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

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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:
 
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:
  
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</asy>
 
</asy>
  
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Draw <math> BE </math> and <math> CF </math>, forming a [[trapezoid]]. Since it's cyclic, this trapezoid must be [[Isosceles trapezoid|isosceles]]. Also, drop [[Altitude|altitudes]] from <math> E </math> to <math> AC </math>, <math> B </math> to <math> DF </math>, and <math> C </math> to <math> DF </math>, and let the feet of these altitudes be <math> G </math>, <math> H </math>, and <math> I </math> respectively. <math> AGED </math> is a [[rectangle]] since it has <math> 4 </math> [[Right angle|right angles]]. Therefore, <math> AG=DE=3 </math>, and <math> GB=4-3=1 </math>. By the same logic, <math> GBHE </math> is also a rectangle, and <math> EH=GB=1 </math>. <math> BH=CI </math> since they're both altitudes to a trapezoid, and <math> BE=CF </math> since the trapezoid is isosceles. Therefore, <math> \triangleBHE\congruent\triangleCIF </math> by HL [[congruence]], so <math> IF=EH=1 </math>. Also, <math> BCIH </math> is a rectangle from <math> 4 </math> right angles, and <math> HI=BC=5 </math>. Therefore, <math> EF=EH+HI+IF=1+5+1=\boxed{7} </math>.
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Draw <math> BE </math> and <math> CF </math>, forming a [[trapezoid]]. Since it's cyclic, this trapezoid must be [[Isosceles trapezoid|isosceles]]. Also, drop [[Altitude|altitudes]] from <math> E </math> to <math> AC </math>, <math> B </math> to <math> DF </math>, and <math> C </math> to <math> DF </math>, and let the feet of these altitudes be <math> G </math>, <math> H </math>, and <math> I </math> respectively. <math> AGED </math> is a [[rectangle]] since it has <math> 4 </math> [[Right angle|right angles]]. Therefore, <math> AG=DE=3 </math>, and <math> GB=4-3=1 </math>. By the same logic, <math> GBHE </math> is also a rectangle, and <math> EH=GB=1 </math>. <math> BH=CI </math> since they're both altitudes to a trapezoid, and <math> BE=CF </math> since the trapezoid is isosceles. Therefore, <math> \triangle BHE \cong \triangle CIF </math> by HL [[congruence]], so <math> IF=EH=1 </math>. Also, <math> BCIH </math> is a rectangle from <math> 4 </math> right angles, and <math> HI=BC=5 </math>. Therefore, <math> EF=EH+HI+IF=1+5+1=\boxed{7} </math>.
  
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1984|num-b=3|num-a=5}}
 
{{AHSME box|year=1984|num-b=3|num-a=5}}
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{{MAA Notice}}

Latest revision as of 12:26, 16 July 2024

Problem

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] defaultpen(linewidth(0.7)+fontsize(10)); pair D=origin, E=(3,0), F=(10,0), X=(12,0), Y=(12,1), A=(0,1), B=(4,1), C=(9,1), O=circumcenter(B,C,F), G=foot(E,A,C), H=foot(B,D,F), I=foot(C,D,F); draw(D--X--Y--A--cycle); draw(Circle(O, abs(O-C))); label("$A$", A, NW); label("$B$", B, N); label("$C$", C, NE); label("$D$", D, SW); label("$E$", E, S); label("$F$", F, S); label("$G$", G, N); label("$H$", H, S); label("$I$", I, S); label("4", (2,0.85), N); label("3", D--E, S); label("5", (6.5,0.85), N); draw(E--G^^H--B^^I--C, linetype("4 4")); [/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, $\triangle BHE \cong \triangle CIF$ 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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