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

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-Arnav Nigam
 
-Arnav Nigam
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==Solution 4 (Similar Triangles)==
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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.
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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 18:8 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 18:8 reduces to 9:4, the ratio of their sides must be 3:2. So, because <math>FC</math> and <math>AF</math> represent sides of these triangles, and they add to 840, <math>AF</math> must equal two-fifths of 840, or <math>\boxed{336}</math>.
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==Video Solution==
 
==Video Solution==

Revision as of 21:32, 27 March 2021

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 $AF$.

Diagram

[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

By angle-chasing, $\triangle AGF$ 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 $\frac{1}{2}(\sin 30)(420\sqrt{3} - \frac{\sqrt{3}}{2}x)(840-x).$ Similarly, because $\angle AFG = 120,$ the area of $\triangle AFG = \frac{1}{2}(\sin 120)(x^2).$

Now we can make an equation:

\[\frac{\triangle AFG}{\triangle BED} = \frac{8}{9}\] \[\frac{\frac{1}{2}(\sin 120)(x^2)}{\frac{1}{2}(\sin 30)(420\sqrt{3} - \frac{\sqrt{3}}{2}x)(840-x)} = \frac{8}{9}\] \[\frac{x^2}{(420 - \frac{x}{2})(840-x)} = \frac{8}{9}\]

To make further calculations easier, we scale everything down by $420$ (while keeping the same variable names, so keep that in mind).

\[\frac{x^2}{(1-\frac{x}{2})(2-x)} = \frac{8}{9}\] \[8(1-\frac{x}{2})(2-x) = 9x^2\] \[16-16x + 4x^2 = 9x^2\] \[5x^2 + 16x -16 = 0\] \[(5x-4)(x+4) = 0\]

Thus $x = \frac{4}{5}.$ Because we scaled down everything by $420,$ the actual value of $AF$ is $(420)\frac{4}{5} = \boxed{336}.$

~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

See also

2021 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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All AIME Problems and Solutions

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