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

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

@@ Line 13: / Line 13: @@
 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 = 5k</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
 <cmath>25k^2 = 6x-8y+25</cmath><cmath>25k = 4x+3y.</cmath>
 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 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==
-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.
+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*}
+x &= 40 \widetilde{x} - 3 \widetilde{x}^2 + 75, \\
+y &= 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)
+=== 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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