Difference between revisions of "1966 AHSME Problems/Problem 11"

Latest revision as of 23:50, 25 June 2018

Problem

The sides of triangle $BAC$ are in the ratio $2:3:4$ . $BD$ is the angle-bisector drawn to the shortest side $AC$ , dividing it into segments $AD$ and $CD$ . If the length of $AC$ is $10$ , then the length of the longer segment of $AC$ is:

$\text{(A)} \ 3\frac{1}{2} \qquad \text{(B)} \ 5 \qquad \text{(C)} \ 5\frac{5}{7} \qquad \text{(D)} \ 6 \qquad \text{(E)} \ 7\frac{1}{2}$

Solution

By the Angle Bisector Theorem, we have $\frac{BA}{AD}=\frac{BC}{CD}$ which implies $\frac{AD}{DC}=\frac{BA}{BC}=\frac{3}{4}$ . So $AC=10=AD+DC=\frac{7}{4}DC$ , and thus $DC=\frac{40}{7}=5\frac{5}{7}$ . Hence $\fbox{C}$

1966 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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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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 == Solution ==
+By the [[Angle Bisector Theorem]], we have <math>\frac{BA}{AD}=\frac{BC}{CD}</math> which implies <math>\frac{AD}{DC}=\frac{BA}{BC}=\frac{3}{4}</math>. So <math>AC=10=AD+DC=\frac{7}{4}DC</math>, and thus <math>DC=\frac{40}{7}=5\frac{5}{7}</math>.
+Hence
 <math>\fbox{C}</math>

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Difference between revisions of "1966 AHSME Problems/Problem 11"

Latest revision as of 23:50, 25 June 2018

Problem

Solution

See also