Difference between revisions of "2014 AMC 12A Problems/Problem 25"
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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 | ||
− | <cmath>25k^2 | + | <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>. | 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>. | ||
Revision as of 21:45, 15 March 2015
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)
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.