[/quote]
,
table filled with natural numbers, we say it is a divisor table if:
-
,
-
,
for every 
.
is given. Determine the smallest natural number 
,
be a right-angled triangle where 
=
be an altitude of that triangle and points 
and 
be the midpoints of 
,
,
and 
,
be a point 
)
be the anticomplement of 
be the 
-
is parallel to the real asymptote of the circular pivotal isocubic with pivot 
and 
.
such that 
.
is a power of a prime number, but for arbitrary 
and, for each 
,
be the unique integer in the range 
for which 
is divisible by 
.
the obtained sequence is 
.
eventually becomes constant.
lie on sides 
,
.
meet at orthocenter 
.
such that 
and 
.
and 
intersect at point 
.
.
.
over 
intersects lines 
and 
and 
.
lies on 
.
such that 
be an infinite, nonconstant sequence of rational numbers, meaning it is not the case that 
Suppose that 
is also an infinite, nonconstant sequence of rational numbers with the property that 
is an integer for all 
such that 
and 
are integers for all 
Y by megarnie
Let 
, 
, and 
be squares that have vertices at lattice points (i.e., points whose coordinates are both integers) in the coordinate plane, together with their interiors. The bottom edge of each square is on the x-axis. The left edge of and the right edge of are on the 
-axis, and contains 
as many lattice points as does . The top two vertices of are in 
, and contains 
of the lattice points contained in . See the figure (not drawn to scale).
![[asy]
//kaaaaaaaaaante314
size(8cm);
import olympiad;
label(scale(.8)*"$y$", (0,60), N);
label(scale(.8)*"$x$", (60,0), E);
filldraw((0,0)--(55,0)--(55,55)--(0,55)--cycle, yellow+orange+white+white);
label(scale(1.3)*"$R$", (55/2,55/2));
filldraw((0,0)--(0,28)--(-28,28)--(-28,0)--cycle, green+white+white);
label(scale(1.3)*"$S$",(-14,14));
filldraw((-10,0)--(15,0)--(15,25)--(-10,25)--cycle, red+white+white);
label(scale(1.3)*"$T$",(3.5,25/2));
draw((0,-10)--(0,60),EndArrow(TeXHead));
draw((-34,0)--(60,0),EndArrow(TeXHead));[/asy]](//latex.artofproblemsolving.com/2/4/d/24da341df5acb1143cb0f2ca5f29ded541559594.png)
The fraction of lattice points in that are in 
is 27 times the fraction of lattice points in that are in 
. What is the minimum possible value of the edge length of plus the edge length of plus the edge length of ?
, 
, and 
be squares that have vertices at lattice points (i.e., points whose coordinates are both integers) in the coordinate plane, together with their interiors. The bottom edge of each square is on the x-axis. The left edge of and the right edge of are on the 
-axis, and contains 
as many lattice points as does . The top two vertices of are in 
, and contains 
of the lattice points contained in . See the figure (not drawn to scale).![[asy]
//kaaaaaaaaaante314
size(8cm);
import olympiad;
label(scale(.8)*"$y$", (0,60), N);
label(scale(.8)*"$x$", (60,0), E);
filldraw((0,0)--(55,0)--(55,55)--(0,55)--cycle, yellow+orange+white+white);
label(scale(1.3)*"$R$", (55/2,55/2));
filldraw((0,0)--(0,28)--(-28,28)--(-28,0)--cycle, green+white+white);
label(scale(1.3)*"$S$",(-14,14));
filldraw((-10,0)--(15,0)--(15,25)--(-10,25)--cycle, red+white+white);
label(scale(1.3)*"$T$",(3.5,25/2));
draw((0,-10)--(0,60),EndArrow(TeXHead));
draw((-34,0)--(60,0),EndArrow(TeXHead));[/asy]](http://latex.artofproblemsolving.com/2/4/d/24da341df5acb1143cb0f2ca5f29ded541559594.png)
The fraction of lattice points in
that are in 
is 27 times the fraction of lattice points in that are in 
. What is the minimum possible value of the edge length of plus the edge length of plus the edge length of ?
