AMC 10 · 2011 · #11

Grade 8 geometry-2d
Archetype Coordinate Plane Figure Computation └ Sub-archetype Pythagorean / Distance Formula for a Length
pythagorean-theoremcoordinate-geometryrotation-isometry convert-to-algebra ↑ Prerequisites: pythagorean-theorem
📏 Medium solution 💡 2 insights
Problem
A tilted square EFGH is inscribed inside a bigger square ABCD, with exactly one vertex of EFGH on each side of ABCD. On side AB the vertex E splits the side so that the piece AE is 7 times the piece EB. Find how the area of the tilted square compares to the area of the big square.

Pick an answer.

(A)
$\frac{49}{64}$
(B)
$\frac{25}{32}$
(C)
$\frac78$
(D)
$\frac{5\sqrt{2}}{8}$
(E)
$\frac{\sqrt{14}}{4}$

AMC 10 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

Tool #1 (Draw a Diagram) is the anchor: a labeled sketch of the tilted square inside the big square makes one corner right triangle appear, with its two legs sitting on one side of ABCD. Tool #4 (Introduce a Variable) picks a convenient side length (8) so that AE = 7 · EB turns into whole-number pieces 7 and 1. Tool #17 (Visualize Spatial Relationships) supplies the key move: a quarter-turn of the picture about the center carries ABCD onto itself and EFGH onto itself, so the neighboring side must be split in the SAME ratio — that fixes the second leg of the corner triangle. Then the Pythagorean theorem gives the tilted square's side, and Tool #3 (Eliminate Possibilities) confirms the resulting fraction against the five choices.

1STEP 1

Pick a friendly side length

Only the ratio matters, so let ABCD have side 8. Then AE + EB = 8 with AE = 7 · EB gives EB = 1 and AE = 7.

AE + EB = 8, AE = 7 EB → 8 EB = 8 → EB = 1, AE = 7
2STEP 2

Turn the picture a quarter-turn

A 90° turn about the center maps ABCD and EFGH onto themselves and sends E to F, so BF = 7 and FC = 1.

rotation 90°: A↦ B, E↦ F → BF = AE = 7
3STEP 3

Find the tilted square's side

Corner triangle EBF has legs EB = 1 and BF = 7 with its right angle at B, and its hypotenuse EF gives EF² = 50.

EF² = EB² + BF² = 1² + 7² = 1 + 49 = 50
4STEP 4

Compare the two areas

Area is side squared, so EFGH has area EF² = 50 and ABCD has 8² = 64, and 50/64 reduces to 25/32.

[EFGH]/[ABCD] = EF²/8² = 50/64 = 25/32
5STEP 5

Match to a choice

25/32 ≈ 0.781, while (A) 49/64 ≈ 0.766 and (C) 7/8 = 0.875, so only (B) matches.

25/32 ≈ 0.781 → (B)
Answer
25/32
The tilted square must be smaller than the big square but not by much, since E sits near corner B (only 1 unit away) — the tilt is gentle. A gentle tilt should give a ratio a little below 1, and 25/32 ≈ 0.78 fits. Boundary check: if E were at the midpoint the ratio would be 1/2; if E were at a corner the tilted square would coincide with the big one (ratio 1). Our 7:1 split is near a corner, so a ratio close to but under 1 is exactly right.
💡Key takeaway

Give the big square a handy side of 8, split it 7 and 1, and notice a quarter-turn forces every side to split the same way — then one corner right triangle (1 and 7) hands you the tilted square's side by the Pythagorean theorem, and 50/64 = 25/32.

  • Pick a friendly side length
  • Turn the picture a quarter-turn
  • Find the tilted square's side
  • Compare the two areas
  • Match to a choice

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