2025 AIME I Problem 9

Question

The parabola with equation

y = x^2 - 4

is rotated

60^\circ

counterclockwise around the origin. The unique point in the fourth quadrant where the original parabola and its image intersect has

y

-coordinate

\frac{a - \sqrt{b}}{c},

where

a,

b,

and

c

are positive integers, and

a

and

c

are relatively prime. Find

a + b + c.

Answer

Answer: 62

Concepts:parabola transformation symmetry

Difficulty rating: 2920

Small hint:

P

lies on both curves exactly when

P

and its rotation by

-60^\circ

both lie on the original parabola, at equal distance from the origin

Big hint:

Try making the rotated image the mirror point

(-x, y):

that forces

y = -\sqrt{3}\,x,

which you can substitute into the parabola

Solution:

A point

P

lies on the image parabola exactly when its rotation by

-60^\circ,

namely

Q = \left(\frac{x + \sqrt{3}y}{2},\ \frac{-\sqrt{3}x + y}{2}\right),

lies on the original parabola. So we need

P

and

Q

both on

y = x^2 - 4.

The parabola is symmetric in

x \mapsto -x,

so we look for

P = (x, y)

whose rotated image is the mirror point

Q = (-x, y).

Matching

y

-coordinates gives

-\frac{\sqrt{3}}{2}x + \frac{y}{2} = y,

i.e.

y = -\sqrt{3}\,x,

and then the

x

-coordinate works automatically:

\frac{x}{2} + \frac{\sqrt{3}}{2}(-\sqrt{3}x) = -x.

Substituting

y = -\sqrt{3}\,x

into

y = x^2 - 4

gives

x^2 + \sqrt{3}\,x - 4 = 0,

whose positive root is

x = \frac{-\sqrt{3} + \sqrt{19}}{2}.

Then

y = -\sqrt{3}\,x = \frac{3 - \sqrt{57}}{2} \lt 0,

so this point is in the fourth quadrant, on both curves.

The problem guarantees the fourth-quadrant intersection is unique, so its

y

-coordinate is

\frac{3 - \sqrt{57}}{2},

giving

a + b + c = 3 + 57 + 2 = 62.