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

(→‎Problem)
(→‎Solution)
Line 20: Line 20:
 
==Solution==
 
==Solution==
 
<asy>
 
<asy>
−
unitsize(2.5cm);
+
defaultpen(linewidth(0.7)+fontsize(10));
−
draw(unitcircle);
+
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((sqrt(2)/2,sqrt(2)/2)--(-2,sqrt(2)/2));
+
draw(D--X--Y--A--cycle);
−
draw((sqrt(2/3),-sqrt(3)/3)--(-2,-sqrt(3)/3));
+
draw(Circle(O, abs(O-C)));
−
draw((-2,sqrt(2)/2)--(-2,-sqrt(3)/3));
+
label("$A$", A, NW);
−
draw((-sqrt(2/3),-sqrt(3)/3)--(-sqrt(2)/2,sqrt(2)/2));
+
label("$B$", B, N);
−
draw((sqrt(2)/2,sqrt(2)/2)--(sqrt(2/3),-sqrt(3)/3));
+
label("$C$", C, NE);
−
draw((-sqrt(2/3),sqrt(2)/2)--(-sqrt(2/3),-sqrt(3)/3));
+
label("$D$", D, SW);
−
draw((-sqrt(2)/2,sqrt(2)/2)--(-sqrt(2)/2,-sqrt(3)/3));
+
label("$E$", E, S);
−
draw((sqrt(2)/2,sqrt(2)/2)--(sqrt(2)/2,-sqrt(3)/3));
+
label("$F$", F, S);
−
label("$A$",(-2,sqrt(2)/2),NW);
+
label("$G$", G, N);
−
label("$B$",(-sqrt(2)/2,sqrt(2)/2),SE);
+
label("$H$", H, S);
−
label("$C$",(sqrt(2)/2,sqrt(2)/2),SW);
+
label("$I$", I, S);
−
label("$D$",(-2,-sqrt(3)/3),SW);
+
 
−
label("$E$",(-sqrt(2/3),-sqrt(3)/3),SW);
+
label("4", (2,0.85), N);
−
label("$F$",(sqrt(2/3),-sqrt(3)/3),SE);
+
label("3", D--E, S);
−
label("$G$",(-sqrt(2/3),sqrt(2)/2),NNW);
+
label("5", (6.5,0.85), N);
−
label("$H$",(-sqrt(2)/2,-sqrt(3)/3),NE);
+
draw(E--G^^H--B^^I--C, linetype("4 4"));
−
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>
 
</asy>
  

Revision as of 15:39, 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] 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, $\triangleBHE\congruent\triangleCIF$ (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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