Question

设 $\triangle ABC$ 为整数边长三角形，且满足 $\angle B = 2\angle A$ 。这种三角形的最小可能周长是多少？

Let $\triangle ABC$ be a triangle with integer side lengths and the property that $\angle B = 2\angle A.$ What is the least possible perimeter of such a triangle?

$13$

$14$

$15$

$16$

$17$

答案：C

知识点：正弦定理丢番图方程三角不等式

难度评级：2230

解答：

当 $\angle B = 2\angle A$ 时，边长满足 $b^2 = a(a + c)$ ，其中 $a = BC$ 、 $b = CA$ 、 $c = AB$ 。所以 $c = \dfrac{b^2 - a^2}{a}$ 必须是正整数，且三边必须构成非退化三角形。

尝试小值， $b\gt a$ 、 $\angle B=2\angle A\gt\angle A$ 时， $a+b+c=b+\dfrac{b^2}{a}\gt2b$ 。边长 $b\le6$ 构成三角形，且 $b=2,3,4,5,6$ 。其周长为 $15$ ，检查可知没有更小的周长可行。 $a\lt b$ $b^2$ $(a,b)=(4,6)$ ， $c=5$ 。 $(4,5,6)$ $15$ 。 $\begin{gathered} (b,a)=(2,1),(3,1),\\ (4,1),(4,2),(5,1),\\ (6,1),(6,2),(6,3),(6,4). \end{gathered}$

所以正确答案是 C。

When $\angle B = 2\angle A,$ the side lengths satisfy $b^2 = a(a + c),$ where $a = BC,$ $b = CA,$ $c = AB.$ So $c = \dfrac{b^2 - a^2}{a}$ must be a positive integer, and the sides must form a valid triangle.

Also $b\gt a,$ because $\angle B=2\angle A\gt\angle A.$ The perimeter is $a+b+c=b+\dfrac{b^2}{a}\gt2b.$ Therefore a perimeter below $15$ would require $b\le6.$ For $b=2,3,4,5,6,$ the divisors $a\lt b$ of $b^2$ give the possible pairs $\begin{gathered} (b,a)=(2,1),(3,1),\\ (4,1),(4,2),(5,1),\\ (6,1),(6,2),(6,3),(6,4). \end{gathered}$ Substitution gives a degenerate or invalid triangle in every case except $(a,b)=(4,6),$ which gives $c=5.$ Thus $(4,5,6)$ is the first valid triangle, and its perimeter is $15.$

Thus, the correct answer is C.

2024 AMC 12B 第 22 题

答案：C

难度评级：2230

解答：

其他年份的第 22 题