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

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==solution easiest outline==  
 
==solution easiest outline==  
Draw the other half of the largest circle and proceed with Descartes Circle Theorem.
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Draw the other half of the largest circle and proceed with [[Descartes' Circle Formula]].
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2017|ab=A|num-b=15|num-a=17}}
 
{{AMC12 box|year=2017|ab=A|num-b=15|num-a=17}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:56, 12 November 2017

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:

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

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,SW); label("$C$",C,S); label("$B$",B,SE); 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 $\frac{a+b+c}{2}$

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

Solution 4

Let $C$, the center of the large semicircle, to be at $(0, 0)$, and $P$ to be at $(h, k)$.

[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]

Therefore $A$ is at $(-1, 0)$ and $B$ is at $(2, 0)$.

Let the radius of circle $P$ be $r$.

Using Distance Formula, we get the following system of three equations:

\[h^2+k^2=(3-r)^2, (h+1)^2+k^2=(r+2)^2, (h-2)^2+k^2=(r+1)^2\]

By simplifying, we get

\[h^2+k^2=r^2-6r+9, h^2+2r+1+k^2=r^2+4r+4, h^2-4r+4+k^2=r^2+2r+1\]

By subtracting the first equation from the second and third equations, we get

\[8r=-4h+12, 10r=2h+6\]

which simplifies to

\[2r=3-h, 5r=h+3\]

When we add these two equations, we get

\[7r=6\]

So \[r = \boxed{\frac{6}{7}} \to \boxed{\textbf{(B)}}\] $w^5$

solution easiest outline

Draw the other half of the largest circle and proceed with Descartes' Circle Formula.

See Also

2017 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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