2001 AIME I Problem 13

Question

In a certain circle, the chord of a

d

-degree arc is

22

centimeters long, and the chord of a

2d

-degree arc is

20

centimeters longer than the chord of a

3d

-degree arc, where

d \lt 120.

The length of the chord of a

3d

-degree arc is

-m + \sqrt{n}

centimeters, where

m

and

n

are positive integers. Find

m + n.

Answer

Answer: 174

Concepts:chord trigonometric identity quadratic

Difficulty rating: 2990

Small hint:

A chord of a

\theta

-degree arc has length

2R\sin\frac{\theta}{2}.

Set

t = \frac{d}{2}

and use

\sin 2t = 2\sin t\cos t,

\sin 3t = \sin t\,(4\cos^2 t - 1).

Big hint:

Divide everything by

2R \sin t = 22

to get

44\cos^2 t - 22\cos t - 1 = 0,

and note the

3d

-degree chord equals

44 \cos t - 20

Solution:

A chord subtending a

\theta

-degree arc in a circle of radius

R

has length

2R\sin\frac{\theta}{2}.

Write

t = \frac{d}{2},

so the three chords are

2R\sin t = 22,

2R \sin 2t,

and

2R \sin 3t.

Using

\sin 2t = 2 \sin t \cos t

and

\sin 3t = \sin t\,(4\cos^2 t - 1),

the chords of the

2d

- and

3d

-degree arcs are

22 \cdot 2\cos t = 44 \cos t

and

22\,(4\cos^2 t - 1).

The condition “the

2d

-chord is

20

longer than the

3d

-chord” becomes

44\cos t = 22\,(4\cos^2 t - 1) + 20,

which simplifies to

44 \cos^2 t - 22 \cos t - 1 = 0.

From this,

4\cos^2 t = 2\cos t + \frac{1}{11},

so the

3d

-chord equals

22\left(2\cos t + \frac{1}{11} - 1\right)

= 44 \cos t - 20.

Solving the quadratic,

\cos t = \frac{22 + \sqrt{484 + 176}}{88} = \frac{11 + \sqrt{165}}{44}

(the

+

root since

d \lt 120

means

t \lt 60^\circ,

so

\cos t \gt \frac{1}{2}

). Then the

3d

-chord is

44\cos t - 20

= 11 + \sqrt{165} - 20

= -9 + \sqrt{165},

giving

m + n = 9 + 165 = 174.

2001 AIME I Problem 13

Answer: 174

Difficulty rating: 2990

Small hint:

Big hint:

Solution:

Problem 13 in other years