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Difference between revisions of "2021 AIME II Problems/Problem 2"

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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>.
−
==Solution==
+
<asy>
−
SOMEBODY DELETED A SOLUTION - PLEASE WAIT FOR THE ORIGINAL AUTHOR TO COME BACK ~ARCTICTURN
+
pair A,B,C,D,E,F,G;
 +
B=origin;
 +
A=5*dir(60);
 +
C=(5,0);
 +
E=0.6*A+0.4*B;
 +
F=0.6*A+0.4*C;
 +
G=rotate(240,F)*A;
 +
D=extension(E,F,B,dir(90));
 +
draw(D--G--A,grey);
 +
draw(B--0.5*A+rotate(60,B)*A*0.5,grey);
 +
draw(A--B--C--cycle,linewidth(1.5));
 +
dot(A^^B^^C^^D^^E^^F^^G);
 +
label("$A$",A,dir(90));
 +
label("$B$",B,dir(225));
 +
label("$C$",C,dir(-45));
 +
label("$D$",D,dir(180));
 +
label("$E$",E,dir(-45));
 +
label("$F$",F,dir(225));
 +
label("$G$",G,dir(0));
 +
label("$\ell$",midpoint(E--F),dir(90));
 +
</asy>
  
−
==See also==
+
==Solution 1 (Area Formulas for Triangles)==
 +
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 the brackets denote areas. We have <cmath>[AFG]=\frac12\cdot AF\cdot FG\cdot\sin{\angle AFG}=\frac12\cdot x\cdot x\cdot\sin{120^\circ}=\frac12\cdot x^2\cdot\frac{\sqrt3}{2}</cmath> and <cmath>[BED]=\frac12\cdot DE\cdot DB=\frac12\cdot\frac{840-x}{2}\cdot\left(\frac{840-x}{2}\cdot\sqrt3\right).</cmath>
 +
 
 +
We set up and solve an equation for <math>x:</math>
 +
<cmath>\begin{align*}
 +
\frac{[AFG]}{[BED]}&=\frac89 \\
 +
\frac{\frac12\cdot x^2\cdot\frac{\sqrt3}{2}}{\frac12\cdot\frac{840-x}{2}\cdot\left(\frac{840-x}{2}\cdot\sqrt3\right)}&=\frac89 \\
 +
\frac{2x^2}{(840-x)^2}&=\frac89 \\
 +
\frac{x^2}{(840-x)^2}&=\frac49.
 +
\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 for both sides:
 +
<cmath>\begin{align*}
 +
\frac{x}{840-x}&=\frac23 \\
 +
3x&=1680-2x \\
 +
5x&=1680 \\
 +
x&=\boxed{336}.
 +
\end{align*}</cmath>
 +
 
 +
~MRENTHUSIASM
 +
 
 +
==Solution 2==
 +
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 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>
 +
 
 +
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:
 +
<cmath>\begin{align*}
 +
\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).
 +
<cmath>\begin{align*}
 +
\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>\frac{4}{5}(420) = \boxed{336}. </math>
 +
 
 +
~JimY
 +
 
 +
==Solution 3 (Pretty Straightforward)==
 +
<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>.
 +
 
 +
Let <math>AF = FG = AE = EF = x</math>
 +
 
 +
So, <math>[\Delta AFG] = \frac{1}{2} \cdot x^{2} \textup{sin} 120^{0} = \frac{\sqrt{3}}{4}x^{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) \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 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 23:09, 3 February 2023

Problem

Equilateral triangle $ABC$ has side length $840$. Point $D$ lies on the same side of line $BC$ as $A$ such that $\overline{BD} \perp \overline{BC}$. The line $\ell$ through $D$ parallel to line $BC$ intersects sides $\overline{AB}$ and $\overline{AC}$ at points $E$ and $F$, respectively. Point $G$ lies on $\ell$ such that $F$ is between $E$ and $G$, $\triangle AFG$ is isosceles, and the ratio of the area of $\triangle AFG$ to the area of $\triangle BED$ is $8:9$. Find $AF$. [asy] pair A,B,C,D,E,F,G; B=origin; A=5*dir(60); C=(5,0); E=0.6*A+0.4*B; F=0.6*A+0.4*C; G=rotate(240,F)*A; D=extension(E,F,B,dir(90)); draw(D--G--A,grey); draw(B--0.5*A+rotate(60,B)*A*0.5,grey); draw(A--B--C--cycle,linewidth(1.5)); dot(A^^B^^C^^D^^E^^F^^G); label("$A$",A,dir(90)); label("$B$",B,dir(225)); label("$C$",C,dir(-45)); label("$D$",D,dir(180)); label("$E$",E,dir(-45)); label("$F$",F,dir(225)); label("$G$",G,dir(0)); label("$\ell$",midpoint(E--F),dir(90)); [/asy]

Solution 1 (Area Formulas for Triangles)

By angle chasing, we conclude that $\triangle AGF$ is a $30^\circ\text{-}30^\circ\text{-}120^\circ$ triangle, and $\triangle BED$ is a $30^\circ\text{-}60^\circ\text{-}90^\circ$ triangle.

