AMC12 2025 B

AMC12 2025 B · Q25

AMC12 2025 B · Q25. It mainly tests Triangles (properties), Circle theorems.

Three concentric circles have radii $1$, $2$, and $3$. An equilateral triangle of side length $s$ has one vertex on each circle. What is $s^{2}$?

三个同心圆的半径分别为$1$、$2$和$3$。边长为$s$的等边三角形有一个顶点在每个圆上。求$s^{2}$。

(A) 6 6

(B) \frac{25}{4} \frac{25}{4}

(D) \frac{27}{4} \frac{27}{4}

(E) 7 7

Answer

Correct choice: (E)

正确答案：（E）

Solution

Let $\triangle ABC$ and center $O$ be such that $OA=1, OB=2, OC=3$. Suppose the length of the side of $ABC$ is $s$. Noticing that $OA + OB = OC$, we suspect that $OACB$ is a cyclic quadrilateral. If it was, we could apply Ptolemy's Theorem, which would say that \[CB \cdot OA + OB \cdot AC = OC \cdot AB\] \[s + 2s = 3s.\] Because Ptolemy's is true, $OACB$ is cyclic. Because it's cyclic, $\angle AOB = 180 - \angle BCA = 120$. Applying Law of Cosines on $AOB$, we get \[s^2 = 1^2 + 2^2 - 2 \cdot 1 \cdot 2 \cdot \cos{120}\] \[s^2 = 1 + 4 +2 = \boxed{7}.\]

设$\triangle ABC$和圆心$O$使得$OA=1$、$OB=2$、$OC=3$。设$ABC$的边长为$s$。注意到$OA + OB = OC$，我们怀疑$OACB$是循环四边形。如果是，我们可以应用托勒密定理，即 \[CB \cdot OA + OB \cdot AC = OC \cdot AB\] \[s + 2s = 3s。\] 因为托勒密定理成立，$OACB$是循环的。因为它是循环的，$\angle AOB = 180 - \angle BCA = 120^\circ$。在$AOB$上应用余弦定律，得到 \[s^2 = 1^2 + 2^2 - 2 \cdot 1 \cdot 2 \cdot \cos{120^\circ}\] \[s^2 = 1 + 4 +2 = \boxed{7}。\]

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