AIME 2021 I, Problem 15

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: Note that $y=x^2-k$ is an upward-opening parabola with the vertex at $(0,-k),$ and $x=2(y-20)^2-k$ is a rightward-opening parabola with the vertex at $(-k,20).$ We consider each condition separately: The two parabolas intersect at four distinct points. By a quick sketch, we have two subconditions: The point $(-k,20)$ is on or below the parabola $y=x^2-k.$ We need $20\leq(-k)^2-k,$ from which $k\geq5.$ Moreover, the point $(-k,20)$ is on the parabola $y=x^2-k$ when $k=5.$ We will prove that the two parabolas intersect at four distinct points at this value of $k:$ Substituting $y=x^2-5$ into $x=2(y-20)^2-5,$ we get $x=2\left(\left(x^2-5\right)-20\right)^2-5.$ Expanding and rearranging give \[2x^4-100x^2-x+1245=0. \hspace{20mm}(\bigstar)\] By either the graphs of the parabolas or the Rational Root Theorem, we conclude that $x=-5$ is a root of $(\bigstar).$ So, we factor its left side: \[(x+5)\left(2x^3-10x^2-50x+249\right)=0.\] By either the graphs of the parabolas or Descartes' Rule of Signs, we conclude that $2x^3-10x^2-50x+249=0$ has two positive roots and one negative root such that $x\neq-5.$ So, $(\bigstar)$ has four distinct real roots, or the two parabolas intersect at four distinct points. For Subcondition A, we deduce that $k\geq5.$ Remark for Subcondition A Recall that if $1\leq k\leq 4,$ then the point $(-k,20)$ is above the parabola $y=x^2-k.$ It follows that for $-k\leq x\leq0:$ The maximum value of $y$ for the parabola $y=x^2-k$ occurs at $x=-k,$ from which $y=k^2-k\leq12.$ The minimum value of $y$ for the parabola $x=2(y-20)^2-k$ occurs at $x=0,$ from which $y=20-\sqrt{\frac k2}>18.$ Clearly, the parabola $x=2(y-20)^2-k$ and the left half of the parabola $y=x^2-k$ do not intersect. Therefore, the two parabolas do not intersect at four distinct points. The point $(0,-k)$ is on or below the parabola $x=2(y-20)^2-k.$ The lower half of the parabola $x=2(y-20)^2-k$ is $y=20-\sqrt{\frac{x+k}{2}}.$ We need $-k\leq20-\sqrt{\frac k2},$ which holds for all values of $k.$ For Subcondition B, we deduce that $k$ can be any positive integer. For Condition 1, we obtain $\boldsymbol{k\geq5}$ by taking the intersection of Subconditions A and B. The four points of intersection lie on a circle with radius at most $21.$ For equations of circles, the coefficients of $x^2$ and $y^2$ must be the same. So, we add the equation $y=x^2-k$ to half the equation $x=2(y-20)^2-k:$ \[y+\frac12x=x^2+(y-20)^2-\frac32k.\] We expand, rearrange, and complete the squares: \begin{align} y+\frac12x&=x^2+y^2-40y+400-\frac32k \\ \frac32k-400&=\left(x^2-\frac12x\right)+\left(y^2-41y\right) \\ \frac32k-400+\frac{1}{16}+\frac{1681}{4}&=\left(x-\frac14\right)^2+\left(y-\frac{41}{2}\right)^2. \end{align} We need $\frac32k-400+\frac{1}{16}+\frac{1681}{4}\leq21^2,$ from which $k\leq\left\lfloor\frac{6731}{24}\right\rfloor=280.$ For Condition 2, we obtain $\boldsymbol{k\leq280.}$ Taking the intersection of Conditions 1 and 2 produces $5\leq k\leq280.$ Therefore, the answer is $5+280=\fbox{285}.$ ~MRENTHUSIASM

Answer: 285