AMC 10 · 2024 · #21

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Circle-Polygon Hybrid

pythagorean-theoremcoordinate-geometrytangent-circlessimilar-triangles identify-subproblemsconvert-to-algebracasework ↑ Prerequisites: pythagorean-theoremcoordinate-geometryarea-circles

📏 Long solution 💡 4 insights 📊 Diagram

Problem

Two straight pipes (circular cylinders), with radii $1$ and $\frac{1}{4}$ , lie parallel and in contact on a flat floor. The figure below shows a head-on view. What is the sum of the possible radii of a third parallel pipe lying on the same floor and in contact with both?

Pick an answer.

(A)

$~\frac{1}{9}$

(B)

(C)

$~\frac{10}{9}$

(D)

$~\frac{11}{9}$

(E)

$~\frac{19}{9}$

AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

View mode:

Toolkit + CCSS Solution

Understand

Restated: Two cylindrical pipes of radii $1$ and $\tfrac{1}{4}$ lie on a flat floor, touching each other. A third pipe of radius $r$ also rests on the floor and touches both of the first two. There are exactly two valid choices for $r$; find their sum.

Givens: Pipe $A$ has radius $r_1 = 1$ and rests on the floor; Pipe $B$ has radius $r_2 = \tfrac{1}{4}$ and rests on the floor; Pipes $A$ and $B$ touch each other; A third pipe of unknown radius $r$ rests on the floor and touches both $A$ and $B$; Answer choices: (A) $\tfrac{1}{9}$, (B) $1$, (C) $\tfrac{10}{9}$, (D) $\tfrac{11}{9}$, (E) $\tfrac{19}{9}$

Unknowns: The sum $r_{\text{small}} + r_{\text{large}}$ of the two possible third radii

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems, #13 Convert to Algebra, #14 Use Cases

The whole problem is a head-on picture, so Tool #1 (Draw a Diagram) starts everything: put the floor on a horizontal line, plant the two given circles, and notice that any circle of radius $\rho$ resting on the floor has its center at height $\rho$. Tool #7 (Identify Subproblems) gives the key reusable fact — the horizontal distance between centers of two floor-resting circles tangent to each other is $2\sqrt{\rho_1 \rho_2}$, derived once from the Pythagorean theorem. Tool #13 (Convert to Algebra) then turns "the new pipe touches both given pipes" into two such horizontal-distance equations in $\sqrt{r}$, and Tool #14 (Use Cases) handles the two geometric placements (new pipe between the originals vs. outside them) as a $\pm$ sign in one equation.

Execute — Answer: C

#1 Draw a Diagram 5.G.A.2 Step 1

Set up coordinates.
Let the floor be the line $y = 0$.
A circle of radius $\rho$ resting on the floor is tangent to $y = 0$, so its center sits at height $\rho$.
Place the large pipe's center at $A = (0, 1)$.
Place the small pipe's center at $B = (x_B, \tfrac{1}{4})$ with $x_B > 0$.

$$A = (0, 1), \quad B = (x_B, \tfrac{1}{4})$$

💡 Grade 5 coordinate-plane graphing: picture the floor as a number line and let each circle's center sit directly above where it touches the floor.

#7 Identify Subproblems 8.G.B.7 Step 2

Use the tangency $|AB| = 1 + \tfrac{1}{4} = \tfrac{5}{4}$ to pin down $x_B$.
The vertical drop from $A$ to $B$ is $1 - \tfrac{1}{4} = \tfrac{3}{4}$, so the Pythagorean theorem gives $x_B^2 + (\tfrac{3}{4})^2 = (\tfrac{5}{4})^2$, hence $x_B^2 = \tfrac{25}{16} - \tfrac{9}{16} = 1$ and $x_B = 1$.

$$x_B^2 = \left(\tfrac{5}{4}\right)^2 - \left(\tfrac{3}{4}\right)^2 = \tfrac{25 - 9}{16} = 1 \;\Rightarrow\; x_B = 1$$

💡 Grade 8 Pythagorean theorem reads the horizontal gap directly off the right triangle whose hypotenuse is the line through the two centers.

#7 Identify Subproblems 8.EE.A.2 Step 3

Derive a reusable shortcut.
The same computation for any two floor-resting circles of radii $\rho_1, \rho_2$ that are externally tangent gives horizontal distance $\sqrt{(\rho_1 + \rho_2)^2 - (\rho_1 - \rho_2)^2} = \sqrt{4\rho_1 \rho_2} = 2\sqrt{\rho_1 \rho_2}$.
(Sanity check: with $\rho_1 = 1, \rho_2 = \tfrac{1}{4}$ this gives $2 \sqrt{1/4} = 1$, matching $x_B = 1$.) We will reuse this formula three times.

$$\Delta x(\rho_1, \rho_2) = \sqrt{(\rho_1+\rho_2)^2 - (\rho_1-\rho_2)^2} = 2\sqrt{\rho_1 \rho_2}$$

💡 Grade 8 square-root and squared-difference algebra packages the same Pythagorean step into a one-line tool we can reuse.

#13 Convert to Algebra 8.EE.C.7 Step 4

Set up the third pipe.
Let its radius be $r$ and its center be $C = (x, r)$.
The horizontal distance from $A = (0, 1)$ to $C$ is $|x|$, and the horizontal distance from $B = (1, \tfrac{1}{4})$ to $C$ is $|x - 1|$.
Tangency with $A$ forces $|x| = 2\sqrt{1 \cdot r} = 2\sqrt{r}$.
Tangency with $B$ forces $|x - 1| = 2\sqrt{\tfrac{1}{4} \cdot r} = \sqrt{r}$.

$$|x| = 2\sqrt{r}, \quad |x - 1| = \sqrt{r}$$

💡 Grade 8 turn-words-into-equations: each tangency between the new pipe and a given pipe is one horizontal-distance equation.

