2025 AMC 12A Problem 10

Question

In the figure shown below, major arc

AD

and minor arc

BC

have the same center,

O.

Also,

A

lies between

O

and

B,

and

D

lies between

O

and

C.

Major arc

AD,

minor arc

BC,

and each of the two segments

AB

and

CD

have length

2\pi.

What is the distance from

O

to

A?

1

1 - \pi + \sqrt{1 + \pi^2}

\dfrac{1}{2}\pi

\dfrac{1}{2}\sqrt{1 + \pi^2}

2

Answer: B

Concepts:arc quadratic

Difficulty rating: 1530

Small hint:

Let

OA = R_1,

OB = R_2,

and

\angle AOD = \alpha

; the arcs give

R_2\alpha = 2\pi

and

R_1(2\pi - \alpha) = 2\pi

Big hint:

The segments give

R_2 - R_1 = 2\pi

; eliminating the radii yields

\alpha^2 - 2(1+\pi)\alpha + 2\pi = 0

Solution:

Let

R_1 = OA = OD

and

R_2 = OB = OC,

and let

\alpha = \angle AOD = \angle BOC

(the rays coincide). The minor arc

BC

has length

R_2\alpha = 2\pi,

and the major arc

AD

is the reflex arc, so

R_1(2\pi - \alpha) = 2\pi.

Each segment

AB = CD = R_2 - R_1 = 2\pi.

From the first two equations,

R_2 = \dfrac{2\pi}{\alpha}

and

R_1 = \dfrac{2\pi}{2\pi - \alpha}.

Substituting into

R_2 - R_1 = 2\pi

and dividing by

2\pi

gives

\frac{1}{\alpha} - \frac{1}{2\pi - \alpha} = 1,

which simplifies to

\alpha^2 - 2(1+\pi)\alpha + 2\pi = 0.

The larger root is greater than

2\pi,

so it cannot be an angle of this minor arc. Thus

\alpha = (1+\pi) - \sqrt{1+\pi^2}.

Then

\begin{aligned} R_1 &= \frac{2\pi}{2\pi - \alpha} \\ &= \frac{2\pi}{\pi - 1 + \sqrt{1+\pi^2}} \\ &= 1 - \pi + \sqrt{1+\pi^2}, \end{aligned}

after rationalizing (the denominator times

\sqrt{1+\pi^2} - (\pi-1)

equals

2\pi

).

Thus, the correct answer is B.