Difference between revisions of "2011 AMC 10A Problems/Problem 25"
(→Solution 4) |
(→Solution 4) |
||
Line 71: | Line 71: | ||
Thus a point <math>X</math> is both 100-rays and 60-rays partitional if the simultaneous solutions <math>(a,b,c,d)</math> to the equations <cmath>a+c=50,b+d=50</cmath> are also positive integral solutions to the equations <cmath>a+c=30,b+d=30</cmath> when <math>a</math>,<math>b</math>,<math>c</math>,<math>d</math> are multiplied by <math>\frac{3}{5}</math>, which occurs when all of <math>a</math>,<math>b</math>,<math>c</math>,<math>d</math> are multiples of 5. | Thus a point <math>X</math> is both 100-rays and 60-rays partitional if the simultaneous solutions <math>(a,b,c,d)</math> to the equations <cmath>a+c=50,b+d=50</cmath> are also positive integral solutions to the equations <cmath>a+c=30,b+d=30</cmath> when <math>a</math>,<math>b</math>,<math>c</math>,<math>d</math> are multiplied by <math>\frac{3}{5}</math>, which occurs when all of <math>a</math>,<math>b</math>,<math>c</math>,<math>d</math> are multiples of 5. | ||
− | In this case, there are 9 solutions for each of <math>a+c=50</math>, <math>b+d=50</math> that follow the condition. There is a total of <math>9 | + | In this case, there are 9 solutions for each of <math>a+c=50</math>, <math>b+d=50</math> that follow the condition. There is a total of <math>9\cdot9=81</math> overlapping points for <math>X</math>. |
Thus, the number of point <math>X</math> that is 100-rays but not 60-rays partitional is <math>2401-81=\boxed{2320 (C)}</math> | Thus, the number of point <math>X</math> that is 100-rays but not 60-rays partitional is <math>2401-81=\boxed{2320 (C)}</math> |
Revision as of 21:23, 6 August 2019
Problem 25
Let be a square region and
an integer. A point
in the interior of
is called
partitional if there are
rays emanating from
that divide
into
triangles of equal area. How many points are 100-ray partitional but not 60-ray partitional?
Solution 1
First, notice that there must be four rays emanating from that intersect the four corners of the square region. Depending on the location of
, the number of rays distributed among these four triangular sectors will vary. We start by finding the corner-most point that is
-ray partitional (let this point be the bottom-left-most point). We first draw the four rays that intersect the vertices. At this point, the triangular sectors with bases as the sides of the square that the point is closest to both do not have rays dividing their areas. Therefore, their heights are equivalent since their areas are equal. The remaining
rays are divided among the other two triangular sectors, each sector with
rays, thus dividing these two sectors into
triangles of equal areas. Let the distance from this corner point to the closest side be
and the side of the square be
. From this, we get the equation
. Solve for
to get
. Therefore, point
is
of the side length away from the two sides it is closest to. By moving
to the right, we also move one ray from the right sector to the left sector, which determines another
-ray partitional point. We can continue moving
right and up to derive the set of points that are
-ray partitional. In the end, we get a square grid of points each
apart from one another. Since this grid ranges from a distance of
from one side to
from the same side, we have a
grid, a total of
-ray partitional points. To find the overlap from the
-ray partitional, we must find the distance from the corner-most
-ray partitional point to the sides closest to it. Since the
-ray partitional points form a
grid, each point
apart from each other, we can deduce that the
-ray partitional points form a
grid, each point
apart from each other. To find the overlap points, we must find the common divisors of
and
which are
and
. Therefore, the overlapping points will form grids with points
,
,
, and
away from each other respectively. Since the grid with points
away from each other includes the other points, we can disregard the other grids. The total overlapping set of points is a
grid, which has
points. Subtract
from
to get
.
Solution 2
We may assume that the square has coordinates
. Suppose that
is
-ray partitional.
