1996 AJHSME Problems/Problem 17

Problem

Figure $OPQR$ is a square. Point $O$ is the origin, and point $Q$ has coordinates (2,2). What are the coordinates for $T$ so that the area of triangle $PQT$ equals the area of square $OPQR$ ?

$[asy] pair O,P,Q,R,T; O = (0,0); P = (2,0); Q = (2,2); R = (0,2); T = (-4,0); draw((-5,0)--(3,0)); draw((0,-1)--(0,3)); draw(P--Q--R); draw((-0.2,-0.8)--(0,-1)--(0.2,-0.8)); draw((-0.2,2.8)--(0,3)--(0.2,2.8)); draw((-4.8,-0.2)--(-5,0)--(-4.8,0.2)); draw((2.8,-0.2)--(3,0)--(2.8,0.2)); draw(Q--T); label("$O$",O,SW); label("$P$",P,S); label("$Q$",Q,NE); label("$R$",R,W); label("$T$",T,S); [/asy]$

NOT TO SCALE

$\text{(A)}\ (-6,0) \qquad \text{(B)}\ (-4,0) \qquad \text{(C)}\ (-2,0) \qquad \text{(D)}\ (2,0) \qquad \text{(E)}\ (4,0)$

Solution

The area of $\square OPQR$ is $2^2 = 4$ .

The area of $\triangle PQT$ is $\frac{1}{2}bh = \frac{1}{2}\cdot PT\cdot PQ$ .

If we set the areas equal, the area of $\traingle PQT$ (Error compiling LaTeX. Unknown error_msg) is $4$ . Also, note that $PQ=2$ . Plugging those in, we get:

$4 = \frac{1}{2} \cdot PT \cdot 2$

$PT = 4$

If $PT = 4$ , and $PO = 2$ , then $OT = 2$ , and $T$ must be $2$ units to the left of the origin. This would be $(-2,0)$ , giving answer $\boxed{C}$ .

1996 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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1996 AJHSME Problems/Problem 17

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