AMC 8 · 2016 · #25

Grade 8 geometry-2d

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

pythagorean-theoremarea-trianglessimilar-triangles identify-subproblemsconvert-to-algebra ↑ Prerequisites: pythagorean-theoremarea-triangles

📏 Medium solution 💡 4 insights 📊 Diagram

Problem

An isosceles triangle has base 16 and height 15. A semicircle sits inside the triangle with its diameter lying along the base, and its curved side touches both slanted legs. Find the radius of that semicircle.

Pick an answer.

(A)

$4\sqrt{3}$

(B)

$\dfrac{120}{17}$

(C)

(D)

$\dfrac{17\sqrt{2}}{2}$

(E)

$\dfrac{17\sqrt{3}}{2}$

AMC 8 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

Tool #1 (Draw a Diagram) is the geometry workhorse: sketch the triangle, drop the height CD from the apex to the midpoint of the base, and mark where the semicircle's edge touches a leg. Two facts pop out of that picture — (i) the height CD splits the isosceles triangle into two right triangles with legs 8 and 15, so the leg of the isosceles triangle is the famous 8-15-17 hypotenuse, and (ii) the radius drawn to the tangent point is perpendicular to the leg, which means r is exactly the altitude from D to the hypotenuse of the small right triangle. Tool #7 (Identify Subproblems) then turns the hard "find the radius" question into the easy subproblem "find the area of a right triangle two different ways" — once with legs 8 × 15 and once with hypotenuse 17 and height r. Setting them equal solves for r in one line, no algebra heavier than a linear equation.

1STEP 1

Drop height CD from apex C to base midpoint D: CD = 15 and AD = 8, centering the semicircle (radius r) at D tangent to leg AC.

AD = 8, CD = 15, ∠ ADC = 90°

2STEP 2

By the Pythagorean Theorem on △ ADC, the slanted leg AC = 17 — the famous 8-15-17 triple.

AC² = AD² + CD² = 8² + 15² = 64 + 225 = 289 → AC = 17

3STEP 3

Since the semicircle is tangent to AC at E, radius DE = r is perpendicular to AC — so r is the altitude to the hypotenuse of △ ADC.

DE = r, DE ⊥ AC

4STEP 4

Area of △ ADC from its legs AD and CD: half of 8 × 15 = 60.

Area =

\frac{1}{2}

× AD × CD =

\frac{1}{2}

× 8 × 15 = 60

5STEP 5

Now take the same area with base AC = 17 and height r: 17r = 120, so r = $\frac{120}{17}$ → (B).

\frac{1}{2}

× 17 × r = 60 → 17r = 120 → r =

\frac{120}{17}

→ (B)

Answer

\frac{120}{17}

Sanity check the size:

\frac{120}{17}

≈ 7.06. The semicircle's diameter would be ≈ 14.1, which fits inside the base of length 16 (with a little gap on each side) — exactly what the picture in the problem shows. The radius is also smaller than the height 15, so the curved top doesn't poke through the apex. Both checks pass, so r =

\frac{120}{17}

is geometrically reasonable.

💡Key takeaway

This AMC 8 problem #25 only needs Grade 8 Pythagorean Theorem and the trick that a triangle's area is the same no matter which side you pick as the base!

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