AMC12 2013 B

AMC12 2013 B · Q24

AMC12 2013 B · Q24. It mainly tests Angle chasing, Triangles (properties).

Let $ABC$ be a triangle where $M$ is the midpoint of $\overline{AC}$, and $\overline{CN}$ is the angle bisector of $\angle ACB$ with $N$ on $\overline{AB}$. Let $X$ be the intersection of the median $\overline{BM}$ and the bisector $\overline{CN}$. In addition $\triangle BXN$ is equilateral and $AC=2$. What is $BN^2$?

设$ABC$为一三角形，$M$为$\overline{AC}$的中点，$\overline{CN}$为$\angle ACB$的角平分线，且$N$在$\overline{AB}$上。设$X$为中线$\overline{BM}$与角平分线$\overline{CN}$的交点。另有$\triangle BXN$为等边三角形，且$AC=2$。求$BN^2$。

(A) \frac{10-6\sqrt{2}}{7} \frac{10-6\sqrt{2}}{7}

(B) \frac{2}{9} \frac{2}{9}

(D) \frac{\sqrt{2}}{6} \frac{\sqrt{2}}{6}

(E) \frac{3\sqrt{3}-4}{5} \frac{3\sqrt{3}-4}{5}

Answer

Correct choice: (A)

正确答案：（A）

Solution

Answer (A): Let $\alpha=\angle ACN=\angle NCB$ and $x=BN$. Because $\triangle BXN$ is equilateral it follows that $\angle BXC=\angle CNA=120^\circ$, $\angle CBX=\angle BAC=60^\circ-\alpha$, and $\angle CBA=\angle BMC=120^\circ-\alpha$. Thus $\triangle ABC\sim\triangle BMC$ and $\triangle ANC\sim\triangle BXC$. Then $$ \frac{BC}{2}=\frac{BC}{AC}=\frac{MC}{BC}=\frac{1}{BC}, $$ so $BC=\sqrt2$; and $$ \frac{CX+x}{2}=\frac{CN}{AC}=\frac{CX}{BC}=\frac{CX}{\sqrt2}, $$ so $CX=(\sqrt2+1)x$. Let $P$ be the midpoint of $XN$. Because $\triangle BXN$ is equilateral, the triangle $BPC$ is a right triangle with $\angle BPC=90^\circ$. Then by the Pythagorean Theorem, $$ 2=BC^2=CP^2+PB^2=(CX+XP)^2+PB^2 $$ $$ =\left(CX+\frac12 BN\right)^2+\left(\frac{\sqrt3}{2}BN\right)^2 $$ $$ =\left(\sqrt2+\frac32\right)^2x^2+\left(\frac{\sqrt3}{2}\right)^2x^2=(5+3\sqrt2)x^2. $$ Therefore $$ x^2=\frac{2}{5+3\sqrt2}=\frac{10-6\sqrt2}{7}. $$ OR Establish as in the first solution that $CX=(\sqrt2+1)x$. Then the Law of Cosines applied to $\triangle BCX$ gives $$ 2=BC^2=BX^2+CX^2-2BX\cdot CX\cdot\cos(120^\circ) $$ $$ =x^2+(1+\sqrt2)^2x^2+(1+\sqrt2)x^2 $$ $$ =(5+3\sqrt2)x^2, $$ and solving for $x^2$ gives the requested answer.

答案（A）：令 $\alpha=\angle ACN=\angle NCB$，并令 $x=BN$。因为 $\triangle BXN$ 是等边三角形，所以有 $\angle BXC=\angle CNA=120^\circ$，$\angle CBX=\angle BAC=60^\circ-\alpha$，且 $\angle CBA=\angle BMC=120^\circ-\alpha$。因此 $\triangle ABC\sim\triangle BMC$ 且 $\triangle ANC\sim\triangle BXC$。于是 $$ \frac{BC}{2}=\frac{BC}{AC}=\frac{MC}{BC}=\frac{1}{BC}, $$ 所以 $BC=\sqrt2$；并且 $$ \frac{CX+x}{2}=\frac{CN}{AC}=\frac{CX}{BC}=\frac{CX}{\sqrt2}, $$ 所以 $CX=(\sqrt2+1)x$。令 $P$ 为 $XN$ 的中点。由于 $\triangle BXN$ 是等边三角形，三角形 $BPC$ 为直角三角形，且 $\angle BPC=90^\circ$。由勾股定理， $$ 2=BC^2=CP^2+PB^2=(CX+XP)^2+PB^2 $$ $$ =\left(CX+\frac12 BN\right)^2+\left(\frac{\sqrt3}{2}BN\right)^2 $$ $$ =\left(\sqrt2+\frac32\right)^2x^2+\left(\frac{\sqrt3}{2}\right)^2x^2=(5+3\sqrt2)x^2。 $$ 因此 $$ x^2=\frac{2}{5+3\sqrt2}=\frac{10-6\sqrt2}{7}. $$ 或者同第一种解法可得 $CX=(\sqrt2+1)x$。对 $\triangle BCX$ 应用余弦定理： $$ 2=BC^2=BX^2+CX^2-2BX\cdot CX\cdot\cos(120^\circ) $$ $$ =x^2+(1+\sqrt2)^2x^2+(1+\sqrt2)x^2 $$ $$ =(5+3\sqrt2)x^2， $$ 解出 $x^2$ 即为所求答案。

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