2025 AMC 10A Problem 22

When two circles touch from the outside, the distance between their centers is the sum of their radii. So every "touching" in the picture becomes a distance between centers. Start there.

1. Name the centers. Call the centers of the circles with radius $1, 2, 3$ by the letters $A, B, C$, and call the center of the small circle $O$. Then
$$AB = 1 + 2 = 3, \qquad AC = 1 + 3 = 4, \qquad BC = 2 + 3 = 5,$$
$$OA = r + 1, \qquad OB = r + 2, \qquad OC = r + 3.$$

2. Spot the right triangle. The sides $3, 4, 5$ satisfy $3^2 + 4^2 = 5^2$, so triangle $ABC$ has a right angle at $A$. That makes coordinates easy: put
$$A = (0, 0), \qquad B = (3, 0), \qquad C = (0, 4),$$
and let the small center be $O = (x, y)$.

3. Write the three distances as equations.
$$x^2 + y^2 = (r + 1)^2 \qquad (1)$$
$$(x - 3)^2 + y^2 = (r + 2)^2 \qquad (2)$$
$$x^2 + (y - 4)^2 = (r + 3)^2 \qquad (3)$$

4. Subtract to get rid of the squares. Every equation contains $x^2 + y^2$, so subtracting (1) from the others cancels those squares and leaves simple equations.
- (2) minus (1): $-6x + 9 = (r+2)^2 - (r+1)^2 = 2r + 3$. So $6x = 6 - 2r$, and $x = 1 - \dfrac{r}{3}$.
- (3) minus (1): $-8y + 16 = (r+3)^2 - (r+1)^2 = 4r + 8$. So $8y = 8 - 4r$, and $y = 1 - \dfrac{r}{2}$.

5. Put $x$ and $y$ back into (1).
$$\left(1 - \frac{r}{3}\right)^2 + \left(1 - \frac{r}{2}\right)^2 = (r + 1)^2.$$
Multiply both sides by $36$ to clear the fractions ($36 \cdot (1 - \frac r3)^2 = (6 - 2r)^2$ and $36 \cdot (1 - \frac r2)^2 = (6 - 3r)^2$):
$$(6 - 2r)^2 + (6 - 3r)^2 = 36(r + 1)^2,$$
$$36 - 24r + 4r^2 + 36 - 36r + 9r^2 = 36r^2 + 72r + 36,$$
$$23r^2 + 132r - 36 = 0.$$

6. Quadratic formula. The number under the square root is $132^2 + 4 \cdot 23 \cdot 36 = 17424 + 3312 = 20736 = 144^2$. So
$$r = \frac{-132 + 144}{2 \cdot 23} = \frac{12}{46} = \frac{6}{23}.$$
The other root, $\frac{-132 - 144}{46}$, is negative, so it is not a radius.

1. Check. With $r = \frac{6}{23}$: $x = 1 - \frac{2}{23} = \frac{21}{23}$ and $y = 1 - \frac{3}{23} = \frac{20}{23}$. Then $x^2 + y^2 = \frac{441 + 400}{529} = \frac{841}{529} = \left(\frac{29}{23}\right)^2$, and $r + 1 = \frac{29}{23}$. It works.

The answer is $\boxed{\textbf{(B)}\ \frac{6}{23}}$.

Circles that touch are really a statement about distances between centers, so the picture becomes a set of distance equations. Subtracting two circle equations cancels the squared terms, which is why $x$ and $y$ came out as simple expressions in $r$, leaving one equation in one unknown. The step that keeps the algebra short is noticing the $3$-$4$-$5$ right triangle and putting its right angle at $(0, 0)$. The answer choices are all between $0.25$ and $0.30$, so guessing from the picture will not work; the exact calculation is needed.

If you have seen Descartes' Circle Theorem, it gives the answer in two lines. For each circle, take its "curvature", which is $1$ divided by its radius. When four circles all touch each other, the fourth curvature is
$$k_4 = k_1 + k_2 + k_3 \pm 2\sqrt{k_1 k_2 + k_2 k_3 + k_3 k_1}.$$
Here $k_1 = 1$, $k_2 = \frac12$, $k_3 = \frac13$. Their sum is $\frac{11}{6}$, and $k_1k_2 + k_2k_3 + k_3k_1 = \frac12 + \frac16 + \frac13 = 1$. So $k_4 = \frac{11}{6} \pm 2$.

The plus sign gives $k_4 = \frac{23}{6}$, the small circle in the middle, so $r = \frac{6}{23}$. (The minus sign gives $-\frac16$, which stands for the big circle of radius $6$ that wraps around all three.) This is a formula to memorize, not one you can rebuild during the test, so the coordinate method above is the safer plan.