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

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==Solution==
 
==Solution==
 
SOMEBODY DELETED A SOLUTION - PLEASE WAIT FOR THE ORIGINAL AUTHOR TO COME BACK ~ARCTICTURN
 
SOMEBODY DELETED A SOLUTION - PLEASE WAIT FOR THE ORIGINAL AUTHOR TO COME BACK ~ARCTICTURN
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 +
 +
 +
==Solution==
 +
We express the ares 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 isoceles and <math>\triangle AEF</math> is equilateral, <math>AF = FG = EF = AE = x. </math>
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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>
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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 = \frac{1}{2}(\sin 120)(x^2). </math>
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 +
Now we can make an equation:
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 +
\begin{align*}
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\frac{\triangle AFG}{\triangle BED} &= \frac{8}{9}\\
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\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}\\
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\frac{x^2}{(420 - \frac{x}{2})(840-x)} &= \frac{8}{9}\\
 +
\end{align*}
 +
 +
To make further calculations easier, we scale everything down by <math>420</math> (while keeping the same variable names, so keep that in mind).
 +
 +
LATEX FIXING IN PROGRESS:
 +
 +
\begin{align*}
 +
\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\\
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5x^2 + 16x -16 &= 0\\
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(5x-4)(x+4) &= 0\\
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\end{align*}
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 +
 +
 +
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>(420)\frac{4}{5} = \boxed{336}. </math>
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 +
~JimY
  
 
==See also==
 
==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}}

Revision as of 18:47, 22 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 $AF$.


[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

SOMEBODY DELETED A SOLUTION - PLEASE WAIT FOR THE ORIGINAL AUTHOR TO COME BACK ~ARCTICTURN


Solution

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

We let $x = AF.$ Because $\triangle AFG$ is isoceles 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:

\begin{align*} \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}\\ \end{align*}

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

LATEX FIXING IN PROGRESS:

\begin{align*} \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\\ \end{align*}


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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