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

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==Solution==
 
==Solution==
  
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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>.
+
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>, which is 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
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and accordingly we find by solving the system that <math>x = \frac{1}{2} (3k^2-3) + 4k</math> and <math>y = -2k^2+3k+2</math>.
 
and accordingly we find by solving the system that <math>x = \frac{1}{2} (3k^2-3) + 4k</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 C-273)
 +
 
 +
*You can do this by noting 2 has to divide <math>(3k^2-3)</math>, so <math>3k^2</math> is congruent to 3 modulo 2. Dividing by 3 (since 2 and 3 are coprime), <math>k^2</math> is odd, so <math>k</math> must be odd. - (Orion 2010)
 +
 
 +
==Solution 2==
 +
 
 +
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
 +
<cmath>\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}</cmath>
 +
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
 +
<cmath>\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}</cmath>
 +
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
 +
<cmath>\begin{align*}
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  50x &= 40 \widetilde{x} - 3 \widetilde{x}^2 + 75, \\
 +
  50y &= 30 \widetilde{x} + 4 \widetilde{x}^2 - 100
 +
\end{align*}</cmath>
 +
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
 +
<cmath>2x = 8a - 3 a^2 + 3</cmath>
 +
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)
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 +
=== Video Solution by Richard Rusczyk ===
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 +
https://artofproblemsolving.com/videos/amc/2014amc12a/384
  
 
==See Also==
 
==See Also==

Latest revision as of 01:19, 19 January 2025

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)

  • You can do this by noting 2 has to divide $(3k^2-3)$, so $3k^2$ is congruent to 3 modulo 2. Dividing by 3 (since 2 and 3 are coprime), $k^2$ is odd, so $k$ must be odd. - (Orion 2010)

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

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.

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