2016 AMC 8 Problems/Problem 25

25. A semicircle is inscribed in an isosceles triangle with base $16$ and height $15$ so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?

$[asy]draw((0,0)--(8,15)--(16,0)--(0,0)); draw(arc((8,0),7.0588,0,180));[/asy]$

$\textbf{(A) }4 \sqrt{3}\qquad\textbf{(B) } \dfrac{120}{17}\qquad\textbf{(C) }10\qquad\textbf{(D) }\dfrac{17\sqrt{2}}{2}\qquad \textbf{(E)} \dfrac{17\sqrt{3}}{2}$

Solution

Draw the altitude from the top of the triangle to its base, dividing the isosceles triangle into two right triangles with height $15$ and base $\frac{16}{2} = 8$ . The Pythagorean triple $8$ - $15$ - $17$ tells us that these triangles have hypotenuses of $17$ .

Now draw an altitude of one of the smaller right triangles, starting from the foot of the first altitude we drew (which is also the center of the circle that contains the semicircle) and going to the hypotenuse of the right triangle. This segment is both an altitude of the right triangle as well as the radius of the semicircle (this is because tangent lines to circles, such as the hypotenuse touching the semicircle, are always perpendicular to the radii of the circles drawn to the point of tangency). Let this segment's length be $r$ .

The area of the entire isosceles triangle is $\frac{(16)(15)}{2} = 120$ , so the area of each of the two congruent right triangles it gets split into is $\frac{120}{2} = 60$ . We can also find the area of one of the two congruent right triangles by using its hypotenuse as its base and the radius of the semicircle, the altitude we drew, as its height. Then the area of the triangle is $\frac{17r}{2}$ . Thus we can write the equation $\frac{17r}{2} = 60$ , so $17r = 120$ , so $r = \boxed{\textbf{(B) }\frac{120}{17}}$ .

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2016 AMC 8 Problems/Problem 25

Solution