2016 AMC 8 Problems/Problem 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 1

First, we draw a line perpendicular to the base of the triangle and cut the triangle in half. The base of the resulting right triangle would be 8, and the height would be 15. Using the Pythagorean theorem, we can find the length of the hypotenuse, which would be 17. Using the two legs of the right angle, we can find the area of the right triangle, $60$ . $\frac{60}{17}$ times $2$ results in the radius, which is the height of the right triangle when using the hypotenuse as the base. The answer is $\boxed{\textbf{(B) }\frac{120}{17}}$ .

Solution 2: Similar Triangles

$[asy] pair A, B, C, D, E; A=(0,0); B=(16,0); C=(8,15); D=B/2; E=(64/17*8/17, 64/17*15/17); draw(A--B--C--cycle); draw(C--D); draw(D--E); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); draw(rightanglemark(A,E,D,25)); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,S); label("$E$",E,NW);[/asy]$ Let's call the triangle $\triangle ABC,$ where $AB=16$ and $AC=BC.$ Let's say that $D$ is the midpoint of $AB$ and $E$ is the point where $AC$ is tangent to the semicircle. We could also use $BC$ instead of $AC$ because of symmetry.

Notice that $\triangle ACD \cong \triangle BCD,$ and are both 8-15-17 right triangles. We also know that we create a right angle with the intersection of the radius and a tangent line of a circle (or part of a circle). So, by $AA$ similarity, $\triangle AED \sim \triangle ADC,$ with $\angle EAD \cong \angle DAC$ and $\angle CDA \cong \angle DEA.$ This similarity means that we can create a proportion: $\frac{AD}{AC}=\frac{DE}{CD}.$ We plug in $AD=\frac{AB}{2}=8, AC=17,$ and $CD=15.$ After we multiply both sides by $15,$ we get $DE=\frac{8}{17} \cdot 15= \boxed{\textbf{(B) }\frac{120}{17}}.$

(By the way, we could also use $\triangle DEC \sim \triangle ADC.$ )

Solution 4: Inscribed Circle

$[asy] pair A, B, C, D, M; B=(0,0); D=(16,0); A=(8,15); C=(8,-15); M=D/2; draw(B--D--A--cycle); draw(A--M); draw(arc(M,120/17,0,180)); draw(rightanglemark(D,M,A,25)); draw(rightanglemark(B,M,25)); label("$B$",B,SW); label("$D$",D,SE); label("$A$",A,N); label("$M$",M,S); label("$C$",C,S); draw((0,0)--(8,-15)--(16,0)--(0,0)); draw(arc((8,0),7.0588,0,360));[/asy]$

We'll call this triangle $\triangle ABD$ . Let the midpoint of base $BD$ be $M$ . Divide the triangle in half by drawing a line from $A$ to $M$ . Half the base of $\triangle ABD$ is $\frac{16}{2} = 8$ . The height is $15$ , which is given in the question. Using the Pythagorean Triple $8$ - $15$ - $17$ , the length of each of the legs ( $AB$ and $DA$ ) is 17.

Reflect the triangle over its base. This will create an inscribed circle in a rhombus $ABCD$ . Because $AB \cong DA$ , $BC \cong CD$ . Therefore $AB = BC = CD = DA$ .

The semiperimeter $s$ of the rhombus is $\frac{AB + BC + CD + DA}{2} = \frac{(17)(4)}{2} = 34$ . Since the area of $\triangle ABD$ is $\frac{bh}{2}$ , the area $[ABCD]$ of the rhombus is twice that, which is $bh = (16)(15) = 240$ .

The Formula for the Incircle of a Quadrilateral is $s$ $r$ = $[ABCD]$ . Substituting the semiperimeter and area into the equation, $34r = 240$ . Solving this, $r = \frac{240}{34}$ = $\boxed{\textbf{(B) }\frac{120}{17}}$ .

Solution 4: Inscribed Circle

Noting that we have a 8-15-17 triangle, we can find $AE$ and $CE.$ Let $AE=x$ , $CE=17-x.$ Then by similar triangles (or "Altitude on Hypotenuse") we have $15^2=x*17.$ Thus, $AE=x=225/17, CE=64/17.$ Now again by "Altitude on Hypotenuse”, $r=\sqrt{AE*CE}.$ Therefore $r=120/17.$