and 
be different prime numbers. We are given an infinite decreasing arithmetic progression in which each of the numbers 
and 
occurs. Show that the numbers 
![\[
\text{Given that } 0 < \theta < 90^\circ,\ \text{solve the equation: } \sin(\theta - 60^\circ)\sin\theta + \sin(54^\circ - \theta)\sin 54^\circ = 0
\]](http://latex.artofproblemsolving.com/b/2/4/b24276c1fbddbcc97e50cef52db03ea9b264d7ee.png)
![\[
\text{What is the value of } \theta\ (\text{in degrees})\ \text{that satisfies the equation?}
\]](http://latex.artofproblemsolving.com/0/0/f/00fbcd5839fedced0e5636c1def5360d65fc0444.png)
.
.
be the orthocenter and 
be the incenter of triangle 
be the midpoint of the arc 
of the circumcircle of triangle 
,
be a point on the arc 
.
be the circumcenter of triangle 
.
and 
meet on 
be a positive integer. In an 
table, an upright path is a sequence of adjacent cells starting from the southwest corner to the northeast corner such that the next cell is either on the top or on the right of the previous cell. Find the smallest number of grids one needs to color in an 
be positive reals. Prove that: 
rabbits and 
rabbit.
and 
be positive rational numbers such that: 
,
,
)
and incircle 
.
at 
.
cuts excircle 
)
cuts the excircle with the diameter of 
)
cuts 
~ 
and 
is the tangent line of the excircle of 
into 
sets of 
and that this chapter is constructive counting, this is not intuitive at all. Am i missing something?
and it is indeed correct), but I don't know how to prove it:
.
can be partitioned into 
pairs such that the sum of the elements in each pair is a power of 
is stylish because the set 
can be partitioned as 
,
,
.
Doing this, we get that 
Now, we can solve this by factoring. By using RRT, we get that 
is a factor. Upon division, we get 
is the factored form. Now, we know that our only positive solutions are 
so our values of 
are 
Summing these gives our answer of 
to 
in or something like that idk my friend explained it pretty wonkily idk if he even solved it right but anyhow we press on. Transformations suck.
and 
Determine the value of 
Furthermore, note that 
Thus, the answer is just 
is on point 
such that 
and 
and 
is a point on 
such that 
and 
Given that 
calculate 
so 
Cross multiplying and solving this quadratic, we get that our solution is 
Thus, 
given that 
We know from 
that 
so substituting this gives us 
Now, plugging this in to 
and 
gives us 
Solving these equations gives us 
Thus, by Vieta's, the sum of the roots is 
thus by median properties (the 2:1 ratio split thingy), we know that 
(For reference, I missed this part in-contest) Now, we can calculate 
by the Pythagorean Theorem to be 
and thus 
Now, by Pythagorean theorem, we know that the altitude from 
is 
so our area is just 
Thus, simplifying and substituting 
we get that 
Simple guess and check gives us 
as our solutions. My dumb self forgot the former solution. Thus, 
giving us 
as our sum.![\[f(x) =
\begin{cases}
x - 9, & \text{if } x > 100 \\
f(f(x + 10)), & \text{if } x \leq 100
\end{cases}\]](http://latex.artofproblemsolving.com/4/e/5/4e5b4486030a59e74df648819a1a59f82622b3b7.png)
.
we get that 
Now if we try finding 
we will realize that 
this 
pattern continues until we once again reach 
thus 
Now return to the problem statement; 
but each inner function will just become 
hence the answer is 
be real numbers such that 
Find 
Thus, simplifying, we get that 
or that 
Plugging this in to the first equation, we get that 
so this means that 
Now, 
thus 
are on circle 
and 
Determine the path length from 
is the diameter, thus we can draw 
and so by Special Right Triangles we do NOT in fact get 
but rather 
is also an integer.
