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1961 AHSME Problems/Problem 21

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Problem

Medians $AD$ and $CE$ of $\triangle ABC$ intersect in $M$. The midpoint of $AE$ is $N$. Let the area of $\triangle MNE$ be $k$ times the area of $\triangle ABC$. Then $k$ equals:

$\textbf{(A)}\ \frac{1}{6}\qquad \textbf{(B)}\ \frac{1}{8}\qquad \textbf{(C)}\ \frac{1}{9}\qquad \textbf{(D)}\ \frac{1}{12}\qquad \textbf{(E)}\ \frac{1}{16}$

Solution

[asy] draw((0,0)--(120,0)--(40,60)--(0,0)); draw((40,60)--(60,0)); draw((120,0)--(20,30)); draw((160/3,20)--(30,45)); dot((40,60)); label("A",(40,60),N); dot((0,0)); label("B",(0,0),SW); dot((120,0)); label("C",(120,0),SE); dot((60,0)); label("D",(60,0),S); dot((20,30)); label("E",(20,30),NW); dot((160/3,20)); label("M",(160/3,20),NE); dot((30,45)); label("N",(30,45),NW); draw((160/3,20)--(0,0),dotted); [/asy]

Let $[NME] = a$ and $[BMD] = b$. The altitudes and bases of $\triangle ANM$ and $\triangle NME$ are the same, so $[NME] = [ANM] = a$. Since the base of $\triangle BEM$ is twice as long as $\triangle NME$ and the altitudes of both triangles are the same, $[BEM] = 2a$.

Since the altitudes and bases of $\triangle BMD$ and $\triangle CMD$ are the same, $[BMD] = [CMD] = b$. Additionally, since $\triangle AME$ and $\triangle EMB$ have the same area, $\triangle AMC$ and $\triangle BMC$ also have the same area, so $[AMC] = 2b$.

Also, the altitudes and bases of $\triangle ABD$ and $\triangle ADC$ are the same, so the area of the two triangles are the same. Thus, \[a+a+2a+b=2b+b\] \[4a=2b\] \[2a = b\] Thus, the area of $\triangle ABC$ is $12a$, so the area of $\triangle NME$ is $\frac{1}{12}$ the area of $\triangle ABC$. The answer is $\boxed{\textbf{(D)}}$.


See Also

1961 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions


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