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

(asy)
m (Solution: minor)
Line 6: Line 6:
 
<center><asy>
 
<center><asy>
 
size(200);
 
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);
 
pair F1=(9,20),F2=(49,55);
 
D(shift((F1+F2)/2)*rotate(41.186)*scale(85/2,10*11^.5)*unitcircle);
 
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);
 
path p = F1--(49,-55);
 
pair X = IP(p,(0,0)--(80,0));
 
pair X = IP(p,(0,0)--(80,0));

Revision as of 20:15, 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 $\boxed{085}$.

See also

1985 AIME (ProblemsAnswer KeyResources)
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