Question

Trapezoid $ABCD$ has $AB \parallel CD,$ $BC = CD = 43,$ and $AD \perp BD.$ Let $O$ be the intersection of the diagonals $AC$ and $BD,$ and let $P$ be the midpoint of $BD.$ Given that $OP = 11,$ the length $AD$ can be written in the form $m\sqrt n,$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. What is $m + n?$

$65$

$132$

$157$

$194$

$215$

Answer: D

Concepts:coordinate geometry trapezoid similarity

Difficulty rating: 2080

Solution:

Place $D = (0,0)$ with $B = (b, 0)$ on one axis and $A = (0, a)$ on the other, so that $AD \perp BD.$ Since $CD \parallel AB,$ write $C = t(b, -a)$ for some $t.$ Then $CD = t\sqrt{a^2+b^2}$ and $BC^2 = b^2(1-t)^2 + t^2a^2.$ Setting $BC = CD$ gives $t^2 = (1-t)^2,$ so $t = \tfrac12.$

Thus $C = \left(\tfrac{b}{2}, -\tfrac{a}{2}\right),$ and $CD = 43$ gives $a^2 + b^2 = 4\cdot 43^2 = 7396.$ The diagonal $AC$ meets $BD$ (the $x$ -axis) at $O = \left(\tfrac{b}{3}, 0\right),$ while $P = \left(\tfrac{b}{2}, 0\right).$ Hence $OP = \tfrac{b}{6} = 11,$ so $b = 66.$

Then $a^2 = 7396 - 66^2 = 3040,$ so $AD = a = \sqrt{3040} = 4\sqrt{190}.$ With $m = 4$ and $n = 190,$ we get $m + n = 194.$

Thus, the correct answer is D.

2021 AMC 12A Spring Problem 17

Answer: D

Difficulty rating: 2080

Solution:

Problem 17 in Other Years