[/quote]
,
with respect to 
is greater than 0 if 
and 
.
and since 
?
be an acute scalene triangle with circumcircle 
.
intersect at 
and 
be the projections of 
and 
,
,
,
.
.
of the acute-angled triangle 
is tangent to its side 
at a point 
.
be an altitude of triangle 
be the midpoint of the segment 
is the common point of the circle 
(distinct from 
are tangent to each other at the point 
is called lucky if it has at least two distinct prime divisors and can be written in the form:![\[
n = p_1^{\alpha_1} + \cdots + p_k^{\alpha_k}
\]](http://latex.artofproblemsolving.com/7/4/4/744a5ccaeb9476ebd7d999c395762cb6e99a7a71.png)
where 
are distinct prime numbers that divide 
,
.
,
,
be a non-isosceles triangle with incenter 
Let 
be the smaller circle through 
tangent to 
and 
(the addition of indices being mod 3). Let 
be the second point of intersection of 
and 
Prove that the circumcentres of the triangles 
are collinear.
such that 
coprime, 
and 
.
such that for every 
is satisfied? Explain the answer.
be a scalene oblique triangle, and 
)
-
,
,
through 
.![\[P=\dfrac{a}{\sqrt{2bc+1}}+\dfrac{b}{\sqrt{2ca+1}}+\dfrac{c}{\sqrt{2ab+1}}.\]](http://latex.artofproblemsolving.com/1/f/3/1f39a54e4cf378bfd4e66d97949350b9b37dc1ec.png)
such that 
is the square of an integer.
.
and let 
be the foot of the perpendicular from 
intersects the circumcircle 
of triangle 
be the point where 
intersects the circle 
,
and 
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 
