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==Problem== | ==Problem== | ||
Equilateral triangle <math>ABC</math> has side length <math>840</math>. Point <math>D</math> lies on the same side of line <math>BC</math> as <math>A</math> such that <math>\overline{BD} \perp \overline{BC}</math>. The line <math>\ell</math> through <math>D</math> parallel to line <math>BC</math> intersects sides <math>\overline{AB}</math> and <math>\overline{AC}</math> at points <math>E</math> and <math>F</math>, respectively. Point <math>G</math> lies on <math>\ell</math> such that <math>F</math> is between <math>E</math> and <math>G</math>, <math>\triangle AFG</math> is isosceles, and the ratio of the area of <math>\triangle AFG</math> to the area of <math>\triangle BED</math> is <math>8:9</math>. Find <math>AF</math>. | Equilateral triangle <math>ABC</math> has side length <math>840</math>. Point <math>D</math> lies on the same side of line <math>BC</math> as <math>A</math> such that <math>\overline{BD} \perp \overline{BC}</math>. The line <math>\ell</math> through <math>D</math> parallel to line <math>BC</math> intersects sides <math>\overline{AB}</math> and <math>\overline{AC}</math> at points <math>E</math> and <math>F</math>, respectively. Point <math>G</math> lies on <math>\ell</math> such that <math>F</math> is between <math>E</math> and <math>G</math>, <math>\triangle AFG</math> is isosceles, and the ratio of the area of <math>\triangle AFG</math> to the area of <math>\triangle BED</math> is <math>8:9</math>. Find <math>AF</math>. | ||
− | |||
− | |||
<asy> | <asy> | ||
pair A,B,C,D,E,F,G; | pair A,B,C,D,E,F,G; | ||
Line 26: | Line 24: | ||
</asy> | </asy> | ||
− | ==Solution 1== | + | ==Solution 1 (Area Formulas for Triangles)== |
− | By angle | + | By angle chasing, we conclude that <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> | 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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\frac{x^2}{(840-x)^2}&=\frac49. | \frac{x^2}{(840-x)^2}&=\frac49. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | Since <math>0<x<840,</math> it is clear that <math>\frac{x}{840-x}>0.</math> Therefore, we take the positive square root | + | Since <math>0<x<840,</math> it is clear that <math>\frac{x}{840-x}>0.</math> Therefore, we take the positive square root for both sides: |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
\frac{x}{840-x}&=\frac23 \\ | \frac{x}{840-x}&=\frac23 \\ | ||
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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> | ||
− | Now we can find <math>DB</math> and <math>BE</math> in terms of <math>x. </math> <math>DB = h - h' = 420\sqrt{3} - \frac{\sqrt{3}}{2}x, </math> <math>BE = AB - AE = 840 - x. </math> Because we are given that <math>\angle DBC = 90, </math> <math>\angle DBE = 30. </math> This allows us to use the sin formula for triangle area: the area of <math>\triangle BED</math> is <math>\frac{1}{2}(\sin 30)(420\sqrt{3} - \frac{\sqrt{3}}{2}x)(840-x). </math> Similarly, because <math>\angle AFG = 120, </math> the area of <math>\triangle AFG | + | Now we can find <math>DB</math> and <math>BE</math> in terms of <math>x. </math> <math>DB = h - h' = 420\sqrt{3} - \frac{\sqrt{3}}{2}x, </math> <math>BE = AB - AE = 840 - x. </math> Because we are given that <math>\angle DBC = 90, </math> <math>\angle DBE = 30. </math> This allows us to use the sin formula for triangle area: the area of <math>\triangle BED</math> is <math>\frac{1}{2}(\sin 30)\left(420\sqrt{3} - \frac{\sqrt{3}}{2}x\right)(840-x). </math> Similarly, because <math>\angle AFG = 120, </math> the area of <math>\triangle AFG</math> is <math>\frac{1}{2}(\sin 120)(x^2). </math> |
Now we can make an equation: | Now we can make an equation: | ||
− | + | <cmath>\begin{align*} | |
− | <cmath>\frac{\triangle AFG}{\triangle BED} = \frac{8}{9} | + | \frac{\triangle AFG}{\triangle BED} &= \frac{8}{9} \\ |
− | + | \frac{\frac{1}{2}(\sin 120)(x^2)}{\frac{1}{2}(\sin 30)\left(420\sqrt{3} - \frac{\sqrt{3}}{2}x\right)(840-x)} &= \frac{8}{9} \\ | |
− | + | \frac{x^2}{\left(420 - \frac{x}{2}\right)(840-x)} &= \frac{8}{9}. | |
− | + | \end{align*}</cmath> | |
To make further calculations easier, we scale everything down by <math>420</math> (while keeping the same variable names, so keep that in mind). | To make further calculations easier, we scale everything down by <math>420</math> (while keeping the same variable names, so keep that in mind). | ||
− | + | <cmath>\begin{align*} | |
− | <cmath>\frac{x^2}{(1-\frac{x}{2})(2-x)} = \frac{8}{9} | + | \frac{x^2}{\left(1-\frac{x}{2}\right)(2-x)} &= \frac{8}{9} \\ |
− | + | 8\left(1-\frac{x}{2}\right)(2-x) &= 9x^2 \\ | |
− | + | 16-16x + 4x^2 &= 9x^2 \\ | |
− | + | 5x^2 + 16x -16 &= 0 \\ | |
− | + | (5x-4)(x+4) &= 0. | |
− | + | \end{align*}</cmath> | |
− | Thus <math>x = \frac{4}{5}. </math> Because we scaled down everything by <math>420, </math> the actual value of <math>AF</math> is <math> | + | Thus <math>x = \frac{4}{5}. </math> Because we scaled down everything by <math>420, </math> the actual value of <math>AF</math> is <math>\frac{4}{5}(420) = \boxed{336}. </math> |
