AMC12 2025 A

AMC12 2025 A · Q14

AMC12 2025 A · Q14. It mainly tests Geometry misc.

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 $e = \tfrac{c}{a}$, where $c$ is the distance from the center to a focus, and $2a$ is the length of the major axis.) What is $e$?

点 $F$、$G$ 和 $H$ 共线，且 $G$ 在 $F$ 和 $H$ 之间。以 $G$ 和 $H$ 为焦点的椭圆内切于以 $F$ 和 $G$ 为焦点的椭圆，如下图所示。两个椭圆具有相同的离心率 $e$，且面积比为 $2025$。（回想椭圆的离心率为 $e = \tfrac{c}{a}$，其中 $c$ 是中心到焦点的距离，$2a$ 是长轴长度。）$e$ 是多少？

(A) \frac35 \frac35

(B) \frac{16}{25} \frac{16}{25}

(D) \frac{22}{23} \frac{22}{23}

(E) \frac{44}{45} \frac{44}{45}

Answer

Correct choice: (D)

正确答案：（D）

Solution

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 $\boxed{\text{(D) }\dfrac{22}{23}}.$

令外椭圆为椭圆 $1$，内椭圆为椭圆 $2$。令 $a_1$、$b_1$ 和 $c_1$ 分别对应椭圆 $1$ 的半长轴、半短轴和焦距。类似地，$a_2$、$b_2$ 和 $c_2$ 对应椭圆 $2$。具有相同离心率的椭圆相似，因此 $\frac{a_1}{a_2} = \sqrt{2025} = 45$，因为相似椭圆的半长轴比等于面积比的平方根。注意到 \[a_1 = c_1 + c_2 + a_2.\] 由离心率公式代入， \[a_1 = ea_1 + ea_2 + a_2.\] 整理得 \[a_1 - ea_1 = ea_2 + a_2\] \[a_1(1-e) = a_2(1+e)\] \[\frac{a_1}{a_2} = \frac{1+e}{1-e} = 45.\] 解得 $e=\boxed{\text{(D) }\dfrac{22}{23}}$。

Topics

GE.020 Geometry misc