Difference between revisions of "1984 AHSME Problems/Problem 4"

m (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==
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A rectangle intersects a circle as shown: <math>AB=4</math>, <math>BC=5</math>, and <math>DE=3</math>. Then <math>EF</math> equals:
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 marked outside the circle such that <math> AB=4 </math> and <math> BC=5 </math>. When <math> EF </math> is extended to the left, point <math> D </math> is marked outside the circle such that <math> DE=3 </math>. <math> AD </math> is [[perpendicular]] to both <math> AC </math> and <math> DF </math>. Find the length of <math> EF </math>.
 
  
{{incomplete|material}}
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<asy>defaultpen(linewidth(0.7)+fontsize(10));
[[Category: Incomplete material]]
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pair D=origin, E=(3,0), F=(10,0), G=(12,0), H=(12,1), A=(0,1), B=(4,1), C=(9,1), O=circumcenter(B,C,F);
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draw(D--G--H--A--cycle);
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draw(Circle(O, abs(O-C)));
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label("$A$", A, NW);
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label("$B$", B, NW);
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label("$C$", C, NE);
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label("$D$", D, SW);
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label("$E$", E, SE);
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label("$F$", F, SW);
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label("4", (2,0.85), N);
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label("3", D--E, S);
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label("5", (6.5,0.85), N);
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</asy>
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<math>\mathbf{(A)}\; 6\qquad \mathbf{(B)}\; 7\qquad \mathbf{(C)}\; \frac{20}3\qquad \mathbf{(D)}\; 8\qquad \mathbf{(E)}\; 9</math>
  
 
==Solution==
 
==Solution==

Revision as of 16:36, 30 August 2011

A rectangle intersects a circle as shown: $AB=4$, $BC=5$, and $DE=3$. Then $EF$ equals:

[asy]defaultpen(linewidth(0.7)+fontsize(10)); pair D=origin, E=(3,0), F=(10,0), G=(12,0), H=(12,1), A=(0,1), B=(4,1), C=(9,1), O=circumcenter(B,C,F); draw(D--G--H--A--cycle); draw(Circle(O, abs(O-C))); label("$A$", A, NW); label("$B$", B, NW); label("$C$", C, NE); label("$D$", D, SW); label("$E$", E, SE); label("$F$", F, SW);  label("4", (2,0.85), N); label("3", D--E, S); label("5", (6.5,0.85), N); [/asy] $\mathbf{(A)}\; 6\qquad \mathbf{(B)}\; 7\qquad \mathbf{(C)}\; \frac{20}3\qquad \mathbf{(D)}\; 8\qquad \mathbf{(E)}\; 9$

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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All AHSME Problems and Solutions