2014 AMC 12A Problems/Problem 25

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The parabola $P$ has focus $(0,0)$ and goes through the points $(4,3)$ and $(-4,-3)$. For how many lattice points $(x,y)\in P$ does $-250 \leq x+\frac{3}{4}y \leq 250$ hold?

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


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

Consider the rotation of axes such that the axes are the lines passing through the origin with slope $\dfrac{3}{4}$ and $-\dfrac{4}{3}$ for x-axis and y-axis, respectively, and let the point on the rotated axis be $(x_1, y_1)$. We can check that $x=\dfrac{4}{5}x_1-\dfrac{3}{5}y_1$ and $y=\dfrac{3}{5}x_1+\dfrac{4}{5}y_1$ by the distance from a point to line formula $\dfrac {ax_0+by_0+c} {\sqrt{a^{2}+b^{2}}}$ where the equation of the line is $ax_0+by_0+c=0$ and $(x_0, y_0)$ is the point. We have the focus as $(0,0)$ and $(5,0)$ and $(-5,0)$ as points on the parabola(on the rotated axes). Therefore, the directrix is $y=\pm 5$, and it doesn't matter which one(due to the absolute value) so WLOG we choose $y_1=-5$. The vertex is the midpoint between the focus and the foot of the altitude from focus to directrix, so the vertex is $(0, -\dfrac{5}{2})$. Therefore, the equation is $y_1=\dfrac{x_1^{2}}{10}-\dfrac{5}{2}$, and from the equations above we have $|3x+4y|=5x_1$, so $|{x_1}|<200$. One can check with $4x+3y$ and $4y-3x$ that the only time $x$ and $y$ can both be integers is when $x_1$ and $y_1$ are both integer multiples of $\dfrac{1}{5}$. Therefore, the only time is when $x_1$ is an odd multiple of 5(otherwise $y_1$ is not a multiple of $\dfrac{1}{5}$, and this is obviously sufficient because $y_1$ is also a multiple of $5$. The values that satisfy thus are $x_1={-195, -185, -175, ..., 195}$, and there are $\boxed{(B) 40}$ such numbers.

(Solution by Shaddoll)

Video Solution by Richard Rusczyk


~ dolphin7

See Also

2014 AMC 12A (ProblemsAnswer KeyResources)
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

1) The line of symmetry is NOT y= -x but 4x + 3y = 0

2) In the expression for x, it is NOT 8 but 8k.

With these minor corrections, the solution still holds good.

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