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 (→‎Being more specific)
m (→‎Missed a hole)
Line 23: Line 23:
 
==Solution 2==
 
==Solution 2==
  
−
Consider the rotation of axes such that the axes are the lines passing through the origin with slope <math>\dfrac{3}{4}</math> and <math>-\dfrac{4}{3}</math> for x-axis and y-axis, respectively, and let the point on the rotated axis be <math>(x_1, y_1)</math>. We can check that <math>x=\dfrac{4}{5}x_1-\dfrac{3}{5}y_1</math> and <math>y=\dfrac{3}{5}x_1+\dfrac{4}{5}y_1</math> by dropping perpendiculars from the rotated axes to the original axes. We have the focus as <math>(0,0)</math> and <math>(5,0)</math> and <math>(-5,0)</math> as points on the parabola(on the rotated axes). Therefore, the directrix is <math>y=\pm 5</math>, and it doesn't matter which one(due to the absolute value) so WLOG we choose <math>y_1=-5</math>. The vertex is the midpoint between the focus and the foot of the altitude from focus to directrix, so the vertex is <math>(0, -\dfrac{5}{2})</math>. Therefore, the equation is <math>y_1=\dfrac{x_1^{2}}{10}-\dfrac{5}{2}</math>, and from the equations above we have <math>|3x+4y|=5x_1</math>, so <math>|{x_1}|<200</math>. One can check with <math>7x+y</math> that the only time <math>x</math> and <math>y</math> can both be integers is when <math>x_1</math> and <math>y_1</math> are both integer multiples of <math>\dfrac{1}{5}</math>. Therefore, the only time is when <math>x_1</math> is an odd multiple of 5, and this is obviously sufficient because <math>y_1</math> is also a multiple of <math>5</math>. The values that satisfy thus are <math>x={-195, -185, -175, ..., 195}</math>, and there are <math>\boxed{(B) 40}</math> such numbers.
+
Consider the rotation of axes such that the axes are the lines passing through the origin with slope <math>\dfrac{3}{4}</math> and <math>-\dfrac{4}{3}</math> for x-axis and y-axis, respectively, and let the point on the rotated axis be <math>(x_1, y_1)</math>. We can check that <math>x=\dfrac{4}{5}x_1-\dfrac{3}{5}y_1</math> and <math>y=\dfrac{3}{5}x_1+\dfrac{4}{5}y_1</math> by dropping perpendiculars from the rotated axes to the original axes. We have the focus as <math>(0,0)</math> and <math>(5,0)</math> and <math>(-5,0)</math> as points on the parabola(on the rotated axes). Therefore, the directrix is <math>y=\pm 5</math>, and it doesn't matter which one(due to the absolute value) so WLOG we choose <math>y_1=-5</math>. The vertex is the midpoint between the focus and the foot of the altitude from focus to directrix, so the vertex is <math>(0, -\dfrac{5}{2})</math>. Therefore, the equation is <math>y_1=\dfrac{x_1^{2}}{10}-\dfrac{5}{2}</math>, and from the equations above we have <math>|3x+4y|=5x_1</math>, so <math>|{x_1}|<200</math>. One can check with <math>4x+3y</math> and <math>4y-3x</math> that the only time <math>x</math> and <math>y</math> can both be integers is when <math>x_1</math> and <math>y_1</math> are both integer multiples of <math>\dfrac{1}{5}</math>. Therefore, the only time is when <math>x_1</math> is an odd multiple of 5, and this is obviously sufficient because <math>y_1</math> is also a multiple of <math>5</math>. The values that satisfy thus are <math>x={-195, -185, -175, ..., 195}</math>, and there are <math>\boxed{(B) 40}</math> such numbers.
  
 
(Solution by Shaddoll)
 
(Solution by Shaddoll)

Revision as of 23:50, 30 January 2016

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 \[25k^2 = 6x-8y+25\]\[25k = 4x+3y.\] and accordingly we find by solving the system that $x = \frac{1}{2} (3k^2-3) + 4k$ 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)

Solution 2

Consider the rotation of axes such that the axes are the lines passing through the origin with slope $\dfrac{3}{4}$ and $-\dfrac{4}{3}$ for x-axis and y-axis, respectively, and let the point on the rotated axis be $(x_1, y_1)$. We can check that $x=\dfrac{4}{5}x_1-\dfrac{3}{5}y_1$ and $y=\dfrac{3}{5}x_1+\dfrac{4}{5}y_1$ by dropping perpendiculars from the rotated axes to the original axes. We have the focus as $(0,0)$ and $(5,0)$ and $(-5,0)$ as points on the parabola(on the rotated axes). Therefore, the directrix is $y=\pm 5$, and it doesn't matter which one(due to the absolute value) so WLOG we choose $y_1=-5$. The vertex is the midpoint between the focus and the foot of the altitude from focus to directrix, so the vertex is $(0, -\dfrac{5}{2})$. Therefore, the equation is $y_1=\dfrac{x_1^{2}}{10}-\dfrac{5}{2}$, and from the equations above we have $|3x+4y|=5x_1$, so $|{x_1}|<200$. One can check with $4x+3y$ and $4y-3x$ that the only time $x$ and $y$ can both be integers is when $x_1$ and $y_1$ are both integer multiples of $\dfrac{1}{5}$. Therefore, the only time is when $x_1$ is an odd multiple of 5, and this is obviously sufficient because $y_1$ is also a multiple of $5$. The values that satisfy thus are $x={-195, -185, -175, ..., 195}$, and there are $\boxed{(B) 40}$ such numbers.

(Solution by Shaddoll)

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.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_12A_Problems/Problem_25&oldid=74937"