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> | + | 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 lattice points <math>(x,y)\in P</math> does <math>-250 \leq x+\frac{3}{4}y \leq 250</math> hold? |
<math>\textbf{(A) }38\qquad | <math>\textbf{(A) }38\qquad |
Revision as of 20:42, 25 September 2020
Problem
The parabola has focus and goes through the points and . For how many lattice points does hold?
Solution
The parabola is symmetric through , and the common distance is , so the directrix is the line through and , which is 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 C-273)
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 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(otherwise is not a multiple of , 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)
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2014amc12a/384
~ dolphin7
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.