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.
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== Solution 2==
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Alternatively, we use shoelace to get:
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<cmath>10=\frac {1}{2}|50\sin (\theta)|</cmath>
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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

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 $C$.

See also

1998 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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