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2024 AMC 12A Problems/Problem 25

Revision as of 21:23, 20 March 2025 by Maa is stupid (talk | contribs)

Problem

The ellipse $2x^{2} + 3y^{2} = 7$ consists of all points $P$ in the coordinate plane satisfying $PF_{1} + PF_{2} = \lambda$, for some points $F_{1}$ and $F_{2}$ and some constant $\lambda$. Let $\mathbf{R}$ denote the set of all points $Q$ in the coordinate plane satisfying \[\sqrt{QF_{1}^{2} + 1}\ + \sqrt{QF_{2}^{2} + 1} = \lambda\] What is the square of the area of the region bounded by $\mathbf{R}$?

$\textbf{(A)}~\frac{147\pi^{2}}{64}\qquad\textbf{(B)}~\frac{8\pi^{2}}{3}\qquad\textbf{(C)}~\frac{27\pi^{2}}{7}\qquad\textbf{(D)}~\frac{25\pi^{2}}{6}\qquad\textbf{(E)}~\frac{49\pi^{2}}{10}$

Solution

Rewrite the equation of the ellipse in question as $\tfrac{x^{2}}{7/2} + \tfrac{y^{2}}{7/3} = 1$. Consider the solid "blimp" (known as an ellipsoid) in three-dimensional space that results when said ellipse is revolved about the $x$-axis: $\tfrac{x^{2}}{7/2} + \tfrac{y^{2}}{7/3} + \tfrac{z^{2}}{7/3} = 1$. It is clear that this solid forms the locus of all points $P$ in space that satisfy $PF_{1} + PF_{2} = \lambda$.

Consider any point $Q$ in the $xy$-plane, and construct point $Q^{\prime}$, located directly above $Q$ on the plane $z = 1$. Then, the condition $\sqrt{QF_{1}^{2} + 1}\ + \sqrt{QF_{2}^{2} + 1} = \lambda$ is equivalent to $Q^{\prime}$ lying on the boundary of this solid because \[\sqrt{QF_{1}^{2} + 1}\ + \sqrt{QF_{2}^{2} + 1} = Q^{\prime}F_{1} + Q^{\prime}F_{2} = \lambda\] whenever $Q^{\prime}$ is on the solid.

Thus, region $\mathbf{R}$ is the closed curve that results when the intersection of $z = 1$ and $\tfrac{x^{2}}{7/2} + \tfrac{y^{2}}{7/3} + \tfrac{z^{2}}{7/3} = 1$ is projected on the $xy$-plane. Note that $z^{2} = 1$ so $\tfrac{x^{2}}{7/2} + \tfrac{y^{2}}{7/3} = 1 - \tfrac{3}{7} = \tfrac{4}{7}$. After multiplying by $\tfrac{7}{4}$, we see that $\tfrac{x^{2}}{2} + \tfrac{y^{2}}{4/3} = 1$.

The region in question is an ellipse with a semimajor axis of $\sqrt{2}$ and a semiminor axis of $\sqrt{\tfrac{4}{3}}$, which has area $\sqrt{\tfrac{8}{3}} ~ \pi$. Therefore, the answer is $\boxed{\textbf{(B)}~\frac{8\pi^{2}}{3}}$.

See also

2024 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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