1995 AIME Problem 7

Attempt Problem 7 of the 1995 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 1995 AIME solutions, or check the answer key.

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7.

Given that

(1+\sin t)(1+\cos t)=\frac54

and

\begin{aligned}(1-\sin t)&(1-\cos t)\\&=\frac mn-\sqrt k,\end{aligned}

where

k,

m,

and

n

are positive integers with

m

and

n

relatively prime, find

k+m+n.

Answer

Answer: 27

Concepts:trigonometric identity symmetry (algebra)radical

Difficulty rating: 1900

Small hint:

Set

u=\sin t+\cos t

and express

\sin t\cos t

in terms of

u

Big hint:

Both given products become half of a perfect square in

u

Solution:

Let

u=\sin t+\cos t.

Since

\sin t\cos t=\frac{u^2-1}{2},

\begin{aligned}(1+\sin t)&(1+\cos t)\\&=\frac{(u+1)^2}{2}=\frac54.\end{aligned}

The feasible sign gives

u+1=\sqrt{\frac{5}{2}}.

Therefore

\begin{aligned}(1-\sin t)&(1-\cos t)\\&=\frac{(1-u)^2}{2}\\&=\frac{13}{4}-\sqrt{10}.\end{aligned}

Thus

k+m+n=10+13+4=27.