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Difference between revisions of "1985 AIME Problems/Problem 11"

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[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]

Revision as of 13:35, 6 May 2007

Problem

An ellipse has foci at $(9,20)$ and $(49,55)$ in the $xy$-plane and is tangent to the $x$-axis. What is the length of its major axis?

Solution

An ellipse is defined to be the locus of points $P$ such that the sum of the distances between $P$ and the two foci is constant. Let $F_1 = (9, 20)$, $F_2 = (49, 55)$ and $X = (x, 0)$ be the point of tangency of the ellipse with the $x$-axis. Then $X$ must be the point on the axis such that the sum $F_1X + F_2X$ is minimal. Finding the optimal location for $X$ is a classic problem: for any path from $F_1$ to $X$ and then back to $F_2$, we can reflect the second leg of this path (from $X$ to $F_2$) across the $x$-axis. Then our path connects $F_1$ to the reflection $F_2'$ of $F_2$ via some point on the $x$-axis, and this path will have shortest length exactly when our original path has shortest length. This occurs exactly when we have a straight-line path. The sum of the two distances $\displaystyle F_1 X$ and $F_2X$ is therefore equal to the length of the segment $F_1F_2'$, which by the distance formula is just $d = \sqrt{(9 - 49)^2 + (20 + 55)^2} = \sqrt{40^2 + 75^2} = 5\sqrt{8^2 + 15^2} = 5\cdot 17 = 85$.

Finally, let $A$ and $B$ be the two endpoints of the major axis of the ellipse. Then by symmetry $AF_1 = F_2B$ so $AB = AF_1 + F_1B = F_2B + F_1B = d$ (because $B$ is on the ellipse), so the answer is $085$.


See also

1985 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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