1996 AMC 8 第 17 题

先试着解答 1996 AMC 8 第 17 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 1996 AMC 8 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

17.

图形 OPQROPQR 是正方形。点 OO 是原点,点 QQ 的坐标是 (2,2)(2, 2)。点 TT 的坐标应是多少,才能使三角形 PQTPQT 的面积等于正方形 OPQROPQR 的面积?
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)
答案:C
知识点:坐标几何三角形面积
小提示:
正方形边长是 22,面积为 44;同时 P=(2,0)P = (2, 0),底边 PQPQ 是长度为 22 的竖直线段。
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
大提示:
若 T=(t,0)T = (t, 0),则它到直线 PQPQ 的水平距离为 ∣2−t∣\lvert 2 - t\rvert。
If T=(t,0),T = (t, 0), its horizontal distance from line PQPQ is ∣2−t∣\lvert 2 - t\rvert
解答:
因为 OPQROPQR 是正方形,且 O=(0,0)O = (0, 0)、Q=(2,2)Q = (2, 2),所以 P=(2,0)P = (2, 0)、R=(0,2)R = (0, 2),正方形面积为 22=42^2 = 4。
三角形 PQTPQT 的竖直底边 PQPQ 长为 22。若 T=(t,0)T = (t, 0) 位于 xx 轴上,它的面积是 12⋅2⋅∣2−t∣=∣2−t∣\tfrac12 \cdot 2 \cdot \lvert 2 - t\rvert = \lvert 2 - t\rvert。令 ∣2−t∣=4\lvert 2 - t\rvert = 4,得到 t=−2t = -2 或 t=6t = 6。选项中只有 t=−2t = -2,所以 T=(−2,0)T = (-2, 0)。
所以正确答案是 C。
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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