2025 AMC 12A Problem 14

Problem 14 of 25IntermediateGeometry

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?

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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),

45(1 - e) = 1 + e,

giving

46e = 44

and

e = \dfrac{22}{23}.

Thus, the correct answer is D.