Difference between revisions of "2021 AIME I Problems/Problem 15"
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| + | ===Solution 2=== | ||
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| + | Make the translation <math>x \rightarrow x+20</math> to obtain <math>20+x=y^2-k , y=2x^2-k</math>. Multiply the second equation by 2 and sum, we see that <math>2(x^2+y^2)=3k+40+2x+y</math>. Completing the square gives us <math>(x- \frac{1}{2})^2+(y - \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 \rightarrow 24k \leq 6731</math>, so <math>k \leq 280</math>. | ||
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| + | 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. | ||
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| + | -Ross Gao | ||
==See also== | ==See also== | ||
{{AIME box|year=2021|n=I|num-b=14|after=Last problem}} | {{AIME box|year=2021|n=I|num-b=14|after=Last problem}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 09:52, 12 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
Solution 1
With binary search you can narrow down the k value. Newton raphson method let you narrow down the x and y solution for that specific k value. With 3 (x,y) pairs you can find radius of the circle.
You end up finding the bounds of 5 and 280. The sum is 285.
~Lopkiloinm
Solution 2
Make the translation 
to obtain 
. Multiply the second 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.
-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. 
