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

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 $C_4$ ?

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

Pick an answer.

(A)

$frac{1}{14}$

(B)

$frac{1}{12}$

(C)

$frac{1}{10}$

(D)

$frac{3}{28}$

(E)

$frac{1}{9}$

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Toolkit + CCSS Solution

Understand

Restated: Two unit circles $C_1$ and $C_2$ overlap, with their centers $\frac{1}{2}$ apart. $C_3$ is the largest circle that fits inside both $C_1$ and $C_2$ (internally tangent to each). $C_4$ is another circle internally tangent to $C_1$ and $C_2$ and also externally tangent to $C_3$ (a smaller circle pinched between $C_3$ and the rims of $C_1, C_2$). Find the radius of $C_4$.

Givens: Radii of $C_1$ and $C_2$ are both $1$; Distance between centers of $C_1$ and $C_2$ is $\frac{1}{2}$; $C_3$ is the largest circle internally tangent to both $C_1$ and $C_2$; $C_4$ is internally tangent to $C_1$ and $C_2$, externally tangent to $C_3$; Answer choices: (A) $\tfrac{1}{14}$, (B) $\tfrac{1}{12}$, (C) $\tfrac{1}{10}$, (D) $\tfrac{3}{28}$, (E) $\tfrac{1}{9}$

Unknowns: The radius $r$ of circle $C_4$

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems, #13 Convert to Algebra

Tangency conditions are easy to mis-set-up in words, so Tool #1 (Draw a Diagram) is the first move — place $C_1$ and $C_2$ 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 $C_3$'s radius (one tangency equation), then find $C_4$'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.

Execute — Answer: D

#1 Draw a Diagram 6.NS.C.8 Step 1

Set up coordinates from the picture.
By symmetry, place the centers of $C_1$ and $C_2$ on the x-axis equally spaced about the origin: $A = (-\tfrac{1}{4}, 0)$ and $B = (\tfrac{1}{4}, 0)$, so $|AB| = \tfrac{1}{2}$.
The symmetry axis is the y-axis.

$$A=(-\tfrac{1}{4},0),\;B=(\tfrac{1}{4},0),\;R_1=R_2=1$$

💡 Grade 6 "place points by their coordinates" — set up the picture so the symmetry is obvious.

#7 Identify Subproblems 7.G.B.4 Step 2

Find the radius of $C_3$.
The largest circle inside both $C_1$ and $C_2$ must be centered on the y-axis (by symmetry) and on the x-axis (it sits at the symmetric center).
So its center is the origin $M=(0,0)$.
Internal tangency to $C_2$ means $|MB|=R_2-r_3$, i.e., $\tfrac{1}{4}=1-r_3$, giving $r_3=\tfrac{3}{4}$.

$$|MB|=\tfrac{1}{4}=1-r_3\;\Rightarrow\;r_3=\tfrac{3}{4}$$

💡 Grade 7 "facts about circles" — for two internally tangent circles the centers are radius-difference apart.

#1 Draw a Diagram 8.G.B.7 Step 3

Set up $C_4$.
By symmetry its center sits on the y-axis at some height $y > 0$: call it $O_4 = (0, y)$ with radius $r$.
Internal tangency to $C_1$ means $|AO_4| = R_1 - r = 1 - r$.
Form the right triangle $\triangle AMO_4$ where $M=(0,0)$ is the foot of the perpendicular from $O_4$ to the x-axis.
Its legs are $|AM|=\tfrac{1}{4}$ and $|MO_4|=y$, and its hypotenuse is $|AO_4|=1-r$.

$$\bigl(\tfrac{1}{4}\bigr)^2 + y^2 = (1-r)^2$$

💡 Grade 8 Pythagorean theorem on the right triangle formed by $A$, the y-axis, and $O_4$.

#7 Identify Subproblems 7.G.B.4 Step 4

External tangency between $C_3$ and $C_4$ pins down $y$.
Distance between their centers equals sum of radii: $|MO_4| = r_3 + r$, i.e., $y = \tfrac{3}{4} + r$.

$$y=\tfrac{3}{4}+r$$

💡 External tangency = "two circles kissing on the outside," so centers are radius-sum apart.

