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

m (Fixed type-o. Changed "coefficients" to "coordinates.")
m (→‎Solution)
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Let <math>m = 4x+3y \in \mathbb Z</math>, <math>\left\lvert m \right\rvert \le 1000</math>.  Put <math>m = 25k</math> to obtain
 
Let <math>m = 4x+3y \in \mathbb Z</math>, <math>\left\lvert m \right\rvert \le 1000</math>.  Put <math>m = 25k</math> to obtain
 
<cmath>25k^2 &= 6x-8y+25</cmath><cmath>25k &= 4x+3y.</cmath>
 
<cmath>25k^2 &= 6x-8y+25</cmath><cmath>25k &= 4x+3y.</cmath>
−
and accordingly we find by solving the system that <math>x = \frac{1}{2} (3k^2-3) + 8</math> and <math>y = -2k^2+3k+2</math>.
+
and accordingly we find by solving the system that <math>x = \frac{1}{2} (3k^2-3) + 8k</math> and <math>y = -2k^2+3k+2</math>.
  
 
One can show that the values of <math>k</math> that make <math>(x,y)</math> an integer pair are precisely odd integers <math>k</math>.  For <math>\left\lvert 25k \right\rvert \le 1000</math> this is <math>k= -39,-37,-35,\dots,39</math>, so <math>40</math> values work and the answer is <math>\boxed{\textbf{(B)}}</math>.
 
One can show that the values of <math>k</math> that make <math>(x,y)</math> an integer pair are precisely odd integers <math>k</math>.  For <math>\left\lvert 25k \right\rvert \le 1000</math> this is <math>k= -39,-37,-35,\dots,39</math>, so <math>40</math> values work and the answer is <math>\boxed{\textbf{(B)}}</math>.

Revision as of 11:19, 4 October 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 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 $(-7,1)$. 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 $x = \frac{1}{2} (3k^2-3) + 8k$ 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
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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. 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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