Question

两个不全等的整边等腰三角形有相同的周长和相同的面积。两个三角形的底边长度之比为 $8 : 7$ 。求它们公共周长的最小可能值。

Two noncongruent integer-sided isosceles triangles have the same perimeter and the same area. The ratio of the lengths of the bases of the two triangles is $8 : 7.$ Find the minimum possible value of their common perimeter.

答案

答案：676

知识点：等腰三角形勾股定理丢番图方程

难度评级：3060

解答：

因为整数底边之比为 $8 : 7$ ，它们为 $8a$ 和 $7a$ ，其中 $a$ 是正整数。面积相等使相应高与底边成反比，设为 $7h$ 和 $8h$ 。两条腰分别为 $\sqrt{16a^2 + 49h^2}$ 和 $\sqrt{\frac{49}{4}a^2 + 64h^2}$ ，周长相等给出 $\begin{aligned} &8a + 2\sqrt{16a^2 + 49h^2} \\ &= 7a + 2\sqrt{\tfrac{49}{4}a^2 + 64h^2}. \end{aligned}$

将 $7a$ 移到左边并平方，得 $a\sqrt{16a^2 + 49h^2} = 15h^2 - 4a^2$ ；再次平方并化简，留下 $225h^4 = 169a^2h^2$ ，所以 $h = \frac{13a}{15}$ 。两条腰分别变为 $\sqrt{16a^2 + 49 \cdot \tfrac{169a^2}{225}} = \frac{109a}{15}$ 和 $\sqrt{\tfrac{49}{4}a^2 + 64 \cdot \tfrac{169a^2}{225}} = \frac{233a}{30}.$

为使所有边都是整数， $30$ 必须整除 $a$ 。取 $a = 30$ 得到三角形 $(218, 218, 240)$ 和 $(233, 233, 210)$ ，它们的周长都为 $676$ ，面积都为 $21840$ 。最小公共周长为 $676$ 。

Since the integer bases are in ratio $8 : 7,$ they are $8a$ and $7a$ for a positive integer $a.$ Equal areas make the corresponding altitudes inversely proportional to the bases, say $7h$ and $8h.$ The legs are then $\sqrt{16a^2 + 49h^2}$ and $\sqrt{\frac{49}{4}a^2 + 64h^2},$ and equal perimeters give $\begin{aligned} &8a + 2\sqrt{16a^2 + 49h^2} \\ &= 7a + 2\sqrt{\tfrac{49}{4}a^2 + 64h^2}. \end{aligned}$

Moving $7a$ to the left and squaring yields $a\sqrt{16a^2 + 49h^2} = 15h^2 - 4a^2;$ squaring again and simplifying leaves $225h^4 = 169a^2h^2,$ so $h = \frac{13a}{15}.$ The legs become $\sqrt{16a^2 + 49 \cdot \tfrac{169a^2}{225}} = \frac{109a}{15}$ and $\sqrt{\tfrac{49}{4}a^2 + 64 \cdot \tfrac{169a^2}{225}} = \frac{233a}{30}.$

For all sides to be integers, $30$ must divide $a.$ Taking $a = 30$ gives the triangles $(218, 218, 240)$ and $(233, 233, 210),$ each with perimeter $676$ and area $21840.$ The minimum common perimeter is $676.$

2010 AIME II 第 12 题

答案：676

难度评级：3060

解答：

其他年份的第 12 题