1964 AHSME Problems/Problem 17

Revision as of 20:57, 23 July 2019 by Talkinaway (talk | contribs)

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 $\boxed{\textbf{(D)}}$

See Also

1964 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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 31 32 33 34 35 36 37 38 39 40
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS