Question

Let $\overline{AB}$ be a diameter in a circle of radius $5\sqrt2.$ Let $\overline{CD}$ be a chord in the circle that intersects $\overline{AB}$ at a point $E$ such that $BE = 2\sqrt5$ and $\angle AEC = 45^\circ.$ What is $CE^2 + DE^2?$

$96$

$98$

$44\sqrt{5}$

$70\sqrt{2}$

$100$

Solution:

Place the center at the origin with $\overline{AB}$ on the $x$ -axis; the radius is $R = 5\sqrt2,$ so $R^2 = 50.$ Then $E = (x_E, 0)$ with $x_E = R - 2\sqrt5$ (its exact value is not needed).

Parametrize the chord as $E + t\left(\tfrac{1}{\sqrt2}, \tfrac{1}{\sqrt2}\right).$ Substituting into $x^2 + y^2 = 50$ gives $t^2 + \sqrt2\,x_E\, t + (x_E^2 - 50) = 0,$ whose roots are the signed distances $t_1, t_2$ to $C$ and $D.$

By Vieta, $t_1 + t_2 = -\sqrt2\,x_E$ and $t_1 t_2 = x_E^2 - 50,$ so $CE^2 + DE^2 = t_1^2 + t_2^2 = (t_1 + t_2)^2 - 2t_1 t_2 = 2x_E^2 - 2(x_E^2 - 50) = 100.$

Thus, the correct answer is E.

2020 AMC 12B Problem 12

Solution:

Problem 12 in Other Years