1998 AHSME Problems/Problem 19

Revision as of 10:08, 4 February 2017 by Hinna (talk | contribs) (Solution 2)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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: \[\frac {1}{2}|(-5*0+5*5\sin (\theta)+5 \cos (\theta) *0)-(0*5+0*5 \cos (\theta)-5*5\sin (\theta))|=10 \implies \frac {1}{2}|50\sin (\theta)|=10\] 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 $C$.

See also

1998 AHSME (ProblemsAnswer KeyResources)
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
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS