Difference between revisions of "2001 AIME I Problems/Problem 5"
(Added Solution 4) |
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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 | ||
+ | <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 . 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
.
Contents
Solution
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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 .
Solution 4
Denote as vertex
as the vertex to the left of the
-axis and
as the vertex to the right of the
-axis. Let
be the intersection of
and the
-axis.
Let be the
-coordinate of
This implies
and
Note that
and
This yields
Re-arranging and squaring, we have
Simplifying and solving for
, we have
As the length of each side is
our desired length is
which means our desired answer is
~ASAB
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.