# Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 28"

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== Solution == | == Solution == | ||

− | First note that <math>\cos{A}, \cos{B}, \sin{A}, \sin{B}>0</math> and <math>A+B>90</math>. | + | |

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+ | First note that <math>\cos{A}, \cos{B}, \sin{A}, \sin{B}>0</math> and <math>A+B>90</math>. | ||

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+ | Since <math>\cos B+\sin A>0</math>, <math>\cos B-\sin A</math> and <math>\cos^2B-\sin^2A</math> have the same sign. Similarly, <math>\sin B-\cos A</math> and <math>\sin^2B-\cos^2A</math> have the same sign. | ||

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+ | Notice that <math>\cos^2B-\sin^2A=-(\sin^2B-\cos^2A)</math>. Then <math>(\cos B-\sin A, \sin B-\cos A)</math> can only lie in the 2nd and 4th quadrants. | ||

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## Latest revision as of 17:29, 7 October 2007

## Problem

Suppose is a triangle with 3 acute angles and . Then the point

(A) can be in the 1st quadrant and can be in the 2nd quadrant only

(B) can be in the 3rd quadrant and can be in the 4th quadrant only

(C) can be in the 2nd quadrant and can be in the 3rd quadrant only

(D) can be in the 2nd quadrant only

(E) can be in any of the 4 quadrants

## Solution

First note that and .

Since , and have the same sign. Similarly, and have the same sign.

Notice that . Then can only lie in the 2nd and 4th quadrants.