, 2012

Solution 1

Since the $x$-intercepts of the parabola with equation $y=2(x-3)(x-5)$ are $x=3$ and $x=5$, then its axis of symmetry is at $x = \frac{1}{2}(3+5)=4$. If a horizontal line intersects the parabola at two points, then these points are symmetric across the axis of symmetry. Since the line $y=k$ intersects the parabola at two points $A$ and $B$ with $AB=6$, then each of $A$ and $B$ must be 3 units from the axis of symmetry. Therefore, the $x$-coordinates of $A$ and $B$ are $4-3=1$ and $4+3=7$. Thus, the coordinates of $A$ and $B$, in some order, are $(1,k)$ and $(7,k)$. Substituting $(1,k)$ into the equation of the parabola gives $k = 2(1-3)(1-5)=16$. (Substituting $(7,k)$ would give the same value of $k$.) Solution 2 Let $x_A$ be the $x$-coordinate of $A$ and $x_B$ be the $x$-coordinate of $B$. We may assume that $A$ is to the left of $B$; that is, we assume that $x_A<x_B$. Since $AB$ is horizontal and $AB=6$, then $x_B-x_A=6$. Since $A$ and $B$ are the points of intersection between the line with equation $y=k$ and the parabola with equation $y=2(x-3)(x-5)$, then we can solve for $x_A$ and $x_B$ by equating values of $y$ to obtain the equation $k = 2(x-3)(x-5)$, which is equivalent to $k = 2(x^2-8x+15)$ or $2x^2-16x+(30-k)=0$. Using the quadratic formula, we obtain $$x_A,x_B = \dfrac{16 \pm \sqrt{(-16)^2-4(2)(30-k)}}{2(2)}$$ Thus, $x_A = \dfrac{16 - \sqrt{16^2-4(2)(30-k)}}{2(2)}$ and $x_B = \dfrac{16 + \sqrt{16^2-4(2)(30-k)}}{2(2)}$. Since $x_B-x_A=6$, then $$\begin{aligned} \dfrac{16 + \sqrt{16^2-4(2)(30-k)}}{2(2)} - \dfrac{16 - \sqrt{16^2-4(2)(30-k)}}{2(2)} & = 6 \\ \dfrac{2\sqrt{16^2-4(2)(30-k)}}{2(2)} & = 6 \\ \sqrt{256-(240-8k)} & = 12 \\ \sqrt{16+8k} & = 12 \\ 16+8k & = 144\\ 8k & = 128 \\ k & = 16\end{aligned}$$ Therefore, $k=16$. We can double check that the line with equation $y=16$ intersects the parabola with equation $y=2(x-3)(x-5)$ at the points $(1,16)$ and $(7,16)$, which are a distance $6$ apart. Let $n = (3a+6a+9a+12a+15a)+(6b+12b+18b+24b+30b)$. First, we simplify the given expression for $n$ to obtain $$n = (3a+6a+9a+12a+15a)+(6b+12b+18b+24b+30b) = 45a+90b$$ We then factor the right side to obtain $n = 45(a+2b) = 3^2 5^1 (a+2b)$. If $a+2b=5$, then $n=3^2 5^2 = (3\times 5)^2$, which is a perfect square. Two pairs of positive integers $(a,b)$ that satisfy $a+2b=5$ are $(a,b)=(3,1)$ and $(a,b)=(1,2)$. Another value of $a+2b$ for which $n$ is a perfect square is $a+2b=20$, since here $n=3^2 5^1 20 = 3^2 5^1 2^2 5^1 = 3^2 2^2 5^2 = (3 \times 2 \times 5)^2$. A pair of positive integers $(a,b)$ that satisfies $a+2b=20$ is $(18,1)$. Therefore, three pairs of positive integers $(a,b)$ with the required property are $(3,1),(1,2),(18,1)$.

(There are infinitely many other pairs with this property.)