Difference between revisions of "2021 AIME II Problems/Problem 2"
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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> | ||
| + | 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== | ==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 
has side length 
. Point 
lies on the same side of line 
as 
such that 
. The line 
through parallel to line intersects sides 
and 
at points 
and 
, respectively. Point 
lies on such that is between and , 
is isosceles, and the ratio of the area of to the area of 
is 
. Find 
.
![[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]](http://latex.artofproblemsolving.com/7/1/5/7154e7a32b3eda0a8ba22787a8b4d10ba8b8dc50.png)
Solution
SOMEBODY DELETED A SOLUTION - PLEASE WAIT FOR THE ORIGINAL AUTHOR TO COME BACK ~ARCTICTURN
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
