Difference between revisions of "1985 AIME Problems/Problem 11"
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== Solution 2 (Calculus) == | == Solution 2 (Calculus) == | ||
− | Ellipse is defined as set of points equal to fixed sum of distances from foci. Length of major axis is equal to 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: | + | 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: |
− | + | <math>k = \sqrt{(x - 9)^2 + 20^2} + \sqrt{(x - 49)^2 + 55^2}</math> | |
− | == See also | + | This is the equation of the ellipse expressed in terms of <math>x</math>. The line tangent to the ellipse at the given point <math>P(x, 0)</math> will thus have slope 0. Taking the derivative gives us the slope of this line. To simplify, let <math>f(x) = (x - 9)^2 + 20^2</math> and <math>g(x) = (x - 49)^2 + 55^2</math>. Then we get: |
+ | |||
+ | <math>0 = \frac{f^\prime(x)}{2\sqrt{f(x)}} + \frac{g^\prime(x)}{2\sqrt{g(x)}}</math> | ||
+ | |||
+ | Next, we multiply by the conjugate to remove square roots. We next move the resulting <math>a^2 - b^2</math> form expression into form <math>a^2 = b^2</math>. | ||
+ | |||
+ | <math>\frac{(f^\prime(x))^2}{4\cdot f(x)} = \frac{(g^\prime(x))^2}{4\cdot g(x)}</math> | ||
+ | |||
+ | We know <math>f^\prime(x) = 2x - 18</math> and <math>g^\prime(x) = 2x - 98</math>. Simplifying yields: | ||
+ | |||
+ | <math>\frac{(x - 9)^2}{(x - 9)^2 + 20^2} = \frac{(x - 49)^2}{(x - 49)^2 + 55^2}</math> | ||
+ | |||
+ | To further simplify, let <math>a = (x - 9)^2 and b = (x - 49)^2</math>. This means <math>\frac{a}{a + 400} = \frac{b}{b + 3025}</math>. Solving yields that <math>16b = 121a</math>. | ||
+ | Substituting back <math>a</math> and <math>b</math> yields: | ||
+ | |||
+ | <math>16 \cdot (x - 49)^2 = 121 \cdot (x - 9)^2</math>. | ||
+ | |||
+ | Solving for <math>x</math> yields <math>x = \frac{59}{3}</math>. Substituting back into our original distance formula, solving for <math>k</math> yields <math>\boxed{85}</math>. | ||
+ | |||
+ | == See also = | ||
{{AIME box|year=1985|num-b=10|num-a=12}} | {{AIME box|year=1985|num-b=10|num-a=12}} | ||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] |
Revision as of 13:36, 2 July 2018
Problem
An ellipse has foci at and
in the
-plane and is tangent to the
-axis. What is the length of its major axis?
Solution 1
An ellipse is defined to be the locus of points such that the sum of the distances between
and the two foci is constant. Let
,
and
be the point of tangency of the ellipse with the
-axis. Then
must be the point on the axis such that the sum
is minimal. (The last claim begs justification: Let
be the reflection of
across the
-axis. Let
be where the line through
and
intersects the ellipse. We will show that
. Note that
since
is on the
-axis. Also, since the entire ellipse is on or above the
-axis and the line through
and
is perpendicular to the
-axis, we must have
with equality if and only if
is on the
-axis. Now, we have
But the right most sum is the straight-line distance from
to
and the left is the distance of some path from
to
., so this is only possible if we have equality and thus
). Finding the optimal location for
is a classic problem: for any path from
to
and then back to
, we can reflect (as above) the second leg of this path (from
to
) across the
-axis. Then our path connects
to the reflection
of
via some point on the
-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
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]](http://latex.artofproblemsolving.com/e/1/7/e17edad4abef8ffedba4e8086dacea0dc6c8c309.png)
The sum of the two distances and
is therefore equal to the length of the segment
, which by the distance formula is just
.
Finally, let and
be the two endpoints of the major axis of the ellipse. Then by symmetry
so
(because
is on the ellipse), so the answer is
.
Solution 2 (Calculus)
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 . Thus if we find sum of distances, we get the answer. Let k be this fixed sum, then we get by distance formula:
This is the equation of the ellipse expressed in terms of . The line tangent to the ellipse at the given point
will thus have slope 0. Taking the derivative gives us the slope of this line. To simplify, let
and
. Then we get:
Next, we multiply by the conjugate to remove square roots. We next move the resulting form expression into form
.
We know and
. Simplifying yields:
To further simplify, let . This means
. Solving yields that
.
Substituting back
and
yields:
.
Solving for yields
. Substituting back into our original distance formula, solving for
yields
.
= 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 |