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

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== Solution ==
 
== Solution ==
−
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>\displaystyle F_1 X</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>.   
+
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.   
−
 
+
<center><asy>
−
Finally, let <math>A</math> and <math>B</math> be the two endpoints of the major axis of the ellipse.  Then by symmetry <math>AF_1 = F_2B</math> so <math>AB = AF_1 + F_1B = F_2B + F_1B = d</math> (because <math>B</math> is on the ellipse), so the answer is <math>085</math>.
+
size(200);
 +
pointpen=black;pathpen=black+linewidth(0.6);pen f = fontsize(8);
 +
pair F1=(9,20),F2=(49,55);
 +
D(shift((F1+F2)/2)*rotate(41.186)*scale(85/2,10*11^.5)*unitcircle);
 +
D((-30,0)--(80,0)--(0,0)--(0,80));
 +
path p = F1--(49,-55);
 +
pair X = IP(p,(0,0)--(80,0));
 +
D(p,dashed);D(F1--X--F2);D(F1);D(F2);D((49,-55));
 +
MP("X",X,SW,f);
 +
MP("F_1",F1,NW,f);
 +
MP("F_2",F2,NW,f);
 +
MP("F_2'",(49,-55),NE,f);
 +
</asy></center>
 +
The sum of the two distances <math>F_1 X</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>.   
  
 +
Finally, let <math>A</math> and <math>B</math> be the two endpoints of the major axis of the ellipse.  Then by symmetry <math>AF_1 = F_2B</math> so <math>AB = AF_1 + F_1B = F_2B + F_1B = d</math> (because <math>B</math> is on the ellipse), so the answer is <math>\boxed{085}</math>.
  
 
== See also ==
 
== See also ==
−
* [[Coordinate geometry]]
 
 
{{AIME box|year=1985|num-b=10|num-a=12}}
 
{{AIME box|year=1985|num-b=10|num-a=12}}
−
* [[AIME Problems and Solutions]]
 
−
* [[American Invitational Mathematics Examination]]
 
−
* [[Mathematics competition resources]]
 
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]

Revision as of 19:14, 9 April 2008

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.

[asy] size(200); pointpen=black;pathpen=black+linewidth(0.6);pen f = fontsize(8); pair F1=(9,20),F2=(49,55); D(shift((F1+F2)/2)*rotate(41.186)*scale(85/2,10*11^.5)*unitcircle); D((-30,0)--(80,0)--(0,0)--(0,80)); path p = F1--(49,-55); pair X = IP(p,(0,0)--(80,0)); D(p,dashed);D(F1--X--F2);D(F1);D(F2);D((49,-55)); MP("X",X,SW,f); MP("F_1",F1,NW,f); MP("F_2",F2,NW,f); MP("F_2'",(49,-55),NE,f); [/asy]

The sum of the two distances $F_1 X$ 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 $\boxed{085}$.

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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