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 coefficients 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 $y=-x$ , and the common distance is $5$ , so the directrix is the line through $(1,7)$ and $(-7,1)$ . That's 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

\[25k^2 &= 6x-8y+25\] (Error compiling LaTeX. Unknown error_msg)

\[25k &= 4x+3y.\] (Error compiling LaTeX. Unknown error_msg)

and accordingly we find by solving the system that $x = \frac{1}{2} (3k^2-3) + 8$ 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 v_Enhance)

2014 AMC 12A (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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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2014 AMC 12A Problems/Problem 25

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