1986 AIME Problem 3

Attempt Problem 3 of the 1986 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 1986 AIME solutions, or check the answer key.

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3.

If tanx+tany=25\tan x+\tan y=25 and cotx+coty=30,\cot x+\cot y=30, what is tan(x+y)?\tan(x+y)?

Answer: 150
Concepts:algebraic manipulationtrigonometric identity
Difficulty rating: 1700
Small Hint:

Express the sum of the cotangents using tanx\tan x and tany\tan y

Big Hint:

Use the tangent addition formula after finding tanxtany\tan x\tan y

Solution:

Since cotx+coty=tanx+tanytanxtany, \cot x+\cot y =\frac{\tan x+\tan y}{\tan x\tan y}, the given equations yield tanxtany=2530=56.\tan x\tan y=\frac{25}{30}=\frac{5}{6}. Therefore tan(x+y)=tanx+tany1tanxtany=25156=150. \begin{aligned} \tan(x+y) &=\frac{\tan x+\tan y} {1-\tan x\tan y}\\ &=\frac{25}{1-\frac56}\\ &=150. \end{aligned}

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