Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution)
m (→‎Solution)
Line 11: Line 11:
 
==Solution==
 
==Solution==
  
−
The parabola is symmetric through <math>y=- \frac{3}{4}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=- \frac{4}{3}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

Revision as of 10:13, 30 July 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 $y=- \frac{4}{3}x$, 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 $x = \frac{1}{2} (3k^2-3) + 8$ 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

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.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_12A_Problems/Problem_25&oldid=62715"