AMC 10 · 2025 · #22

Grade 8 geometry-2d

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

tangent-circlesdescartes-circle-theoremcoordinate-geometry convert-to-algebra ↑ Prerequisites: tangent-circles

📏 Long solution 💡 3 insights 📊 Diagram

Problem

A small circle of radius r sits in the middle, touching three larger circles of radii 1, 2, and 3. Every pair of the four circles touches from the outside, and the three big circles also touch each other. From the picture the small circle is the one nestled snugly in the gap between the three big ones. Find r.

Pick an answer.

(A)

$\frac{1}{4}$

(B)

$\frac{6}{23}$

(C)

$\frac{3}{11}$

(D)

$\frac{5}{17}$

(E)

$\frac{3}{10}$

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

How to solve

Strategy Introduce a Variable

Working with the radii directly is messy because tangency ties the centers together through square roots of sums. The clean move is to name a new variable for each circle: its curvature k = 1/r. For four circles that all touch one another, the curvatures obey Descartes' Circle Theorem, a single tidy equation. So I introduce k for every circle, split the theorem into two easy pieces (a plain sum and a sum of products under a square root), compute each, and read off the two possible circles. The picture then eliminates the wrong one, and flipping the curvature back gives r.

1STEP 1

Switch from radius to curvature

Radii tangle, but their reciprocals do not. Give each circle a curvature k = 1/r: the known three are 1, 1/2, 1/3, and the unknown is 1/r.

k₁=1/1=1, k₂=1/2, k₃=1/3, k₄=1/r

2STEP 2

Write Descartes' Circle Theorem

Descartes' Circle Theorem: k4 is the sum of the other three curvatures, plus or minus twice the root of their pairwise products.

k₄=k₁+k₂+k₃± 2√(k₁k₂+k₂k₃+k₃k₁)

3STEP 3

Add the three curvatures

The easy piece first: over denominator 6, 1 + 1/2 + 1/3 = 6/6 + 3/6 + 2/6 = 11/6.

k₁+k₂+k₃=1+1/2+1/3=6/6+3/6+2/6=11/6

4STEP 4

Add the three pairwise products

Under the root, the pairwise products give 1/2 + 1/6 + 1/3 = 3/6 + 1/6 + 2/6 = exactly 1, a perfect square that keeps this clean.

k₁k₂+k₂k₃+k₃k₁=1/2+1/6+1/3=3/6+1/6+2/6=1

5STEP 5

Combine the two pieces

The root of 1 is 1, so the correction term is 2, and k4 = 11/6 ± 2 splits into 23/6 or -1/6.

k₄=11/6± 2√(1)=11/6± 2=23/6 or -1/6

6STEP 6

Pick the small circle and flip back

A bigger curvature means a smaller circle, so the nestled one is 23/6 and -1/6 is the radius-6 wrapper; flipping back, r = 6/23.

r=1/k₄=1/23/6=6/23

Answer

6/23

The value r = 6/23 is about 0.26, a small positive length smaller than all three given radii, exactly the size of the tiny circle drawn in the central gap, so it passes the eye test. The rejected sign gave curvature -1/6, i.e. radius 6, the giant circle circling the other three, which the picture rules out. Note the answer choices 1/4 = 0.25, 6/23 approx 0.261, and 3/11 approx 0.273 are all close, so the exact fraction matters; the perfect cancellation to 6/23 confirms choice (B) rather than a near neighbor.

💡Key takeaway

When circles all touch, stop juggling radii and switch to curvature (1/r); Descartes' Circle Theorem then turns the whole picture into one short fraction sum.

Switch from radius to curvature
Write Descartes' Circle Theorem
Add the three curvatures
Add the three pairwise products
Combine the two pieces
Pick the small circle and flip back