2025 AMC 12A Problem 14

Attempt Problem 14 of the 2025 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2025 AMC 12A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

14.

Points

F,

G,

and

H

are collinear with

G

between

F

and

H.

The ellipse with foci at

G

and

H

is internally tangent to the ellipse with foci at

F

and

G,

as shown below.

The two ellipses have the same eccentricity

e,

and the ratio of their areas is

2025.

(Recall that the eccentricity of an ellipse is

e = \dfrac{c}{a},

where

c

is the distance from the center to a focus, and

2a

is the length of the major axis.) What is

e?

\dfrac{3}{5}

\dfrac{16}{25}

\dfrac{4}{5}

\dfrac{22}{23}

\dfrac{44}{45}

Answer: D

Concepts:ellipse area ratio

Difficulty rating: 1730

Small hint:

Equal eccentricity makes area proportional to

a^2,

so

\dfrac{a_1}{a_2} = \sqrt{2025} = 45

Big hint:

The ellipses share focus

G

; equate their right vertices to get

a_1(1 - e) = a_2(1 + e)

Solution:

With the same eccentricity,

b = a\sqrt{1 - e^2},

so the area

\pi a b \propto a^2.

The area ratio

2025

gives

\dfrac{a_1}{a_2} = \sqrt{2025} = 45,

where

a_1, a_2

are the semi-major axes.

Both ellipses share focus

G.

On the large ellipse

G

is the right focus, so its right vertex lies

a_1 - c_1

to the right of

G.

On the small ellipse

G

is the left focus, so its right vertex lies

a_2 + c_2

to the right of

G.

Internal tangency makes these coincide:

a_1 - c_1 = a_2 + c_2.

Using

c = ea,

a_1(1 - e) = a_2(1 + e),

so

45(1 - e) = 1 + e,

giving

46e = 44

and

e = \dfrac{22}{23}.

Thus, the correct answer is D.

Problem 14 in other years