Question

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:

$\dfrac12\lt r\lt1$ depending upon the position of $P$

$r=\dfrac12$ independent of the position of $P$

$\dfrac12\leq r\lt1$ depending upon the position of $P$

$\dfrac13\lt r\lt\dfrac23$ depending upon the position of $P$

$r=\dfrac13$ independent of the position of $P$

Answer: B

Concepts:area ratio median (geometry)parallel lines

Difficulty rating: 1650

Small Hint:

Triangles $MDP$ and $MDC$ have equal bases on parallel lines

Big Hint:

Add their areas to $[BMD]$ and use that $CM$ is a median

Solution:

Since $MD\parallel PC,$ triangles $MDP$ and $MDC$ have the same base $MD$ and equal altitudes, so their areas are equal. Hence $\begin{aligned} [BPD] &=[BMD]+[MDP]\\ &=[BMD]+[MDC]\\ &=[BMC]. \end{aligned}$ Because $CM$ is a median, $[BMC]=\frac{[ABC]}{2}.$ Thus $r=\frac{1}{2}$ for every allowed $P.$

Therefore, the correct answer is B.

1966 AMC 12 Problem 32

Answer: B

Difficulty rating: 1650

Small Hint:

Big Hint:

Solution:

Problem 32 in Other Years