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

Revision as of 13:42, 12 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

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 $x \rightarrow x+20$ to obtain $20+x=y^2-k , y=2x^2-k$ . Multiply the first equation by 2 and sum, we see that $2(x^2+y^2)=3k+40+2x+y$ . Completing the square gives us $(x- \frac{1}{2})^2+(y - \frac{1}{4})^2 = \frac{325+24k}{16}$ ; 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

@@ Line 16: / Line 16: @@
 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 <math>ax^2+by^2=f(x,y)</math> and the other as <math>cx^2+dy^2=g(x,y)</math> for f,g polynomials of degree at most 1, whenever <math>(a,b),(c,d)</math> are linearly independent, we can combine the two equations and then complete the square. We can also combine these two equations to form a parabola, or a hyperbola, or an ellipse. When <math>(a,b),(c,d)</math> 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.
+*In general, (Assuming four intersections exist) when two conics intersect, if one conic can be written as <math>ax^2+by^2=f(x,y)</math> and the other as <math>cx^2+dy^2=g(x,y)</math> for f,g polynomials of degree at most 1, whenever <math>(a,b),(c,d)</math> are linearly independent, we can combine the two equations and then complete the square to achieve <math>(x-p)^2+(y-q)^2=r^2</math>. We can also combine these two equations to form a parabola, or a hyperbola, or an ellipse. When <math>(a,b),(c,d)</math> 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

2021 AIME I (Problems • Answer Key • Resources)
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