# Difference between revisions of "2017 AMC 12A Problems/Problem 16"

## Problem

In the figure below, semicircles with centers at $A$ and $B$ and with radii 2 and 1, respectively, are drawn in the interior of, and sharing bases with, a semicircle with diameter $JK$. The two smaller semicircles are externally tangent to each other and internally tangent to the largest semicircle. A circle centered at $P$ is drawn externally tangent to the two smaller semicircles and internally tangent to the largest semicircle. What is the radius of the circle centered at $P$? $[asy] size(5cm); draw(arc((0,0),3,0,180)); draw(arc((2,0),1,0,180)); draw(arc((-1,0),2,0,180)); draw((-3,0)--(3,0)); pair P = (-1,0)+(2+6/7)*dir(36.86989); draw(circle(P,6/7)); dot((-1,0)); dot((2,0)); dot(P); [/asy]$ $\textbf{(A)}\ \frac{3}{4} \qquad \textbf{(B)}\ \frac{6}{7} \qquad\textbf{(C)}\ \frac{1}{2}\sqrt{3} \qquad\textbf{(D)}\ \frac{5}{8}\sqrt{2} \qquad\textbf{(E)}\ \frac{11}{12}$

## Solution 1

Connect the centers of the tangent circles! (call the center of the large circle $C$) $[asy] size(5cm); draw(arc((0,0),3,0,180)); draw(arc((2,0),1,0,180)); draw(arc((-1,0),2,0,180)); draw((-3,0)--(3,0)); pair P = (9/7,12/7); pair A = (-1,0); pair C = (0,0); pair B = (2,0); draw(circle(P,6/7)); dot((-1,0)); dot((2,0)); dot((0,0)); dot(P); draw((-1,0)--P); draw((2,0)--P); draw((0,0)--(9/5,12/5)); label("A",A,SW); label("C",C,S); label("B",B,SE); label("P",P,N); [/asy]$

Notice that we don't even need the circles anymore; thus, draw triangle $\Delta ABP$ with cevian $PC$: $[asy] size(5cm); draw((-1,0)--(2,0)); pair P = (9/7,12/7); pair A = (-1,0); pair C = (0,0); pair B = (2,0); dot((-1,0)); dot((2,0)); dot((0,0)); dot(P); draw((-1,0)--P); draw((2,0)--P); draw((0,0)--P); label("A",A,SW); label("C",C,S); label("B",B,SE); label("P",P,N); [/asy]$

and use Stewart's Theorem: $$AB \cdot AC \cdot BC + AB \cdot {CP}^2 = AC \cdot {BP}^2 + BC \cdot {AP}^2$$

From what we learned from the tangent circles, we have $AB = 3$, $AC = 1$, $BC = 2$, $AP = 2 + r$, $BP = 1 + r$, and $CP = 3 - r$, where $r$ is the radius of the circle centered at $P$ that we seek.

Thus: $$3 \cdot 1 \cdot 2 + 3 {\left(3-r\right)}^2 = 1 {\left(1+r\right)}^2 + 2 {\left(2+r\right)}^2$$ $$6 + 3\left(9 - 6r + r^2\right) = \left(1 + 2r + r^2\right) + 2\left(4 + 4r + r^2\right)$$ $$33 - 18r + 3r^2 = 9 + 10r + 3r^2$$ $$28r = 24$$ $$r = \boxed{\frac{6}{7}} \to \boxed{\textbf{(B)}}$$

NOTICE to proficient editors: please label the points on the diagrams, thanks!

