AMC 10 · 2023 · #18

Grade 8 geometry-2d
tangent-circlespythagorean-theoremcoordinate-geometry identify-subproblemsconvert-to-algebra ↑ Prerequisites: tangent-circlespythagorean-theorem
📏 Medium solution 💡 3 insights 📊 Diagram
Problem
Two unit circles overlap with their centres half a unit apart. A third circle is the largest that fits inside both. A fourth circle is internally tangent to both unit circles and externally tangent to the third. Find the radius of the fourth circle.

Pick an answer.

(A)
$\frac{1}{14}$
(B)
$\frac{1}{12}$
(C)
$\frac{1}{10}$
(D)
$\frac{3}{28}$
(E)
$\frac{1}{9}$
How to solve
Strategy Draw a Diagram

Tangency conditions are easy to mis-set-up in words, so Tool #1 (Draw a Diagram) is the first move — place C₁ and C₂ symmetrically on a horizontal axis, mark the centers A, B, and the symmetry axis. Tool #7 (Identify Subproblems) splits the work into two clean pieces: first find C₃'s radius (one tangency equation), then find C₄'s radius (a right triangle plus one tangency equation). Tool #13 (Convert to Algebra) finishes by solving a linear equation in r once the right triangle is set up — the Pythagorean theorem is the workhorse.

1STEP 1

Place the two centres

Place them symmetrically.

A=(-1/4,0), B=(1/4,0), R₁=R₂=1
2STEP 2

The third circle's radius

Internal tangency gives its radius.

|MB|=1/4=1-r₃ → r₃=3/4
3STEP 3

The fourth circle's internal tangency

Tangency becomes a centre distance.

(1/4)² + y² = (1-r)²
4STEP 4

Write the external tangency

The external tangency is a distance too.

y=3/4+r
5STEP 5

Combine the equations

Combine the two equations.

5/8+3/2r=1-2r
6STEP 6

Solve for the radius

Solving gives three twenty-eighths.

r=3/8·2/7=3/28 → (D)
Answer
3/28
Sanity check the sizes. C₃ has radius 3/4 and sits centered between C₁, C₂; C₄ should be tiny because it has to squeeze between C₃ (radius 3/4) and the top arc of C₁∪ C₂. The height of C₄'s center is y=3/4+3/28=21/28+3/28=24/28=6/7. Pythagorean check: (1/4)²+(6/7)²=1/16+36/49. And (1-r)²=(25/28)²=625/784. Common denominator: 1/16=49/784 and 36/49=576/784; sum =625/784. Matches exactly — r=3/28 is correct.
💡Key takeaway

This AMC 12 problem only needs Grade 8 Pythagorean theorem plus the two simple circle-tangency rules you already know — drop in one right triangle, the squared terms cancel, and r=3/28 falls out of a single linear equation.