Question

在四边形 $ABCD$ 中， $\angle BAD \cong \angle ADC$ 且 $\angle ABD \cong \angle BCD$ ，并且 $AB = 8$ 、 $BD = 10$ 、 $BC = 6$ 。长度 $CD$ 可写成 $\frac{m}{n}$ 的形式，其中 $m$ 和 $n$ 是互质正整数。求 $m + n$ 。

In quadrilateral $ABCD,$ $\angle BAD \cong \angle ADC$ and $\angle ABD \cong \angle BCD,$ $AB = 8,$ $BD = 10,$ and $BC = 6.$ The length $CD$ may be written in the form $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

答案：69

知识点：相似等腰三角形导角

难度评级：2990

解答：

将 $\overline{AB}$ 越过 $B$ 延长，将 $\overline{DC}$ 越过 $C$ 延长，交于 $P$ 。因为 $\angle PAD = \angle PDA$ ，三角形 $APD$ 是等腰三角形，且 $PA = PD$ 。另外， $\angle PBD = 180^\circ - \angle ABD$ ， $\angle PCB = 180^\circ - \angle BCD$ ，所以 $\angle PBD = \angle PCB$ 。

三角形 $PCB$ 与 $PBD$ 共用角 $P$ ，并且 $\angle PCB = \angle PBD$ ，所以它们相似，得到 $\frac{PC}{PB} = \frac{PB}{PD} = \frac{CB}{BD} = \frac{3}{5}.$ 因为 $PB = PA - 8 = PD - 8$ ，中间的比例为 $\frac{PD - 8}{PD} = \frac{3}{5}$ ，所以 $PD = 20$ ，且 $PB = 12$ 。于是 $PC = \frac{3}{5} \cdot 12 = \frac{36}{5}$ 。

最后 $CD = PD - PC$ $= 20 - \frac{36}{5}$ $= \frac{64}{5}$ ，已经是最简形式，所以 $m + n = 64 + 5 = 69$ 。

Extend $\overline{AB}$ beyond $B$ and $\overline{DC}$ beyond $C$ to meet at $P.$ Since $\angle PAD = \angle PDA,$ triangle $APD$ is isosceles with $PA = PD.$ Also $\angle PBD = 180^\circ - \angle ABD$ and $\angle PCB = 180^\circ - \angle BCD,$ so $\angle PBD = \angle PCB.$

Triangles $PCB$ and $PBD$ share angle $P$ and have $\angle PCB = \angle PBD,$ so they are similar, giving $\frac{PC}{PB} = \frac{PB}{PD} = \frac{CB}{BD} = \frac{3}{5}.$ Since $PB = PA - 8 = PD - 8,$ the middle ratio reads $\frac{PD - 8}{PD} = \frac{3}{5},$ so $PD = 20$ and $PB = 12.$ Then $PC = \frac{3}{5} \cdot 12 = \frac{36}{5}.$

Finally $CD = PD - PC$ $= 20 - \frac{36}{5}$ $= \frac{64}{5},$ which is in lowest terms, so $m + n = 64 + 5 = 69.$

2001 AIME II 第 13 题

答案：69

难度评级：2990

解答：

其他年份的第 13 题