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Difference between revisions of "2014 AMC 12A Problems/Problem 25"

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1)  The line of symmetry is NOT y= -x but 4x + 3y = 0
 +
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2)  In the expression for x, it is NOT 8 but 8k.
 +
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With these minor corrections, the solution still holds good.

Revision as of 18:45, 15 February 2014

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)

See Also

2014 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Question
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the 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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