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

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== Solution ==
 
== Solution ==
  
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First note that <math>\cos{A}, \cos{B}, \sin{A}, \sin{B}>0</math> and <math>A+B>90</math>. From these, we deduce that the <math>x</math> value and the <math>y</math> value of the coordinates will always be positive and thus the point can only be in the first quadrant.
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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 $\triangle ABC$ is a triangle with 3 acute angles $A, B,$ and $C$. Then the point $( \cos B - \sin A, \sin B - \cos A)$

(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 $\cos{A}, \cos{B}, \sin{A}, \sin{B}>0$ and $A+B>90$.

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

Notice that $\cos^2B-\sin^2A=-(\sin^2B-\cos^2A)$. Then $(\cos B-\sin A, \sin B-\cos A)$ can only lie in the 2nd and 4th quadrants.


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