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1984 AHSME Problems/Problem 4

Revision as of 21:28, 8 July 2011 by Admin25 (talk | contribs) (Created page with "==Problem== Points <math> B, C, F, E </math> are picked on a circle such that <math> BC||EF </math>. When <math> BC </math> is extended to the left, point <math> A </math> is...")
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Problem

Points $B, C, F, E$ are picked on a circle such that $BC||EF$. When $BC$ is extended to the left, point $A$ is marked outside the circle such that $AB=4$ and $BC=5$. When $EF$ is extended to the left, point $D$ is marked outside the circle such that $DE=3$. $AD$ is perpendicular to both $AC$ and $DF$. Find the length of $EF$.

Template:Incomplete

Solution

[asy] unitsize(2.5cm); draw(unitcircle); draw((sqrt(2)/2,sqrt(2)/2)--(-2,sqrt(2)/2)); draw((sqrt(2/3),-sqrt(3)/3)--(-2,-sqrt(3)/3)); draw((-2,sqrt(2)/2)--(-2,-sqrt(3)/3)); draw((-sqrt(2/3),-sqrt(3)/3)--(-sqrt(2)/2,sqrt(2)/2)); draw((sqrt(2)/2,sqrt(2)/2)--(sqrt(2/3),-sqrt(3)/3)); draw((-sqrt(2/3),sqrt(2)/2)--(-sqrt(2/3),-sqrt(3)/3)); draw((-sqrt(2)/2,sqrt(2)/2)--(-sqrt(2)/2,-sqrt(3)/3)); draw((sqrt(2)/2,sqrt(2)/2)--(sqrt(2)/2,-sqrt(3)/3)); label("$A$",(-2,sqrt(2)/2),NW); label("$B$",(-sqrt(2)/2,sqrt(2)/2),SE); label("$C$",(sqrt(2)/2,sqrt(2)/2),SW); label("$D$",(-2,-sqrt(3)/3),SW); label("$E$",(-sqrt(2/3),-sqrt(3)/3),SW); label("$F$",(sqrt(2/3),-sqrt(3)/3),SE); label("$G$",(-sqrt(2/3),sqrt(2)/2),NNW); label("$H$",(-sqrt(2)/2,-sqrt(3)/3),NE); label("$I$",(sqrt(2)/2,-sqrt(3)/3),NW); label("$5$",(0,sqrt(2)/2),S); label("$4$",((-4-sqrt(2))/4,sqrt(2)/2),N); label("$3$",((-2-sqrt(2/3))/2,-sqrt(3)/3),S); [/asy]

Draw $BE$ and $CF$, forming a trapezoid. Since it's cyclic, this trapezoid must be isosceles. Also, drop altitudes from $E$ to $AC$, $B$ to $DF$, and $C$ to $DF$, and let the feet of these altitudes be $G$, $H$, and $I$ respectively. $AGED$ is a rectangle since it has $4$ right angles. Therefore, $AG=DE=3$, and $GB=4-3=1$. By the same logic, $GBHE$ is also a rectangle, and $EH=GB=1$. $BH=CI$ since they're both altitudes to a trapezoid, and $BE=CF$ since the trapezoid is isosceles. Therefore, $\triangleBHE\congruent\triangleCIF$ (Error compiling LaTeX. Unknown error_msg) by HL congruence, so $IF=EH=1$. Also, $BCIH$ is a rectangle from $4$ right angles, and $HI=BC=5$. Therefore, $EF=EH+HI+IF=1+5+1=\boxed{7}$.

See Also

1984 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions
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