2014 AMC 12A Problems/Problem 25

Problem

The parabola $P$ has focus $(0,0)$ and goes through the points $(4,3)$ and $(-4,-3)$ . For how many points $(x,y)\in P$ with integer coordinates is it true that $|4x+3y|\leq 1000$ ?

$\textbf{(A) }38\qquad \textbf{(B) }40\qquad \textbf{(C) }42\qquad \textbf{(D) }44\qquad \textbf{(E) }46\qquad$

Solution

The parabola is symmetric through $y=- \frac{4}{3}x$ , and the common distance is $5$ , so the directrix is the line through $(1,7)$ and $(-7,1)$ , which is the line $3x-4y = -25.$ Using the point-line distance formula, the parabola is the locus $x^2+y^2 = \frac{\left\lvert 3x-4y+25 \right\rvert^2}{3^2+4^2}$ which rearranges to $(4x+3y)^2 = 25(6x-8y+25)$ .

Let $m = 4x+3y \in \mathbb Z$ , $\left\lvert m \right\rvert \le 1000$ . Put $m = 25k$ to obtain $25k^2 = 6x-8y+25$ $25k = 4x+3y.$ and accordingly we find by solving the system that $x = \frac{1}{2} (3k^2-3) + 4k$ and $y = -2k^2+3k+2$ .

One can show that the values of $k$ that make $(x,y)$ an integer pair are precisely odd integers $k$ . For $\left\lvert 25k \right\rvert \le 1000$ this is $k= -39,-37,-35,\dots,39$ , so $40$ values work and the answer is $\boxed{\textbf{(B)}}$ .

(Solution by C-273)

Solution 2

The axis of $P$ is inclined at an angle $\theta$ relative to the coordinate axis, where $\tan\theta = \tfrac 34$ . We rotate the coordinate axis by angle $\theta$ anti-clockwise, so that the parabola now has a vertical symmetry axis relative to the rotated coordinates. Let $(\widetilde{x}, \widetilde{y})$ be the coordinates in the rotated system. Then $(x,y)$ and $(\widetilde{x}, \widetilde{y})$ are related by $\begin{align} \nonumber x = \widetilde{x}\cos\theta -\widetilde{y}\sin\theta &= \tfrac 45 \widetilde{x} - \tfrac 35 \widetilde{y}, \\ y = \widetilde{x}\sin\theta +\widetilde{y}\cos\theta &= \tfrac 35 \widetilde{x} + \tfrac 45 \widetilde{y} \end{align}$ In the rotated coordinate system, the parabola has focus at $(0,0)$ and the two points on it are at $(5,0)$ and $(-5,0)$ . Therefore, the directrix is $\widetilde{y}=\pm 5$ ; we can, WLOG, choose $\widetilde{y}=-5$ . For a point on the parabola, it is equidistant from the focus and directrix, so the equation of the parabola is $\begin{align}\tag{2} \widetilde{x}^2+\widetilde{y}^2 = (\widetilde{y}+5)^2 \qquad &\Longrightarrow\qquad \widetilde{y} = \tfrac{1}{10}(\widetilde{x}^2-25) \end{align}$ From $(1)$ we have $|4x+3y|=5\widetilde{x}$ , so we need $|\widetilde{x}|<200$ . Substituting $(2)$ in $(1)$ , we get $\begin{align*} 50x &= 40 \widetilde{x} - 3 \widetilde{x}^2 + 75, \\ 50y &= 30 \widetilde{x} + 4 \widetilde{x}^2 - 100 \end{align*}$ For $x$ to be an integer $\widetilde{x}$ must be a multiple of 5; setting $\widetilde{x}=5a$ we get $2x = 8a - 3 a^2 + 3$ Now we need $a$ to be odd, i.e. $\widetilde{x}=5a$ is an odd multiple of $5$ , in which case we get $y = 3 a + 2 a^2 - 2$ , which is also an integer. The values that satisfy the given conditions correspond to $\widetilde{x}= \{\pm 5\cdot (2k-1)\mid k = 0, 1, \ldots , 19 \}={-195, -185, -175, ..., 195}$ , and there are $\boxed{\textbf{(B)} \ 40}$ such numbers.

(Solution by Shaddoll)

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2014amc12a/384

~ dolphin7

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2014 AMC 12A Problems/Problem 25

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Problem

Solution

Solution 2

Video Solution by Richard Rusczyk

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