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

(Created page with "==Problem== The parabola <math>P</math> has focus <math>(0,0)</math> and goes through the points <math>(4,3)</math> and <math>(-4,-3)</math>. For how many points <math>(x,y)\in...")
 
(→‎Solution)
Line 13: Line 13:
 
The parabola is symmetric through <math>y=-x</math>, and the common distance is <math>5</math>, so the directrix is the line through <math>(1,7)</math> and <math>(-7,1)</math>. That's the line <cmath> 3x-4y = -25. </cmath> Using the point-line distance formula, the parabola is the locus <cmath> x^2+y^2 = \frac{\left\lvert 3x-4y+25 \right\rvert^2}{3^2+4^2} </cmath> which rearranges to <math>(4x+3y)^2 = 25(6x-8y+25)</math>.
 
The parabola is symmetric through <math>y=-x</math>, and the common distance is <math>5</math>, so the directrix is the line through <math>(1,7)</math> and <math>(-7,1)</math>. That's the line <cmath> 3x-4y = -25. </cmath> Using the point-line distance formula, the parabola is the locus <cmath> x^2+y^2 = \frac{\left\lvert 3x-4y+25 \right\rvert^2}{3^2+4^2} </cmath> which rearranges to <math>(4x+3y)^2 = 25(6x-8y+25)</math>.
  
−
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
−
\begin{align*}
+
<cmath>25k^2 &= 6x-8y+25</cmath><cmath>25k &= 4x+3y.</cmath>
−
25k^2 &= 6x-8y+25 \\
+
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>.
−
25k &= 4x+3y.
 
−
\end{align*}
 
−
\] 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>.
 
  
 
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>.
  
 
(Solution by v_Enhance)
 
(Solution by v_Enhance)

Revision as of 19:39, 7 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)

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