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

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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 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>.
 
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>.
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== Solution 1 Simplified ==
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This assumes you know properties of ellipses.
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Like above, X is the point on the x-axis minimizing <math>XF_1+XF_2</math>.
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By reflection, this equals <math>F_1F_2'</math>
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...But this just equals the length of the major axis.
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Solving like above, we get <math>\boxed{085}</math>
  
 
== Solution 2 (Calculus) ==
 
== Solution 2 (Calculus) ==

Latest revision as of 23:57, 5 December 2024

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 $x$-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 1 Simplified

This assumes you know properties of ellipses. Like above, X is the point on the x-axis minimizing $XF_1+XF_2$. By reflection, this equals $F_1F_2'$ ...But this just equals the length of the major axis. Solving like above, we get $\boxed{085}$

Solution 2 (Calculus)

An ellipse is defined as the set of points where the sum of the distances from the foci to the point is fixed. The 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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