AMC 10 · 2014 · #22

Grade 8 geometry-2d
Archetype Coordinate Plane Figure Computation └ Sub-archetype Pythagorean / Distance Formula for a Length
tangent-circlespythagorean-theoremspatial-visualization symmetry-argument ↑ Prerequisites: tangent-circlespythagorean-theorem
📏 Long solution 💡 3 insights 📊 Diagram
Problem
A square has side length 2. Along the inside of the square sit eight congruent semicircles, two flat against each side, their diameters covering the side. One circle is drawn in the middle so that it just touches every semicircle. Find the radius of that middle circle.

Pick an answer.

(A)
$\dfrac{1+\sqrt2}4$
(B)
$\dfrac{\sqrt5-1}2$
(C)
$\dfrac{\sqrt3+1}4$
(D)
$\dfrac{2\sqrt3}5$
(E)
$\dfrac{\sqrt5}3$

AMC 10 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

Tangency problems collapse to one clean fact: two circles that touch on the outside have their centers a distance apart equal to the sum of their radii. So Tool #1 (Draw a Diagram) sets up coordinates with the square centered at the origin, turning the picture into exact points. Tool #7 (Identify Subproblems) splits the work into three small pieces: find a semicircle's radius, find where its center is, and measure the distance from the square's center to that center. Tool #4 (Introduce a Variable) names the unknown radius r and writes the tangency condition as an equation. Tool #17 (Visualize Spatial Relationships) is the key sighting: the touching point, the two centers all line up, so the center-to-center distance is exactly r plus the semicircle radius.

1STEP 1

Put the square on coordinates

Center the square at the origin: its sides lie on x=±1 and y=±1, and by symmetry the middle circle sits at (0,0).

corners=(±1,±1); central circle center=(0,0)
2STEP 2

Find one semicircle's radius and center

Two semicircles split the bottom side of length 2, so each has diameter 1 and radius 1/2; the right one is centered at (1/2,-1).

diameter=2/2=1 → radius=1/2; center=(1/2,-1)
3STEP 3

Write the tangency condition

Call the middle radius r. The touch point lies on the segment joining the two centers, so that distance is exactly r+1/2.

dist((0,0),(1/2,-1))=r+1/2
4STEP 4

Measure the distance between centers

From (0,0) to (1/2,-1) the gaps are 1/2 across and 1 down, so Pythagoras gives √(1/4+1)=√(5/4)=√5/2.

√((1/2)²+1²)=√(5/4)=√5/2
5STEP 5

Solve for the radius

Set √5/2=r+1/2 and subtract 1/2 from both sides: r=(√5-1)/2, which is choice (B).

r=√5/2-1/2=(√5-1)/2 → (B)
Answer
(√5-1)/2
Numerically r=(√5-1)/2≈(2.236-1)/2≈0.618. That is larger than a semicircle's radius 0.5 (the central circle should be the biggest disk in the picture) yet smaller than the half-side 1 (it must fit inside the square without poking through a side). Both checks pass. Testing the near-miss choices as decimals, (A) ≈0.604, (C) ≈0.683, (D) ≈0.693, (E) ≈0.745; only the exact Pythagorean value matches (B), and (E) would already be too wide to sit clear of the semicircles.
💡Key takeaway

Two circles that just touch on the outside sit a sum-of-radii apart, so measure the center-to-center distance √5/2 and subtract the semicircle's 1/2 to get (√5-1)/2.

  • Put the square on coordinates
  • Find one semicircle's radius and center
  • Write the tangency condition
  • Measure the distance between centers
  • Solve for the radius

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