.
to be an integer. Plugging in factors of 
we see that there are 
is a multiple of 
Plugging in numbers gives us 
as our smallest value.
are real numbers such that 
and 
Calculate 
Furthermore, 
so 
meaning that 
.
and thus 
giving us 
will follow the rules such that![\[
p_1 = (0,0), \quad p_{i+1} = (x_i + 1, y_i) \text{ or } (x_i, y_i + 1), \quad p_{10} = (4,5).
\]](http://latex.artofproblemsolving.com/6/0/3/6036ef2619ffb1e91db0ab310388217e472490f2.png)
How many sequences 
are possible such that 
is the only point with equal coordinates?
Furthermore, we know that 
so our answer is just 
in the summation. Essentially, we have 
.
In other words, we want to find 
.
.
in such a polynomial is 
.
,
.
,
.
and 
,
and divide by 
.
and 
in our sum, subtract by 
.
term look kind of weird? Well, we actually notice 
,
.
,
,
is our answer!
,
terms. Drawing it out on the unit circle (which I heavily advise), it cycles between 
and 
.
,
,
.
,
for when 
terms! Anyways, the answer is 
.
so our coefficient is just 
Find 
Thus, our answer is just 
on the positive integers using the rule that for 
For all prime 
and for all other 
Find the smallest possible value of 
can be written as the sum of two distinct, non-negative integer powers of 
etc.
we get that it is 
Thus, we note that for an integer 
Since 
we know that all values of 
satisfy our conditions. Thus, our answer is just 
be the set of positive integers of 
for some other positive integer 
Find the only three-digit value of 
is our solution. For these Pell Equations, all solutions are of the form 
for 
Thus, we plug in 
giving us 
as our answer.
Find the value of 
and let 
be our three-digit number that we remove. Thus, we can write 
By our condition, we know that 
or that 
Furthermore, we know that 
does not work, so we try 
We know that 
by the fact that it must be divisible by a 
Noting that 
doesn't work, we try 
This, miraculously, works, and thus our value of 
Y by mkhayech, Mathuzb, Davi-8191, tiendung2006, Mathlover_1, Adventure10, Rounak_iitr
Let 
be the circumcircle of triangle 
. A circle 
is tangent to the line segment 
and is tangent to 
at a point lying on the same side of the line as 
. The angle bisector of 
intersects at two different points 
and 
.
Prove that
.
be the circumcircle of triangle 
. A circle 
is tangent to the line segment 
and is tangent to 
at a point lying on the same side of the line as 
. The angle bisector of 
intersects at two different points 
and 
.Prove that
.This post has been edited 2 times. Last edited by djmathman, Apr 23, 2018, 1:41 PM
.
,
is tangent to the circumcircle of 
,
then the inversion at 
fixes 
.
and 
,
.
are inverses, we get 
as desired.
be 
respectively,then obviously
Which implies that 
in triangle 
.
at 
.
and 
are colinear and 
.
and line 
,
and 
Pencil from B gives involution swapping 
but the first two pairs are isogonal w.r.t. 
so we are done.
(directed lenghts), 
and 
.
,
be tangent to 
.
.
are collinear, we know that 
are concyclic. So 
,
;
;
and 
are tangent at 
meet 
,
and the result follows.
are collinear . Because 
is cyclic we have 
,but 
.
, so 
.
, but 
and 
, so 
.
.
= 
- 
= 
- 
=
=
of 
and 
As desired.
be the second intersection of 
.
.
,
.
so 
and 
are isogonal. But 
and 
are isogonal, so 
and 
are isogonal, so 
.
,
,
.
and 
;
,
lie on a line, and so do 
.
is tangent to the line 
,
.
.
is tangent to 
and 
are colinear. Also note 
as they both subtend arc 
hence using sine rule and angle bisector theorem on 
we see:
Now by power of a point:
But it's well-known this means 
are isogonal in 
and hence by symmetry 
as desired.
be the tangency point of 
.
collinear.
and 
denotes the circumcenters of 
collinear.
and 
implies 
.
cyclic.
is tangent to 
is tangent to 
bisects 
and the angle bisector of 
other than 
at 
and 
and 
which means 
be the tangency point
1: 
: 
.
is similar to 
where 
And, 
collinear. So, 
intersect 
at 
. 
as desired.
a
w
then with angle chasing we get the solution.
are collinear. 
