Difference between revisions of "2021 AIME I Problems/Problem 15"
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For the lower bound, we need to ensure there are 4 intersections to begin with. (Here I'm using the un-translated coordinates.) | For the lower bound, we need to ensure there are 4 intersections to begin with. (Here I'm using the un-translated coordinates.) | ||
| − | <math>k=4</math> does not work because the "leftmost" point of <math>x=2(y-20)^2-k</math> is <math>(-4,20)</math> which lies to the right of <math>(-\sqrt{24}, 20)</math>, which is on the graph <math>y=x^2-k</math>. | + | <math>k=4</math> does not work because the "leftmost" point of <math>x=2(y-20)^2-k</math> is <math>(-4,20)</math> which lies to the right of <math>(-\sqrt{24}, 20)</math>, which is on the graph <math>y=x^2-k</math>. Clearly, no k less than 4 works either. |
<math>k=5</math> does work because the two graphs intersect at <math>(-5,20)</math>, and by drawing the graph, you realize this is not a tangent point and there is in fact another intersection, due to slope. Therefore, the answer is <math>5+280=285</math>. | <math>k=5</math> does work because the two graphs intersect at <math>(-5,20)</math>, and by drawing the graph, you realize this is not a tangent point and there is in fact another intersection, due to slope. Therefore, the answer is <math>5+280=285</math>. | ||
Revision as of 10:17, 16 March 2021
Problem
Let 
be the set of positive integers 
such that the two parabolas![\[y=x^2-k~~\text{and}~~x=2(y-20)^2-k\]](http://latex.artofproblemsolving.com/6/f/9/6f9e25390a9657644dc69c3cd2ca6d154cd6e9be.png)
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. (Here I'm using the un-translated coordinates.)
does not work because the "leftmost" point of 
is 
which lies to the right of 
, which is on the graph 
. 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, 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.
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 -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. 
