Difference between revisions of "1974 AHSME Problems/Problem 25"

Latest revision as of 11:44, 5 July 2013

Problem

In parallelogram $ABCD$ of the accompanying diagram, line $DP$ is drawn bisecting $BC$ at $N$ and meeting $AB$ (extended) at $P$ . From vertex $C$ , line $CQ$ is drawn bisecting side $AD$ at $M$ and meeting $AB$ (extended) at $Q$ . Lines $DP$ and $CQ$ meet at $O$ . If the area of parallelogram $ABCD$ is $k$ , then the area of the triangle $QPO$ is equal to

$[asy] size((400)); draw((0,0)--(5,0)--(6,3)--(1,3)--cycle); draw((6,3)--(-5,0)--(10,0)--(1,3)); label("A", (0,0), S); label("B", (5,0), S); label("C", (6,3), NE); label("D", (1,3), NW); label("P", (10,0), E); label("Q", (-5,0), W); label("M", (.5,1.5), NW); label("N", (5.65, 1.5), NE); label("O", (3.4,1.75));[/asy]$

$\mathrm{(A)\ } k \qquad \mathrm{(B) \ }\frac{6k}{5} \qquad \mathrm{(C) \ } \frac{9k}{8} \qquad \mathrm{(D) \ } \frac{5k}{4} \qquad \mathrm{(E) \ }2k$

Solution

Note that $[QPO]=[QAM]+[PBN]+[AMONB]=[AMONB]+[MDC]+[NCD]$

$=[AMONB]+[MDC]+[NOC]+[DOC]=[ABCD]+[DOC]=k+[DOC].$

Also, note that $DCNM$ is a parallelogram, and so $[DOC]=\frac{1}{4}[DCNM]=\frac{1}{8}[ABCD]=\frac{k}{8}$ .

Therefore, $[QPO]=k+\frac{k}{8}=\frac{9k}{8}, \boxed{\text{C}}$ .

1974 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Problem 26
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.

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 {{AHSME box|year=1974|num-b=24|num-a=26}}
 [[Category:Introductory Geometry Problems]]
+{{MAA Notice}}

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Difference between revisions of "1974 AHSME Problems/Problem 25"

Latest revision as of 11:44, 5 July 2013

Problem

Solution

See Also