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

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==Solution 2==
 
==Solution 2==
  
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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.
+
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 the distance from a point to line formula <math>\dfrac{ax_0+by_0+c}{\sqrt{a^{2}+b^{2}}</math> where the equation of the line is <math>ax_0+by_0+c=0</math> and <math>(x_0, y_0)</math> is the point. 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 23:09, 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 the distance from a point to line formula $\dfrac{ax_0+by_0+c}{\sqrt{a^{2}+b^{2}}$ (Error compiling LaTeX. Unknown error_msg) where the equation of the line is $ax_0+by_0+c=0$ and $(x_0, y_0)$ is the point. 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
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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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