AMC12 2017 B

AMC12 2017 B · Q24

AMC12 2017 B · Q24. It mainly tests Triangles (properties), Similarity.

Quadrilateral ABCD has right angles at B and C, $\triangle ABC \sim \triangle BCD$, and AB > BC. There is a point E in the interior of ABCD such that $\triangle ABC \sim \triangle CEB$ and the area of $\triangle AED$ is 17 times the area of $\triangle CEB$. What is $\frac{AB}{BC}$?

四边形ABCD在B和C处有直角，$\triangle ABC \sim \triangle BCD$，且AB > BC。存在点E在ABCD内部，使得$\triangle ABC \sim \triangle CEB$，且$\triangle AED$的面积是$\triangle CEB$面积的17倍。求$\frac{AB}{BC}$？

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

(B) 2 + \sqrt{2} 2 + \sqrt{2}

(D) 2 + \sqrt{5} 2 + \sqrt{5}

(E) 1 + 2\sqrt{3} 1 + 2\sqrt{3}

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Let $F$ lie on $\overline{AB}$ so that $\overline{DF}\perp\overline{AB}$. Because $BCDF$ is a rectangle, $\angle FCB\cong\angle DBC\cong\angle CAB\cong\angle BCE$, so $E$ lies on $CF$ and it is the foot of the altitude to the hypotenuse in $\triangle CBF$. Therefore $\triangle BEF\sim\triangle CBF\cong\triangle BCD\sim\triangle ABC$. Because \[ \overline{DF}\perp\overline{AB},\qquad \overline{FE}\perp\overline{EB},\qquad \text{and}\qquad \frac{AB}{DF}=\frac{AB}{BC}=\frac{BE}{FE}, \] it follows that $\triangle ABE\sim\triangle DFE$. Thus $\angle DEA=\angle DEF-\angle AEF=\angle AEB-\angle AEF=\angle FEB=90^\circ$. Furthermore, \[ \frac{AE}{ED}=\frac{BE}{EF}=\frac{AB}{BC}, \] so $\triangle AED\sim\triangle ABC$. Assume without loss of generality that $BC=1$, and let $AB=r>1$. Because $\frac{AB}{BC}=\frac{BC}{CD}$, it follows that $BF=CD=\frac{1}{r}$. Then \[ 17=\frac{\text{Area}(\triangle AED)}{\text{Area}(\triangle CEB)}=AD^2=FD^2+AF^2=1+\left(r-\frac{1}{r}\right)^2, \] and because $r>1$ this yields $r^2-4r-1=0$, with positive solution $r=2+\sqrt{5}$.

答案（D）：设点 $F$ 在 $\overline{AB}$ 上，使得 $\overline{DF}\perp\overline{AB}$。由于 $BCDF$ 是矩形，$\angle FCB\cong\angle DBC\cong\angle CAB\cong\angle BCE$，所以点 $E$ 在 $CF$ 上，并且它是 $\triangle CBF$ 中斜边上的高的垂足。因此 $\triangle BEF\sim\triangle CBF\cong\triangle BCD\sim\triangle ABC$。又因为 \[ \overline{DF}\perp\overline{AB},\qquad \overline{FE}\perp\overline{EB},\qquad \text{且}\qquad \frac{AB}{DF}=\frac{AB}{BC}=\frac{BE}{FE}, \] 可得 $\triangle ABE\sim\triangle DFE$。于是 $\angle DEA=\angle DEF-\angle AEF=\angle AEB-\angle AEF=\angle FEB=90^\circ$。此外， \[ \frac{AE}{ED}=\frac{BE}{EF}=\frac{AB}{BC}, \] 所以 $\triangle AED\sim\triangle ABC$。不妨设 $BC=1$，令 $AB=r>1$。由于 $\frac{AB}{BC}=\frac{BC}{CD}$，可得 $BF=CD=\frac{1}{r}$。那么 \[ 17=\frac{\text{Area}(\triangle AED)}{\text{Area}(\triangle CEB)}=AD^2=FD^2+AF^2=1+\left(r-\frac{1}{r}\right)^2, \] 且因为 $r>1$，得到 $r^2-4r-1=0$，其正根为 $r=2+\sqrt{5}$。

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