2025 AMC 12A Problems/Problem 14

Problem

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.

$[asy] import graph; unitsize(0.15 inch); draw(ellipse((0,0), 12, 4)); draw(ellipse((8,0), 4, 1.45)); pair F = (-6,0); pair G = (6,0); pair H = (10,0); dot(F^^G^^H); label("$F$", F, 1.5*S, p=fontsize(8pt)); label("$G$", G, 1.5*S, p=fontsize(8pt)); label("$H$", H, 1.5*S, p=fontsize(8pt)); [/asy]$

The two ellipses have the same eccentricity $e$ , and the ratio of their areas is $2025$ . (Recall that the eccentricity of an ellipse is $e = \tfrac{c}{a}$ , where $c$ is the distance from the center to a focus, and $2a$ is the length of the major axis.) What is $e$ ?

$\textbf{(A)}~\frac35\qquad\textbf{(B)}~\frac{16}{25}\qquad\textbf{(C)}~\frac45\qquad\textbf{(D)}~\frac{22}{23}\qquad\textbf{(E)}~\frac{44}{45}$

Solution 1(Simple)

Let the outer ellipse be ellipse 1, and the inner ellipse be ellipse 2. Let $a_1$ , $b_1$ , and $c_1$ , correspond to the semimajor axis, semiminor axis, and focal distance of ellipse $1$ , respectively. Similarly, let $a_2$ , $b_2$ , and $c_2$ correspond to the semimajor axis, semiminor axis, and focal distance of ellipse $2$ , respectively.

Ellipses with the same eccentricity are similar, so $\frac{a_1}{a_2} = \sqrt{2025} = 45$ because the ratio of semimajor axes between similar ellipses is equal to the square root of the ratio between their areas. Notice how $a_1 = c_1 + c_2 + a_2.$ Substituting from the eccentricity equation, $a_1 = ea_1 + ea_2 + a_2.$ Rearranging gives us $a_1 - ea_1 = ea_2 + a_2$ $a_1(1-e) = a_2(1+e)$ $\frac{a_1}{a_2} = \frac{1+e}{1-e} = 45.$ Solving for $e$ then yields $\boxed{\text{(D) }\dfrac{22}{23}}.$

~Kevin Wang ~Edits by pythagoreanMath

Solution 2

We label the outer ellipse as Ellipse $1$ and the inner ellipse as Ellipse $2$ . Let the focal distance of Ellipse $1$ be $2c_1$ and the major axis be $2a_1$ . Similarly, Ellipse $2$ has a focal distance of $2c_2$ and a major axis of $2a_2$ .

We label the rightmost vertex of both ellipses as $I$ . Because $I$ lies on Ellipse $1$ , the sum of the lengths of $FI$ and $GI$ equals $2a_1$ . The length of $FI$ can be written as $c_2+a_2+2c_1$ , and the length of $GI$ can be written as $c_2+a_2$ . Therefore, $(c_2+a_2+2c_1) + (c_2+a_2) = 2a_1$ $2c_2+2a_2+2c_1 = 2a_1$ $c_2+a_2+c_1=a_1.$ Substituting $a_1 = c_1/e$ and $a_2 = c_2/e$ , we get $c_2 + \frac{c_2}{e} + c_1=\frac{c_1}{e}$ $c_2e + c_2 + c_1e = c_1$ $c_2(e+1) = c_1(1-e)$ $\frac{c_1}{c_2} = \frac{1+e}{1-e}$ We now take into account the information regarding the ratio of the ellipses' areas. Because the area of an ellipse with semi-major axis $2a$ and eccentricity $e$ can be written as $\pi a^2\sqrt{1-e^2}$ , we get that: $\frac{\text{Area}_1}{{\text{Area}_2}} = \frac{a_1^2\sqrt{1-e^2}}{a_2^2\sqrt{1-e^2}} = \frac{a_1^2}{a_2^2} = 2025.$

Therefore, $\frac{a_1}{a_2} = 45$ . Because $a_1 = c_1/e$ and $a_2 = c_2/e$ , then $\frac{c_1/e}{c_2/e} = \frac{c_1}{c_2}=45$ as well.

Substituting this into the equation found earlier, we get that $\frac{c_1}{c_2} = \frac{1+e}{1-e} = 45$ $1+e = 45-45e$ $46e = 44$ $e = \boxed{\frac{22}{23}}.$

~lprado

Solution 3 (No area formulas!)

In the larger ellipse, let $a$ be half the major axis (also called the semi-major axis) and $b$ be half the minor axis (also called the semi-minor axis), and $c$ be the distance between its center and a focus. Since we are given that $2025 = 45^2$ is the ratio of the areas of the two ellipses, it follows that the linear scale factor of the two is $45:1$ . Therefore, the semi-major axis of the smaller ellipse is $\frac{1}{45}a$ and its semi-minor axis is $\frac{1}{45}b$ . Furthermore, the distance between a focus of the smaller ellipse and its center is $\frac{1}{45}c$ . There are now two ways to express the length of the major axis of the large ellipse: $2a$ and $a + c + \frac{1}{45}c + \frac{1}{45}a$ . Setting these equal, we now have an expression of degree 1 entirely in terms of $a$ and $c$ . Some algebra gives us:

$a-\frac{1}{45}a=c+\frac{1}{45}c$

$\frac{44}{45}a=\frac{46}{45}c$

$44a=46c$

$e=\frac{c}{a}=\boxed{\text{(D) }\dfrac{22}{23}}$ .

Note that this method did not require you to know the area formula of an ellipse (which, incidentally, is $\pi a b$ ).

~dg6665

Solution 4 (Similarity)

Parametrize the bigger ellipse with $a,c$ as defined in the problem, and the smaller ellipse with $a',c'$ . Note that the ellipses have the same eccentricity, so they are similar with ratio $k^2=2025$ or $k=45$ . We wish to find $e=\frac{c}{a}=\frac{c'}{a'}$ .

Looking at the figure, we can decompose the semi-major axis $a$ of the bigger ellipse as $a=c+c'+a'$ Using $c=kc'=45c'$ and $a=45a'$ and rearranging gives us: $45a'-a'=44a'=45c'+c'=46c'$ Thus $e=\frac{c'}{a'}=\frac{44}{46}=\boxed{\textbf{(D)}~\frac{22}{23}}.$

~imosilver

Video Solution by StressedPineapple

https://youtube.com/watch?v=NWBPm3lThH4&t=28s

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=dAeyV60Hu5c

2025 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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