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

Revision as of 19:15, 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.

$[asy] size(200); pointpen=black;pathpen=black+linewidth(0.6);pen f = fontsize(10); pair F1=(9,20),F2=(49,55); D(shift((F1+F2)/2)*rotate(41.186)*scale(85/2,10*11^.5)*unitcircle); D((-20,0)--(80,0)--(0,0)--(0,80)--(0,-60)); path p = F1--(49,-55); pair X = IP(p,(0,0)--(80,0)); D(p,dashed);D(F1--X--F2);D(F1);D(F2);D((49,-55)); MP("X",X,SW,f); MP("F_1",F1,NW,f); MP("F_2",F2,NW,f); MP("F_2'",(49,-55),NE,f); [/asy]$

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 $\boxed{085}$ .

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

Revision as of 19:15, 9 April 2008

Problem

Solution

See also

@@ Line 6: / Line 6: @@
 <center><asy>
 size(200);
-pointpen=black;pathpen=black+linewidth(0.6);pen f = fontsize(8);
+pointpen=black;pathpen=black+linewidth(0.6);pen f = fontsize(10);
 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));
+D((-20,0)--(80,0)--(0,0)--(0,80)--(0,-60));
 path p = F1--(49,-55);
 pair X = IP(p,(0,0)--(80,0));