AMC 10 · 2025 · #20

Grade 8 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity

tangent-circlesthirty-sixty-ninety-trianglepythagorean-theorem convert-to-algebra ↑ Prerequisites: pythagorean-theorem

📏 Medium solution 💡 4 insights 📊 Diagram

Problem

Four congruent semicircles sit inside a square of side length

1

, with each semicircle's diameter lying on a side of the square and one endpoint of every diameter at a vertex of the square, so that adjacent semicircles are tangent to each other. A small circle centered at the center of the square is tangent to all four semicircles. The diameter of that small circle can be written as

(\sqrt{a}+b)(\sqrt{c}+d)

, where

a

b

c

d

are integers. Find

a+b+c+d

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

Everything here is about where the circles sit, so put the picture on a coordinate grid and mark each semicircle's center. Once the centers have coordinates, two facts turn the picture into equations: two circles that just touch have their centers a distance of (sum of radii) apart. Call the semicircle radius r, use tangency to pin down r, then use tangency again for the small circle. Each tangency is a right-triangle distance, so the Pythagorean theorem does all the heavy lifting.

1STEP 1

Put the picture on a grid

Place the square's corners at $(0,0),(1,0),(1,1),(0,1)$ ; with semicircle radius $r$ the four centers are $(r,0),(1,r),(1-r,1),(0,1-r)$ .

E=(r,0), F=(1,r), G=(1-r,1), H=(0,1-r)

2STEP 2

Neighbors touch: centers are 2r apart

The bottom and right semicircles touch, so their centers are $2r$ apart; the gap between them is $1-r$ across and $r$ up.

√((1-r)²+r²)=2r

3STEP 3

Solve for the semicircle radius

Squaring gives $(1-r)^2+r^2=4r^2$ , i.e. $2r^2+2r-1=0$ , whose positive root is $r=\frac{\sqrt{3}-1}{2}$ .

2r²+2r-1=0 ⟹ r=(√(3)-1)/2

4STEP 4

The small circle touches too

The small circle has center $(\frac{1}{2},\frac{1}{2})$ and radius $R$ ; tangency to the bottom semicircle gives $(r+R)^2=(\frac{1}{2}-r)^2+\frac{1}{4}$ , which is $2-\sqrt{3}$ .

√((1/2-r)²+(1/2)²)=r+R, ((2-√3)/2)²+1/4=(7-4√3)/4+1/4=2-√3

5STEP 5

Simplify and read off the answer

Rewrite $2-\sqrt{3}$ as $\frac{(\sqrt{3}-1)^2}{2}$ , so $r+R=\frac{\sqrt{6}-\sqrt{2}}{2}$ and $2R=\sqrt{6}-\sqrt{2}-\sqrt{3}+1$ , which factors as $(\sqrt{2}-1)(\sqrt{3}-1)$ : $a+b+c+d=3$ .

2R=√6-√2-√3+1=(√2-1)(√3-1), a+b+c+d=2-1+3-1=3

Answer

Numerically r≈0.366 (under 1/2, so the semicircles fit) and the diameter 2R=(√2-1)(√3-1)≈0.414×0.732≈0.30, a small circle radius of about 0.15 that comfortably sits in the middle without overlapping the semicircles. Both the shape and the size look right, and a+b+c+d=3 is one of the offered choices, matching (A).

💡Key takeaway

When two circles just touch, the straight line between their centers equals their two radii added together—write that as a right-triangle distance and the algebra falls right out.

Put the picture on a grid
Neighbors touch: centers are 2r apart
Solve for the semicircle radius
The small circle touches too
Simplify and read off the answer

Put the picture on a grid

Neighbors touch: centers are 2r apart

Solve for the semicircle radius

The small circle touches too

Simplify and read off the answer

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