Difference between revisions of "1989 AHSME Problems/Problem 4"
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<math>XY=AB=4</math> and <math>DX=YC=3</math>, hence <math>DY=7\implies DF=14\implies CF=\boxed{4.0}</math>. | <math>XY=AB=4</math> and <math>DX=YC=3</math>, hence <math>DY=7\implies DF=14\implies CF=\boxed{4.0}</math>. | ||
| + | |||
| + | == See also == | ||
| + | {{AHSME box|year=1989|num-b=3|num-a=5}} | ||
| + | |||
| + | [[Category: Introductory Geometry Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 06:42, 22 October 2014
Problem
In the figure, 
is an isosceles trapezoid with side lengths 
, 
, and 
. The point 
is on 
and 
is the midpoint of hypotenuse 
in right triangle 
. Then 
![[asy] defaultpen(fontsize(10)); pair D=origin, A=(3,4), B=(7,4), C=(10,0), E=(14,8), F=(14,0); draw(B--C--F--E--B--A--D--B^^C--D, linewidth(0.7)); dot(A^^B^^C^^D^^E^^F); pair point=(5,3); label("$A$", A, N); label("$B$", B, N); label("$C$", C, S); label("$D$", D, S); label("$E$", E, dir(point--E)); label("$F$", F, dir(point--F)); markscalefactor=0.05; draw(rightanglemark(E,F,D), linewidth(0.7));[/asy]](http://latex.artofproblemsolving.com/8/5/3/853469c3a6d4a7177bed8ab01d73e9d527b8f5ab.png)
Solution
Drop perpendiculars from 
and ; then the triangle 
is similar to but with corresponding sides of half the length.
![[asy] defaultpen(fontsize(10)); pair D=origin, A=(3,4), B=(7,4), C=(10,0), E=(14,8), F=(14,0); draw(B--C--F--E--B--A--D--B^^C--D, linewidth(0.7)); dot(A^^B^^C^^D^^E^^F); pair point=(5,3); label("$A$", A, N); label("$B$", B, N); label("$C$", C, S); label("$D$", D, S); label("$E$", E, dir(point--E)); label("$F$", F, dir(point--F)); markscalefactor=0.05; draw(rightanglemark(E,F,D), linewidth(0.7)); pair X=(3,0), Y=(7,0); draw(A--X^^B--Y,dotted); dot(X^^Y); label("$X$", X, S); label("$Y$", Y, S); [/asy]](http://latex.artofproblemsolving.com/f/7/5/f7569f21adae4ca88e5a41a8abd568a5e6b88685.png)
and 
, hence 
.
See also
| 1989 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
