AMC10 2017 B

AMC10 2017 B · Q24

AMC10 2017 B · Q24. It mainly tests Polygons, Coordinate geometry.

The vertices of an equilateral triangle lie on the hyperbola $xy = 1$, and a vertex of this hyperbola is the centroid of the triangle. What is the square of the area of the triangle?

一个正三角形的顶点位于双曲线$xy=1$上，且该双曲线的一个顶点是该三角形的质心。求该三角形面积的平方。

(A) 48 48

(B) 60 60

(D) 120 120

(E) 169 169

Answer

Correct choice: (C)

正确答案：（C）

Solution

Assume without loss of generality that two of the vertices of the triangle are on the branch of the hyperbola in the first quadrant. This forces the centroid of the triangle to be the vertex (1, 1) of the hyperbola. Because the vertices of the triangle are equidistant from the centroid, the first-quadrant vertices must be (a, 1/a) and (1/a, a) for some positive number a. By symmetry, the third vertex must be (−1, −1). The distance between the vertex (−1, −1) and the centroid (1, 1) is 2√2, so the altitude of the triangle must be (3/2) · 2√2 = 3√2, which makes the side length of the triangle s = (2/√3) · 3√2 = 2√6. The required area is (√3/4) s² = 6√3. The requested value is (6√3)² = 108.

不失一般性，假设三角形的两个顶点位于第一象限的双曲线支上。这迫使三角形的质心为双曲线顶点(1,1)。因为三角形的顶点到质心的距离相等，第一象限的顶点必须是(a,1/a)和(1/a,a)，其中a为某个正数。由对称性，第三个顶点必须是(-1,-1)。顶点(-1,-1)与质心(1,1)的距离是$2\sqrt{2}$，因此三角形的高必须是(3/2)·$2\sqrt{2}$=$3\sqrt{2}$，这使得边长s=(2/\sqrt{3})·$3\sqrt{2}$=$2\sqrt{6}$。所需的面积是(\sqrt{3}/4)s²=$6\sqrt{3}$。所求值为($6\sqrt{3}$)^2=108。

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