AMC 10 · 2020 · #21

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Polygon Triangulation

area-trianglessimilar-trianglesisosceles-trianglepythagorean-theorem identify-subproblemsconvert-to-algebracasework ↑ Prerequisites: area-trianglessimilar-trianglespythagorean-theorem

📏 Long solution 💡 3 insights 📊 Diagram

Problem

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 ⊥ EH and GJ ⊥ EH. The four regions — triangle AEH, quadrilateral BFIE, quadrilateral DHJG, pentagon FCGJI — each have area 1. Find FI².

Pick an answer.

(A)

$\frac{7}{3}$

(B)

$8-4\sqrt2$

(C)

$1+\sqrt2$

(D)

$\frac{7}{4}\sqrt2$

(E)

$2\sqrt2$

AMC 10 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

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 ∥ 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.

1STEP 1

Four area-1 pieces tile ABCD, so its area is 4 and side is 2; the right isosceles △AEH of area 1 gives AE = AH = √(2) and EH = 2.

s = 2, AE = AH = √(2), EH = 2

2STEP 2

AE = AH makes the figure symmetric across diagonal AC, so CF = CG and FI = GJ, and line FG is perpendicular to AC.

CF = CG, FI = GJ

3STEP 3

FI ∥ GJ (both ⊥ EH) and FI = GJ, so FGJI has equal parallel opposite sides — a parallelogram — and with FI ⊥ IJ it is a rectangle.

FGJI is a rectangle with sides FI and IJ

4STEP 4

Diagonal FG splits pentagon FCGJI into right isosceles △FCG ( $\frac{c²}{2}$ , FG = c√(2) = IJ) plus rectangle FGJI, so $\frac{c²}{2}$ + c√(2)·FI = 1.

\frac{c²}{2}

+ c√(2) FI = 1

5STEP 5

Coordinates make EH: x+y=√(2), FG: x+y=4-c; the C-to-EH distance 2√(2)-1 splits as $\frac{c}{√(2)}$ +FI, so c + FI√(2) = 4 - √(2).

c + FI √(2) = 4 - √(2)

6STEP 6

Squaring gives (c+FI√(2))²=18-8√(2); doubling the pentagon eq gives c²+2c√(2)·FI=2; subtracting, 2FI²=16-8√(2), so FI²=8-4√(2) → (B).

FI² = 8 - 4√(2) → (B)

Answer

8-4√2

Numerical check: FI² ≈ 2.343, so FI ≈ 1.531. Constraints: FI ＜ distance C-to-EH = 2√(2) - 1 ≈ 1.828 ✓. Recover c = (4 - √(2)) - FI√(2) ≈ 4 - 1.414 - 2.165 = 0.421, inside (0, 2) ✓. Verify pentagon:

\frac{c²}{2}

≈ 0.089, rectangle c√(2) FI ≈ 0.595 · 1.531 ≈ 0.911, sum ≈ 1.000 ✓. Choice (B) confirmed.

💡Key takeaway

This AMC 10 problem only needs Grade 8 geometry — spot that FGJI is a rectangle (because FI ∥ GJ), then the pentagon equation plus the diagonal-distance equation snap together: square the distance, subtract the area, and 2 FI² = 16 - 8√(2) gives FI² = 8 - 4√(2).

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