Competition · AMC preparation · step 4 of 4

AMC 10 · 2024B · #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 cylindrical pipes of radii 1 and

\frac{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.

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.

How to solve

Strategy Draw a Diagram

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 ρ resting on the floor has its center at height ρ. 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√(ρ₁ ρ₂), 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 √(r), and Tool #14 (Use Cases) handles the two geometric placements (new pipe between the originals vs. outside them) as a ± sign in one equation.

1STEP 1

Set up coordinates

Floor at y = 0; a resting circle's center sits at its radius height, so put centers at A = (0, 1) and B = (x_B, $\frac{1}{4}$ ).

A = (0, 1), B = (x_B, 1/4)

2STEP 2

Find B's x-coordinate

Tangency gives |AB| = $\frac{5}{4}$ with vertical drop $\frac{3}{4}$ , so the Pythagorean theorem yields x_B = 1.

x_B² = (5/4)² - (3/4)² = (25 - 9)/16 = 1 → x_B = 1

3STEP 3

Make a reusable shortcut

The same triangle for any two floor-resting tangent circles gives horizontal gap 2√(ρ₁ ρ₂) — a shortcut we reuse.

Δ x(ρ₁, ρ₂) = √((ρ₁+ρ₂)² - (ρ₁-ρ₂)²) = 2√(ρ₁ ρ₂)

4STEP 4

Set up the third pipe

For the new pipe C = (x, r), tangency with each given pipe gives |x| = 2√(r) and |x - 1| = √(r).

|x| = 2√(r), |x - 1| = √(r)

5STEP 5

Split by the new pipe's spot

Case between (0 ＜ x ＜ 1) gives 3√(r) = 1, so r = $\frac{1}{9}$ ; case outside (x ＞ 1) gives √(r) = 1, so r = 1.

(i) 3√(r) = 1 → r = 1/9, (ii) √(r) = 1 → r = 1

6STEP 6

Add the two radii

Add the two radii: $\frac{1}{9}$ + 1 = $\frac{10}{9}$ , choice (C).

r_small + r_large = 1/9 + 1 = 10/9 → (C)

Answer

10/9

Each value passes a sanity check. For r =

\frac{1}{9}

the new pipe is much smaller than the small pipe (

\frac{1}{9}

＜

\frac{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

\frac{10}{9}

≈ 1.11 matches choice (C). The nearby distractor (B) 1 would mean only the large outside solution counts; (A)

\frac{1}{9}

would mean only the small wedge solution counts; neither matches the problem's "two possible radii".

💡Key takeaway

When two same-floor circles touch, the horizontal gap between their centers is 2√(ρ₁ ρ₂) — 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 $\frac{1}{9}$ and 1 pop right out.

Set up coordinates
Find B's x-coordinate
Make a reusable shortcut
Set up the third pipe
Split by the new pipe's spot
Add the two radii

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