1999 AIME Problem 11

Attempt Problem 11 of the 1999 AIME below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1999 AIME solutions, or check the answer key.

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11.

Given that

\sum_{k=1}^{35} \sin 5k = \tan \frac{m}{n},

where angles are measured in degrees, and

m

and

n

are relatively prime positive integers that satisfy

\frac{m}{n} \lt 90,

find

m + n.

Answer

Answer: 177

Concepts:trigonometric identity telescoping

Difficulty rating: 2560

Small hint:

Multiply the sum by

2 \sin 2.5^\circ

and use

2 \sin\alpha \sin\beta

= \cos(\alpha - \beta)

- \cos(\alpha + \beta)

to telescope

Big hint:

The two surviving terms combine because

\cos(180^\circ - \theta) = -\cos\theta;

finish with

\cot\theta = \tan(90^\circ - \theta)

Solution:

Multiply the sum by

2 \sin 2.5^\circ

and apply

2 \sin 5k^\circ \sin 2.5^\circ

= \cos(5k - 2.5)^\circ

- \cos(5k + 2.5)^\circ,

so the sum telescopes:

\begin{aligned} &2 \sin 2.5^\circ \sum_{k=1}^{35} \sin 5k^\circ \\ &= \cos 2.5^\circ - \cos 177.5^\circ \\ &= 2\cos 2.5^\circ. \end{aligned}

Hence the sum equals

\frac{\cos 2.5^\circ}{\sin 2.5^\circ}

= \cot 2.5^\circ

= \tan 87.5^\circ

= \tan \frac{175}{2}^\circ.

Since

\gcd(175, 2) = 1

and

\frac{175}{2} \lt 90,

we get

m + n = 175 + 2 = 177.