Difference between revisions of "2014 AMC 12A Problems/Problem 25"
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Revision as of 13:40, 8 February 2014
Problem
The parabola 
has focus 
and goes through the points 
and 
. For how many points 
with integer coefficients 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 ![\[3x-4y = -25.\]](http://latex.artofproblemsolving.com/6/a/4/6a4d38879d2c5db2afd19b7e9154d1935fea2661.png)
Using the point-line distance formula, the parabola is the locus ![\[x^2+y^2 = \frac{\left\lvert 3x-4y+25 \right\rvert^2}{3^2+4^2}\]](http://latex.artofproblemsolving.com/2/a/5/2a5479a13d0f971d8bdfebc9c4fbe79416af5157.png)
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. 
