Difference between revisions of "2008 AIME I Problems/Problem 13"
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As the answer format implies that the <math>x</math>-coordinate of the root is non-integral, <math>x(x - 1)\left(x + 1 - \frac{3}{2}y\right) = 0 \iff x + 1 - \frac{3}{2}y = 0 \iff y = \frac{2}{3}(x + 1)\ (1)</math>. The format also implies that <math>y</math> is positive, so <math>y\left(y^2 - 1 - \frac{3}{2}x(x - 1)\right) = 0 \iff y^2 - 1 - \frac{3}{2}x(x - 1) = 0\ (2)</math>. Substituting <math>(1)</math> into <math>(2)</math> and reducing to a quadratic yields <math>(19x - 5)(x - 2) = 0</math>, in which the only non-integral root is <math>x = \frac{5}{19}</math>, so <math>y = \frac{16}{19}</math>. | As the answer format implies that the <math>x</math>-coordinate of the root is non-integral, <math>x(x - 1)\left(x + 1 - \frac{3}{2}y\right) = 0 \iff x + 1 - \frac{3}{2}y = 0 \iff y = \frac{2}{3}(x + 1)\ (1)</math>. The format also implies that <math>y</math> is positive, so <math>y\left(y^2 - 1 - \frac{3}{2}x(x - 1)\right) = 0 \iff y^2 - 1 - \frac{3}{2}x(x - 1) = 0\ (2)</math>. Substituting <math>(1)</math> into <math>(2)</math> and reducing to a quadratic yields <math>(19x - 5)(x - 2) = 0</math>, in which the only non-integral root is <math>x = \frac{5}{19}</math>, so <math>y = \frac{16}{19}</math>. | ||
− | The answer is <math>5 + 16 + 19 = \boxed{ | + | The answer is <math>5 + 16 + 19 = \boxed{040}</math>. |
=== Solution 2 === | === Solution 2 === |
Revision as of 13:30, 3 January 2021
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,
, or equivalently
.
Substituting these equations into the original polynomial , we find that at
,
.
The remaining coefficients
and
are now completely arbitrary because the original equations impose no more restrictions on them. Hence, for the final equation to hold for all possible
, we must have
.
As the answer format implies that the -coordinate of the root is non-integral,
. The format also implies that
is positive, so
. Substituting
into
and reducing to a quadratic yields
, in which the only non-integral root is
, so
.
The answer is .
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.