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Difference between revisions of "2014 AMC 12A Problems/Problem 25"
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==Problem==
 
==Problem==
  
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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 lattice points <math>(x,y)\in P</math> does <math>-250 \leq x+\frac{3}{4}y \leq 250</math> hold?
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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 P</math> with integer coordinates is it true that <math>|4x+3y|\leq 1000</math>?
  
 
<math>\textbf{(A) }38\qquad
 
<math>\textbf{(A) }38\qquad
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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 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(otherwise <math>y_1</math> is not a multiple of <math>\dfrac{1}{5}</math>, 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_1={-195, -185, -175, ..., 195}</math>, and there are <math>\boxed{(B) 40}</math> such numbers.
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The axis of <math>P</math> is inclined at an angle <math>\theta</math> relative to the coordinate axis, where <math>\tan\theta = \tfrac 34</math>. We rotate the coordinate axis by angle <math>\theta</math> anti-clockwise, so that the parabola now has a vertical symmetry axis relative to the rotated coordinates. Let <math>(\widetilde{x}, \widetilde{y})</math> be the coordinates in the rotated system. Then <math>(x,y)</math> and <math>(\widetilde{x}, \widetilde{y})</math> are related by
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<cmath>\begin{align}
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  \nonumber x = \widetilde{x}\cos\theta -\widetilde{y}\sin\theta &= \tfrac 45 \widetilde{x} - \tfrac 35 \widetilde{y}, \\
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  y = \widetilde{x}\sin\theta +\widetilde{y}\cos\theta &= \tfrac 35 \widetilde{x} + \tfrac 45 \widetilde{y}
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\end{align}</cmath>
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In the rotated coordinate system, the parabola has focus at <math>(0,0)</math> and the two points on it are at <math>(5,0)</math> and <math>(-5,0)</math>. Therefore, the directrix is <math>\widetilde{y}=\pm 5</math>; we can, WLOG, choose <math>\widetilde{y}=-5</math>. For a point on the parabola, it is equidistant from the focus and directrix, so the equation of the parabola is  
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<cmath>\begin{align}\tag{2}
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  \widetilde{x}^2+\widetilde{y}^2 = (\widetilde{y}+5)^2 \qquad  &\Longrightarrow\qquad \widetilde{y} = \tfrac{1}{10}(\widetilde{x}^2-25)
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\end{align}</cmath>
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From <math>(1)</math> we have <math>|4x+3y|=5\widetilde{x}</math>, so we need <math>|\widetilde{x}|<200</math>. Substituting <math>(2)</math> in <math>(1)</math>, we get
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<cmath>\begin{align*}
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  50x &= 40 \widetilde{x} - 3 \widetilde{x}^2 + 75, \\
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  50y &= 30 \widetilde{x} + 4 \widetilde{x}^2 - 100
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\end{align*}</cmath>
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For <math>x</math> to be an integer <math>\widetilde{x}</math> must be a multiple of 5; setting <math>\widetilde{x}=5a</math> we get
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<cmath>2x = 8a - 3 a^2 + 3</cmath>
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Now we need <math>a</math> to be odd, i.e. <math>\widetilde{x}=5a</math> is an odd multiple of <math>5</math>, in which case we get <math>y = 3 a + 2 a^2 - 2</math>, which is also an integer. The values that satisfy the given conditions correspond to <math>\widetilde{x}= \{\pm 5\cdot (2k-1)\mid k = 0, 1, \ldots , 19 \}={-195, -185, -175, ..., 195}</math>, and there are <math>\boxed{\textbf{(B)} \ 40}</math> such numbers.
  
 
(Solution by Shaddoll)
 
(Solution by Shaddoll)
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https://artofproblemsolving.com/videos/amc/2014amc12a/384
 
https://artofproblemsolving.com/videos/amc/2014amc12a/384
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~ dolphin7
 
  
 
==See Also==
 
==See Also==

Latest revision as of 15:18, 2 September 2022

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 $(-7,1)$, which is 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 C-273)

Solution 2

The axis of $P$ is inclined at an angle $\theta$ relative to the coordinate axis, where $\tan\theta = \tfrac 34$. We rotate the coordinate axis by angle $\theta$ anti-clockwise, so that the parabola now has a vertical symmetry axis relative to the rotated coordinates. Let $(\widetilde{x}, \widetilde{y})$ be the coordinates in the rotated system. Then $(x,y)$ and $(\widetilde{x}, \widetilde{y})$ are related by \begin{align} \nonumber x = \widetilde{x}\cos\theta -\widetilde{y}\sin\theta &= \tfrac 45 \widetilde{x} - \tfrac 35 \widetilde{y}, \\ y = \widetilde{x}\sin\theta +\widetilde{y}\cos\theta &= \tfrac 35 \widetilde{x} + \tfrac 45 \widetilde{y} \end{align} In the rotated coordinate system, the parabola has focus at $(0,0)$ and the two points on it are at $(5,0)$ and $(-5,0)$. Therefore, the directrix is $\widetilde{y}=\pm 5$; we can, WLOG, choose $\widetilde{y}=-5$. For a point on the parabola, it is equidistant from the focus and directrix, so the equation of the parabola is \begin{align}\tag{2} \widetilde{x}^2+\widetilde{y}^2 = (\widetilde{y}+5)^2 \qquad &\Longrightarrow\qquad \widetilde{y} = \tfrac{1}{10}(\widetilde{x}^2-25) \end{align} From $(1)$ we have $|4x+3y|=5\widetilde{x}$, so we need $|\widetilde{x}|<200$. Substituting $(2)$ in $(1)$, we get \begin{align*} 50x &= 40 \widetilde{x} - 3 \widetilde{x}^2 + 75, \\ 50y &= 30 \widetilde{x} + 4 \widetilde{x}^2 - 100 \end{align*} For $x$ to be an integer $\widetilde{x}$ must be a multiple of 5; setting $\widetilde{x}=5a$ we get \[2x = 8a - 3 a^2 + 3\] Now we need $a$ to be odd, i.e. $\widetilde{x}=5a$ is an odd multiple of $5$, in which case we get $y = 3 a + 2 a^2 - 2$, which is also an integer. The values that satisfy the given conditions correspond to $\widetilde{x}= \{\pm 5\cdot (2k-1)\mid k = 0, 1, \ldots , 19 \}={-195, -185, -175, ..., 195}$, and there are $\boxed{\textbf{(B)} \ 40}$ such numbers.

(Solution by Shaddoll)

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2014amc12a/384

See Also

2014 AMC 12A (Problems • Answer Key • Resources)
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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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