2019 AMC 12A Problem 22

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

Answer: E

Concepts:coordinate geometry equilateral triangle tangent line

Difficulty rating: 2310

Small hint:

Let

s = AB.

Since

\tfrac{BP}{CP} = 3,

we have

BP = \tfrac{3s}{4}

and

CP = \tfrac{s}{4}

Big hint:

Place

P

at the origin with

BC

horizontal;

O

lies on line

AP

at distance

17

from

BC

and

20

from

A

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 has

O

and

A

on opposite sides of

P,

so

|t-1|s=|t|s+s.

Therefore

AB = s = \dfrac{80}{\sqrt{13}} - \dfrac{34}{\sqrt{3}}.