2017 AMC 8 Problem 22

Attempt Problem 22 of the 2017 AMC 8 below, then check your answer against the video solution and professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2017 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

22.

In the right triangle ABC,ABC, AC=12,AC=12, BC=5,BC=5, and angle CC is a right angle. A semicircle is inscribed in the triangle as shown. What is the radius of the semicircle?
76 \dfrac{7}{6}
135 \dfrac{13}{5}
5918 \dfrac{59}{18}
103 \dfrac{10}{3}
6013 \dfrac{60}{13}
Answer: D
Concepts:tangent linesimilarity
Small hint:
Draw the radius to the point where the semicircle touches the hypotenuse.
Big hint:
Use tangent lengths from the same external point, then set up similar triangles.
Video solution:
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Written solution:
Let OO be the center of the inscribed semicircle and DD be the tangent point of the semicircle on AB‾.\overline{AB}. Then BD=5BD = 5 since BD‾\overline{BD} and BC‾\overline{BC} are tangents to the semicircle. Then AD=8AD = 8 and OD=r.OD = r. OD‾\overline{OD} is perpendicular to AB‾\overline{AB} so △ADO∼△BCA,\triangle ADO \sim \triangle BCA, so r8=512.\dfrac{r}{8} = \dfrac{5}{12}. Solving this, we get r=103.r = \dfrac{10}{3}.
Thus, D is the correct answer.
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