2026 AIME I Problem 5

Question

A plane contains points

A

and

B

with

AB = 1.

Point

A

is rotated in the plane counterclockwise through an acute angle

\theta

around point

B

to point

A'.

Then

B

is rotated in the plane clockwise through angle

\theta

around point

A'

to point

B'.

Suppose

AB' = \frac{4}{3}.

The value of

\cos\theta

can be written as

\frac{m}{n},

where

m

and

n

are relatively prime positive integers. Find

m + n.

Answer

Answer: 65

Concepts:complex number transformation trigonometry

Difficulty rating: 2400

Small hint:

Put

B = 0

and

A = 1

in the complex plane; rotating

z

about

P

through

\varphi

counterclockwise gives

P + e^{i\varphi}(z - P)

Big hint:

Compute

B' = e^{i\theta} - 1,

so

AB'^2 = \left|e^{i\theta} - 2\right|^2 = 5 - 4\cos\theta

Solution:

Work in the complex plane with

B = 0

and

A = 1.

Rotating

z

about

P

through angle

\varphi

counterclockwise gives

P + e^{i\varphi}(z - P).

So

A' = e^{i\theta},

and rotating

B

clockwise through

\theta

about

A'

gives

\begin{aligned} &B' = A' + e^{-i\theta}(0 - A') \\ &= e^{i\theta} - e^{-i\theta}e^{i\theta} \\ &= e^{i\theta} - 1. \end{aligned}

Then

\begin{aligned} &AB'^2 = \left|e^{i\theta} - 2\right|^2 \\ &= (\cos\theta - 2)^2 + \sin^2\theta \\ &= 5 - 4\cos\theta. \end{aligned}

Setting this equal to

\left(\frac{4}{3}\right)^2 = \frac{16}{9}

gives

4\cos\theta = 5 - \frac{16}{9} = \frac{29}{9},

so

\cos\theta = \frac{29}{36}

(indeed positive, consistent with

\theta

acute). Thus

m + n = 29 + 36 = 65.

2026 AIME I Problem 5

Answer: 65

Difficulty rating: 2400

Small hint:

Big hint:

Solution:

Problem 5 in other years