2023 AMC 10 A Problem 22

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To help you master 2023 AMC 10 A Problem 22, this page includes:

The original AMC problem to try, plus the correct answer

Clear, step-by-step solutions using different methods

Key ideas, useful formulas, and takeaways you can apply to similar problems

Related problems from past AMC exams, plus new practice problems that build the same skills

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2023 AMC10 A Problem 22

Solution

Solution 1

Pythagorean Decomposition Method

The distance between the centers of two tangent circles is either the sum of their radii (external tangency) or the difference of their radii (internal tangency). Using the problem’s symmetry, we can form right triangles whose side lengths are expressed in terms of the unknown radii, allowing us to solve for them with the Pythagorean theorem.

For two tangent circles with radii $R$ and $r$ (where $R > r$ ): If they are externally tangent, the distance between their centers is $R + r$ . If they are internally tangent, the distance between their centers is $R - r$ .

Step 1: Identify the line of symmetry

Because $C_{1}$ and $C_{2}$ have the same radius and we seek circles tangent to both, the centers of $C_{3}$ and $C_{4}$ must lie on the perpendicular bisector of the segment connecting the centers of $C_{1}$ and $C_{2}$ . Let $O_{1}$ and $O_{2}$ be the centers of $C_{1}$ and $C_{2}$ , and let $M$ be their midpoint. Since the distance between $O_{1}$ and $O_{2}$ is $\frac{1}{2}$ , the distance from $O_{1}$ to $M$ is half of that.

O_{1} M = \frac{1}{4}

Step 2: Find the radius and center of $C_{3}$

Let $C_{3}$ have center $O_{3}$ and radius $r_{3}$ . Because it is internally tangent to $C_{1}$ , the distance between their centers is $1 - r_{3}$ . To make $C_{3}$ as large as possible (maximizing $r_{3}$ ), we must minimize the distance $O_{1} O_{3}$ . The distance $O_{1} O_{3}$ forms the hypotenuse of right $△ O_{1} M O_{3}$ . This distance is minimized when the leg $M O_{3}$ is $0$ , meaning $O_{3}$ is located exactly at $M$ . Setting the expressions equal gives us the radius.

O_{1} O_{3} = O_{1} M^{2} + M O_{3}^{2}

O_{1} O_{3} = 1 - r_{3}

1 - r_{3} = \frac{1}{4} ⟹ r_{3} = \frac{3}{4}

Step 3: Set up the right triangle for $C_{4}$

Let $C_{4}$ have center $O_{4}$ and radius $r_{4}$ . Because $O_{4}$ lies on the perpendicular bisector, $△ O_{1} M O_{4}$ is a right triangle. Circle $C_{4}$ is internally tangent to $C_{1}$ (radius $1$ ), so the hypotenuse is $1 - r_{4}$ . Circle $C_{4}$ is externally tangent to $C_{3}$ (centered at $M$ with radius $\frac{3}{4}$ ), so the vertical leg is the sum of their radii. The horizontal leg remains $O_{1} M = \frac{1}{4}$ .

O_{1} O_{4} = 1 - r_{4}

M O_{4} = \frac{3}{4} + r_{4}

Step 4: Apply the Pythagorean theorem

In right $△ O_{1} M O_{4}$ , we use the Pythagorean theorem to relate the legs to the hypotenuse.

O_{1} M^{2} + M O_{4}^{2} = O_{1} O_{4}^{2}

(\frac{1}{4})^{2} + (\frac{3}{4} + r_{4})^{2} = (1 - r_{4})^{2}

Step 5: Solve for the radius of $C_{4}$

Expand both squared binomials and collect like terms. Notice that the $r_{4}^{2}$ terms cancel out, leaving a linear equation that we can easily solve for $r_{4}$ .

\frac{1}{16} + \frac{9}{16} + \frac{3}{2} r_{4} + r_{4}^{2} = 1 - 2 r_{4} + r_{4}^{2}

\frac{10}{16} + \frac{3}{2} r_{4} = 1 - 2 r_{4}

\frac{5}{8} + \frac{3}{2} r_{4} = 1 - 2 r_{4}

\frac{7}{2} r_{4} = \frac{3}{8}

r_{4} = \frac{3}{8} \cdot \frac{2}{7} = \frac{3}{28}

Correct answer: D. $\frac{3}{28}$
The radius of

C_{4}

is

\frac{3}{28}

, which matches answer choice D. We can verify this using the distance values: if

r_{4} = \frac{3}{28}

, then the vertical leg is

\frac{3}{4} + \frac{3}{28} = \frac{24}{28} = \frac{6}{7}

. The horizontal leg is

\frac{1}{4}

. The sum of their squares is

\frac{1}{16} + \frac{36}{49} = \frac{49}{784} + \frac{576}{784} = \frac{625}{784}

. The square root of this sum is

\frac{25}{28}

, which perfectly matches the required hypotenuse length

1 - r_{4} = 1 - \frac{3}{28} = \frac{25}{28}

.

Solution 2

Cartesian Coordinate Method

By placing the configuration on a coordinate plane, geometric tangency constraints can be translated directly into algebraic distance formulas. This method mirrors the geometric logic but uses explicit coordinates, which some students find more mechanical and easier to manage.

The distance between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ on a Cartesian plane is given by $d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$ .

Step 1: Set up the coordinate plane

Place the origin $(0, 0)$ at the midpoint of the segment connecting the centers of $C_{1}$ and $C_{2}$ . Since the distance between the centers is $\frac{1}{2}$ , they lie on the x-axis symmetrically.

