AMC 8 · 2017 · #22

Grade 8 geometry-2dalgebra

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

similar-trianglespythagorean-theoremratio-proportionarea-triangles identify-subproblemsconvert-to-algebra ↑ Prerequisites: pythagorean-theoremsimilar-trianglesratio-proportion

📏 Medium solution 💡 4 insights 📊 Diagram

Problem

Right triangle ABC has a right angle at C with legs AC = 12 and BC = 5. A semicircle is drawn inside the triangle so that its flat diameter sits on leg AC (with one endpoint at C) and its curved edge just touches the hypotenuse AB. Find the radius r of that semicircle.

Pick an answer.

(A)

$frac{7}{6}$

(B)

$frac{13}{5}$

(C)

$frac{59}{18}$

(D)

$frac{10}{3}$

(E)

$frac{60}{13}$

AMC 8 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

The asy figure is the seed, but the key move is Tool #1 (Draw a Diagram): add the unseen pieces — the center O on AC, the radius OC = r to leg BC, and the radius OT drawn perpendicular to the hypotenuse AB. Once those are on the picture, Tool #7 (Identify Subproblems) splits the work into two clean pieces — first find AB with the Pythagorean theorem, then notice the small right triangle △ AOT tucked inside the big right triangle △ ABC. The two triangles share angle A and both have a right angle, so they are similar. The similar-triangle proportion gives a single linear equation in r, which Tool #13 (Convert to Algebra) finishes. (We could lean on Tool #6 Guess & Check against the answer choices as a fast verification, and we do exactly that in the Review.)

1STEP 1

Pythagoras on the legs 12 and 5 (a classic triple) gives hypotenuse AB = 13.

AB = √(12² + 5²) = √(144 + 25) = √(169) = 13

2STEP 2

Put center O on AC with OC = r; mark tangent point T, so OT ⊥ AB, OT = r, and AO = 12 - r.

OC = r, OT = r, OT ⊥ AB, AO = 12 - r

3STEP 3

Triangles AOT and ABC share angle A and each have a right angle, so by AA they are similar: OT/BC = AO/AB.

△ AOT ∼ △ ABC → OT/BC = AO/AB

4STEP 4

Substitute into r/5 = (12 - r)/13, cross-multiply to 18r = 60, so r = $\frac{10}{3}$ .

r/5 = (12 - r)/13 ⟹ 13r = 5(12 - r) ⟹ 13r = 60 - 5r ⟹ 18r = 60 ⟹ r =

\frac{60}{18}

\frac{10}{3}

5STEP 5

The value r = $\frac{10}{3}$ is exactly choice (D).

r =

\frac{10}{3}

→ (D)

Answer

\frac{10}{3}

The semicircle must fit inside a triangle whose shorter leg is 5, so the radius must be less than 5. Our answer r =

\frac{10}{3}

≈ 3.33 is comfortably below 5, and it's also large enough that the semicircle visibly reaches across most of the figure — matching the asy picture (where the arc spans roughly from x = 5.33 to x = 12, giving radius (12 - 5.33)/2 ≈ 3.33). The value is consistent with the area cross-check too: splitting △ ABC (area 30) into △ BCO (area 5r/2) and △ ABO (area 13r/2) gives 18r/2 = 30 → r =

\frac{10}{3}

. Same answer, different path.

💡Key takeaway

This AMC 8 problem only needs Grade 8 Pythagorean theorem and similar-triangle reasoning you already know!

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