AMC 10 · 2005 · #19

Grade 8 geometry-2d
Archetype Coordinate Plane Figure Computation └ Sub-archetype Pythagorean / Distance Formula for a Length
pythagorean-theoremspatial-visualization identify-subproblemsphysical-representation ↑ Prerequisites: pythagorean-theoremspatial-visualizationisosceles-triangle
📏 Long solution 💡 3 insights 📊 Diagram
Problem
Two unit squares sit on a line with a one-inch gap between them where a third square used to be. That third square is turned 45° into a diamond and lowered into the gap until it rests on the two squares beside it. Find how high the top corner B of the diamond ends up above the base line.

Pick an answer.

(A)
1
(B)
$\sqrt{2}$
(C)
$\frac{3}{2}$
(D)
$\sqrt{2}+\frac{1}{2}$
(E)
2

AMC 10 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

This is a picture problem, so Tool #1 (Draw a Diagram) carries it: mark the two resting corners and the diamond's own corners to see exactly where things touch. Tool #17 (Visualize Spatial Relationships) is needed to see that the tilted square is too wide to drop through and instead hangs on the two top corners of its neighbours. Tool #7 (Identify Subproblems) then splits the height of B into two easy pieces — how high the diamond's bottom corner sits, plus the diamond's vertical diagonal.

1STEP 1

Mark the gap and its corners

Lifting the middle square out leaves a gap exactly 1 inch wide, and the two inner top corners become ledges at height 1.

gap width=1, corner height=1
2STEP 2

The diamond is too wide to fall through

Turned 45°, the square's width becomes its diagonal, √(2) ≈ 1.41 inch — wider than the 1-inch gap, so it catches on the two corners.

diamond width=√(2) ＞ 1=gap width
3STEP 3

Spot the 45-45-90 triangle

Join the bottom corner to both resting corners: the edges meet at 90° and each slopes 45°, giving a 45-45-90 triangle with hypotenuse 1.

right angle at bottom corner, hypotenuse=1
4STEP 4

Find the height of the bottom corner

In a 45-45-90 triangle the drop from the right angle to the hypotenuse is half of it, so the bottom corner sits at height 1/2.

drop=1/2, bottom corner height=1-1/2=1/2
5STEP 5

Add the vertical diagonal to reach B

B sits straight above that corner, one vertical diagonal √(1²+1²)=√(2) higher, so B is at height √(2)+1/2 — choice (D).

height of B=1/2+√(2)=√(2)+1/2 → (D)
Answer
√(2)+1/2
Check with coordinates. Put the gap between x=1 and x=2, both resting corners at height 1: points (1,1) and (2,1). The diamond is centred at x=1.5; its lower-right edge has slope -1 through (2,1) and its lower-left edge has slope +1 through (1,1). These meet at the bottom corner (1.5, 0.5), matching the 1/2 height. Adding the vertical diagonal √(2) puts B at (1.5, 0.5+√(2))≈(1.5, 1.91). That is a bit under 2, so it beats choices (A) 1, (B) √2≈1.41, (C) 1.5 and stays under (E) 2 — only (D) √2+1/2≈1.91 fits.
💡Key takeaway

A square turned on its point is wider than it looks, so it hangs on the corners — then just stack the little drop and the straight-up diagonal to find the top.

  • Mark the gap and its corners
  • The diamond is too wide to fall through
  • Spot the 45-45-90 triangle
  • Find the height of the bottom corner
  • Add the vertical diagonal to reach B

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