~JimY | ~JimY | ||
− | ==Solution 3== | + | ==Solution 3 (Pretty Straightforward)== |
<math>\angle AFE = \angle AEF = \angle EAF = 60^{0} \Rightarrow \angle AFG = 120^{0}</math> | <math>\angle AFE = \angle AEF = \angle EAF = 60^{0} \Rightarrow \angle AFG = 120^{0}</math> | ||
So, If <math>\Delta AFG</math> is isosceles, it means that <math>AF = FG</math>. | So, If <math>\Delta AFG</math> is isosceles, it means that <math>AF = FG</math>. | ||
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In <math>\Delta BED</math>, <math>BE = 840 - x</math>, Hence <math>DE = \frac{840 - x}{2}</math> (because <math>\textup{sin} 30^{0} = \frac{1}{2}</math>) | In <math>\Delta BED</math>, <math>BE = 840 - x</math>, Hence <math>DE = \frac{840 - x}{2}</math> (because <math>\textup{sin} 30^{0} = \frac{1}{2}</math>) | ||
− | Therefore, <math>[\Delta BED] = \frac{1}{2} (840 - x) | + | Therefore, <math>[\Delta BED] = \frac{1}{2} (840 - x) \left (\frac{840-x}{2} \right) \textup{sin} 60^{0}</math> |
+ | |||
+ | So, <math>[\Delta BED] = \frac{\sqrt{3}}{4} (840 - x) \left (\frac{840-x}{2} \right) = \frac{\sqrt{3}}{8} (840 - x)^{2}</math> | ||
+ | |||
+ | |||
+ | Now, as we know that the ratio of the areas of <math>\Delta AFG</math> and <math>\Delta BED</math> is <math>8:9</math> | ||
+ | |||
+ | Substituting the values, we get | ||
+ | |||
+ | <math>\frac{\frac{\sqrt{3}}{4}x^{2}}{\frac{\sqrt{3}}{8} (840 - x)^{2}} = \frac{8}{9} \Rightarrow \left (\frac{x}{840 - x} \right)^{2} = \frac{4}{9}</math> | ||
+ | Hence, <math>\frac{x}{840 - x} = \frac{2}{3}</math>. Solving this, we easily get <math>x = 336</math> | ||
+ | |||
+ | We have taken <math>AF = x</math>, Hence, <math>AF = \boxed{336}</math> | ||
+ | |||
+ | -Arnav Nigam | ||
+ | |||
+ | ==Solution 4 (Similar Triangles)== | ||
+ | |||
+ | Since <math>\triangle AFG</math> is isosceles, <math>AF = FG</math>, and since <math>\triangle AEF</math> is equilateral, <math>AF = EF</math>. Thus, <math>EF = FG</math>, and since these triangles share an altitude, they must have the same area. | ||
+ | |||
+ | Drop perpendiculars from <math>E</math> and <math>F</math> to line <math>BC</math>; call the meeting points <math>P</math> and <math>Q</math>, respectively. <math>\triangle BEP</math> is clearly congruent to both <math>\triangle BED</math> and <math>\triangle FQC</math>, and thus each of these new triangles has the same area as <math>\triangle BED</math>. But we can "slide" <math>\triangle BEP</math> over to make it adjacent to <math>\triangle FQC</math>, thus creating an equilateral triangle whose area has a ratio of <math>18:8</math> when compared to <math>\triangle AEF</math> (based on our conclusion from the first paragraph). Since these triangles are both equilateral, they are similar, and since the area ratio <math>18:8</math> reduces to <math>9:4</math>, the ratio of their sides must be <math>3:2</math>. So, because <math>FC</math> and <math>AF</math> represent sides of these triangles, and they add to <math>840</math>, <math>AF</math> must equal two-fifths of <math>840</math>, or <math>\boxed{336}</math>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=ol-Nl-t9X04 | ||
+ | ==Video Solution by Interstigation (Similar Triangles)== | ||
+ | https://youtu.be/qjiOhBEfpWY | ||
+ | |||
+ | ~Interstigation | ||
− | ==See | + | ==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}} |
Latest revision as of 00:09, 4 February 2023
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
.
Solution 1 (Area Formulas for Triangles)
By angle chasing, we conclude that is a
triangle, and
is a
triangle.
Let It follows that
and
By the side-length ratios in
we have
and
Let the brackets denote areas. We have and
We set up and solve an equation for
Since
it is clear that
Therefore, we take the positive square root for both sides:
~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
is
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
Solution 3 (Pretty Straightforward)
So, If
is isosceles, it means that
.
Let
So,
In ,
, Hence
(because
)
Therefore,
So,
Now, as we know that the ratio of the areas of and
is
Substituting the values, we get
Hence,
. Solving this, we easily get
We have taken , Hence,
-Arnav Nigam
Solution 4 (Similar Triangles)
Since is isosceles,
, and since
is equilateral,
. Thus,
, and since these triangles share an altitude, they must have the same area.
Drop perpendiculars from and
to line
; call the meeting points
and
, respectively.
is clearly congruent to both
and
, and thus each of these new triangles has the same area as
. But we can "slide"
over to make it adjacent to
, thus creating an equilateral triangle whose area has a ratio of
when compared to
(based on our conclusion from the first paragraph). Since these triangles are both equilateral, they are similar, and since the area ratio
reduces to
, the ratio of their sides must be
. So, because
and
represent sides of these triangles, and they add to
,
must equal two-fifths of
, or
.
Video Solution
https://www.youtube.com/watch?v=ol-Nl-t9X04
Video Solution by Interstigation (Similar Triangles)
~Interstigation
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.