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2025 AMC 12A 第 14 题

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所有题目均经美国数学协会（MAA）官方合法授权使用。

14.

点 FF、GG 和 HH 共线，且 GG 在 FF 和 HH 之间。以 GG 和 HH 为焦点的椭圆与以 FF 和 GG 为焦点的椭圆内切，如下图所示。

这两个椭圆的离心率相同，均为 ee，且它们的面积之比为 20252025。（回忆：椭圆的离心率为 e=cae = \dfrac{c}{a}，其中 cc 是从中心到焦点的距离，2a2a 是长轴长度。）ee 是多少？

Points F,F, G,G, and HH are collinear with GG between FF and H.H. The ellipse with foci at GG and HH is internally tangent to the ellipse with foci at FF and G,G, as shown below.

The two ellipses have the same eccentricity e,e, and the ratio of their areas is 2025.2025. (Recall that the eccentricity of an ellipse is e=ca,e = \dfrac{c}{a}, where cc is the distance from the center to a focus, and 2a2a is the length of the major axis.) What is e?e?

35\dfrac{3}{5}

1625\dfrac{16}{25}

45\dfrac{4}{5}

2223\dfrac{22}{23}

4445\dfrac{44}{45}

答案：D
知识点：椭圆面积比
难度评级：1730
小提示：

离心率相同使面积与 a2a^2 成正比，所以 a1a2=2025=45\dfrac{a_1}{a_2} = \sqrt{2025} = 45。

Equal eccentricity makes area proportional to a2,a^2, so a1a2=2025=45\dfrac{a_1}{a_2} = \sqrt{2025} = 45

大提示：

两个椭圆共用焦点 GG；令它们的右顶点重合，得到 a1(1−e)=a2(1+e)a_1(1 - e) = a_2(1 + e)

The ellipses share focus GG; equate their right vertices to get a1(1−e)=a2(1+e)a_1(1 - e) = a_2(1 + e)

解答：

离心率相同，则 b=a1−e2b = a\sqrt{1 - e^2}，所以面积 πab∝a2\pi a b \propto a^2。面积之比为 20252025，因此 a1a2=2025=45\dfrac{a_1}{a_2} = \sqrt{2025} = 45，其中 a1,a2a_1, a_2 是两个半长轴。

两个椭圆共用焦点 GG。在大椭圆中，GG 是右焦点，所以右顶点位于 GG 右侧 a1−c1a_1 - c_1 处。在小椭圆中，GG 是左焦点，所以右顶点位于 GG 右侧 a2+c2a_2 + c_2 处。内切使这两个点重合：a1−c1=a2+c2。a_1 - c_1 = a_2 + c_2\text{。}

利用 c=eac = ea，得 a1(1−e)=a2(1+e)a_1(1 - e) = a_2(1 + e)，所以 45(1−e)=1+e45(1 - e) = 1 + e，从而 46e=4446e = 44，e=2223e = \dfrac{22}{23}。

因此，正确答案是 D。

With the same eccentricity, b=a1−e2,b = a\sqrt{1 - e^2}, so the area πab∝a2.\pi a b \propto a^2. The area ratio 20252025 gives a1a2=2025=45,\dfrac{a_1}{a_2} = \sqrt{2025} = 45, where a1,a2a_1, a_2 are the semi-major axes.

Both ellipses share focus G.G. On the large ellipse GG is the right focus, so its right vertex lies a1−c1a_1 - c_1 to the right of G.G. On the small ellipse GG is the left focus, so its right vertex lies a2+c2a_2 + c_2 to the right of G.G. Internal tangency makes these coincide: a1−c1=a2+c2.a_1 - c_1 = a_2 + c_2.

Using c=ea,c = ea, a1(1−e)=a2(1+e),a_1(1 - e) = a_2(1 + e), so 45(1−e)=1+e,45(1 - e) = 1 + e, giving 46e=4446e = 44 and e=2223.e = \dfrac{22}{23}.

Thus, the correct answer is D.

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