By definition, there exist
rays from
which divide
into
triangles of equal area, and
of these rays intersect the vertices of
. Let
be the number of triangles that share a side with
, respectively. Then
Let the common area of the triangles be
. Each of the triangles that share a side with
have common height
; since they have the same area, they must have the same base
. Hence
. Similarly,
Substituting gives
and
Since
and
and
are integers,
are rational. Let
be integers such that
and
Thus we have derived a necessary condition for
to be
-ray partitional:
Conversely, if
satisfies the above condition, then it is
-ray partitional since we can define
in terms of
as above.
To count the points that are -ray partitional, it suffices to count the ordered pairs of rationals
in the interior of
such that
divides
and
divides
. This is just
which has
points.
We need to subtract the points that are -ray partitional and
-ray partitional; those are the ordered pairs of rationals
in the interior of
such that
divides
and
divides
. This is just
which has
points.
Hence the answer is .
Solution 3
As with solution , place the square on the coordinate plane with coordinates
. Let the point
be
-ray partitional. Note that we must have four rays going to each corner of the square to have all triangles, so we gave four "big" triangles of area
We need to split each one into an integral number of triangles with area
, so we must have
be an integer. Letting
equal
and
, we now have that we need to find all points
such that
and
are integers, but
and
aren't.
We will use complementary counting. I claim that that there are
fractions less than one such that for every fraction
,
is an integer, where
is Euler's totient function. This is easy to see: we want to count the number of fractions with a denominator that divides
but is simplified, so a numerator that is relatively prime to the denominator. However, we must exclude one since our fraction is less than
. This is precisely what our sum counts. However, this sum is very well known and evaluates to
. Thus there are
that are
-ray partitional (I believe that this is for even
only but I haven't checked the odd case yet).
Letting
gives us
points that are
-ray partitional. Since we are counting the complement, we want to find all points that are both
-ray partitional and
-ray partitional, or counting all pairs of fractions
such that
are all integers. Since the GCD of the two numbers is
, it suffices to find the number of pairs of fractions such that
and
are integers, which is simply
. Thus, our answer is simply
. -franchester
Solution 4
![[asy] unitsize(36); draw((0,0)--(3,0)--(3,3)--(0,3)--cycle); draw((0,0)--(1.8,0.8)); draw((3,0)--(1.8,0.8)); draw((3,3)--(1.8,0.8)); draw((0,3)--(1.8,0.8)); label("$R_1$",(-0.25,3.5),S); label("$R_2$",(3.5,3.25),W); label("$R_3$",(3.5,-0.25),W); label("$R_4$",(0,-0.25),W); label("$X$",(2,1.25),W); label("$a$",(1.5,3.25),W); label("$b$",(3.5,1.5),W); label("$c$",(1.5,-0.25),W); label("$d$",(-0.25,1.5),W); [/asy]](http://latex.artofproblemsolving.com/0/5/c/05c0356f1ae7635cee95d934827162aab2201f35.png)
Let ,
,
,
be divided into
,
,
,
equal segments, respectively, to be the bases of the triangles.
For triangles with equal area, the ratio between their heights is equivalent to the reciprocal of the ratio between their bases, which in this case, refers to the ratio between the numbers of segments that the sides of are divided into.
Let the altitudes from to
,
,
,
be
,
,
,
, respectively. Since
, we have the equivalent equation
.
When is 100-rays partitional, we have
, or
Since are positive integers, there are
solutions for each of the two Diophantine equations. Thus there is a total of
points for
to be 100-rays partitional.
Now when is 60-rays partitional, we have
, or
Thus a point
is both 100-rays and 60-rays partitional if the simultaneous solutions
to the equations
are also positive integral solutions to the equations
when
,
,
,
are multiplied by
, which occurs when all of
,
,
,
are multiples of 5.
In this case, there are 9 solutions for each of ,
that follow the condition. There is a total of
overlapping points for
.
Thus, the number of point that is 100-rays but not 60-rays partitional is
~ Nafer
See Also
2011 AMC 10A (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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.