Difference between revisions of "2021 AIME I Problems/Problem 15"
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− | 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>. | + | 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>\left(y- \frac{1}{2}\right)^2+\left(x - \frac{1}{4}\right)^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. (Here I'm using the un-translated coordinates.) Draw up a graph, and realize that two intersections are guaranteed, on the so called "right branch" of <math>y=x^2-k</math>. As we increase the value of k, two more intersections appear on the "left branch." | For the lower bound, we need to ensure there are 4 intersections to begin with. (Here I'm using the un-translated coordinates.) Draw up a graph, and realize that two intersections are guaranteed, on the so called "right branch" of <math>y=x^2-k</math>. As we increase the value of k, two more intersections appear on the "left branch." |
Revision as of 17:09, 31 May 2021
Contents
Problem
Let be the set of positive integers such that the two parabolasintersect 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. (Here I'm using the un-translated coordinates.) Draw up a graph, and realize that two intersections are guaranteed, on the so called "right branch" of . As we increase the value of k, two more intersections appear on the "left branch."
does not work because the "leftmost" point of is which lies to the right of , which is on the graph . While technically speaking this doesn't prove that there are no intersections (why?), drawing the graph should convince you that this is the case. Clearly, no k less than 4 works either.
does work because the two graphs intersect at , and by drawing the graph, you realize this is not a tangent point and there is in fact another intersection nearby, due to slope. Therefore, the answer is .
- 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.