#14 Use Cases 8.EE.A.2 Step 5

Split into cases by the new pipe's position relative to pipe $B$.
Case (i) — the new pipe sits in the wedge to the right of $A$ and to the left of $B$: $0 < x < 1$, so $|x| = x$ and $|x - 1| = 1 - x$, giving $x = 2\sqrt{r}$ and $1 - x = \sqrt{r}$.
Add: $1 = 3\sqrt{r}$, so $\sqrt{r} = \tfrac{1}{3}$ and $r = \tfrac{1}{9}$.
Case (ii) — the new pipe sits to the right of $B$: $x > 1$, so $|x| = x$ and $|x - 1| = x - 1$, giving $x = 2\sqrt{r}$ and $x - 1 = \sqrt{r}$.
Subtract: $1 = \sqrt{r}$, so $r = 1$.

$$(\text{i}) \; 3\sqrt{r} = 1 \Rightarrow r = \tfrac{1}{9}, \qquad (\text{ii}) \; \sqrt{r} = 1 \Rightarrow r = 1$$

💡 Grade 8 use-cases: the $\pm$ in $|x - 1|$ packages "new pipe between vs. outside" into one clean sign choice.

#13 Convert to Algebra 7.NS.A.1 Step 6

Add the two possible radii to answer the problem.

$$r_{\text{small}} + r_{\text{large}} = \tfrac{1}{9} + 1 = \tfrac{10}{9} \;\Rightarrow\; \textbf{(C)}$$

💡 Grade 7 fraction addition closes it: $\tfrac{1}{9} + \tfrac{9}{9} = \tfrac{10}{9}$.

[1] #1 5.G.A.2 Set up coordinates. Let the floor be the line $y = 0$. A circle of radius $\rho$

[2] #7 8.G.B.7 Use the tangency $|AB| = 1 + \tfrac{1}{4} = \tfrac{5}{4}$ to pin down $x_B$. The

[3] #7 8.EE.A.2 Derive a reusable shortcut. The same computation for any two floor-resting circl

[4] #13 8.EE.C.7 Set up the third pipe. Let its radius be $r$ and its center be $C = (x, r)$. The

[5] #14 8.EE.A.2 Split into cases by the new pipe's position relative to pipe $B$. Case (i) — the

[6] #13 7.NS.A.1 Add the two possible radii to answer the problem.

Review

Reasonableness: Each value passes a sanity check. For $r = \tfrac{1}{9}$ the new pipe is much smaller than the small pipe ($\tfrac{1}{9} < \tfrac{1}{4}$), which is exactly what fits in the tiny wedge between the two given pipes — the picture supports this. For $r = 1$ the new pipe is the same size as the large pipe, sitting on the opposite side of the small pipe; symmetry makes this plausible too. The sum $\tfrac{10}{9} \approx 1.11$ matches choice (C). The nearby distractor (B) $1$ would mean only the large outside solution counts; (A) $\tfrac{1}{9}$ would mean only the small wedge solution counts; neither matches the problem's "two possible radii".

Alternative: Tool #3 (Eliminate Possibilities): the small wedge radius is clearly less than $\tfrac{1}{4}$, and $\tfrac{1}{9}$ is the only choice that even looks like a wedge radius — so $r_{\text{small}} = \tfrac{1}{9}$ is forced. Then the sum must be $\tfrac{1}{9} +$ something, which immediately rules out (A), and leaves (C) $\tfrac{10}{9}$, (D) $\tfrac{11}{9}$, (E) $\tfrac{19}{9}$, corresponding to $r_{\text{large}} = 1, \tfrac{10}{9}, 2$. A quick sketch shows the outside pipe is roughly as wide as the large pipe (both rest on the floor and span the small pipe), so $r_{\text{large}} = 1$, giving (C). This avoids the full algebra.

CCSS standards used (min grade 8)

5.G.A.2 Represent real-world and mathematical problems by graphing points in the first quadrant of the coordinate plane (Placing the floor as the $x$-axis and the pipe centers at heights equal to their radii, $A = (0, 1)$ and $B = (1, \tfrac{1}{4})$.)
8.G.B.7 Apply the Pythagorean Theorem to determine unknown side lengths in right triangles in real-world and mathematical problems (Computing the horizontal gap between two floor-resting circles' centers as $\sqrt{(\rho_1+\rho_2)^2 - (\rho_1-\rho_2)^2}$ from the right triangle of centers.)
8.EE.A.2 Use square root and cube root symbols to represent solutions to equations of the form $x^2 = p$ and $x^3 = p$; evaluate square roots of small perfect squares (Simplifying $\sqrt{(\rho_1+\rho_2)^2 - (\rho_1-\rho_2)^2} = 2\sqrt{\rho_1 \rho_2}$ and packaging the casework as a $\pm$ inside $|x - 1|$.)
8.EE.C.7 Solve linear equations in one variable (Solving $3\sqrt{r} = 1$ and $\sqrt{r} = 1$ — each a linear equation in $u = \sqrt{r}$ — to extract the two possible radii.)
7.NS.A.1 Apply and extend previous understandings of addition and subtraction to add and subtract rational numbers (Adding $\tfrac{1}{9} + 1 = \tfrac{10}{9}$ for the final answer.)

⭐ When two same-floor circles touch, the horizontal gap between their centers is $2\sqrt{\rho_1 \rho_2}$ — a one-line shortcut from the Pythagorean theorem. Use that shortcut twice for the new pipe, split into "between" and "outside" cases, and the two answers $\tfrac{1}{9}$ and $1$ pop right out.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 8)

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