Difference between revisions of "2014 AMC 12A Problems/Problem 25"
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==Solution 2== | ==Solution 2== | ||
− | + | The axis of <math>P</math> is inclined at an angle <math>\theta</math> relative to the coordinate axis, where <math>\tan\theta = \tfrac 34</math>. We rotate the coordinate axis by angle <math>\theta</math> anti-clockwise, so that the parabola now has a vertical symmetry axis relative to the rotated coordinates. Let <math>(\widetilde{x}, \widetilde{y})</math> be the coordinates in the rotated system. Then <math>(x,y)</math> and <math>(\widetilde{x}, \widetilde{y})</math> are related by | |
+ | <cmath>\begin{align} | ||
+ | \nonumber x = \widetilde{x}\cos\theta -\widetilde{y}\sin\theta &= \tfrac 45 \widetilde{x} - \tfrac 35 \widetilde{y}, \\ | ||
+ | y = \widetilde{x}\sin\theta +\widetilde{y}\cos\theta &= \tfrac 35 \widetilde{x} + \tfrac 45 \widetilde{y} | ||
+ | \end{align}</cmath> | ||
+ | In the rotated coordinate system, the parabola has focus at <math>(0,0)</math> and the two pints on it are at <math>(5,0)</math> and <math>(-5,0)</math>. Therefore, the directrix is <math>\widetilde{y}=\pm 5</math>; we can, WLOG, choose <math>\widetilde{y}=-5</math>. For a point on the parabola, it is equidistant from the focus and directrix, so the equation of the parabola is | ||
+ | <cmath>\begin{align}\tag{2} | ||
+ | \widetilde{x}^2+\widetilde{y}^2 = (\widetilde{y}+5)^2 \qquad &\Longrightarrow\qquad \widetilde{y} = \tfrac{1}{10}(\widetilde{x}^2-25) | ||
+ | \end{align}</cmath> | ||
+ | From <math>(1)</math> we have <math>|4x+3y|=5\widetilde{x}</math>, so we need <math>|\widetilde{x}|<200</math>. Substituting <math>(2)</math> in <math>(1)</math>, we get | ||
+ | <cmath>\begin{align*} | ||
+ | 50x &= 40 \widetilde{x} - 3 \widetilde{x}^2 + 75, \\ | ||
+ | 50y &= 30 \widetilde{x} + 4 \widetilde{x}^2 - 100 | ||
+ | \end{align*}</cmath> | ||
+ | For <math>x</math> to be an integer <math>\widetilde{x}</math> must be a multiple of 5; setting <math>\widetilde{x}=5a</math> we get | ||
+ | <cmath>2x = 8a - 3 a^2 + 3</cmath> | ||
+ | Now we need <math>a</math> to be odd, i.e. <math>\widetilde{x}=5a</math> is an odd multiple of <math>5</math>, in which case we get <math>y = 3 a + 2 a^2 - 2</math>, which is also an integer. The values that satisfy the given conditions correspond to <math>\widetilde{x}= \{\pm 5\cdot (2k-1)\mid k = 0, 1, \ldots , 19 \}={-195, -185, -175, ..., 195}</math>, and there are <math>\boxed{\textbf{(B)} \ 40}</math> such numbers. | ||
(Solution by Shaddoll) | (Solution by Shaddoll) |
Revision as of 18:44, 22 November 2021
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
The axis of is inclined at an angle
relative to the coordinate axis, where
. We rotate the coordinate axis by angle
anti-clockwise, so that the parabola now has a vertical symmetry axis relative to the rotated coordinates. Let
be the coordinates in the rotated system. Then
and
are related by
In the rotated coordinate system, the parabola has focus at
and the two pints on it are at
and
. Therefore, the directrix is
; we can, WLOG, choose
. For a point on the parabola, it is equidistant from the focus and directrix, so the equation of the parabola is
From
we have
, so we need
. Substituting
in
, we get
For
to be an integer
must be a multiple of 5; setting
we get
Now we need
to be odd, i.e.
is an odd multiple of
, in which case we get
, which is also an integer. The values that satisfy the given conditions correspond to
, 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.