2021 AMC 10A Problem 17

Draw $CP$. Because $CB = CD$, triangle $BCD$ is isosceles and its median $CP$ is also an altitude: $CP \perp BD$. We are also told $AD \perp BD$, so $CP \parallel AD$.

By alternate interior angles ($AB \parallel CD$), $\angle CDB = \angle ABD$. So right triangles $CPD$ (right angle at $P$) and $ADB$ (right angle at $D$) share an acute angle and are similar, with
$$\frac{CD}{AB} = \frac{DP}{DB} = \frac12 \quad\Longrightarrow\quad AB = 2 \cdot 43 = 86.$$

In a trapezoid the diagonals cut each other in the ratio of the parallel sides, so $\frac{DO}{OB} = \frac{CD}{AB} = \frac12$, giving $DO = \frac13 BD$. Since $DP = \frac12 BD$,
$$OP = DP - DO = \frac{BD}{2} - \frac{BD}{3} = \frac{BD}{6} = 11 \quad\Longrightarrow\quad BD = 66.$$

Finally, in right triangle $ADB$,
$$AD^2 = AB^2 - BD^2 = 86^2 - 66^2 = (86-66)(86+66) = 20 \cdot 152 = 3040,$$
so $AD = \sqrt{3040} = \sqrt{16 \cdot 190} = 4\sqrt{190}$, and $m + n = 4 + 190 = 194$.

The answer is $\boxed{\textbf{(D)}\ 194}$.

Two perpendiculars to the same line are parallel, and isosceles triangles hand you a perpendicular for free through the midpoint of the base. The diagonal-ratio fact for trapezoids converts the odd-looking datum $OP = 11$ into the full diagonal length. Difference of squares keeps the last computation clean.

Coordinates: $D = (0,0)$, $B = (2x, 0)$, $A = (0, h)$. Requiring $CD = CB$ with $C$ on the line through $D$ parallel to $AB$ forces $C = (x, -h/2)$. Then $O = (2x/3, 0)$, so $OP = x/3 = 11$ and $x = 33$; $CD^2 = x^2 + h^2/4 = 43^2$ gives $h^2 = 4(1849 - 1089) = 3040$.

• Assuming O is the midpoint of BD (it is not; O splits BD in ratio AB:CD = 2:1), or mixing up which segment OP = 11 measures.
• Simplifying $\sqrt{3040}$ incorrectly (e.g. $2\sqrt{760}$) and reporting $762$ or another non-choice, or forgetting to reduce at all.