Difference between revisions of "1998 AHSME Problems/Problem 19"
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Visually, the set of points of the form <math>(5\cos \theta, 5\sin \theta)</math> is a circle centered at <math>(0,0)</math> with radius 5. The missing vertex of the triangle must lie on this circle. At the same time, its distance from the <math>x</math> axis must be 2. The set of all such points are precisely the lines <math>y=2</math> and <math>y=-2</math>, and each of these lines intersects the circle in two points. | Visually, the set of points of the form <math>(5\cos \theta, 5\sin \theta)</math> is a circle centered at <math>(0,0)</math> with radius 5. The missing vertex of the triangle must lie on this circle. At the same time, its distance from the <math>x</math> axis must be 2. The set of all such points are precisely the lines <math>y=2</math> and <math>y=-2</math>, and each of these lines intersects the circle in two points. | ||
| + | |||
| + | == Solution 2== | ||
| + | Alternatively, we use shoelace to get: | ||
| + | <cmath>10=\frac {1}{2}|50\sin (\theta)|</cmath> | ||
| + | This means <math>\sin (\theta)=\pm \frac {2}{5}</math>. We see that if it equals <math>\frac {2}{5}</math>, then <math>\cos (\theta)=\pm \frac {\sqrt {21}}{5}</math>. Likewise, we see that if <math>\sin (\theta)=-\frac {2}{5}</math>, then <math>\cos (\theta)</math> has <math>2</math> solutions. Thus, there are <math>\boxed {4}</math> unique points such that the triangle has an area of <math>10</math>, or <math>C</math>. | ||
== See also == | == See also == | ||
{{AHSME box|year=1998|num-b=18|num-a=20}} | {{AHSME box|year=1998|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 10:04, 4 February 2017
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:
![\[10=\frac {1}{2}|50\sin (\theta)|\]](http://latex.artofproblemsolving.com/7/6/1/761d5adafa434afffbe5418b61084d5f731636ad.png)
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 | |
| 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. 
