Difference between revisions of "1998 AHSME Problems/Problem 8"
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== Solution 3 == | == Solution 3 == | ||
− | Divide the | + | Divide the pentagon into 2 small congruent trapezoids by extending the common shorter base of the 2 larger trapezoids. |
− | Since each of the smaller trapezoids has its area half each of the larger trapezoids, and each of them has a base | + | Since each of the smaller trapezoids has its area half each of the larger trapezoids, and each of them has a base <math>\frac{1}{2}</math>, we have |
− | <cmath>b_{large}+\frac{1}{2}=2(b_{small}+ \frac{1}{2}</cmath> | + | <cmath>b_{large}+\frac{1}{2}=2(b_{small}+ \frac{1}{2})</cmath> |
− | <cmath>x+\frac{1}{2}=2((1-x)+ | + | <cmath>x+\frac{1}{2}=2((1-x)+\frac{1}{2})</cmath> |
− | <cmath>x=\frac{5}{6}</cmath> | + | <cmath>x=\frac{5}{6}\boxed{D}</cmath> |
+ | |||
+ | ~ Nafer | ||
== See also == | == See also == |
Latest revision as of 11:28, 2 December 2019
Problem
A square with sides of length is divided into two congruent trapezoids and a pentagon, which have equal areas, by joining the center of the square with points on three of the sides, as shown. Find
, the length of the longer parallel side of each trapezoid.
![[asy] pointpen = black; pathpen = black; D(unitsquare); D((0,0)); D((1,0)); D((1,1)); D((0,1)); D(D((.5,.5))--D((1,.5))); D(D((.17,1))--(.5,.5)--D((.17,0))); MP("x",(.58,1),N); [/asy]](http://latex.artofproblemsolving.com/6/d/6/6d64e011eb1d7812da21a5f3cd62791c42186299.png)
Solution
Solution 1
![[asy] pointpen = black; pathpen = black; D(unitsquare); D((0,0)); D((1,0)); D((1,1)); D((0,1)); D(D((.5,.5))--D((1,.5))); D(D((.17,1))--(.5,.5)--D((.5,1)));D(D((1-.17,1))--(.5,.5)--D((.17,0))); D((.17,1)--(.17,0));D((1-.17,1)--(1-.17,.5));D((0,.5)--(.5,.5)); MP("x",(.58,1),N); MP("I",(.17/2,.25),(0,0));MP("I",(.17/2,.75),(0,0));MP("I",(1-.17/2,.75),(0,0));MP("II",(.5-.17,.4),(0,0));MP("II",(.5-.17,.6),(0,0));MP("II",(.5-.17,.9),(0,0));MP("II",(.5+.17,.9),(0,0));MP("II",(.5+.17,.6),(0,0)); [/asy]](http://latex.artofproblemsolving.com/9/1/f/91f3434c146d8b8a3545dd6f748035fcf194f9f1.png)
Then . Let the shorter side of
be
and the base of
be
such that
; then
implies that
, and since
it follows that
and
.
Solution 2
The area of the trapezoid is , and the shorter base and height are both
. Therefore,
Solution 3
Divide the pentagon into 2 small congruent trapezoids by extending the common shorter base of the 2 larger trapezoids.
Since each of the smaller trapezoids has its area half each of the larger trapezoids, and each of them has a base , we have
~ Nafer
See also
1998 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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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