Difference between revisions of "1998 AHSME Problems/Problem 19"

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.

$[asy] 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), (-6,2) -- (6,2) ); pair ip2[] = intersectionpoints( circle((0,0),5), (-6,-2) -- (6,-2) ); draw ( (-6,0) -- (6,0) ); draw ( A -- ip1[1] -- B, Dotted ); draw ( circle((0,0),5), red ); draw ( (-6,2) -- (6,2), blue ); draw ( (-6,-2) -- (6,-2), blue ); dot(ip1[0]); dot(ip1[1]); dot(ip2[0]); dot(ip2[1]); label("$$(-5,0)$$", A, SW ); label("$$(5,0)$$", B, SE ); label("$$(0,0)$$", (0,0), SE ); label("$$y=2$$", (6,2), N, blue ); label("$$y=-2$$", (6,-2), S, blue ); label("$$y=-2$$", (6,-2), S, blue ); label("$$(5\cos\theta,5\sin\theta)$$", 5*dir(-45), SE, red ); [/asy]$

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.

Solution 2

Alternatively, we use shoelace to get: $$10=\frac {1}{2}|50\sin (\theta)|$$ This means $\sin (\theta)=\pm \frac {2}{5}$. We see that if it equals $\frac {2}{5}$, then $\cos (\theta)=\pm \frac {\sqrt {21}}{5}$. Likewise, we see that if $\sin (\theta)=-\frac {2}{5}$, then $\cos (\theta)$ has $2$ solutions. Thus, there are $\boxed {4}$ unique points such that the triangle has an area of $10$, or $C$.