1986 AMC 12 第 26 题

先试着解答 1986 AMC 12 第 26 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 1986 AMC 12 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。

26.

要在坐标平面内构造一个直角三角形,使它的两条直角边分别平行于 xx 轴和 yy 轴,并使连接直角顶点与两条直角边中点的中线分别位于直线 y=3x+1y=3x+1 和 y=mx+2y=mx+2 上。能使这样的三角形存在的不同常数 mm 的个数是
It is desired to construct a right triangle in the coordinate plane so that its legs are parallel to the xx and yy axes and so that the medians to the midpoints of the legs lie on the lines y=3x+1y=3x+1 and y=mx+2.y=mx+2. The number of different constants mm for which such a triangle exists is
00
11
22
33
多于 33
more than 33
答案:C
知识点:坐标几何median斜率construction
小提示:
将直角边平行于坐标轴的直角三角形放在便于计算的位置,并求出通向两条直角边中点的两条中线的斜率
Place an axis-aligned right triangle at convenient coordinates and compute the slopes of the two medians to its legs
大提示:
两条中线斜率的比值为 44;注意给定的任一条直线都可能是斜率绝对值较大的那一条
The two slopes differ by a factor of 4;4; remember that either given line could be the steeper one
解答:
将直角顶点置于 (0,0)(0,0),另两个顶点分别置于 (a,0)(a,0) 和 (0,b)(0,b)。通向两条直角边中点的中线斜率分别为 −2ba-\frac{2b}{a} 和 −b2a-\frac{b}{2a},二者之比为 44。因此,若一条中线的斜率为 33,另一条的斜率可以为 1212 或 34\frac{3}{4}。两种情况都能实现:先选取具有所需两条斜率的三角形,再将它的重心平移到两条给定直线的交点即可。因此 mm 有两个可能值。
因此正确答案是 C。
Place the right-angle vertex at (0,0)(0,0) and the other vertices at (a,0)(a,0) and (0,b).(0,b). The medians to the legs have slopes −2ba-\frac{2b}{a} and −b2a,-\frac{b}{2a}, whose ratio is 4.4. Therefore, if one median has slope 3,3, the other can have slope 1212 or 34.\frac{3}{4}. Both occur: choose a triangle with the required pair of slopes and translate its centroid to the intersection of the two specified lines. Thus there are two possible values of m.m.
Therefore the correct answer is C.
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