AMC12 2024 B

AMC12 2024 B · Q21

AMC12 2024 B · Q21. It mainly tests Triangles (properties), Pythagorean theorem.

The measures of the smallest angles of three different right triangles sum to $90^\circ$. All three triangles have side lengths that are primitive Pythagorean triples. Two of them are $3-4-5$ and $5-12-13$. What is the perimeter of the third triangle?

三个不同的直角三角形的最小角度的度量和为$90^\circ$。这三个三角形都有原始勾股三元组作为边长。其中两个是$3-4-5$和$5-12-13$。第三条三角形的周长是多少？

(A) 40 40

(B) 126 126

(D) 176 176

(E) 208 208

Answer

Correct choice: (C)

正确答案：（C）

Solution

Let $\alpha$ and $\beta$ be the smallest angles of the $3-4-5$ and $5-12-13$ triangles respectively. We have \[\tan(\alpha)=\frac{3}{4} \text{ and } \tan(\beta)=\frac{5}{12}\] Then \[\tan(\alpha+\beta)=\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4}\cdot\frac{5}{12}}=\frac{56}{33}\] Let $\theta$ be the smallest angle of the third triangle. Consider \[\tan{90^\circ}=\tan((\alpha+\beta)+\theta)=\frac{\frac{56}{33}+\tan{\theta}}{1-\frac{56}{33}\cdot\tan{\theta}}\] In order for this to be undefined, we need \[1-\frac{56}{33}\cdot\tan{\theta}=0\] so \[\tan{\theta}=\frac{33}{56}\] Hence the base side lengths of the third triangle are $33$ and $56$. By the Pythagorean Theorem, the hypotenuse of the third triangle is $65$, so the perimeter is $33+56+65=\boxed{\textbf{(C) }154}$.

设$\alpha$和$\beta$分别为$3-4-5$和$5-12-13$三角形的最小角度。我们有 \[\tan(\alpha)=\frac{3}{4} \text{ and } \tan(\beta)=\frac{5}{12}\] 则 \[\tan(\alpha+\beta)=\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4}\cdot\frac{5}{12}}=\frac{56}{33}\] 设$\theta$为第三条三角形的最小角度。考虑 \[\tan{90^\circ}=\tan((\alpha+\beta)+\theta)=\frac{\frac{56}{33}+\tan{\theta}}{1-\frac{56}{33}\cdot\tan{\theta}}\] 为了使之未定义，我们需要 \[1-\frac{56}{33}\cdot\tan{\theta}=0\] 所以 \[\tan{\theta}=\frac{33}{56}\] 因此第三条三角形的底边长为$33$和$56$。由勾股定理，斜边为$65$，周长为$33+56+65=\boxed{\textbf{(C) }154}$。

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