By converse alternate segment theorem 
and again by POP we have 
hence 
and from here the result follows.
and 
where 
so 
which implies the result on angle chase.
and let 
implies 
are collinear and ![[asy]/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(14cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
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draw(arc((2.7730186914271573,-1.7657842667556372),0.6,155.87081438240853,189.67467443386437)--(2.7730186914271573,-1.7657842667556372)--cycle, linewidth(1) + qqwuqq);
draw(arc((-4.76,-3.05),0.6,62.29386147827237,96.09772152972822)--(-4.76,-3.05)--cycle, linewidth(1) + qqwuqq);
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draw((-4.76,-3.05)--(-2.7794754967753703,0.7213614896243091), linewidth(1));
draw((-4.76,-3.05)--(2.7730186914271573,-1.7657842667556372), linewidth(1));
draw((-2.7794754967753703,0.7213614896243091)--(2.7730186914271573,-1.7657842667556372), linewidth(1));
draw((2.7730186914271573,-1.7657842667556372)--(5.64,-3.05), linewidth(1));
draw((-5.282674200057174,1.8426269644214222)--(-2.7794754967753703,0.7213614896243091), linewidth(1));
draw((-5.282674200057174,1.8426269644214222)--(-4.76,-3.05), linewidth(1) + qqffff);
draw((-5.282674200057174,1.8426269644214222)--(-1.98,5.49), linewidth(1) + qqffff);
draw((-5.282674200057174,1.8426269644214222)--(-2.751424150756261,3.1202294073890404), linewidth(1));
draw((-2.751424150756261,3.1202294073890404)--(2.3402005304231963,5.69013439688057), linewidth(1));
draw(circle((-5.282674200057174,1.8426269644214222), 4.920466129584573), linewidth(1) + linetype("2 2") + blue);
/* dots and labels */
dot((-1.98,5.49),dotstyle);
label("$A$", (-1.9,5.69), NE * labelscalefactor);
dot((-4.76,-3.05),dotstyle);
label("$B$", (-5.06,-3.43), NE * labelscalefactor);
dot((5.64,-3.05),dotstyle);
label("$C$", (5.72,-2.85), NE * labelscalefactor);
dot((-2.751424150756261,3.1202294073890404),dotstyle);
label("$D$", (-2.68,3.33), NE * labelscalefactor);
dot((2.3402005304231963,5.69013439688057),linewidth(4pt) + dotstyle);
label("$K$", (2.42,5.85), NE * labelscalefactor);
dot((2.7730186914271573,-1.7657842667556372),linewidth(4pt) + dotstyle);
label("$Q$", (2.86,-1.61), NE * labelscalefactor);
dot((-2.7794754967753703,0.7213614896243091),linewidth(4pt) + dotstyle);
label("$P$", (-2.7,0.89), NE * labelscalefactor);
dot((-5.282674200057174,1.8426269644214222),linewidth(4pt) + dotstyle);
label("$M$", (-5.2,2.01), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */[/asy]](http://latex.artofproblemsolving.com/5/f/5/5f5df6886b09dd6b696a7b731b6ae8efb1015259.png)
)
is kept in mind then it becomes clear that considering 
fixes a lot of things I don't bother to write now,Anyways one thing to note that 
.
,
.Also, 
.
be incenter, 
and isogonal conjugate of 
is an involution fixing 
and hence 
as desired.
,
.
.
points 
are concyclic.
points 
are concyclic.
.
are collinear.
so 
as desired. 
.
is tangent to 
.
are collinear(by homothety), now by shooting lemma 
is tangent to 
and we are done 
.
and 
are collinear. 
so 
.
,
,
is tangent to 
lie on 
in that order, so
which is evident. 
tangent 
.
and 
,
are collinear and also 
As a result, 
.
which suffices.
collinear from shooting lemma. We have 
from PoP. So from similarities if we chasing angle we get desired result.
![[asy]
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Converted from GeoGebra by User:