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2021 AIME II Problems/Problem 2

Revision as of 22:02, 22 March 2021 by MRENTHUSIASM (talk | contribs) (→‎Solution 1)

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 $DB=\frac{840-x}{2}\cdot\sqrt3.$

The area of $\triangle AFG$ is \[\frac12\cdot AF\cdot FG\cdot\sin{\angle AFG}=\frac12\cdot x\cdot x\cdot\sin{120^\circ}=\frac12\cdot x^2\cdot\frac{\sqrt3}{2},\] and the area of $\triangle BED$ is \[\frac12\cdot DE\cdot DB=\frac12\cdot\frac{840-x}{2}\cdot\frac{840-x}{2}\cdot\sqrt3.\]

~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

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

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