2021 AIME II Problems/Problem 2

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 (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 $\frac{1}{2}(\sin 30)(420\sqrt{3} - \frac{\sqrt{3}}{2}x)(840-x).$ 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

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