Difference between revisions of "2014 AMC 12A Problems/Problem 25"
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The parabola is symmetric through <math>y=- \frac{4}{3}x</math>, and the common distance is <math>5</math>, so the directrix is the line through <math>(1,7)</math> and <math>(-7,1)</math>, which is the line <cmath> 3x-4y = -25. </cmath> Using the point-line distance formula, the parabola is the locus <cmath> x^2+y^2 = \frac{\left\lvert 3x-4y+25 \right\rvert^2}{3^2+4^2} </cmath> which rearranges to <math>(4x+3y)^2 = 25(6x-8y+25)</math>. | The parabola is symmetric through <math>y=- \frac{4}{3}x</math>, and the common distance is <math>5</math>, so the directrix is the line through <math>(1,7)</math> and <math>(-7,1)</math>, which is the line <cmath> 3x-4y = -25. </cmath> Using the point-line distance formula, the parabola is the locus <cmath> x^2+y^2 = \frac{\left\lvert 3x-4y+25 \right\rvert^2}{3^2+4^2} </cmath> which rearranges to <math>(4x+3y)^2 = 25(6x-8y+25)</math>. | ||
− | Let <math>m = 4x+3y \in \mathbb Z</math>, <math>\left\lvert m \right\rvert \le 1000</math>. Put <math>m = | + | 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 = 6x-8y+25</cmath><cmath>25k = 4x+3y.</cmath> | <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 12:51, 25 January 2020
Contents
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
, 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)
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.