Let $AF=x.$ It follows that $FG=x$ and $EB=FC=840-x.$ By the side-length ratios in $\triangle BED,$ we have $DE=\frac{840-x}{2}$ and $DB=\frac{840-x}{2}\cdot\sqrt3.$

Let the brackets denote areas. We have \[[AFG]=\frac12\cdot AF\cdot FG\cdot\sin{\angle AFG}=\frac12\cdot x\cdot x\cdot\sin{120^\circ}=\frac12\cdot x^2\cdot\frac{\sqrt3}{2}\] and \[[BED]=\frac12\cdot DE\cdot DB=\frac12\cdot\frac{840-x}{2}\cdot\left(\frac{840-x}{2}\cdot\sqrt3\right).\]

We set up and solve an equation for $x:$ \begin{align*} \frac{[AFG]}{[BED]}&=\frac89 \\ \frac{\frac12\cdot x^2\cdot\frac{\sqrt3}{2}}{\frac12\cdot\frac{840-x}{2}\cdot\left(\frac{840-x}{2}\cdot\sqrt3\right)}&=\frac89 \\ \frac{2x^2}{(840-x)^2}&=\frac89 \\ \frac{x^2}{(840-x)^2}&=\frac49. \end{align*} Since $0<x<840,$ it is clear that $\frac{x}{840-x}>0.$ Therefore, we take the positive square root for both sides: \begin{align*} \frac{x}{840-x}&=\frac23 \\ 3x&=1680-2x \\ 5x&=1680 \\ x&=\boxed{336}. \end{align*}

~MRENTHUSIASM

Solution 2

We express the areas of $\triangle BED$ and $\triangle AFG$ in terms of $AF$ in order to solve for $AF.$

We let $x = AF.$ Because $\triangle AFG$ is isosceles and $\triangle AEF$ is equilateral, $AF = FG = EF = AE = x.$

Let the height of $\triangle ABC$ be $h$ and the height of $\triangle AEF$ be $h'.$ Then we have that $h = \frac{\sqrt{3}}{2}(840) = 420\sqrt{3}$ and $h' = \frac{\sqrt{3}}{2}(EF) = \frac{\sqrt{3}}{2}x.$

Now we can find $DB$ and $BE$ in terms of $x.$ $DB = h - h' = 420\sqrt{3} - \frac{\sqrt{3}}{2}x,$ $BE = AB - AE = 840 - x.$ Because we are given that $\angle DBC = 90,$ $\angle DBE = 30.$ This allows us to use the sin formula for triangle area: the area of $\triangle BED$ is $\frac{1}{2}(\sin 30)\left(420\sqrt{3} - \frac{\sqrt{3}}{2}x\right)(840-x).$ Similarly, because $\angle AFG = 120,$ the area of $\triangle AFG$ is $\frac{1}{2}(\sin 120)(x^2).$

Now we can make an equation: \begin{align*} \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*} To make further calculations easier, we scale everything down by $420$ (while keeping the same variable names, so keep that in mind). \begin{align*} \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*} Thus $x = \frac{4}{5}.$ Because we scaled down everything by $420,$ the actual value of $AF$ is $\frac{4}{5}(420) = \boxed{336}.$

~JimY

Solution 3 (Pretty Straightforward)

$\angle AFE = \angle AEF = \angle EAF = 60^{0} \Rightarrow \angle AFG = 120^{0}$ So, If $\Delta AFG$ is isosceles, it means that $AF = FG$.

Let $AF = FG = AE = EF = x$

So, $[\Delta AFG] = \frac{1}{2} \cdot x^{2} \textup{sin} 120^{0} = \frac{\sqrt{3}}{4}x^{2}$

In $\Delta BED$, $BE = 840 - x$, Hence $DE = \frac{840 - x}{2}$ (because $\textup{sin} 30^{0} = \frac{1}{2}$)

Therefore, $[\Delta BED] = \frac{1}{2} (840 - x) \left (\frac{840-x}{2} \right) \textup{sin} 60^{0}$

So, $[\Delta BED] = \frac{\sqrt{3}}{4} (840 - x) \left (\frac{840-x}{2} \right) = \frac{\sqrt{3}}{8} (840 - x)^{2}$


Now, as we know that the ratio of the areas of $\Delta AFG$ and $\Delta BED$ is $8:9$

Substituting the values, we get

$\frac{\frac{\sqrt{3}}{4}x^{2}}{\frac{\sqrt{3}}{8} (840 - x)^{2}} = \frac{8}{9} \Rightarrow \left (\frac{x}{840 - x} \right)^{2} = \frac{4}{9}$ Hence, $\frac{x}{840 - x} = \frac{2}{3}$. Solving this, we easily get $x = 336$

We have taken $AF = x$, Hence, $AF = \boxed{336}$

-Arnav Nigam

Solution 4 (Similar Triangles)

Since $\triangle AFG$ is isosceles, $AF = FG$, and since $\triangle AEF$ is equilateral, $AF = EF$. Thus, $EF = FG$, and since these triangles share an altitude, they must have the same area.

Drop perpendiculars from $E$ and $F$ to line $BC$; call the meeting points $P$ and $Q$, respectively. $\triangle BEP$ is clearly congruent to both $\triangle BED$ and $\triangle FQC$, and thus each of these new triangles has the same area as $\triangle BED$. But we can "slide" $\triangle BEP$ over to make it adjacent to $\triangle FQC$, thus creating an equilateral triangle whose area has a ratio of $18:8$ when compared to $\triangle AEF$ (based on our conclusion from the first paragraph). Since these triangles are both equilateral, they are similar, and since the area ratio $18:8$ reduces to $9:4$, the ratio of their sides must be $3:2$. So, because $FC$ and $AF$ represent sides of these triangles, and they add to $840$, $AF$ must equal two-fifths of $840$, or $\boxed{336}$.

Video Solution

https://www.youtube.com/watch?v=ol-Nl-t9X04

Video Solution by Interstigation (Similar Triangles)

https://youtu.be/qjiOhBEfpWY

~Interstigation

See Also

2021 AIME II (Problems • Answer Key • Resources)
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