AMC12 2006 B

AMC12 2006 B · Q23

AMC12 2006 B · Q23. It mainly tests Angle chasing, Triangles (properties).

Isosceles $\triangle ABC$ has a right angle at $C$. Point $P$ is inside $\triangle ABC$, such that $PA=11$, $PB=7$, and $PC=6$. Legs $\overline{AC}$ and $\overline{BC}$ have length $s=\sqrt{a+b\sqrt{2}}$, where $a$ and $b$ are positive integers. What is $a+b$?

等腰 $\triangle ABC$ 在 $C$ 处有直角。点 $P$ 在 $\triangle ABC$ 内部，且 $PA=11$，$PB=7$，$PC=6$。直角边 $\overline{AC}$ 和 $\overline{BC}$ 的长度为 $s=\sqrt{a+b\sqrt{2}}$，其中 $a$ 和 $b$ 为正整数。求 $a+b$。

(A) 85 85

(B) 91 91

(D) 121 121

(E) 127 127

Answer

Correct choice: (E)

正确答案：（E）

Solution

Using the Law of Cosines on $\triangle PBC$, we have: \begin{align*} PB^2&=BC^2+PC^2-2\cdot BC\cdot PC\cdot \cos(\alpha) \Rightarrow 49 = 36 + s^2 - 12s\cos(\alpha) \Rightarrow \cos(\alpha) = \dfrac{s^2-13}{12s}. \end{align*} Using the Law of Cosines on $\triangle PAC$, we have: \begin{align*} PA^2&=AC^2+PC^2-2\cdot AC\cdot PC\cdot \cos(90^\circ-\alpha) \Rightarrow 121 = 36 + s^2 - 12s\sin(\alpha) \Rightarrow \sin(\alpha) = \dfrac{s^2-85}{12s}. \end{align*} Now we use $\sin^2(\alpha) + \cos^2(\alpha) = 1$. \begin{align*} \sin^2(\alpha)+\cos^2(\alpha) = 1 &\Rightarrow \frac{s^4-26s^2+169}{144s^2} + \frac{s^4-170s^2+7225}{144s^2} = 1 \\ &\Rightarrow s^4-170s^2+3697 = 0 \\ &\Rightarrow s^2 = \dfrac{170 \pm 84\sqrt{2}}{2} = 85 \pm 42\sqrt2 \end{align*} Note that we know that we want the solution with $s^2 > 85$ since we know that $\sin(\alpha) > 0$. Thus, $a+b=85+42=\boxed{127}$.

在 $\triangle PBC$ 上使用余弦定理，有： \begin{align*} PB^2&=BC^2+PC^2-2\cdot BC\cdot PC\cdot \cos(\alpha) \Rightarrow 49 = 36 + s^2 - 12s\cos(\alpha) \Rightarrow \cos(\alpha) = \dfrac{s^2-13}{12s}. \end{align*} 在 $\triangle PAC$ 上使用余弦定理，有： \begin{align*} PA^2&=AC^2+PC^2-2\cdot AC\cdot PC\cdot \cos(90^\circ-\alpha) \Rightarrow 121 = 36 + s^2 - 12s\sin(\alpha) \Rightarrow \sin(\alpha) = \dfrac{s^2-85}{12s}. \end{align*} 现在使用 $\sin^2(\alpha) + \cos^2(\alpha) = 1$。 \begin{align*} \sin^2(\alpha)+\cos^2(\alpha) = 1 &\Rightarrow \frac{s^4-26s^2+169}{144s^2} + \frac{s^4-170s^2+7225}{144s^2} = 1 \\ &\Rightarrow s^4-170s^2+3697 = 0 \\ &\Rightarrow s^2 = \dfrac{170 \pm 84\sqrt{2}}{2} = 85 \pm 42\sqrt2 \end{align*} 注意我们需要满足 $s^2 > 85$ 的解，因为我们知道 $\sin(\alpha) > 0$。因此 $a+b=85+42=\boxed{127}$。

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