2021 AMC 10A Problem 17

Draw the trapezoid with the long side $AB$ at the bottom and $CD$ at the top, and draw the diagonal $BD$.

1. Triangle $BCD$ has $CB = CD = 43$, so it is isosceles. In an isosceles triangle, the segment from the top corner to the midpoint of the base is at right angles to the base. So draw $CP$: it is perpendicular to $BD$.
2. Compare triangles $CPD$ and $ADB$.
- Each has a right angle: at $P$, and at $D$ (given $AD \perp BD$).
- Since $AB \parallel CD$, the angles $\angle CDB$ and $\angle ABD$ are equal (they form a "Z" across the diagonal).
So the two triangles have the same angles, and the same shape.
3. Match the sides. The long side $CD$ matches the long side $AB$, and $DP$ matches $BD$. Since $DP$ is half of $BD$, triangle $CPD$ is exactly half the size of triangle $ADB$. So $AB = 2 \times CD = 86$.
4. Now find where $O$ sits on $BD$. Triangles $ODC$ and $OBA$ also have the same angles (again "Z" angles from the parallel sides), and $AB$ is twice $CD$. So $OB$ is twice $OD$. That means $OD$ is $\frac13$ of $BD$.
5. $P$ is the midpoint, so $DP$ is $\frac12$ of $BD$. Then
$$OP = DP - DO = \tfrac12 BD - \tfrac13 BD = \tfrac16 BD.$$
Since $OP = 11$, we get $BD = 66$.
6. Triangle $ADB$ has a right angle at $D$, long side $AB = 86$ and leg $BD = 66$. By the Pythagorean theorem,
$$AD^2 = 86^2 - 66^2 = (86 - 66)(86 + 66) = 20 \times 152 = 3040.$$
7. $3040 = 16 \times 190$, and $190 = 2 \times 5 \times 19$ has no square factor. So $AD = 4\sqrt{190}$ and $m + n = 4 + 190 = 194$.

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

An isosceles triangle gives you a free right angle at the midpoint of its base. Two lines at right angles to the same line are parallel, and parallel lines give equal "Z" angles, which is how the matching triangles appear. In any trapezoid the diagonals cut each other in the same ratio as the two parallel sides; that turns the odd-looking fact $OP = 11$ into the full length of $BD$.

Use coordinates. Put $D = (0, 0)$, $B = (2x, 0)$ and $A = (0, h)$, so $AD \perp BD$. Point $C$ lies on the line through $D$ parallel to $AB$, and $CD = CB$ means $C$ has the same $x$-coordinate as the midpoint of $DB$, so $C = (x, -\tfrac{h}{2})$. The diagonal $AC$ crosses the $x$-axis at $O = (\tfrac{2x}{3}, 0)$, and $P = (x, 0)$, so $OP = \tfrac{x}{3} = 11$ and $x = 33$. Then $CD^2 = x^2 + \tfrac{h^2}{4} = 43^2$ gives $h^2 = 4(1849 - 1089) = 3040$, so $AD = h = 4\sqrt{190}$.

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 $\boxed{\textbf{(D)}\ 194}$.