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"

(Added Solution 4)
(→‎Solution 5)
 
(2 intermediate revisions by 2 users not shown)
Line 62: Line 62:
 
Moving stuff around and solving:
 
Moving stuff around and solving:
  
−
<math>y = \frac{-11}{3}, 1</math>
+
<math>y = \frac{-11}{13}, 1</math>
  
 
The second is found to be extraneous, so, when we go back and figure out <math>x</math> and then <math>2x</math>  (which is the side length), we find it to be:
 
The second is found to be extraneous, so, when we go back and figure out <math>x</math> and then <math>2x</math>  (which is the side length), we find it to be:
Line 92: Line 92:
  
 
~ASAB
 
~ASAB
 +
 +
===Solution 5===
 +
 +
Notice that <math>x^2+4y^2=4</math> can be rewritten as <math>(x)^2+(2y)^2=2^2</math>. The points of the triangle are <math>(0, 1)</math>, <math>(-x, 1-x\sqrt{3})</math>, and <math>(x, 1-x\sqrt{3})</math>. When plugging the second coordinate into the equation, we get <math>x^2+4-8x\sqrt{3}+12x^2=4</math>, which equals <math>13x^2-8x\sqrt{3}=0</math>.
 +
This yields <math>x(13x-8\sqrt{3})=0</math>. Obviously x can't be 0, so <math>x=\frac{8\sqrt{3}}{13}</math>. The side length of the equilateral triangle is twice of this, so <math>\frac{16\sqrt{3}}{13}</math>.
 +
This can be rewritten as <math>\sqrt{\frac{256\cdot3}{169}}=\sqrt{\frac{768}{169}}</math>.
 +
<math>768+169=\boxed{937}</math>.
 +
~ MC413551
 +
 +
===Solution 6===
 +
Consider the transformation <math>(x,y)</math> to <math>(x/2, y).</math> This sends the ellipse to the unit circle. If we let <math>n</math> be one-fourth of the side length of the triangle, the equilateral triangle is sent to an isosceles triangle with side lengths <math>2n, n\sqrt{13}, n\sqrt{13}.</math> Let the triangle be <math>ABC</math> such that <math>AB=AC.</math> Let the foot of the altitude from A be <math>X.</math> Then <math>BX=n,</math> and <math>AX=2n\sqrt{3}.</math> Let <math>C</math> be a point such that <math>AC</math> is a diameter of the unit circle. Then <math>XC=2-2n\sqrt{3}.</math> Using power of a point on X,
 +
<cmath>n^2=2n\sqrt{3}(2-2n\sqrt{3})</cmath>
 +
Simplifying gets us to <math>13n^2=4n\sqrt{3}.</math> Then, <math>n=\dfrac{4\sqrt{3}}{13}</math> which means the side length is <math>\dfrac{16\sqrt{3}}{13}=\sqrt{\dfrac{768}{169}}.</math>
 +
Thus, the answer is <math>768+169=\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}}

Latest revision as of 09:09, 15 January 2024

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:

$y = \frac{-11}{13}, 1$

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

Solution 5

Notice that $x^2+4y^2=4$ can be rewritten as $(x)^2+(2y)^2=2^2$. The points of the triangle are $(0, 1)$, $(-x, 1-x\sqrt{3})$, and $(x, 1-x\sqrt{3})$. When plugging the second coordinate into the equation, we get $x^2+4-8x\sqrt{3}+12x^2=4$, which equals $13x^2-8x\sqrt{3}=0$. This yields $x(13x-8\sqrt{3})=0$. Obviously x can't be 0, so $x=\frac{8\sqrt{3}}{13}$. The side length of the equilateral triangle is twice of this, so $\frac{16\sqrt{3}}{13}$. This can be rewritten as $\sqrt{\frac{256\cdot3}{169}}=\sqrt{\frac{768}{169}}$. $768+169=\boxed{937}$. ~ MC413551

Solution 6

Consider the transformation $(x,y)$ to $(x/2, y).$ This sends the ellipse to the unit circle. If we let $n$ be one-fourth of the side length of the triangle, the equilateral triangle is sent to an isosceles triangle with side lengths $2n, n\sqrt{13}, n\sqrt{13}.$ Let the triangle be $ABC$ such that $AB=AC.$ Let the foot of the altitude from A be $X.$ Then $BX=n,$ and $AX=2n\sqrt{3}.$ Let $C$ be a point such that $AC$ is a diameter of the unit circle. Then $XC=2-2n\sqrt{3}.$ Using power of a point on X, \[n^2=2n\sqrt{3}(2-2n\sqrt{3})\] Simplifying gets us to $13n^2=4n\sqrt{3}.$ Then, $n=\dfrac{4\sqrt{3}}{13}$ which means the side length is $\dfrac{16\sqrt{3}}{13}=\sqrt{\dfrac{768}{169}}.$ Thus, the answer is $768+169=\boxed{937}.$

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_I_Problems/Problem_5&oldid=210800"