Difference between revisions of "2001 AIME I Problems/Problem 5"
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== Problem == | == Problem == | ||
− | An equilateral triangle is inscribed in the ellipse whose equation is <math>x^2+4y^2=4</math>. One vertex of the triangle is <math>(0,1)</math>, one altitude is contained in the y-axis, and the length of each side is <math>\sqrt{\frac mn}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | + | An [[equilateral triangle]] is inscribed in the [[ellipse]] whose equation is <math>x^2+4y^2=4</math>. One vertex of the triangle is <math>(0,1)</math>, one altitude is contained in the y-axis, and the length of each side is <math>\sqrt{\frac mn}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. |
+ | __TOC__ | ||
== Solution == | == Solution == | ||
− | Solving for y in terms of x gives <math>y=\sqrt{4-x^2}/2</math>, so the two other points of the triangle are <math>(x,\sqrt{4-x^2}/2)</math> and <math>(-x,\sqrt{4-x^2}/2)</math>, which are a distance of 2x apart. Thus 2x equals the distance between <math>(x,\sqrt{4-x^2}/2)</math> and (0,1), so by the distance formula we have < | + | <center><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></center> | ||
+ | === Solution 1 === | ||
+ | Denote the vertices of the triangle <math>A,B,</math> and <math>C,</math> where <math>B</math> is in [[quadrant]] 4 and <math>C</math> is in quadrant <math>3.</math> | ||
+ | |||
+ | Note that the slope of <math>\overline{AC}</math> is <math>\tan 60^\circ = \sqrt {3}.</math> Hence, the equation of the line containing <math>\overline{AC}</math> is | ||
+ | <cmath> | ||
+ | y = x\sqrt {3} + 1. | ||
+ | </cmath> | ||
+ | This will intersect the ellipse when | ||
+ | <cmath> | ||
+ | \begin{eqnarray*}4 = x^{2} + 4y^{2} & = & x^{2} + 4(x\sqrt {3} + 1)^{2} \\ | ||
+ | & = & x^{2} + 4(3x^{2} + 2x\sqrt {3} + 1) \implies x = \frac { - 8\sqrt {3}}{13}. \end{eqnarray*} | ||
+ | </cmath> | ||
+ | Since the triangle is symmetric with respect to the y-axis, the coordinates of <math>B</math> and <math>C</math> are now <math>\left(\frac {8\sqrt {3}}{13},y_{0}\right)</math> and <math>\left(\frac { - 8\sqrt {3}}{13},y_{0}\right),</math> respectively, for some value of <math>y_{0}.</math> | ||
+ | |||
+ | Since we're going to use the distance formula, the value of <math>y_{0}</math> is irrelevant. Our answer is | ||
+ | <cmath> | ||
+ | BC = \sqrt {2\left(\frac {8\sqrt {3}}{13}\right)^{2}} = \sqrt {\frac {768}{169}}\implies m + n = \boxed{937}. | ||
+ | </cmath> | ||
+ | |||
+ | === Solution 2 === | ||
+ | Solving for <math>y</math> in terms of <math>x</math> gives <math>y=\sqrt{4-x^2}/2</math>, so the two other points of the triangle are <math>(x,\sqrt{4-x^2}/2)</math> and <math>(-x,\sqrt{4-x^2}/2)</math>, which are a distance of <math>2x</math> apart. Thus <math>2x</math> equals the distance between <math>(x,\sqrt{4-x^2}/2)</math> and <math>(0,1)</math>, so by the distance formula we have | ||
+ | |||
+ | <cmath>2x=\sqrt{x^2+(1-\sqrt{4-x^2}/2)^2}.</cmath> | ||
+ | |||
+ | Squaring both sides and simplifying through algebra yields <math>x^2=192/169</math>, so <math>2x=\sqrt{768/169}</math> and the answer is <math>\boxed{937}</math>. | ||
== See also == | == See also == | ||
{{AIME box|year=2001|n=I|num-b=4|num-a=6}} | {{AIME box|year=2001|n=I|num-b=4|num-a=6}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] |
Revision as of 18:11, 11 June 2008
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
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
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
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 |