AMC12 2022 B

AMC12 2022 B · Q19

AMC12 2022 B · Q19. It mainly tests Triangles (properties), Trigonometry (basic).

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 $m+n+p$?

在$\triangle ABC$中，中线$\overline{AD}$与$\overline{BE}$相交于$G$，且$\triangle AGE$是等边三角形。则$\cos(C)$可表示为$\dfrac{m\sqrt{p}}{n}$，其中$m,n$为互质的正整数，$p$为不被任何素数的平方整除的正整数。求$m+n+p$。

(A) 44 44

(B) 48 48

(D) 56 56

(E) 60 60

Answer

Correct choice: (C)

正确答案：（C）

Solution

Let $AG=AE=EG=2x$. Since $E$ is the midpoint of $\overline{AC}$, we must have $EC=2x$. Since the centroid splits the median in a $2:1$ ratio, $GD=x$ and $BG=4x$. Applying Law of Cosines on $\triangle ADC$ and $\triangle{}AGB$ yields $AB=\sqrt{28}x$ and $CD=BD=\sqrt{13}x$. Finally, applying Law of Cosines on $\triangle ABC$ yields $\cos(C)=\frac{5}{2\sqrt{13}}=\frac{5\sqrt{13}}{26}$. The requested sum is $5+13+26=44$.

设 $AG=AE=EG=2x$。由于 $E$ 是 $\overline{AC}$ 的中点，故 $EC=2x$。由于质心将中线按 $2:1$ 比例分割，故 $GD=x$ 和 $BG=4x$。在 $\triangle ADC$ 和 $\triangle{}AGB$ 上应用余弦定律，得 $AB=\sqrt{28}x$ 和 $CD=BD=\sqrt{13}x$。最后，在 $\triangle ABC$ 上应用余弦定律，得 $\cos(C)=\frac{5}{2\sqrt{13}}=\frac{5\sqrt{13}}{26}$。所求和为 $5+13+26=44$。

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