# Difference between revisions of "1964 AHSME Problems/Problem 17"

## Problem 17

Given the distinct points $P(x_1, y_1), Q(x_2, y_2)$ and $R(x_1+x_2, y_1+y_2)$. Line segments are drawn connecting these points to each other and to the origin $O$. Of the three possibilities: (1) parallelogram (2) straight line (3) trapezoid, figure $OPRQ$, depending upon the location of the points $P, Q$, and $R$, can be: $\textbf{(A)}\ \text{(1) only}\qquad \textbf{(B)}\ \text{(2) only}\qquad \textbf{(C)}\ \text{(3) only}\qquad \textbf{(D)}\ \text{(1) or (2) only}\qquad \textbf{(E)}\ \text{all three}$

# Solution

Using vector addition can help solve this problem quickly. Note that algebraically, adding $\overrightarrow{OP}$ to $\overrightarrow{OQ}$ will give $\overrightarrow{OR}$. One method of vector addition is literally known as the "parallelogram rule" - if you are given $\overrightarrow{OP}$ and $\overrightarrow{OQ}$, to find $\overrightarrow{OR}$, you can literally draw a parallelogram, making a line though $P$ parallel to $OQ$, and a line through $Q$ parallel to $OP$. The intersection of those lines will give the fourth point $R$, and that fourth point will form a parallelogram with $O, P, Q$.

Thus, $1$ is a possibility. Case $2$ is also a possibility, if $O, P, Q$ are collinear, then $R$ is also on that line.

Since $OP \parallel QR$ and $PQ \parallel RO$, which can be seen from either the prior reasoning or by examining slopes, the figure can never be a trapezoid, which requires exactly one of parallel sides.

Thus, the answer is $\boxed{\textbf{(D)}}$

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