[/quote]
,
be a circle with centre 
,
a convex quadrilateral such that each of the segments 
and 
is tangent to 
be the circumcircle of the triangle 
.
beyond 
meets 
,
beyond 
meets 
.
and 
beyond 
meet 
and ![\[A D+D T+T X+X A=C D+D Y+Y Z+Z C.\]](http://latex.artofproblemsolving.com/8/2/6/8261276053071cc6cc06dda824a883426bf1d4ac.png)
such that for every 
is satisfied? Explain the answer.
be a random point on the smaller arc 
of the circumcircle of square 
,
be the intersection point of segments 
and 
.
are 
and 
, where 
is the center of the square. Prove that points 
and 
such that
such that: 
be an integer, and let 
be positive real numbers such that 
.![\[(1 + a_2)^2 (1 + a_3)^3 \dotsm (1 + a_n)^n > n^n.\]](http://latex.artofproblemsolving.com/7/7/f/77fb719e975225b59c64f1f3aa102c98a6c25f50.png)
colors, there exists a monochromatic regular hexagon.
is the smallest number such that if 
,
with two colors there exists a monochromatic arithmetic progression of length 
.
.
and 
.
points in plane. Prove that the number of isosceles triangles having their vertices among these 
.
.
be an integer. Find all positive real solutions to the following system of 
equations:
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 
satisfying both conditions is 
.![\[y = m^2 = 25\]](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 
