Difference between revisions of "1966 AHSME Problems/Problem 32"

Latest revision as of 01:34, 15 September 2014

Problem

Let $M$ be the midpoint of side $AB$ of triangle $ABC$ . Let $P$ be a point on $AB$ between $A$ and $M$ , and let $MD$ be drawn parallel to $PC$ and intersecting $BC$ at $D$ . If the ratio of the area of triangle $BPD$ to that of triangle $ABC$ is denoted by $r$ , then

$\text{(A) } \frac{1}{2}<r<1 \text{, depending upon the position of P} \\ \text{(B) } r=\frac{1}{2} \text{, independent of the position of P} \\ \text{(C) } \frac{1}{2} \le r <1 \text{, depending upon the position of P} \\ \text{(D) } \frac{1}{3}<r<\frac{2}{3} \text{, depending upon the position of P}\\ \text{(E) } r=\frac{1}{3} \text{, independent of the position of P}$

Solution

$\fbox{B}$

1966 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 31	Followed by Problem 33
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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 == Solution ==
+<math>\fbox{B}</math>
 == See also ==

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Difference between revisions of "1966 AHSME Problems/Problem 32"

Latest revision as of 01:34, 15 September 2014

Problem

Solution

See also