This post has been edited 3 times. Last edited by Spectator, Dec 24, 2022, 12:54 PM
Reason: Asy
Reason: Asy
lattice points (i.e. has coordinates ranging from 
to 
)
lattice point (i.e. has coordinates ranging from 
)
,
and 
for some positive integer 
.
lattice points since they share 
lattice points of the form 
.![\[t^2 = \frac{r^2 + s^2 - s}{4} = \frac{13x^2 - 2x}{4}\]](http://latex.artofproblemsolving.com/f/c/4/fc4ae53b2d489a26b13843529f2daaecfff7018b.png)
and so:![\[13x^2 - 2x = 4t^2\]](http://latex.artofproblemsolving.com/8/a/c/8ac5a8b441d2bae80fc97d93c942c9c5938bfd53.png)
Since the right hand side is even, 
is even, and so we can write 
for some integer ![\[52y^2 - 4y = 4t^2\]](http://latex.artofproblemsolving.com/e/4/2/e425485796440f591b8ab28cd66677ec5881f23d.png)
![\[13y^2 - y = y(13y - 1) = t^2\]](http://latex.artofproblemsolving.com/6/3/e/63ee5a94b3c8502a2382b770917e7f36070f534f.png)
But by the Euclidean algorithm, 
are relatively prime, and so both ![\[y = m^2\]](http://latex.artofproblemsolving.com/1/a/f/1af351f43ddedd512c7184c859f3c06899d742c1.png)
![\[13y - 1 = 13m^2 - 1 = n^2\]](http://latex.artofproblemsolving.com/c/a/8/ca82753b7e4bfd38438e0b728024c0d83b299c5c.png)
Suppose that the top left corner of 
.
.![\[\frac{|S \cap T|}{|S|} = 27 \cdot \frac{|R \cap T|}{|R|}\]](http://latex.artofproblemsolving.com/7/6/6/766a454e168dc6d4a641899975557ffe3b00c223.png)
where 
denotes the number of lattice points in a region 
.![\[\frac{k \cdot t}{s^2} = 27 \cdot \frac{(t - k + 1) \cdot t}{r^2}\]](http://latex.artofproblemsolving.com/2/0/2/202c443c7dba6554cca4168a193724a693f5bf97.png)
Cross-multiplying, we get:![\[k \cdot t \cdot r^2 = 27 \cdot (t - k + 1) \cdot t \cdot s^2\]](http://latex.artofproblemsolving.com/a/6/2/a624b4117c01383bedffb06e29fd3507d8309c69.png)
Simplifying and using 
,![\[9 \cdot k = (t - k + 1) \cdot 4 \cdot 27\]](http://latex.artofproblemsolving.com/c/0/f/c0f98e5d34ac5ba3cf03f5def675ad46ba77d95b.png)
![\[12 \cdot (t - k + 1) = k\]](http://latex.artofproblemsolving.com/f/2/4/f248b6d164f1c15e457dc392151474c0f35875df.png)
![\[12 \cdot (t + 1) = 13k\]](http://latex.artofproblemsolving.com/c/7/6/c76f38c04724afbdd7aa6cb2e11640898654ba3a.png)
Therefore, for any 
,![\[y(13y - 1) \equiv -y \equiv t^2 \equiv 1 \pmod{13}\]](http://latex.artofproblemsolving.com/d/a/4/da427eea26324c6193ab7d249f2a52de8740b500.png)
and so 
.
,![\[m^2 + 1 \equiv 0 \pmod{13}\]](http://latex.artofproblemsolving.com/4/3/c/43c89990cea60fa78cdb38188ab5a657c04cd427.png)
and that 
is a perfect square. The smallest 
.![\[y = m^2 = 25\]](http://latex.artofproblemsolving.com/8/0/6/806eef5119615591ba80797a650f109097659b6f.png)
![\[x = 2y = 50\]](http://latex.artofproblemsolving.com/b/9/d/b9deb69882dd10a43c4437f9116e6a3eaf4f7db7.png)
and so 
,
,
,
,![\[(r - 1) + (s - 1) + (t - 1) = 150 + 100 + 90 - 3 = 340 - 3 = 337\]](http://latex.artofproblemsolving.com/5/c/2/5c294416cabb27240c17e09203bf3076311d4d3c.png)
so the answer is B.
larger than what it's supposed to be.
then 
lattice points in 
lattice points in 
so side length of 
from there just use the other two given conditions are you get modular congruences to give side lengths 
.
.
and 
be the lengths of the portions of the sides of 
by the third condition, so 
,
for some 
,
— thus only 
works
Notice that 
,
for some positive integers 
.
At this point, there isn't really a better way than to test out values for 
,
is a solution. Then, 
,
which yields 
.
,
.
as 
respectively. According to the first condition, we have 
.
,
simplify this to 
.
.
,
.
while only 
satisfies.
,
.
,
.
,
,
.
where all corners are on lattice points has 
lattice points on+in it.
Cancelling and rearranging the last equation, we have 
and so 
then taking 
gives 
.
Since 
is an integer, it must be true that 
,
,
Now we just test 
which doesn't work, and 
works which gives us 
and 
which gives a sum of 
