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: | ||
Line 20: | Line 22: | ||
==Solution== | ==Solution== | ||
<asy> | <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); | |
− | draw( | + | label("$D$", D, SW); |
− | + | label("$E$", E, S); | |
− | draw(( | + | label("$F$", F, S); |
− | label("$A$", | + | label("$G$", G, N); |
− | label("$B$", | + | label("$H$", H, S); |
− | label("$C$", | + | label("$I$", I, S); |
− | label("$D$", | + | |
− | label("$E$", | + | label("4", (2,0.85), N); |
− | label("$F$", | + | label("3", D--E, S); |
− | label("$G$", | + | label("5", (6.5,0.85), N); |
− | label("$H$", | + | draw(E--G^^H--B^^I--C, linetype("4 4")); |
− | label("$I$", | ||
− | label(" | ||
− | label(" | ||
− | label(" | ||
</asy> | </asy> | ||
− | 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> \ | + | 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}} | ||
+ | {{MAA Notice}} |
Latest revision as of 13:26, 16 July 2024
Problem
A rectangle intersects a circle as shown: ,
, and
. Then
equals:
Solution
Draw and
, forming a trapezoid. Since it's cyclic, this trapezoid must be isosceles. Also, drop altitudes from
to
,
to
, and
to
, and let the feet of these altitudes be
,
, and
respectively.
is a rectangle since it has
right angles. Therefore,
, and
. By the same logic,
is also a rectangle, and
.
since they're both altitudes to a trapezoid, and
since the trapezoid is isosceles. Therefore,
by HL congruence, so
. Also,
is a rectangle from
right angles, and
. Therefore,
.
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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