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

(→‎Solution 2)
(→‎Solution 2)
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   y = \widetilde{x}\sin\theta +\widetilde{y}\cos\theta &= \tfrac 35 \widetilde{x} + \tfrac 45 \widetilde{y}
 
   y = \widetilde{x}\sin\theta +\widetilde{y}\cos\theta &= \tfrac 35 \widetilde{x} + \tfrac 45 \widetilde{y}
 
\end{align}</cmath>
 
\end{align}</cmath>
−
In the rotated coordinate system, the parabola has focus at <math>(0,0)</math> and the two pints 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  
+
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}
 
<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)
 
   \widetilde{x}^2+\widetilde{y}^2 = (\widetilde{y}+5)^2 \qquad  &\Longrightarrow\qquad \widetilde{y} = \tfrac{1}{10}(\widetilde{x}^2-25)

Revision as of 17:46, 22 November 2021

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

~ dolphin7

See Also

2014 AMC 12A (Problems • Answer Key • Resources)
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Problem 24 		Followed by
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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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