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>\ | + | 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 square of the length of each side is <math>\frac{m}{n}</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(13x+8\sqrt 3)=0\implies x = \frac { - 8\sqrt {3}}{13}. \end{eqnarray*} | ||
+ | </cmath> | ||
+ | We ignore the <math>x=0</math> solution because it is not in quadrant 3. | ||
+ | |||
+ | 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> | ||
+ | |||
+ | It is clear that the value of <math>y_{0}</math> is irrelevant to the length of <math>BC</math>. Our answer is | ||
+ | <cmath> | ||
+ | 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}. | ||
+ | </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>. | ||
+ | |||
+ | === Solution 3 === | ||
+ | Since the altitude goes along the <math>y</math> axis, this means that the base is a horizontal line, which means that the endpoints of the base are <math>(x,y)</math> and <math>(-x,y)</math>, and WLOG, we can say that <math>x</math> 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): | ||
+ | |||
+ | <math>\sqrt{x^2 + (y-1)^2} = 2x</math> | ||
+ | |||
+ | Square both sides, | ||
+ | |||
+ | <math>x^2 + (y-1)^2 = 4x^2\implies (y-1)^2 = 3x^2</math> | ||
+ | |||
+ | Now, with the equation of the ellipse: | ||
+ | <math>x^2 + 4y^2 = 4</math> | ||
+ | |||
+ | <math>x^2 = 4-4y^2</math> | ||
+ | |||
+ | <math>3x^2 = 12-12y^2</math> | ||
+ | |||
+ | Substituting, | ||
+ | |||
+ | <math>12-12y^2 = y^2 - 2y +1</math> | ||
+ | |||
+ | Moving stuff around and solving: | ||
+ | |||
+ | <math>y = \frac{-11}{3}, 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: | ||
+ | |||
+ | <math>\sqrt{\frac{768}{169}}</math> | ||
+ | |||
+ | and so we get the desired answer of <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]] | ||
+ | {{MAA Notice}} |
Latest revision as of 13:32, 30 July 2020
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 square of the length of each side is , where and are relatively prime positive integers. Find .
Solution
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 We ignore the solution because it is not in quadrant 3.
Since the triangle is symmetric with respect to the y-axis, the coordinates of and are now and respectively, for some value of
It is clear that the value of is irrelevant to the length of . 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 .
Solution 3
Since the altitude goes along the axis, this means that the base is a horizontal line, which means that the endpoints of the base are and , and WLOG, we can say that 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):
Square both sides,
Now, with the equation of the ellipse:
Substituting,
Moving stuff around and solving:
The second is found to be extraneous, so, when we go back and figure out and then (which is the side length), we find it to be:
and so we get the desired answer of .
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.