Difference between revisions of "2021 AIME II Problems/Problem 2"
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==Solution 1== | ==Solution 1== | ||
− | + | By angle-chasing, <math>\triangle AGF</math> is a <math>30^\circ\text{-}30^\circ\text{-}120^\circ</math> triangle, and <math>\triangle BED</math> is a <math>30^\circ\text{-}60^\circ\text{-}90^\circ</math> triangle. | |
+ | |||
+ | Let <math>AF=x.</math> It follows that <math>FG=x</math> and <math>EB=FC=840-x.</math> By the side-length ratios in <math>\triangle BED,</math> we have <math>DE=\frac{840-x}{2}</math> and <math>DB=\frac{840-x}{2}\cdot\sqrt3.</math> | ||
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+ | The area of <math>\triangle AFG</math> is <cmath>\frac12\cdot AF\cdot FG\cdot\sin{\angle AFG}=\frac12\cdot x\cdot x\sin{120^\circ}=\frac12\cdot x^2\cdot\frac{\sqrt3}{2},</cmath> and the area of <math>\triangle BED</math> is <cmath>\frac12\cdot DE\cdot DB=\frac12\cdot\frac{840-x}{2}\frac{840-x}{2}\cdot\sqrt3.</cmath> | ||
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+ | ~MRENTHUSIASM | ||
==Solution 2== | ==Solution 2== | ||
− | We express the | + | We express the areas of <math>\triangle BED</math> and <math>\triangle AFG</math> in terms of <math>AF</math> in order to solve for <math>AF. </math> |
− | We let <math>x = AF. </math> Because <math>\triangle AFG</math> is | + | We let <math>x = AF. </math> Because <math>\triangle AFG</math> is isosceles and <math>\triangle AEF</math> is equilateral, <math>AF = FG = EF = AE = x. </math> |
Let the height of <math>\triangle ABC</math> be <math>h</math> and the height of <math>\triangle AEF</math> be <math>h'. </math> Then we have that <math>h = \frac{\sqrt{3}}{2}(840) = 420\sqrt{3}</math> and <math>h' = \frac{\sqrt{3}}{2}(EF) = \frac{\sqrt{3}}{2}x. </math> | Let the height of <math>\triangle ABC</math> be <math>h</math> and the height of <math>\triangle AEF</math> be <math>h'. </math> Then we have that <math>h = \frac{\sqrt{3}}{2}(840) = 420\sqrt{3}</math> and <math>h' = \frac{\sqrt{3}}{2}(EF) = \frac{\sqrt{3}}{2}x. </math> | ||
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~JimY | ~JimY | ||
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==See also== | ==See also== | ||
{{AIME box|year=2021|n=II|num-b=1|num-a=3}} | {{AIME box|year=2021|n=II|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 23:01, 22 March 2021
Contents
Problem
Equilateral triangle has side length
. Point
lies on the same side of line
as
such that
. The line
through
parallel to line
intersects sides
and
at points
and
, respectively. Point
lies on
such that
is between
and
,
is isosceles, and the ratio of the area of
to the area of
is
. Find
.
Diagram
Solution 1
By angle-chasing, is a
triangle, and
is a
triangle.
Let It follows that
and
By the side-length ratios in
we have
and
The area of is
and the area of
is
~MRENTHUSIASM
Solution 2
We express the areas of and
in terms of
in order to solve for
We let Because
is isosceles and
is equilateral,
Let the height of be
and the height of
be
Then we have that
and
Now we can find and
in terms of
Because we are given that
This allows us to use the sin formula for triangle area: the area of
is
Similarly, because
the area of
Now we can make an equation:
To make further calculations easier, we scale everything down by (while keeping the same variable names, so keep that in mind).
Thus Because we scaled down everything by
the actual value of
is
~JimY
See also
2021 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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.