Difference between revisions of "2014 AMC 12A Problems/Problem 25"
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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 | + | 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
Contents
[hide]Problem
The parabola has focus and goes through the points and . For how many points with integer coordinates is it true that ?
Solution
The parabola is symmetric through , and the common distance is , so the directrix is the line through and . That's the line Using the point-line distance formula, the parabola is the locus which rearranges to .
Let , . Put to obtain and accordingly we find by solving the system that and .
One can show that the values of that make an integer pair are precisely odd integers . For this is , so values work and the answer is .
(Solution by v_Enhance)
Solution 2
Consider the rotation of axes such that the axes are the lines passing through the origin with slope and for x-axis and y-axis, respectively, and let the point on the rotated axis be . We can check that and 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 and is the point. We have the focus as and and as points on the parabola(on the rotated axes). Therefore, the directrix is , and it doesn't matter which one(due to the absolute value) so WLOG we choose . The vertex is the midpoint between the focus and the foot of the altitude from focus to directrix, so the vertex is . Therefore, the equation is , and from the equations above we have , so . One can check with and that the only time and can both be integers is when and are both integer multiples of . Therefore, the only time is when is an odd multiple of 5, and this is obviously sufficient because is also a multiple of . The values that satisfy thus are , and there are such numbers.
(Solution by Shaddoll)
See Also
2014 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Question |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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.