1998 AHSME Problems/Problem 19
Contents
Problem
How many triangles have area and vertices at
and
for some angle
?
Solution
The triangle can be seen as having the base on the axis and height
. The length of the base is
, thus the height must be
. The equation
has
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]](http://latex.artofproblemsolving.com/c/6/e/c6e55834b32f7a0ec73bf97c9d22c44ab8d50c75.png)
Visually, the set of points of the form is a circle centered at
with radius 5. The missing vertex of the triangle must lie on this circle. At the same time, its distance from the
axis must be 2. The set of all such points are precisely the lines
and
, and each of these lines intersects the circle in two points.
Solution 2
Alternatively, we use shoelace to get:
This means
. We see that if it equals
, then
. Likewise, we see that if
, then
has
solutions. Thus, there are
unique points such that the triangle has an area of
, or
.
See also
1998 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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