2020 AMC 12B Problem 12

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

Put the center at the origin with

\overline{AB}

on the

x

-axis and parametrize the chord from

E

at

45^\circ

If

t_1, t_2

are the signed distances from

E

to

C, D,

then

CE^2 + DE^2 = t_1^2 + t_2^2

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

\begin{gathered} 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. \end{gathered}

Thus, the correct answer is E.

Problem 12 in other years