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

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(Problem)
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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>.
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<asy>
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pair A,B,C,D,E,F,G;
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B=origin;
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A=5*dir(60);
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C=(5,0);
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E=0.6*A+0.4*B;
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F=0.6*A+0.4*C;
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G=rotate(240,F)*A;
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D=extension(E,F,B,dir(90));
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draw(D--G--A,grey);
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draw(B--0.5*A+rotate(60,B)*A*0.5,grey);
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draw(A--B--C--cycle,linewidth(1.5));
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dot(A^^B^^C^^D^^E^^F^^G);
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label("$A$",A,dir(90));
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label("$B$",B,dir(225));
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label("$C$",C,dir(-45));
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label("$D$",D,dir(180));
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label("$E$",E,dir(-45));
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label("$F$",F,dir(225));
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label("$G$",G,dir(0));
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label("$\ell$",midpoint(E--F),dir(90));
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</asy>
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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

Revision as of 19:46, 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

See also

2021 AIME II (ProblemsAnswer KeyResources)
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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