1961 AHSME Problems/Problem 36

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Problem

In $\triangle ABC$ the median from $A$ is given perpendicular to the median from $B$. If $BC=7$ and $AC=6$, find the length of $AB$.

$\textbf{(A)}\ 4\qquad \textbf{(B)}\ \sqrt{17} \qquad \textbf{(C)}\ 4.25\qquad \textbf{(D)}\ 2\sqrt{5} \qquad \textbf{(E)}\ 4.5$

Solution

[asy]  draw((-16,0)--(8,0)); draw((-16,0)--(16,-24)); draw((16,-24)--(0,24)--(0,-12)); draw((-16,0)--(0,24)); draw((0,2)--(2,2)--(2,0)); draw((0,-12)--(8,0),dotted);  dot((16,-24)); label("C",(16,-24),SE); dot((-16,0)); label("A",(-16,0),W); dot((0,24)); label("B",(0,24),N);  label("3",(8,-18),SW); label("3",(-8,-6),SW); label("3.5",(12,-12),NE); label("3.5",(4,12),NE);  dot((0,-12)); label("M",(0,-12),SW); dot((8,0)); label("N",(8,0),NE); dot((0,0)); label("G",(0,0),NW);  [/asy] By SAS Similarity, $\triangle ABC \sim \triangle MNC$, so $AB \parallel MN$. Thus, by AA Similarity, $\triangle AGB \sim \triangle NGM$.

Let $a = GN$ and $b = GM$, so $AG = 2a$ and $BG = 2b$. By the Pythagorean Theorem, \[4a^2 + b^2 = 9\] \[a^2 + 4b^2 = \frac{49}{4}\] Adding the two equations yields $5a^2 + 5b^2 = \frac{85}{4}$, so $a^2 + b^2 = \frac{17}{4}$. Thus, $MN = \frac{\sqrt{17}}{2}$, so $AB = \sqrt{17}$, which is answer choice $\boxed{\textbf{(B)}}$.

See Also

1961 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 35
Followed by
Problem 37
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