Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2001 AIME I Problems/Problem 5"
(solution by 4everwise + asy)
Line 1: Line 1:
 
== 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 <math>2x=\sqrt{x^2+(1-\sqrt{4-x^2}/2)^2}</math>. Squaring both sides, letting <math>x^2=n</math>, and simplifying gives <math>n=192/169</math>, so <math>2x=\sqrt{768/169}</math> and the answer is 937.
+
<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 17:11, 11 June 2008

Problem

An equilateral triangle is inscribed in the ellipse whose equation is $x^2+4y^2=4$. One vertex of the triangle is $(0,1)$, one altitude is contained in the y-axis, and the length of each side is $\sqrt{\frac mn}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

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]

Solution 1

Denote the vertices of the triangle $A,B,$ and $C,$ where $B$ is in quadrant 4 and $C$ is in quadrant $3.$

Note that the slope of $\overline{AC}$ is $\tan 60^\circ = \sqrt {3}.$ Hence, the equation of the line containing $\overline{AC}$ is \[y = x\sqrt {3} + 1.\] This will intersect the ellipse when \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*} Since the triangle is symmetric with respect to the y-axis, the coordinates of $B$ and $C$ are now $\left(\frac {8\sqrt {3}}{13},y_{0}\right)$ and $\left(\frac { - 8\sqrt {3}}{13},y_{0}\right),$ respectively, for some value of $y_{0}.$

Since we're going to use the distance formula, the value of $y_{0}$ is irrelevant. Our answer is \[BC = \sqrt {2\left(\frac {8\sqrt {3}}{13}\right)^{2}} = \sqrt {\frac {768}{169}}\implies m + n = \boxed{937}.\]

Solution 2

Solving for $y$ in terms of $x$ gives $y=\sqrt{4-x^2}/2$, so the two other points of the triangle are $(x,\sqrt{4-x^2}/2)$ and $(-x,\sqrt{4-x^2}/2)$, which are a distance of $2x$ apart. Thus $2x$ equals the distance between $(x,\sqrt{4-x^2}/2)$ and $(0,1)$, so by the distance formula we have

\[2x=\sqrt{x^2+(1-\sqrt{4-x^2}/2)^2}.\]

Squaring both sides and simplifying through algebra yields $x^2=192/169$, so $2x=\sqrt{768/169}$ and the answer is $\boxed{937}$.

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_I_Problems/Problem_5&oldid=26471"