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

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== Solution 2 (Calculus) ==
 
== Solution 2 (Calculus) ==
  
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Ellipse is defined as set of points equal to fixed sum of distances from foci. Length of major axis is equal to the sum of these distances <math>(2a)</math>. Thus if we find sum of distances, we get the answer. Let k be this fixed sum, then we get by distance formula:
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An ellipse is defined as the set of points equal to fixed sum of distances from foci. Length of major axis is equal to the sum of these distances <math>(2a)</math>. Thus if we find the sum of the distances, we get the answer. Let k be this fixed sum; then we get, by the distance formula:
  
 
<math>k = \sqrt{(x - 9)^2 + 20^2} + \sqrt{(x - 49)^2 + 55^2}</math>
 
<math>k = \sqrt{(x - 9)^2 + 20^2} + \sqrt{(x - 49)^2 + 55^2}</math>

Revision as of 21:05, 2 July 2018

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 1

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. (The last claim begs justification: Let $F'_2$ be the reflection of $F_2$ across the $y$-axis. Let $Y$ be where the line through $F_1$ and $F’_2$ intersects the ellipse. We will show that $X=Y$. Note that $X F_2 = X F’_2$ since $X$ is on the $x$-axis. Also, since the entire ellipse is on or above the $x$-axis and the line through $F_2$ and $F’_2$ is perpendicular to the $x$-axis, we must have $F_2 Y \leq F’_2 Y$ with equality if and only if $Y$ is on the $x$-axis. Now, we have \[F_1 X + F'_2 X = F_1 X + F_2 X = F_1 Y + F_2 Y \leq F_1 Y + F’_2 Y\] But the right most sum is the straight-line distance from $F_1$ to $F’_2$ and the left is the distance of some path from $F_1$ to $F_2$., so this is only possible if we have equality and thus $X = Y$). 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 (as above) 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, and by the above argument, this path passes through the point of tangency of the ellipse with the $x-$axis.

[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 $\boxed{085}$.

Solution 2 (Calculus)

An ellipse is defined as the set of points equal to fixed sum of distances from foci. Length of major axis is equal to the sum of these distances $(2a)$. Thus if we find the sum of the distances, we get the answer. Let k be this fixed sum; then we get, by the distance formula:

$k = \sqrt{(x - 9)^2 + 20^2} + \sqrt{(x - 49)^2 + 55^2}$

This is the equation of the ellipse expressed in terms of $x$. The line tangent to the ellipse at the given point $P(x, 0)$ will thus have slope $0$. Taking the derivative gives us the slope of this line. To simplify, let $f(x) = (x - 9)^2 + 20^2$ and $g(x) = (x - 49)^2 + 55^2$. Then we get:

$0 = \frac{f^\prime(x)}{2\sqrt{f(x)}} + \frac{g^\prime(x)}{2\sqrt{g(x)}}$

Next, we multiply by the conjugate to remove square roots. We next move the resulting $a^2 - b^2$ form expression into form $a^2 = b^2$.

$\frac{(f^\prime(x))^2}{4\cdot f(x)} = \frac{(g^\prime(x))^2}{4\cdot g(x)}$

We know $f^\prime(x) = 2x - 18$ and $g^\prime(x) = 2x - 98$. Simplifying yields:

$\frac{(x - 9)^2}{(x - 9)^2 + 20^2} = \frac{(x - 49)^2}{(x - 49)^2 + 55^2}$

To further simplify, let $a = (x - 9)^2$ and $b = (x - 49)^2$. This means $\frac{a}{a + 400} = \frac{b}{b + 3025}$. Solving yields that $16b = 121a$. Substituting back $a$ and $b$ yields:

$16 \cdot (x - 49)^2 = 121 \cdot (x - 9)^2$.

Solving for $x$ yields $x = \frac{59}{3}$. Substituting back into our original distance formula, solving for $k$ yields $\boxed{085}$.

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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