2017 AMC 12A 第 19 题

先试着解答 2017 AMC 12A 第 19 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 2017 AMC 12A 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。

19.

一个边长为 xx 的正方形内接于边长为 33,44,55 的直角三角形中,使得正方形的一个顶点与三角形的直角顶点重合。另一个边长为 yy 的正方形内接于另一个边长为 33,44,55 的直角三角形中,使得正方形的一条边落在三角形的斜边上。xy\dfrac{x}{y} 是多少?
A square with side length xx is inscribed in a right triangle with sides of length 3,3, 4,4, and 55 so that one vertex of the square coincides with the right-angle vertex of the triangle. A square with side length yy is inscribed in another right triangle with sides of length 3,3, 4,4, and 55 so that one side of the square lies on the hypotenuse of the triangle. What is xy?\dfrac{x}{y}?
1213\dfrac{12}{13}
3537\dfrac{35}{37}
11
3735\dfrac{37}{35}
1312\dfrac{13}{12}
答案:D
知识点:相似直角三角形
小提示:
顶点在直角处的正方形边长为 x=3⋅43+4x=\dfrac{3\cdot4}{3+4}。
A square with its corner at the right angle has side x=3⋅43+4x=\dfrac{3\cdot4}{3+4}
大提示:
若正方形一边在长度为 bb 的底边上,而到底边的高为 hh,则正方形边长为 bhb+h\dfrac{bh}{b+h};用斜边作底
A square with a side on a base of length bb and height hh to that base has side bhb+h;\dfrac{bh}{b+h}; use the hypotenuse as the base
解答:
对第一个正方形,它截出的两个小三角形与整个三角形相似,得到 x3−x=4−xx\dfrac{x}{3-x}=\dfrac{4-x}{x},所以 x=127x=\dfrac{12}{7}。(等价地,直角处内接正方形的边长为 3⋅43+4\dfrac{3\cdot4}{3+4}。)
对第二个正方形,以长度为 55 的斜边作底;到斜边的高为 h=3⋅45=125h=\dfrac{3\cdot4}{5}=\dfrac{12}{5}。若正方形一边在底边 bb 上,高为 hh,则正方形边长为 bhb+h\dfrac{bh}{b+h},所以 y=5⋅1255+125=12375=6037。 y=\dfrac{5\cdot\tfrac{12}{5}}{5+\tfrac{12}{5}}=\dfrac{12}{\tfrac{37}{5}}=\dfrac{60}{37}\text{。}
因此 xy=127⋅3760=3735\dfrac{x}{y}=\dfrac{12}{7}\cdot\dfrac{37}{60}=\dfrac{37}{35}。
所以正确答案是 D。
For the first square, the two smaller triangles it cuts off are similar to the whole triangle, giving x3−x=4−xx,\dfrac{x}{3-x}=\dfrac{4-x}{x}, so x=127.x=\dfrac{12}{7}. (Equivalently, a square in the right angle has side 3⋅43+4.\dfrac{3\cdot4}{3+4}.)
For the second square, take the hypotenuse of length 55 as base; the altitude to it is h=3⋅45=125.h=\dfrac{3\cdot4}{5}=\dfrac{12}{5}. A square with a side on a base bb and height hh has side bhb+h,\dfrac{bh}{b+h}, so y=5⋅1255+125=12375=6037. y=\dfrac{5\cdot\tfrac{12}{5}}{5+\tfrac{12}{5}}=\dfrac{12}{\tfrac{37}{5}}=\dfrac{60}{37}.
Therefore xy=127⋅3760=3735.\dfrac{x}{y}=\dfrac{12}{7}\cdot\dfrac{37}{60}=\dfrac{37}{35}.
Thus, the correct answer is D.
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