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Difference between revisions of "2001 AIME I Problems/Problem 5"
(Added Solution 4)
Line 70: Line 70:
 
and so we get the desired answer of <math>\boxed{937}</math>.
 
and so we get the desired answer of <math>\boxed{937}</math>.
  
 +
===Solution 4===
 +
 +
Denote <math>(0,1)</math> as vertex <math>A,</math> <math>B</math> as the vertex to the left of the <math>y</math>-axis and <math>C</math> as the vertex to the right of the <math>y</math>-axis. Let <math>D</math> be the intersection of <math>BC</math> and the <math>y</math>-axis.
 +
 +
Let <math>x_0</math> be the <math>x</math>-coordinate of <math>C.</math> This implies
 +
<cmath>C=\left(x_0 , \sqrt{\frac{4-x_0^2}{4}}\right)</cmath>
 +
and
 +
<cmath>B=\left(-x_0 , \sqrt{\frac{4-x_0^2}{4}}\right).</cmath>
 +
Note that <math>BC=2x_0</math> and
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<cmath>\frac{BC}{\sqrt3}=AD=1-\sqrt{\frac{4-x_0^2}{4}}.</cmath>
 +
This yields
 +
<cmath>\frac{2x_0}{\sqrt3}=1-\sqrt{\frac{4-x_0^2}{4}}.</cmath>
 +
Re-arranging and squaring, we have
 +
<cmath>\frac{4-x_0^2}{4}=\frac{4x_0^2}{3}-\frac{4x_0}{\sqrt3} +1.</cmath>
 +
Simplifying and solving for <math>x_0</math>, we have
 +
<cmath>x_0=\frac{48}{13\sqrt 3}.</cmath>
 +
As the length of each side is <math>2x_0,</math> our desired length is
 +
<cmath>4x_0^2=\frac{768}{169}</cmath>
 +
which means our desired answer is
 +
<cmath>768+169=\boxed{937}</cmath>
 +
 +
~ASAB
 
== 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}}

Revision as of 20:19, 11 March 2022

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 square of the length of each side is $\frac{m}{n}$, 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(13x+8\sqrt 3)=0\implies x = \frac { - 8\sqrt {3}}{13}. \end{eqnarray*} We ignore the $x=0$ solution because it is not in quadrant 3.

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 $y_{0}.$

It is clear that the value of $y_{0}$ is irrelevant to the length of $BC$. Our answer is \[BC = 2*\frac {8\sqrt {3}}{13}=\sqrt {4\left(\frac {8\sqrt {3}}{13}\right)^{2}} = \sqrt {\frac {768}{169}}\implies m + n = \boxed{937}.\]

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 $\boxed{937}$.

Solution 3

Since the altitude goes along the $y$ axis, this means that the base is a horizontal line, which means that the endpoints of the base are $(x,y)$ and $(-x,y)$, and WLOG, we can say that $x$ is positive.

Now, since all sides of an equilateral triangle are the same, we can do this (distance from one of the endpoints of the base to the vertex and the length of the base):

$\sqrt{x^2 + (y-1)^2} = 2x$

Square both sides,

$x^2 + (y-1)^2 = 4x^2\implies (y-1)^2 = 3x^2$

Now, with the equation of the ellipse: $x^2 + 4y^2 = 4$

$x^2 = 4-4y^2$

$3x^2 = 12-12y^2$

Substituting,

$12-12y^2 = y^2 - 2y +1$

Moving stuff around and solving:

$y = \frac{-11}{3}, 1$

The second is found to be extraneous, so, when we go back and figure out $x$ and then $2x$ (which is the side length), we find it to be:

$\sqrt{\frac{768}{169}}$

and so we get the desired answer of $\boxed{937}$.

Solution 4

Denote $(0,1)$ as vertex $A,$ $B$ as the vertex to the left of the $y$-axis and $C$ as the vertex to the right of the $y$-axis. Let $D$ be the intersection of $BC$ and the $y$-axis.

Let $x_0$ be the $x$-coordinate of $C.$ This implies \[C=\left(x_0 , \sqrt{\frac{4-x_0^2}{4}}\right)\] and \[B=\left(-x_0 , \sqrt{\frac{4-x_0^2}{4}}\right).\] Note that $BC=2x_0$ and \[\frac{BC}{\sqrt3}=AD=1-\sqrt{\frac{4-x_0^2}{4}}.\] This yields \[\frac{2x_0}{\sqrt3}=1-\sqrt{\frac{4-x_0^2}{4}}.\] Re-arranging and squaring, we have \[\frac{4-x_0^2}{4}=\frac{4x_0^2}{3}-\frac{4x_0}{\sqrt3} +1.\] Simplifying and solving for $x_0$, we have \[x_0=\frac{48}{13\sqrt 3}.\] As the length of each side is $2x_0,$ our desired length is \[4x_0^2=\frac{768}{169}\] which means our desired answer is \[768+169=\boxed{937}\]

~ASAB

See also

2001 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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