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Difference between revisions of "2021 AIME I Problems/Problem 15"

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===Solution 1===
 
===Solution 1===
  
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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 first 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 21^2=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>(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 $S$ be the set of positive integers $k$ such that the two parabolas\[y=x^2-k~~\text{and}~~x=2(y-20)^2-k\]intersect in four distinct points, and these four points lie on a circle with radius at most $21$. Find the sum of the least element of $S$ and the greatest element of $S$.

Solution 1

Make the translation $y \rightarrow y+20$ to obtain $20+y=x^2-k , x=2y^2-k$. Multiply the first equation by 2 and sum, we see that $2(x^2+y^2)=3k+40+2y+x$. Completing the square gives us $(y- \frac{1}{2})^2+(x - \frac{1}{4})^2 = \frac{325+24k}{16}$; this explains why the two parabolas intersect at four points that lie on a circle*. For the upper bound, observe that $LHS \leq 21^2=441 \rightarrow 24k \leq 6731$, so $k \leq 280$.

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 $ax^2+by^2=f(x,y)$ and the other as $cx^2+dy^2=g(x,y)$ for f,g polynomials of degree at most 1, whenever $(a,b),(c,d)$ are linearly independent, we can combine the two equations and then complete the square to achieve $(x-p)^2+(y-q)^2=r^2$. We can also combine these two equations to form a parabola, or a hyperbola, or an ellipse. When $(a,b),(c,d)$ 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)
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