AIME 2026 II · 第 6 题

Solution 1

The parabola has equation $y=\frac{1}{2}(x-2)(x-6)$, which has vertex $(4,-2)$ and axis of symmetry $x=4$. Suppose the circle is tangent to the parabola at a point $(a,b)$. Note that the slope of the tangent line to the parabola at $(a,b)$ has slope $y'=a-4$. In the case where $a=4$ ($b=-2$), this tangent is horizontal and $r=41$ is the only real number that works.

Assume $a\neq 4$. The line between $(4,39)$ and $(a,b)$ must be perpendicular to the tangent line of slope $a-4$, so it turns out that

$$\frac{b-39}{a-4} = -\frac{1}{a-4}$$ and $b=38$. Solving, we find $a=4\pm 4\sqrt{5}$, and we have

$$r = \sqrt{1+(\pm4\sqrt{5})^2} = 9.$$ The only working real numbers are $41$ and $9$, and they have sum $\boxed{050}$. ~TThB0501

Solution 2

We start with the parabola $2y = x^2 - 8x + 12$, which can be rewritten in vertex form as:

$$y = \frac{1}{2}(x - 4)^2 - 2.$$ The vertex of the parabola is at $(4, -2)$.

The circle is centered at $(4, 39)$ with radius $r$, so its equation is:

$$(x - 4)^2 + (y - 39)^2 = r^2.$$ First, we note an obvious case: the distance from the center $(4, 39)$ to the vertex $(4, -2)$ of the parabola is $39 - (-2) = 41$. Thus, $r = 41$ is a solution where the circle is tangent to the parabola at its vertex.

Next, we consider the general case by substituting the parabola equation into the circle equation. From $2y = x^2 - 8x + 12$, we have $x^2 - 8x = 2y - 12$. Substituting this into the circle equation gives:

$$(x^2 - 8x + 16) + (y - 39)^2 = r^2,$$ $$(2y - 12 + 16) + y^2 - 78y + 39^2 = r^2,$$ $$y^2 - 76y + (39^2 + 4 - r^2) = 0.$$ For the circle and parabola to be tangent, this quadratic equation in $y$ must have exactly one solution, so its discriminant must be zero:

$$\Delta = (-76)^2 - 4 \cdot 1 \cdot (39^2 + 4 - r^2) = 0.$$ We calculate the discriminant:

$$76^2 - 4(39^2 + 4 - r^2) = 0,$$ $$38^2 - (39^2 + 4 - r^2) = 0,$$ $$r^2 = 39^2 + 4 - 38^2.$$ Using the difference of squares $a^2 - b^2 = (a - b)(a + b)$:

$$r^2 = (39 - 38)(39 + 38) + 4 = 1 \cdot 77 + 4 = 81,$$ $$r = 9.$$ Thus, the real numbers $r$ are $9$ and $41$. The sum of all such $r$ is:

$$9 + 41 = \boxed{050}.$$ ~Steven Zheng

Solution 3

We first substitute $2y = x^2 - 8x + 12$ into the circle $(x-4)^2 + (y-39)^2 = r^2$:

$$(x-4)^2 + (\frac{1}{2}x^2-4x-33)^2 = r^2$$ .

$$x^2-8x+16 + \frac{1}{4}x^4-4x^3-17x^2+264x+1089 = r^2$$ $$\frac{1}{4}x^4-4x^3-16x^2+256x+1105-r^2=0$$ Since we are finding the tangent points, we should find the values of r so that the $y$ values of the critical points is $0$. We know that the function would not go through the $x$ axis at the critical point, since we can imagine the circle expanding to the parabola, and the tangency point would not cross the parabola, either the circle tangency completely being in the parabola, or completely outside. Since we are finding the tangency point, we can take the derivative of $\frac{1}{4}x^4-4x^3-16x^2+256x+1105-r^2$:

$$\frac{d}{dx}(\frac{1}{4}x^4-4x^3-16x^2+256x+1105-r^2)$$ $$=x^3-12x^2-32x+256$$ Using Rational Root Theorem, we can factor the expression to get $(x-4)(x^2-8x-64)$. The solutions are $x=4,x=4+4\sqrt5,$ and $x=4-4\sqrt5$. Now we plug in the numbers back into $\frac{1}{4}x^4-4x^3-16x^2+256x+1105-r^2=0$.

• Plug in $x=4$ we get $1681-r^2=0$, so $r=41$.
• Plug in $x=4\pm 4\sqrt5$ we get $81-r^2=0$, so $r=9$. (they are the same because it's symmetrical, the AoS passes through the center of circle)

Solution is $9+41=\boxed{050}$

~TheRealMee

Solution 4 (Not a great solution, but simple)

Since the center of the circle is $(4,39),$ we know that the circle must be in the form $(x-4)^2+(y-39)^2=r^2.$ The vertex of the parabola $2y=x^2-8x+12$ is $(4,-2),$ so the circle must have a radius of $39-(-2)=41$ to be tangent to the vertex. Since the line from a circle's center to a tangent is always perpendicular to that tangent, and the slope of the parabola is very close to a vertical line as $x$ becomes further from $4,$ we can find the $x-$value of the horizontal line from the circle's center to where it intersects the parabola. This means that the $x-$value of the position where the parabola reaches $39$ is a great estimate for the $x-$value of the tangent point. Plugging $39$ in for $y,$ we get that the $x$ value is equal to $4\pm{\sqrt{82}}.$ This means that the radius of the circle must be very close to $\sqrt{82}.$ Since the radius must be an integer since this is an AIME problem, we find that $9=\sqrt{81},$ giving $9$ as a second radius. Adding these two values up, $41+9=50,$ so our answer is $\boxed{50}.$

~Nodskin Goble (feel free to change the wording of this solution)