2021 AMC 10A Problem 17

Because $BC=CD,$ the median from $C$ to $BD$ is perpendicular to $BD.$ Thus $\triangle BPC$ is a right triangle. Let $\angle DBC=\alpha.$ Since $AB\parallel CD,$ we also have $\angle ABD=\alpha,$ so $\triangle BPC\sim\triangle BDA.$ Since $P$ is the midpoint of $BD,$ we have $\frac{BD}{BP}=2.$ In the similarity, $BC$ corresponds to $AB,$ so $$\begin{aligned} \frac{AB}{BC} &=2, \\ AB &=2\cdot43=86. \end{aligned}$$ Also, $\triangle ABO\sim\triangle CDO,$ so $$\frac{BO}{OD}=\frac{AB}{CD}=2.$$ Since $OP=11$ and $P$ is the midpoint of $BD,$ write $BP=PD=t.$ Then $BO=t+11$ and $OD=t-11,$ so $$\frac{t+11}{t-11}=2.$$ This gives $t=33,$ hence $BD=66.$ Finally, $\triangle ABD$ is right, so $$\begin{aligned} AD &= \sqrt{AB^2-BD^2} \\ &= \sqrt{86^2-66^2} \\ &= 4\sqrt{190}. \end{aligned}$$ Thus $m+n=4+190=194.$ Thus, D is the correct answer.