AMC 10 · 2025 · #14

Grade 10 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity

ellipse-eccentricityarea-ellipsesimilar-figureslinear-equations-one-var identify-subproblemsconvert-to-algebra ↑ Prerequisites: similar-figures

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

Three points F, G, H sit on one line with G between F and H. One ellipse has foci F and G; a second, smaller ellipse has foci G and H. The second ellipse sits inside the first and touches it at exactly one point. Both ellipses have the same eccentricity e, and the larger ellipse has 2025 times the area of the smaller one. Find e.

Pick an answer.

(A)

$\frac35$

(B)

$\frac{16}{25}$

(C)

$\frac45$

(D)

$\frac{22}{23}$

(E)

$\frac{44}{45}$

How to solve

Strategy Introduce a Variable

Nothing in the problem is a number except the area ratio, so the first move is to name the pieces: a semi-major axis and a focal half-distance for each ellipse. Once those four lengths have names, the two given facts split cleanly into two independent conditions. The area ratio fixes the size relationship between the two ellipses, and the tangency-plus-collinearity fixes how their axes line up. Both conditions turn into equations in the same lengths, and because eccentricity ties c to a, everything collapses to a single equation in e.

1STEP 1

Name the axes and set the eccentricities equal

Name a₁, c₁ for the outer ellipse and a₂, c₂ for the inner; then FG=2c₁, GH=2c₂, and equal eccentricity gives c₁=e a₁, c₂=e a₂.

e = c₁/a₁ = c₂/a₂ → c₁ = e a₁, c₂ = e a₂

2STEP 2

Split the givens into two conditions

An ellipse of semi-major axis a and eccentricity e has area π a² √(1-e²); treat the area ratio and the tangency as two separate conditions.

Area = π a² √(1 - e²)

3STEP 3

Turn the area ratio into a length ratio

Same e makes both areas share √(1-e²), so it cancels: $\frac{Area₁}{Area₂}$ =( $\frac{a₁}{a₂}$ )²=2025, giving $\frac{a₁}{a₂}$ =45.

Area₁/Area₂ = a₁²/a₂² = 2025 → a₁/a₂ = 45

4STEP 4

Write the tangency as a line-up of vertices

With F at 0, the two right vertices c₁+a₁ and 2c₁+c₂+a₂ must coincide, giving a₁=c₁+c₂+a₂.

c₁ + a₁ = 2c₁ + c₂ + a₂ → a₁ = c₁ + c₂ + a₂

5STEP 5

Substitute c = e a and solve for e

Substituting c₁=e a₁, c₂=e a₂ gives a₁(1-e)=a₂(1+e), so $\frac{1+e}{1-e}$ =45; then 46e=44 and e= $\frac{22}{23}$ , choice (D).

a₁(1-e) = a₂(1+e) → (1+e)/(1-e) = 45 → 46e = 44 → e = 22/23

Answer

22/23

Check e =

\frac{22}{23}

against both conditions. Then 1 - e =

\frac{1}{23}

and 1 + e =

\frac{45}{23}

, so

\frac{1 + e}{1 - e}

= 45, which matches the length ratio

\frac{a₁}{a₂}

= 45. Squaring gives an area ratio of 45² = 2025, exactly as required. Also

\frac{22}{23}

is between 0 and 1, so it is a legitimate ellipse eccentricity, and it is close to 1, matching a very stretched ellipse whose two foci are far apart relative to its width. Everything is consistent.

💡Key takeaway

Give every length a name, let one letter carry the shape, and two separate facts about size and touching will meet in a single equation you can solve.

Name the axes and set the eccentricities equal
Split the givens into two conditions
Turn the area ratio into a length ratio
Write the tangency as a line-up of vertices
Substitute c = e a and solve for e

Name the axes and set the eccentricities equal

Split the givens into two conditions

Turn the area ratio into a length ratio

Write the tangency as a line-up of vertices

Substitute c = e a and solve for e

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