#13 Convert to Algebra 8.EE.C.7 Step 5

Substitute $y$ from step 4 into the Pythagorean equation from step 3 and solve for $r$.
Expand both sides and watch the $r^2$ terms cancel: $\tfrac{1}{16} + (\tfrac{3}{4}+r)^2 = (1-r)^2$ becomes $\tfrac{1}{16}+\tfrac{9}{16}+\tfrac{3}{2}r+r^2 = 1-2r+r^2$, so $\tfrac{10}{16}+\tfrac{3}{2}r = 1 - 2r$.

$$\tfrac{5}{8}+\tfrac{3}{2}r=1-2r$$

💡 Grade 8 "solve a linear equation in one variable" — quadratic terms cancel, leaving a single equation in $r$.

#13 Convert to Algebra 8.EE.C.7 Step 6

Collect $r$ terms and constants: $\tfrac{3}{2}r + 2r = 1 - \tfrac{5}{8}$, i.e., $\tfrac{7}{2}r = \tfrac{3}{8}$.
Multiply both sides by $\tfrac{2}{7}$: $r = \tfrac{3}{8}\cdot\tfrac{2}{7} = \tfrac{6}{56} = \tfrac{3}{28}$, which is choice (D).

$$r=\tfrac{3}{8}\cdot\tfrac{2}{7}=\tfrac{3}{28}\;\Rightarrow\;\textbf{(D)}$$

💡 One-step linear solve in $r$ — the unique answer drops out.

[1] #1 6.NS.C.8 Set up coordinates from the picture. By symmetry, place the centers of $C_1$ and

[2] #7 7.G.B.4 Find the radius of $C_3$. The largest circle inside both $C_1$ and $C_2$ must be

[3] #1 8.G.B.7 Set up $C_4$. By symmetry its center sits on the y-axis at some height $y > 0$:

[4] #7 7.G.B.4 External tangency between $C_3$ and $C_4$ pins down $y$. Distance between their

[5] #13 8.EE.C.7 Substitute $y$ from step 4 into the Pythagorean equation from step 3 and solve f

[6] #13 8.EE.C.7 Collect $r$ terms and constants: $\tfrac{3}{2}r + 2r = 1 - \tfrac{5}{8}$, i.e.,

Review

Reasonableness: Sanity check the sizes. $C_3$ has radius $\tfrac{3}{4}$ and sits centered between $C_1, C_2$; $C_4$ should be tiny because it has to squeeze between $C_3$ (radius $\tfrac{3}{4}$) and the top arc of $C_1\cup C_2$. The height of $C_4$'s center is $y=\tfrac{3}{4}+\tfrac{3}{28}=\tfrac{21}{28}+\tfrac{3}{28}=\tfrac{24}{28}=\tfrac{6}{7}$. Pythagorean check: $\bigl(\tfrac{1}{4}\bigr)^2+\bigl(\tfrac{6}{7}\bigr)^2=\tfrac{1}{16}+\tfrac{36}{49}$. And $(1-r)^2=\bigl(\tfrac{25}{28}\bigr)^2=\tfrac{625}{784}$. Common denominator: $\tfrac{1}{16}=\tfrac{49}{784}$ and $\tfrac{36}{49}=\tfrac{576}{784}$; sum $=\tfrac{625}{784}$. Matches exactly — $r=\tfrac{3}{28}$ is correct.

Alternative: Tool #3 (Eliminate Possibilities) on the answer choices: $r$ must satisfy $\bigl(\tfrac{1}{4}\bigr)^2+\bigl(\tfrac{3}{4}+r\bigr)^2=(1-r)^2$. Plug each choice in and check. $r=\tfrac{3}{28}$ matches as shown; $r=\tfrac{1}{12}$ gives left side $\approx 0.755$, right side $\approx 0.840$, mismatch. The other choices similarly miss.

CCSS standards used (min grade 8)

6.NS.C.8 Solve real-world and mathematical problems by graphing points in all four quadrants of the coordinate plane (Placing the centers of $C_1, C_2$ on the x-axis symmetrically about the origin to exploit the picture's symmetry.)
7.G.B.4 Know the formulas for the area and circumference of a circle; use facts about circles (Using the rules for internally tangent circles (distance = radius difference) and externally tangent circles (distance = radius sum).)
8.G.B.7 Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Setting up $(\tfrac{1}{4})^2 + y^2 = (1-r)^2$ on the right triangle $\triangle AMO_4$.)
8.EE.C.7 Solve linear equations in one variable (Substituting $y = \tfrac{3}{4}+r$ and reducing the resulting equation (the $r^2$ terms cancel) to a one-step linear solve for $r$.)

⭐ 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 $r=\tfrac{3}{28}$ falls out of a single linear equation.