O_{1} = (- \frac{1}{4}, 0)

O_{2} = (\frac{1}{4}, 0)

Step 2: Determine the center and radius of $C_{3}$

Any circle tangent to both $C_{1}$ and $C_{2}$ must have its center on the y-axis (the line $x = 0$ ) due to symmetry. Let $C_{3}$ be centered at $(0, y_{3})$ with radius $r_{3}$ . Because it is internally tangent to $C_{1}$ (which has radius $1$ ), the distance to $O_{1}$ must equal $1 - r_{3}$ . To maximize $r_{3}$ , we must minimize this distance, which occurs exactly when $y_{3} = 0$ .

(- \frac{1}{4} - 0)^{2} + (0 - y_{3})^{2} = 1 - r_{3}

y_{3} = 0 ⟹ \frac{1}{4} = 1 - r_{3}

r_{3} = \frac{3}{4}

Step 3: Define the coordinates for $C_{4}$

Let $C_{4}$ be centered at $(0, y_{4})$ with radius $r_{4}$ . The condition that it is internally tangent to $C_{1}$ means the distance from $(0, y_{4})$ to $(- \frac{1}{4}, 0)$ is $1 - r_{4}$ . We can write this using the distance formula squared.

(- \frac{1}{4} - 0)^{2} + (0 - y_{4})^{2} = (1 - r_{4})^{2}

\frac{1}{16} + y_{4}^{2} = (1 - r_{4})^{2}

Step 4: Apply the external tangency condition

Circle $C_{4}$ is externally tangent to $C_{3}$ . Since $C_{3}$ is centered at $(0, 0)$ with radius $\frac{3}{4}$ , the distance between their centers is simply $∣ y_{4} ∣$ (we can assume $C_{4}$ is in the upper half-plane without loss of generality). The external tangency rule states this distance is the sum of their radii.

y_{4} = \frac{3}{4} + r_{4}

Step 5: Substitute and solve for $r_{4}$

Substitute the expression for $y_{4}$ into the squared distance equation and solve for $r_{4}$ . The algebra identically matches the geometric method.

\frac{1}{16} + (\frac{3}{4} + r_{4})^{2} = (1 - r_{4})^{2}

\frac{1}{16} + \frac{9}{16} + \frac{3}{2} r_{4} + r_{4}^{2} = 1 - 2 r_{4} + r_{4}^{2}

\frac{10}{16} + \frac{3}{2} r_{4} = 1 - 2 r_{4}

\frac{7}{2} r_{4} = \frac{3}{8}

r_{4} = \frac{3}{28}

Correct answer: D. $\frac{3}{28}$
The radius is

\frac{3}{28}

(Choice D). Both the pure geometric right triangle model and the coordinate-based algebra lead to the exact same linear equation, confirming that our model securely represents the problem conditions.

Adapted Problems to Practice

1 / 10

What Formulas Do You Need for This Problem?

Distance Between Centers of Tangent Circles:

For two circles with radii $r_{1}$ and $r_{2}$ : 1. If they are externally tangent, the distance between their centers is $r_{1} + r_{2}$ . 2. If they are internally tangent, the distance between their centers is $∣ r_{1} - r_{2} ∣$ .

$r_{1}$ and $r_{2}$ are the radii of the two respective circles. The theorem applies universally in planar geometry to find the exact distance between centers of tangent circles. Circle $C_{4}$ is internally tangent to $C_{1}$ and externally tangent to $C_{3}$ . We used these rules to express the distances between their centers precisely in terms of their unknown radii.

Line of Symmetry for Centers:

If a geometric configuration of circles is symmetric with respect to a line, any circle uniquely defined by symmetric constraints must have its center on that line.

Applies when dealing with geometric configurations featuring reflectional symmetry. Because circles $C_{1}$ and $C_{2}$ have the same radius, the entire setup is symmetric across the perpendicular bisector of the segment connecting their centers. This guarantees the centers of $C_{3}$ and $C_{4}$ lie exactly on that bisector, significantly simplifying the problem into a 2D right triangle.

Key Takeaway: Remember This Strategy for Next Time

A problem asking for the radius of a circle that is tangent to two or more other circles.

Try: Connect the centers of the circles to form a triangle. Express the side lengths of this triangle using the sums and differences of the circle radii.

Why: The centers and points of tangency are collinear, making the distance between centers a simple arithmetic combination of their radii ( $R + r$ or $R - r$ ). This transforms a pure geometry problem into an algebraic equation.

A symmetrical arrangement of equally sized circles (like $C_{1}$ and $C_{2}$ ).

Try: Identify the axis of symmetry (usually the perpendicular bisector between their centers) and place the centers of newly formed tangent circles along this axis.

Why: Symmetry dictates that to remain tangent to two identical circles simultaneously, a new circle cannot lean closer to one than the other. Placing its center on the line of symmetry reduces the problem to calculating distances along a single axis, allowing the creation of right triangles.

The phrase "largest circle internally tangent to..."

Try: Set up an algebraic expression for the radius in terms of an unknown distance, and determine which value of that distance maximizes the radius.

Why: For an internally tangent circle, the radius $r$ relates to the outer radius $R$ and the center-to-center distance $d$ via $d = R - r$ . Rearranging gives $r = R - d$ . To maximize $r$ , you must minimize $d$ , which often forces the center of the inner circle to be as close to the center of the outer circle as the other constraints allow.

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2023 AMC 10 A Problem 22

In This Page:

2023 AMC10 A Problem 22