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2019 AMC 12A Problems/Problem 22

Problem

Circles $\omega$ and $\gamma$, both centered at $O$, have radii $20$ and $17$, respectively. Equilateral triangle $ABC$, whose interior lies in the interior of $\omega$ but in the exterior of $\gamma$, has vertex $A$ on $\omega$, and the line containing side $\overline{BC}$ is tangent to $\gamma$. Segments $\overline{AO}$ and $\overline{BC}$ intersect at $P$, and $\dfrac{BP}{CP} = 3$. Then $AB$ can be written in the form $\dfrac{m}{\sqrt{n}} - \dfrac{p}{\sqrt{q}}$ for positive integers $m$, $n$, $p$, $q$ with $\text{gcd}(m,n) = \text{gcd}(p,q) = 1$. What is $m+n+p+q$? $\phantom{}$

$\textbf{(A) } 42 \qquad \textbf{(B) }86 \qquad \textbf{(C) } 92 \qquad \textbf{(D) } 114 \qquad \textbf{(E) } 130$

Solution

[asy] size(20cm); draw(circle((0,0), 20)); label("$\omega$", (0,0), 4.05*20*dir(149)*20/21); draw(circle((0,0), 17)); label("$\gamma$", (0,0), 4.05*17*dir(149)*20/21); dot((0,0)); label("$O$", (0,0), E); pair aa = (-20, 0); dot(aa); label("$A$", aa, W); draw((-20,0)--(0,0)); real a = (-20 + (80/sqrt(13) - 34/sqrt(3))*(sqrt(13)/sqrt(12))*(sqrt(3)/2)); real ans = (80/sqrt(13) - 34/sqrt(3)); dot((a,0)); label("$P$", (a, 0), dir(290)*0.58); pair s = ((12*a + 0)/13, 0-sqrt(12)*a/13); dot(s); label("$S$", s, dir(135)); pair c = (a + 1/4*ans*1/sqrt(13), 0 + 1/4*ans*sqrt(12)/sqrt(13)); dot(c); label("$C$", c, dir(110)); pair m = (a - 1/4*ans*1/sqrt(13), 0 - 1/4*ans*sqrt(12)/sqrt(13)); dot(m); label("$M$", m, dir(285)); pair b = (a - 3/4*ans*1/sqrt(13), 0 - 3/4*ans*sqrt(12)/sqrt(13)); dot(b); label("$B$", b, S); draw(b--s); draw(s--(0,0)); draw(aa--b); draw(aa--c); draw(aa--m); markscalefactor=0.1; draw(rightanglemark(s,m,aa,3.4)); draw(rightanglemark((0,0),s,m,3.4)); [/asy]

Let $S$ be the point of tangency between $\overline{BC}$ and $\gamma$, and $M$ be the midpoint of $\overline{BC}$. Note that $AM \perp BS$ and $OS \perp BS$. This implies that $\angle OAM \cong \angle AOS$, and $\angle AMP \cong \angle OSP$. Thus, $\triangle PMA \sim \triangle PSO$.

If we let $s$ be the side length of $\triangle ABC$, then it follows that $AM = \frac{\sqrt{3}}{2}s$ and $PM = \frac{s}{4}$. This implies that $AP = \frac{\sqrt{13}}{4}s$, so $\frac{AM}{AP} = \frac{2\sqrt{3}}{\sqrt{13}}$. Furthermore, $\frac{AM + SO}{AO} = \frac{AM}{AP}$ (because $\triangle PMA \sim \triangle PSO$) so this gives us the equation \[\frac{\frac{\sqrt{3}}{2}s + 17}{20} = \frac{2\sqrt{3}}{\sqrt{13}}\] to solve for the side length $s$, or $AB$. Thus, \[\frac{\sqrt{39}}{2}s + 17\sqrt{13} = 40\sqrt{3}\] \[\frac{\sqrt{39}}{2}s = 40\sqrt{3} - 17\sqrt{13}\] \[s = \frac{80}{\sqrt{13}} - \frac{34}{\sqrt{3}} = AB\] The problem asks for $m + n + p + q = 80 + 13 + 34 + 3 = \boxed{\textbf{(E) } 130}$.

Video Solution

https://www.youtube.com/watch?v=2eASfdhEyUE

Video Solution by Richard Rusczyk - https://www.youtube.com/watch?v=CaYgfNEUBwA&list=PLyhPcpM8aMvLgfgbaTLDaV_jRYfn1A-x_&index=2 - AMBRIGGS

See Also

2019 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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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