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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.

See also

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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