AMC 8 1996 Problem 17

AMC8 1996 年第 17 题

坐标几何★★★☆☆
Figure OPQROPQR is a square. Point OO is the origin, and point QQ has coordinates (2,2). What are the coordinates for TT so that the area of triangle PQTPQT equals the area of square OPQROPQR ?
NOT TO SCALE
图形 OPQROPQR 是一个正方形。点 OO 是原点,点 QQ 的坐标为 (2,2)。点 TT 的坐标应为何值,才能使三角形 PQTPQT 的面积等于正方形 OPQROPQR 的面积?
(图形未按比例绘制)
AMC 8 1996 Problem 17 figure
  1. A.(-6,0)
  2. B.(-4,0)
  3. C.(-2,0)
  4. D.(2,0)
  5. E.(4,0)

正确答案 · Correct Answer

C

解析 · Solution

Solution 1

The area of □OPQR\square OPQR is 22=42^2 = 4.
The area of △PQT\triangle PQT is 12bh=12⋅PT⋅PQ\displaystyle \frac{1}{2} bh = \frac{1}{2} \cdot PT\cdot PQ.
If we set the areas equal, the area of △PQT\triangle PQT is 4. Also, note that PQ=2. Plugging those in, we get:
4=12⋅PT⋅2\displaystyle 4 = \frac{1}{2} \cdot PT \cdot 2, PT = 4
If PT = 4, and PO = 2, then OT = 2, and TT must be 2 units to the left of the origin. This would be (-2,0), giving answer CC.

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