Question

在 $\triangle ABC$ 中，边长均为整数，且 $AB = AC$ 圆 $\omega$ 的圆心为 $\triangle ABC$ 的内心。 $\triangle ABC$ 的一个旁切圆是指位于 $\triangle ABC$ 外部、与三角形一边相切并与另外两边的延长线相切的圆。设与 $\overline{BC}$ 相切的旁切圆与 $\omega$ 内切，另外两个旁切圆都与 $\omega$ 外切。求 $\triangle ABC$ 周长的最小可能值。

In $\triangle ABC,$ the sides have integer lengths and $AB = AC.$ Circle $\omega$ has its center at the incenter of $\triangle ABC.$ An excircle of $\triangle ABC$ is a circle in the exterior of $\triangle ABC$ that is tangent to one side of the triangle and tangent to the extensions of the other two sides. Suppose that the excircle tangent to $\overline{BC}$ is internally tangent to $\omega,$ and the other two excircles are both externally tangent to $\omega.$ Find the minimum possible value of the perimeter of $\triangle ABC.$

答案

答案：20

知识点：内切圆、内心与内切圆半径相切圆坐标几何

难度评级：3160

解答：

设 $BC = a$ ， $AB = AC = b$ 。取 $B = \left(-\frac{a}{2}, 0\right)$ 、 $C = \left(\frac{a}{2}, 0\right)$ 、 $A = (0, h)$ ，其中 $h = \sqrt{b^2 - \frac{a^2}{4}}$ ，于是半周长为 $s = b + \frac{a}{2}$ 面积为 $K = \frac{ah}{2}$ 内切圆半径与旁切圆半径为 $r = \frac{K}{s} = \frac{ah}{a + 2b}$ 、 $r_A = \frac{K}{s - a} = \frac{ah}{2b - a}$ ，以及 $r_B = \frac{K}{s - b} = h$ 内心为 $I = (0, r)$ ， $A$ -旁切圆圆心为 $(0, -r_A)$ 。 $B$ -旁切圆在距离 $B$ 为 $s$ 的位置与直线 $BC$ 相切，即在 $x = b$ 处相切，所以其圆心为 $(b, h)$ 。

与 $A$ -旁切圆内切时，圆心距为 $r + r_A$ ，所以 $\omega$ 的半径 $\rho$ 满足 $\rho - r_A = r + r_A$ ，即 $\rho = r + 2r_A$ 与 $B$ -旁切圆外切要求 $b^2 + (h - r)^2 = (\rho + h)^2$ $= (h + r + 2r_A)^2$ ，整理得 $b^2 = 4(r + r_A)(h + r_A)$ 。由于且并且 $h^2 = \frac{4b^2 - a^2}{4}$ ，条件化为 $b^2 = \frac{8ab^2}{2b - a}$ ，也就是 $2b - a = 8a$ ，所以 $2b = 9a$ 。 $r + r_A = \frac{4abh}{4b^2 - a^2}$ $h + r_A = \frac{2bh}{2b - a},$

若边长为整数，则 $a = 2t$ 、 $b = 9t$ ，其中 $t$ 为正整数，周长为 $20t$ 最小值为 $20$ ，由边长 $9, 9, 2$ 的三角形达到。

Let $BC = a$ and $AB = AC = b.$ Place $B = \left(-\frac{a}{2}, 0\right),$ $C = \left(\frac{a}{2}, 0\right),$ $A = (0, h)$ with $h = \sqrt{b^2 - \frac{a^2}{4}},$ so the semiperimeter is $s = b + \frac{a}{2}$ and the area is $K = \frac{ah}{2}.$ The inradius and exradii are $r = \frac{K}{s} = \frac{ah}{a + 2b},$ $r_A = \frac{K}{s - a} = \frac{ah}{2b - a},$ and $r_B = \frac{K}{s - b} = h.$ The incenter is $I = (0, r)$ and the $A$ -excircle has center $(0, -r_A).$ The $B$ -excircle touches line $BC$ at distance $s$ from $B,$ that is, at $x = b,$ so its center is $(b, h).$

Internal tangency with the $A$ -excircle: the center distance is $r + r_A,$ so the radius $\rho$ of $\omega$ satisfies $\rho - r_A = r + r_A,$ i.e. $\rho = r + 2r_A.$ External tangency with the $B$ -excircle requires $b^2 + (h - r)^2 = (\rho + h)^2$ $= (h + r + 2r_A)^2,$ which rearranges to $b^2 = 4(r + r_A)(h + r_A).$ Since $r + r_A = \frac{4abh}{4b^2 - a^2}$ and $h + r_A = \frac{2bh}{2b - a},$ and $h^2 = \frac{4b^2 - a^2}{4},$ the condition becomes $b^2 = \frac{8ab^2}{2b - a},$ that is, $2b - a = 8a,$ so $2b = 9a.$

For integer sides, $a = 2t$ and $b = 9t$ for a positive integer $t,$ giving perimeter $20t.$ The minimum is $20,$ achieved by the triangle with sides $9, 9, 2.$

2019 AIME I 第 11 题

答案：20

难度评级：3160

解答：

其他年份的第 11 题