2001 AIME I Problems/Problem 5
Problem
An equilateral triangle is inscribed in the ellipse whose equation is 
. One vertex of the triangle is 
, one altitude is contained in the y-axis, and the length of each side is 
, where 
and 
are relatively prime positive integers. Find 
.
Solution
![[asy] pointpen = black; pathpen = black + linewidth(0.7); path e = xscale(2)*unitcircle; real x = -8/13*3^.5; D((-3,0)--(3,0)); D((0,-2)--(0,2)); /* axes */ D(e); D(D((0,1))--(x,x*3^.5+1)--(-x,x*3^.5+1)--cycle); [/asy]](http://latex.artofproblemsolving.com/c/1/c/c1c17659e426bf52cb1df3d3e5f87476ced5b24c.png)
Solution 1
Denote the vertices of the triangle 
and 
where 
is in quadrant 4 and 
is in quadrant 
Note that the slope of 
is 
Hence, the equation of the line containing is
![\[y = x\sqrt {3} + 1.\]](http://latex.artofproblemsolving.com/6/4/7/647cf5f03f729a69f551bc8dfd47692937940287.png)
This will intersect the ellipse when
Since the triangle is symmetric with respect to the y-axis, the coordinates of and are now 
and 
respectively, for some value of 
Since we're going to use the distance formula, the value of 
is irrelevant. Our answer is
![\[BC = \sqrt {2\left(\frac {8\sqrt {3}}{13}\right)^{2}} = \sqrt {\frac {768}{169}}\implies m + n = \boxed{937}.\]](http://latex.artofproblemsolving.com/a/e/d/aedb862dfb0696d379eb4da678c729e1369ec040.png)
Solution 2
Solving for 
in terms of 
gives 
, so the two other points of the triangle are 
and 
, which are a distance of 
apart. Thus equals the distance between and , so by the distance formula we have
![\[2x=\sqrt{x^2+(1-\sqrt{4-x^2}/2)^2}.\]](http://latex.artofproblemsolving.com/c/d/9/cd9820889aa97371bcb588d88e45e003d74177ed.png)
Squaring both sides and simplifying through algebra yields 
, so 
and the answer is 
.
See also
| 2001 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