## Solution 2 $[asy] size(5cm); draw(arc((0,0),3,0,180)); draw(arc((2,0),1,0,180)); draw(arc((-1,0),2,0,180)); draw((-3,0)--(3,0)); pair P = (9/7,12/7); draw(circle(P,6/7)); dot((-1,0)); dot((2,0)); dot((0,0)); dot(P); draw((-1,0)--P); draw((2,0)--P); draw((0,0)--(9/5,12/5)); [/asy]$

Like the solution above, connecting the centers of the circles results in triangle $\Delta ABP$ with cevian $PC$. The two triangles $\Delta APC$ and $\Delta ABP$ share angle $A$, which means we can use Law of Cosines to set up a system of 2 equations that solve for $r$ respectively: $(2 + r)^2 + 1^2 - 2(2 + r)(1)cosA = (3 - r)^2$ (notice that the diameter of the largest semicircle is 6, so its radius is 3 and $PC$ is 3 - r) $(2 + r)^2 + 3^2 - 2(2 + r)(3)cosA = (r+1)^2$

We can eliminate the extra variable of angle $A$ by multiplying the first equation by 3 and subtracting the second from it. Then, expand to find $r$: $2(r^2 + 4r + 4) - 6 = 2r^2 - 20r + 26$ $8r + 2 = -20r + 26$ $28r = 24$, so $r$ = $6/7$ $\boxed{(B)}$

## Solution 3 $[asy] size(7.5cm); draw(arc((0,0),3,0,180)); draw(arc((2,0),1,0,180)); draw(arc((-1,0),2,0,180)); draw((-3,0)--(3,0)); pair P = (9/7,12/7); pair A = (-1,0); pair C = (0,0); pair B = (2,0); draw(circle(P,6/7)); dot((-1,0)); dot((2,0)); dot((0,0)); dot(P); draw((-1,0)--P); draw((-2.45,12/7)--P); draw((2.45,12/7)--P); draw((2,0)--P); draw((0,0)--(9/5,12/5)); label("A",A,SW); label("C",C,S); label("B",B,SE); label("P",P,N); [/asy]$

Let $C$ be the center of the largest semicircle and $r$ be the radius of $\circ P$. We know that $AC = 1$, $CB = 2$, $AP = r + 2$, $BP = r + 1$, and $CP = 3 - r$. Notice that $\Delta ACP$ and $\Delta CBP$ are bounded by the same two parallel lines, so these triangles have the same heights. Because the bases of these two triangles (that have the same heights) differ by a factor of 2, the area of $\Delta CBP$ must be twice that of $\Delta ACP$, since the area of a triangle is $\frac{1}{2} Base \cdot Height$.

Again, we don't need to look at the circle and the semicircles anymore; just focus on the triangles. $[asy] size(7.5cm); pair A = (-1,0); pair C = (0,0); pair B = (2,0); draw((-1,0)--(2,0)); pair P = (9/7,12/7); dot((-1,0)); dot((2,0)); dot((0,0)); dot(P); draw((-1,0)--P); draw((2,0)--P); draw((0,0)--P); label("A",A,W); label("C",C,S); label("B",B,E); label("P",P,N); [/asy]$

Let $A_1$ equal to the area of $\Delta ACP$ and $A_2$ equal to the area of $\Delta CBP$. Heron's Formula states that the area of an triangle with sides $a$ $b$ and $c$ is $$\sqrt{s(s-a)(s-b)(s-c)}$$ where $s$, or the semiperimeter, is half of the triangle's perimeter.

The semiperimeter $s_1$ of $\Delta ACP$ is $$\frac{[(r + 2) + (3 - r) + 1]}{2} = \frac{6}{2} = 3$$ Use Heron's Formula to obtain $$A_1 = \sqrt{3(2)(3-2-r)(3-3+r)} = \sqrt{6r(1-r)} = \sqrt{6r-6r^2}$$

Using Heron's Formula again, find the area of $\Delta CBP$ with sides $r+1$, $2$, and $3-r$. $$s_2 = \frac{(r + 1) + 2 + (3 - r)}{2} = 3$$ $$A_2 = \sqrt{3(3-2)(3-1-r)(3-3+r)} = \sqrt{3(2r-r^2)} = \sqrt{6r-3r^2}$$

Now, $$2 \cdot A_1 = A_2$$ $$2\sqrt{6r-6r^2} = \sqrt{6r-3r^2}$$ $$4(6r-6r^2) = 6r-3r^2$$ $$24r-24r^2 = 6r-3r^2$$ $$18r = 21r^2$$ $$r = \frac{18}{21} = \boxed{\frac{6}{7}} \to \boxed{\textbf{(B)}}$$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 