AMC 10 · 2023 · #22

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 C₁ and C₂ overlap, with their centers 12\frac{1}{2} apart. C₃ is the largest circle that fits inside both C₁ and C₂ (internally tangent to each). C₄ is another circle internally tangent to C₁ and C₂ and also externally tangent to C₃ (a smaller circle pinched between C₃ and the rims of C₁, C₂). Find the radius of C₄.

Pick an answer.

(A)
$frac{1}{14}$
(B)
$frac{1}{12}$
(C)
$frac{1}{10}$
(D)
$frac{3}{28}$
(E)
$frac{1}{9}$

AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

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 C₁, C₂ symmetrically on the x-axis about the origin: A=(−14-\frac{1}{4},0), B=(14\frac{1}{4},0), so |AB|=12\frac{1}{2} and the y-axis is the symmetry axis.

A=(−14-\frac{1}{4},0), B=(14\frac{1}{4},0), R₁=R₂=1
2STEP 2

The largest inner circle is centered at the origin M=(0,0); internal tangency |MB|=1-r₃ with |MB|=14\frac{1}{4} gives r₃=34\frac{3}{4}.

|MB|=14\frac{1}{4}=1-r₃ → r₃=34\frac{3}{4}
3STEP 3

C₄'s center is O₄=(0,y), radius r; internal tangency to C₁ gives hypotenuse |AO₄|=1-r in right triangle △AMO₄ with legs 14\frac{1}{4} and y.

(14\frac{1}{4})² + y² = (1-r)²
4STEP 4

External tangency of C₃ and C₄ sets centers a radius-sum apart: |MO₄|=r₃+r, so y=34\frac{3}{4}+r.

y=34\frac{3}{4}+r
5STEP 5

Substitute y=34\frac{3}{4}+r into (14\frac{1}{4})²+y²=(1-r)²; the r² terms cancel, leaving 58\frac{5}{8}+32\frac{3}{2}r=1-2r.

58\frac{5}{8}+32\frac{3}{2}r=1-2r
6STEP 6

Gather terms: 32\frac{3}{2}r+2r=1-58\frac{5}{8} gives 72\frac{7}{2}r=38\frac{3}{8}, so r=38\frac{3}{8}·27\frac{2}{7}=328\frac{3}{28} — choice (D).

r=38\frac{3}{8}·27\frac{2}{7}=328\frac{3}{28} → (D)
Answer
frac{3}{28}
Sanity check the sizes. C₃ has radius 34\frac{3}{4} and sits centered between C₁, C₂; C₄ should be tiny because it has to squeeze between C₃ (radius 34\frac{3}{4}) and the top arc of C₁∪ C₂. The height of C₄'s center is y=34\frac{3}{4}+328\frac{3}{28}=2128\frac{21}{28}+328\frac{3}{28}=2428\frac{24}{28}=67\frac{6}{7}. Pythagorean check: (14\frac{1}{4})²+(67\frac{6}{7})²=116\frac{1}{16}+3649\frac{36}{49}. And (1-r)²=(2528\frac{25}{28})²=625784\frac{625}{784}. Common denominator: 116\frac{1}{16}=49784\frac{49}{784} and 3649\frac{36}{49}=576784\frac{576}{784}; sum =625784\frac{625}{784}. Matches exactly — r=328\frac{3}{28} is correct.
💡Key takeaway

This AMC 10 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=328\frac{3}{28} falls out of a single linear equation.