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

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== Problem ==
 
== Problem ==
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An [[ellipse]] has [[focus | foci]] at <math>(9,20)</math> and <math>(49,55)</math> in the <math>xy</math>-plane and is [[tangent line | tangent]] to the <math>x</math>-axis. What is the length of its [[major axis]]?
  
 
== Solution ==
 
== Solution ==
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{{solution}}
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An ellipse is defined to be the [[locus]] of points <math>P</math> such that the sum of the distances between <math>P</math> and the two foci is constant.  Let <math>F_1 = (9, 20)</math>, <math>F_2 = (49, 55)</math> and <math>X = (x, 0)</math> be the point of tangency of the ellipse with the <math>x</math>-axis.  Then <math>X</math> must be the point on the axis such that the sum <math>F_1X + F_2X</math> is minimal.  Finding the optimal location for <math>X</math> is a classic problem: for any path from <math>F_1</math> to <math>X</math> and then back to <math>F_2</math>, we can reflect the second leg of this path (from <math>X</math> to <math>F_2</math>) across the <math>x</math>-axis.  Then our path connects <math>F_1</math> to the reflection <math>F_2'</math> of <math>F_2</math> via some point on the <math>x</math>-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 <math>F_1X</math> and <math>F_2X</math> is therefore equal to the length of the segment <math>F_1F_2'</math>, which by the [[distance formula]] is just <math>d = \sqrt{(9 - 49)^2 + (20 + 55)^2} = \sqrt{40^2 + 75^2} = 5\sqrt{8^2 + 15^2} = 5\cdot 17 = 85</math>. 
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Finally, let <math>A</math> and <math>B</math> be the two endpoints of the major axis of the ellipse.  Then <math>AF_1 = BF_2</math> so <math>AB = AF_1 + F_1B = BF_2 + F_1B = d</math> (because <math>B</math> is on the ellipse), so the answer is <math>085</math>.
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== See also ==
 
== See also ==
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* [[1985 AIME Problems/Problem 10 | Previous problem]]
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* [[1985 AIME Problems/Problem 12 | Next problem]]
 
* [[1985 AIME Problems]]
 
* [[1985 AIME Problems]]
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* [[Coordinate geometry]]
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[[Category:Intermediate Geometry Problems]]

Revision as of 18:41, 1 December 2006

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 $F_1X$ 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 $AF_1 = BF_2$ so $AB = AF_1 + F_1B = BF_2 + F_1B = d$ (because $B$ is on the ellipse), so the answer is $085$.


See also

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