1998 AHSME Problems/Problem 19

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Problem

How many triangles have area $10$ and vertices at $(-5,0),(5,0)$ and $(5\cos \theta, 5\sin \theta)$ for some angle $\theta$ ?

$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ }4 \qquad \mathrm{(D) \ }6 \qquad \mathrm{(E) \ } 8$

Solution

The triangle can be seen as having the base on the $x$ axis and height $|5\sin\theta|$ . The length of the base is $10$ , thus the height must be $2$ . The equation $|\sin\theta| = \frac 25$ has $\boxed{4}$ solutions, one in each quadrant.

size(250);defaultpen(0.8);pair A=(-5,0), B=(5,0);dot(A); dot(B); dot((0,0));pair ip1[] = intersectionpoints( circle((0,0),5),...

Visually, the set of points of the form $(5\cos \theta, 5\sin \theta)$ is a circle centered at $(0,0)$ with radius 5. The missing vertex of the triangle must lie on this circle. At the same time, its distance from the $x$ axis must be 2. The set of all such points are precisely the lines $y=2$ and $y=-2$ , and each of these lines intersects the circle in two points.

1998 AHSME (Problems)
Preceded by Problem 18	Followed by Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30

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1998 AHSME Problems/Problem 19

From AoPSWiki

Problem

Solution

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