AMC12 2002 B

AMC12 2002 B · Q23

AMC12 2002 B · Q23. It mainly tests Triangles (properties), Trigonometry (basic).

In $\triangle ABC$, we have $AB = 1$ and $AC = 2$. Side $\overline{BC}$ and the median from $A$ to $\overline{BC}$ have the same length. What is $BC$?

在 $\triangle ABC$ 中，$AB = 1$ 且 $AC = 2$。边 $\overline{BC}$ 与从 $A$ 到 $\overline{BC}$ 的中线长度相同。求 $BC$。

(A) \frac{1+\sqrt{2}}{2} \frac{1+\sqrt{2}}{2}

(B) \frac{1+\sqrt{3}}{2} \frac{1+\sqrt{3}}{2}

(D) \frac{3}{2} \frac{3}{2}

(E) \sqrt{3} \sqrt{3}

Answer

Correct choice: (C)

正确答案：（C）

Solution

Let $D$ be the foot of the altitude from $A$ to $\overline{BC}$ extended past $B$. Let $AD = x$ and $BD = y$. Using the Pythagorean Theorem, we obtain the equations \begin{align*} x^2 + y^2 = 1 \hspace{0.5cm}(1)\\ x^2 + y^2 + 2ya + a^2 = 4a^2 \hspace{0.5cm}(2)\\ x^2 + y^2 + 4ya + 4a^2 = 4 \hspace{0.5cm}(3) \end{align*} Subtracting $(1)$ equation from $(2)$ and $(3)$, we get \begin{align*} 2ya + a^2 = 4a^2 - 1 \hspace{0.5cm}(4)\\ 4ya + 4a^2 = 3 \hspace{0.5cm}(5) \end{align*} Then, subtracting $2 \times (4)$ from $(5)$ and rearranging, we get $10a^2 = 5$, so $BC = 2a = \sqrt{2}\Rightarrow \boxed{\mathrm{(C)}}$ Error creating thumbnail: Unable to save thumbnail to destination

设 $D$ 为从 $A$ 向 $\overline{BC}$ 作高的垂足，并将 $\overline{BC}$ 延长过 $B$。令 $AD = x$，$BD = y$。利用勾股定理，得到方程 \begin{align*} x^2 + y^2 = 1 \hspace{0.5cm}(1)\\ x^2 + y^2 + 2ya + a^2 = 4a^2 \hspace{0.5cm}(2)\\ x^2 + y^2 + 4ya + 4a^2 = 4 \hspace{0.5cm}(3) \end{align*} 用 $(2)$ 与 $(3)$ 分别减去 $(1)$，得 \begin{align*} 2ya + a^2 = 4a^2 - 1 \hspace{0.5cm}(4)\\ 4ya + 4a^2 = 3 \hspace{0.5cm}(5) \end{align*} 再用 $(5)$ 减去 $2 \times (4)$ 并整理，得 $10a^2 = 5$，所以 $BC = 2a = \sqrt{2}\Rightarrow \boxed{\mathrm{(C)}}$ Error creating thumbnail: Unable to save thumbnail to destination

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