AMC 8 2023 Problem 24

AMC8 2023 年第 24 题

三角形★★★★☆
Isosceles triangle ABCABC has equal side lengths ABAB and BCBC. In the figures below, segments are drawn parallel to AC‾\overline{AC} so that the shaded portions of △ABC\triangle ABC have the same area. The heights of the two unshaded portions are 11 and 5 units, respectively. What is the height hh of △ABC\triangle ABC?
在等腰三角形ABC中，边AB和BC的长度相等．在下图中，由平行于AC的线段形成的△ABC中的两个阴影部分具有相同的面积. 两个空白部分的高度分别为 11和5. 问△ABC的高h是多少?
AMC 8 2023 Problem 24 figure
  1. A.14.6
  2. B.14.8
  3. C.15
  4. D.15.2
  5. E.15.4

正确答案 · Correct Answer

A

解析 · Solution

解法1

常规解法 · 相似面积比列方程 小三角形与大三角形相似，面积按高的平方比。第一图阴影 =[ABC](1−(11h)2)\displaystyle =[ABC]\left(1-(\tfrac{11}{h})^2\right)；第二图阴影 =[ABC](h−5h)2\displaystyle =[ABC]\left(\tfrac{h-5}{h}\right)^2。两者相等：
1−121h2=(h−5)2h21-\frac{121}{h^2}=\frac{(h-5)^2}{h^2}
两边乘 h2h^2：h2−121=h2−10h+25⇒10h=146⇒h=14.6h^2-121=h^2-10h+25\Rightarrow 10h=146\Rightarrow h=14.6。

Solution 1

Elementary · similar-triangle area ratios The smaller triangles are similar to the larger ones, so areas scale as the square of the height. The first gray region is [ABC](1−(11h)2)\displaystyle [ABC]\left(1-(\tfrac{11}{h})^2\right) and the second is [ABC](h−5h)2\displaystyle [ABC]\left(\tfrac{h-5}{h}\right)^2. Setting them equal,
1−121h2=(h−5)2h2.1-\frac{121}{h^2}=\frac{(h-5)^2}{h^2}.
Multiplying by h2h^2: h2−121=h2−10h+25⇒10h=146⇒h=14.6h^2-121=h^2-10h+25\Rightarrow 10h=146\Rightarrow h=14.6.

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