Question

在 $\triangle ABC$ 中，中线 $\overline{AD}$ 和 $\overline{BE}$ 相交于 $G$ ，且 $\triangle AGE$ 是等边三角形。那么 $\cos(C)$ 可写为 $\dfrac{m\sqrt p}{n}$ ，其中 $m$ 和 $n$ 是互质正整数， $p$ 是不被任何质数平方整除的正整数。求 $m + n + p$ 。

In $\triangle ABC$ medians $\overline{AD}$ and $\overline{BE}$ intersect at $G$ and $\triangle AGE$ is equilateral. Then $\cos(C)$ can be written as $\dfrac{m\sqrt p}{n},$ where $m$ and $n$ are relatively prime positive integers and $p$ is a positive integer not divisible by the square of any prime. What is $m + n + p?$

$44$

$48$

$52$

$56$

$60$

答案：A

知识点：中线（几何）重心余弦定理

难度评级：2020

解答：

令 $a = BC$ $b = CA$ $c = AB$ 由于 $E$ 是 $AC$ 的中点， $AE = \tfrac{b}{2}$ 重心性质给出 $AG = \tfrac23 m_a$ 和 $GE = \tfrac13 m_b$ 其中 $m_a, m_b$ 分别是从 $A$ 和 $B$ 出发的中线长。

等边三角形 $\triangle AGE$ 说明 $AG = GE = AE$ 。由 $\tfrac23 m_a = \tfrac{b}{2}$ 得 $m_a = \tfrac34 b$ ，结合 $m_a^2 = \tfrac{2b^2 + 2c^2 - a^2}{4}$ 得 $2c^2 - a^2 = \tfrac{b^2}{4}$ 。由 $\tfrac13 m_b = \tfrac{b}{2}$ 得 $m_b = \tfrac32 b$ ，所以 $a^2 + c^2 = 5b^2$ 。

解这两个关系得 $c^2 = \tfrac{7b^2}{4}$ 、 $a^2 = \tfrac{13b^2}{4}$ 。取 $b = 2$ 则 $a^2 = 13$ ， $c^2 = 7$ ， $\begin{aligned} \cos C &= \dfrac{a^2 + b^2 - c^2}{2ab} \\ &= \dfrac{13 + 4 - 7}{2 \cdot \sqrt{13} \cdot 2} \\ &= \dfrac{5\sqrt{13}}{26}. \end{aligned}$

因此 $m + n + p = 5 + 26 + 13 = 44$ 。

所以正确答案是 A。

Let $a = BC,$ $b = CA,$ $c = AB.$ Since $E$ is the midpoint of $AC,$ $AE = \tfrac{b}{2}.$ The centroid gives $AG = \tfrac23 m_a$ and $GE = \tfrac13 m_b,$ where $m_a, m_b$ are the medians from $A$ and $B.$

Equilateral $\triangle AGE$ means $AG = GE = AE.$ From $\tfrac23 m_a = \tfrac{b}{2}$ we get $m_a = \tfrac34 b,$ which with $m_a^2 = \tfrac{2b^2 + 2c^2 - a^2}{4}$ gives $2c^2 - a^2 = \tfrac{b^2}{4}.$ From $\tfrac13 m_b = \tfrac{b}{2}$ we get $m_b = \tfrac32 b,$ giving $a^2 + c^2 = 5b^2.$

Solving, $c^2 = \tfrac{7b^2}{4}$ and $a^2 = \tfrac{13b^2}{4}.$ Taking $b = 2$ gives $a^2 = 13,$ $c^2 = 7,$ so $\begin{aligned} \cos C &= \dfrac{a^2 + b^2 - c^2}{2ab} \\ &= \dfrac{13 + 4 - 7}{2 \cdot \sqrt{13} \cdot 2} \\ &= \dfrac{5\sqrt{13}}{26}. \end{aligned}$

Then $m + n + p = 5 + 26 + 13 = 44.$

Thus, the correct answer is A.

2022 AMC 12B 第 19 题

答案：A

难度评级：2020

解答：

其他年份的第 19 题