1996 AMC 8 Problem 17

Attempt Problem 17 of the 1996 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1996 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

17.

Figure OPQROPQR is a square. Point OO is the origin, and point QQ has coordinates (2,2).(2, 2). What are the coordinates for TT so that the area of triangle PQTPQT equals the area of square OPQR?OPQR?
(−6,0)(-6, 0)
(−4,0)(-4, 0)
(−2,0)(-2, 0)
(2,0)(2, 0)
(4,0)(4, 0)
Answer: C
Concepts:coordinate geometrytriangle area
Small hint:
The square has side 2,2, so its area is 4;4; also P=(2,0)P = (2, 0) and the base PQPQ is vertical with length 22
Big hint:
If T=(t,0),T = (t, 0), its horizontal distance from line PQPQ is ∣2−t∣\lvert 2 - t\rvert
Solution:
Since OPQROPQR is a square with O=(0,0)O = (0, 0) and Q=(2,2),Q = (2, 2), we have P=(2,0)P = (2, 0) and R=(0,2),R = (0, 2), so the area is 22=4.2^2 = 4.
Triangle PQTPQT has vertical base PQPQ of length 2,2, and T=(t,0)T = (t, 0) lies on the xx-axis. Its area is 12⋅2⋅∣2−t∣=∣2−t∣.\tfrac12 \cdot 2 \cdot \lvert 2 - t\rvert = \lvert 2 - t\rvert. Setting ∣2−t∣=4\lvert 2 - t\rvert = 4 gives t=−2t = -2 or t=6.t = 6. Only t=−2t = -2 appears among the choices, so T=(−2,0).T = (-2, 0).
Thus, the correct answer is C .
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