AMC 8 2016 Problem 25: Geometry Solution

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Answer: B — Radius 120/17.

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Hint 1 of 3

The big idea: a radius drawn to the point where the semicircle touches a slant side is PERPENDICULAR to that side (radius ⊥ tangent). So the radius is just the perpendicular distance from the semicircle's center to the slant — and that perpendicular is an altitude of a triangle hiding inside the figure.

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Hint 2 of 3

Isolate the triangle made by the center O (midpoint of the base), a base corner, and the apex. Its area can be written TWO ways — using the base×height you know, and using the slant side with the unknown radius as its height. Setting them equal frees the radius.

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Hint 3 of 3

You'll need the slant length: half the base is 8 and the height is 15, so the slant is the hypotenuse of an 8-15-17 right triangle — a clean 17, no square roots.

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Approach: area of a sub-triangle computed two ways (radius = its altitude to the slant)

Let O be the midpoint of the base (the semicircle's center). The semicircle is tangent to the slant side, and a radius to a tangent point is perpendicular to it — so the radius equals the perpendicular distance from O to that slant, i.e. the altitude of ▵BOC dropped onto side BC.
Compute ▵BOC's area the easy way: base OB = 8 (half the base) and height 15 (the full triangle height), so area = (1/2)(8)(15) = 60.
Find the slant BC: with legs 8 and 15 it's an 8-15-17 right triangle, so BC = 17. Now write the SAME area using BC as base and the radius as height: 60 = (1/2)(17)(radius).
Solve: radius = 120/17 = 120/17.
Why this transfers: "area two ways" turns any perpendicular-distance-to-a-side into algebra — compute a triangle's area with the convenient base, then re-express it with the side you care about, and the unknown height drops out. (Tangent ⊥ radius is the standard hook for inscribed-circle problems.)

Another way — perpendicular distance via coordinates:

Put the base's midpoint at (0, 0), apex at (0, 15), right base corner at (8, 0). The right slant has equation 15x + 8y = 120.
The radius is the distance from (0, 0) to that line: 120 / √(152 + 82) = 120/17.