Difference between revisions of "2021 AIME I Problems/Problem 15"
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− | Make the translation <math> | + | Make the translation <math>y \rightarrow y+20</math> to obtain <math>20+y=x^2-k , x=2y^2-k</math>. Multiply the first equation by 2 and sum, we see that <math>2(x^2+y^2)=3k+40+2y+x</math>. Completing the square gives us <math>(y- \frac{1}{2})^2+(x - \frac{1}{4})^2 = \frac{325+24k}{16}</math>; this explains why the two parabolas intersect at four points that lie on a circle*. For the upper bound, observe that <math>LHS \leq 21^2=441 \rightarrow 24k \leq 6731</math>, so <math>k \leq 280</math>. |
For the lower bound, we need to ensure there are 4 intersections to begin with. A quick check shows k=5 works while k=4 does not. Therefore, the answer is 5+280=285. | For the lower bound, we need to ensure there are 4 intersections to begin with. A quick check shows k=5 works while k=4 does not. Therefore, the answer is 5+280=285. |
Revision as of 22:49, 14 March 2021
Problem
Let be the set of positive integers
such that the two parabolas
intersect in four distinct points, and these four points lie on a circle with radius at most
. Find the sum of the least element of
and the greatest element of
.
Solution 1
Make the translation to obtain
. Multiply the first equation by 2 and sum, we see that
. Completing the square gives us
; this explains why the two parabolas intersect at four points that lie on a circle*. For the upper bound, observe that
, so
.
For the lower bound, we need to ensure there are 4 intersections to begin with. A quick check shows k=5 works while k=4 does not. Therefore, the answer is 5+280=285.
- In general, (Assuming four intersections exist) when two conics intersect, if one conic can be written as
and the other as
for f,g polynomials of degree at most 1, whenever
are linearly independent, we can combine the two equations and then complete the square to achieve
. We can also combine these two equations to form a parabola, or a hyperbola, or an ellipse. When
are not L.I., the intersection points instead lie on a line, which is a circle of radius infinity. When the two conics only have 3,2 or 1 intersection points, the statement that all these points lie on a circle is trivially true.
-Ross Gao
See also
2021 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last problem | |
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.