Printable

2023 AMC 10A Problem 22

2023 AMC 10A Problem 22AMC 10 Step by Step

Name: Date: September 13, 2026

Circles · difficulty 4 of 5 · about 6 min · position P21-25

Circle $C_1$ and $C_2$ each have radius $1$ , and the distance between their centers is $\frac{1}{2}$ . Circle $C_3$ is the largest circle internally tangent to both $C_1$ and $C_2$ . Circle $C_4$ is internally tangent to both $C_1$ and $C_2$ and externally tangent to $C_3$ . What is the radius of $C_4$ ?

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

Answer

Key insight

C3 is centered at the midpoint with radius 3/4; C4's center is on the perpendicular bisector, so a right triangle with legs 1/4, 3/4+r and hypotenuse 1-r gives r.

Solution

Let $O_1$ and $O_2$ be the centers of $C_1$ and $C_2$ , and $M$ the midpoint of $O_1O_2$ , so $MO_1 = MO_2 = \frac14$ .

The largest circle inside both $C_1$ and $C_2$ is centered at $M$ : it must fit inside the lens-shaped overlap, and the circle centered at $M$ of radius $1 - \frac14 = \frac34$ touches both big circles internally. So $C_3$ has center $M$ and radius $\frac34$ .

By symmetry the center $O_4$ of $C_4$ lies on the perpendicular bisector of $O_1O_2$ , directly "above" $M$ . Let $r$ be its radius. External tangency with $C_3$ gives $MO_4 = \frac34 + r$ . Internal tangency with $C_1$ gives $O_1O_4 = 1 - r$ .

Triangle $O_1 M O_4$ has a right angle at $M$ , so
$\left(\tfrac14\right)^2 + \left(\tfrac34 + r\right)^2 = (1 - r)^2.$
Expanding: $\frac{1}{16} + \frac{9}{16} + \frac32 r + r^2 = 1 - 2r + r^2$ , so $\frac{7}{2} r = 1 - \frac{10}{16} = \frac{3}{8}$ and $r = \frac{3}{28}$ .

The answer is $\boxed{\textbf{(D)}\ \frac{3}{28}}$ .

Why this works

Tangency conditions are distance conditions between centers: difference of radii for internal, sum for external. Once the centers are located (here two of them by symmetry), the problem is a single right triangle. Look for the line of symmetry first; it usually turns a tangent-circle configuration into one Pythagorean equation.

Alternative approach

Coordinates: $O_1 = (-\frac14, 0)$ , $O_4 = (0, y)$ with $y = \frac34 + r$ . Then $\frac{1}{16} + y^2 = (1 - r)^2$ , the same equation.

The trap

Using external tangency (distance = 1 + r) between C4 and C1, or taking C3's radius as 1/2 instead of 3/4.

Common mistakes

Using external tangency (distance = 1 + r) between C4 and C1, or taking C3's radius as 1/2 instead of 3/4.
Squaring $\frac34 + r$ as $\frac{9}{16} + r^2$ and losing the cross term $\frac32 r$ .

Techniques

Add construction lines/points (drop altitudes, extend segments, connect centers) · Set up the equation/formula and compute; no special trick needed