Difference between revisions of "2008 AIME I Problems/Problem 13"
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At this point, we see that <math>p_1</math> and <math>p_2</math> both must have <math>\left(\frac{a}{c}, \frac{b}{c}\right)</math> as a zero. A quick graph of the 4 lines and the ellipse used to create <math>p_1</math> and <math>p_2</math> gives nine intersection points. Eight of them are the given ones, and the ninth is <math>\left(\frac{5}{9}, \frac{16}{9}\right)</math>. The last intersection point can be found by finding the intersection points of <math>y = \frac{2}{3}x + \frac{2}{3}</math> and <math>x^2 + xy + y^2 = 1</math>. | At this point, we see that <math>p_1</math> and <math>p_2</math> both must have <math>\left(\frac{a}{c}, \frac{b}{c}\right)</math> as a zero. A quick graph of the 4 lines and the ellipse used to create <math>p_1</math> and <math>p_2</math> gives nine intersection points. Eight of them are the given ones, and the ninth is <math>\left(\frac{5}{9}, \frac{16}{9}\right)</math>. The last intersection point can be found by finding the intersection points of <math>y = \frac{2}{3}x + \frac{2}{3}</math> and <math>x^2 + xy + y^2 = 1</math>. | ||
− | Finally, just add the values of <math>a</math>, <math>b</math>, and <math>c</math> to get <math>5 + 16 + | + | Finally, just add the values of <math>a</math>, <math>b</math>, and <math>c</math> to get <math>5 + 16 + 19 = \boxed{040}</math> |
== See also == | == See also == |
Revision as of 00:30, 21 March 2018
Problem
Let
Suppose that
There is a point for which
for all such polynomials, where
,
, and
are positive integers,
and
are relatively prime, and
. Find
.
Solution
Solution 1
Adding the above two equations gives , and so we can deduce that
.
Similarly, plugging in and
gives
and
. Now,
Therefore
and
. Finally,
So
.
Now
.
In order for the above to be zero, we must have
![$x(1-x)(1+x) = y(1-y)(1+y)$](http://latex.artofproblemsolving.com/8/4/4/8441c3c87aea15dfd987d273cd22864e0ce424ea.png)
and
![$x(1-x)(1+x) = 1.5 xy (1-x).$](http://latex.artofproblemsolving.com/e/9/c/e9ca0f9bd6511fccd9398392a365b7f5c3cb7aef.png)
Canceling terms on the second equation gives us . Plugging that into the first equation and solving yields
, and
.
Solution 2
Consider the cross section of on the plane
. We realize that we could construct the lines/curves in the cross section such that their equations multiply to match the form of
and they go over the eight given points. A simple way to do this would be to use the equations
,
, and
, giving us
.
Another way to do this would to use the line and the ellipse,
. This would give
.
At this point, we see that and
both must have
as a zero. A quick graph of the 4 lines and the ellipse used to create
and
gives nine intersection points. Eight of them are the given ones, and the ninth is
. The last intersection point can be found by finding the intersection points of
and
.
Finally, just add the values of
,
, and
to get
See also
2008 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.