Question

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

Solution:

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 Problem 13

Solution:

Problem 13 in Other Years