Question

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 $BC$ is tangent to $\gamma.$ Segments $AO$ and $BC$ 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 $\gcd(m, n) = \gcd(p, q) = 1.$ What is $m + n + p + q?$

$42$

$86$

$92$

$114$

$130$

Solution:

Let $s = AB.$ Since $\dfrac{BP}{CP} = 3,$ we have $BP = \dfrac{3s}{4}$ and $CP = \dfrac{s}{4}.$ Put $P$ at the origin with $BC$ on the $x$ -axis, $B = \left(-\tfrac{3s}{4}, 0\right),$ $C = \left(\tfrac{s}{4}, 0\right),$ and apex $A = \left(-\tfrac{s}{4}, \tfrac{s\sqrt{3}}{2}\right).$

Points $P, O, A$ are collinear, so $O = t \cdot A$ for some scalar $t.$ Two conditions pin it down: $O$ is at distance $17$ from line $BC,$ giving $|t| \cdot \dfrac{s\sqrt{3}}{2} = 17,$ and $A$ is on $\omega,$ giving $|t - 1| \cdot \dfrac{s\sqrt{13}}{4} = 20$ since $|A| = \dfrac{s\sqrt{13}}{4}.$

Solving, $|t| s = \dfrac{34}{\sqrt{3}}$ and $|t - 1| s = \dfrac{80}{\sqrt{13}}.$ The valid configuration gives $AB = s = \dfrac{80}{\sqrt{13}} - \dfrac{34}{\sqrt{3}}.$

Then $m + n + p + q = 80 + 13 + 34 + 3 = 130.$

Thus, the correct answer is E.

2019 AMC 12A Problem 22

Solution:

Problem 22 in Other Years