by graphing points (Coordinate setup $A = (0,0), B = (2,0), C = (2,2), D = (0,2)$ to compute distances and line equations.)
  • 6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing (Pentagon $FCGJI$ = right triangle $FCG$ + rectangle $FGJI$ (decompose-and-add).)
  • 8.EE.C.7 Solve linear equations in one variable (After squaring the distance equation and subtracting the pentagon equation, $2 FI^2 = 16 - 8\sqrt{2}$ is a linear equation in $FI^2$.)
  • 8.G.B.7 Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Recovering $AE = AH = \sqrt{2}$ and $EH = 2$ from the right-isosceles $\triangle AEH$ of area $1$.)
  • 8.G.B.8 Apply the Pythagorean theorem to find distance between two points in a coordinate system (Point-to-line distance from $C$ to line $EH$ and from line $FG$ to line $EH$ via the standard formula.)

⭐ This AMC 10 problem only needs Grade 8 geometry — spot that $FGJI$ is a rectangle (because $FI \parallel GJ$), then the pentagon equation plus the diagonal-distance equation snap together: square the distance, subtract the area, and $2\,FI^2 = 16 - 8\sqrt{2}$ gives $FI^2 = 8 - 4\sqrt{2}$.

⭐ This AMC 10 problem only needs Grade 8 geometry — spot that $FGJI$ is a rectangle (because $FI \parallel GJ$), then the pentagon equation plus the diagonal-distance equation snap together: square the distance, subtract the area, and $2\,FI^2 = 16 - 8\sqrt{2}$ gives $FI^2 = 8 - 4\sqrt{2}$.

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