Difference between revisions of "2014 AMC 12A Problems/Problem 25"
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==Problem== | ==Problem== | ||
− | The parabola <math>P</math> has focus <math>(0,0)</math> and goes through the points <math>(4,3)</math> and <math>(-4,-3)</math>. For how many points <math>(x,y)\in P</math> with integer | + | The parabola <math>P</math> has focus <math>(0,0)</math> and goes through the points <math>(4,3)</math> and <math>(-4,-3)</math>. For how many points <math>(x,y)\in P</math> with integer coordinates is it true that <math>|4x+3y|\leq 1000</math>? |
<math>\textbf{(A) }38\qquad | <math>\textbf{(A) }38\qquad |
Revision as of 19:01, 1 September 2014
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
\[25k^2 &= 6x-8y+25\] (Error compiling LaTeX. Unknown error_msg)
\[25k &= 4x+3y.\] (Error compiling LaTeX. Unknown error_msg)
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.