2021 AMC 12A Spring Problem 24

Question

Semicircle

\Gamma

has diameter

AB

of length

14.

Circle

\Omega

lies tangent to

AB

at a point

P

and intersects

\Gamma

at points

Q

and

R.

If

QR = 3\sqrt3

and

\angle QPR = 60^\circ,

then the area of

\triangle PQR

is

\dfrac{a\sqrt b}{c},

where

a

and

c

are relatively prime positive integers and

b

is a positive integer not divisible by the square of any prime. What is

a + b + c?

110

114

118

122

126

Answer: D

Concepts:law of sines radical axis coordinate geometry

Difficulty rating: 2760

Small hint:

In circle

\Omega,

QR = 2r\sin\angle QPR,

which gives the radius

r = 3

Big hint:

Set

A = (-7,0),

B = (7,0),

P = (p, 0),

center

(p, 3);

the radical axis of the two circles locates

QR,

and the area is

\tfrac12\, QR\cdot d(P, QR)

Solution:

In circle

\Omega,

the chord

QR

subtends the inscribed angle

\angle QPR = 60^\circ,

so

QR = 2r\sin 60^\circ,

giving

3\sqrt3 = r\sqrt3,

hence

r = 3.

Place

A = (-7, 0),

B = (7, 0),

with

\Gamma: x^2 + y^2 = 49

(upper half). Since

\Omega

is tangent to

AB

at

P = (p, 0),

its center is

(p, 3).

Subtracting the two circle equations gives the line

QR,

and the distance from the center

(p,3)

to

QR

must equal

r\cos 60^\circ = \tfrac32.

This yields

(p^2 - 31)^2 = 9p^2 + 81,

so

p^2 = 16

(the root

p^2 = 55

places

P

outside

AB

).

With

p^2 = 16,

the distance from

P

to line

QR

is

\dfrac{49 - p^2}{\sqrt{4p^2 + 36}} = \dfrac{33}{10}.

Thus

\begin{aligned} [\triangle PQR] &= \tfrac12\cdot QR\cdot d \\ &= \tfrac12\cdot 3\sqrt3\cdot\tfrac{33}{10} \\ &= \frac{99\sqrt3}{20}. \end{aligned}

So

a = 99,

b = 3,

c = 20,

and

a + b + c = 122.

Thus, the correct answer is D.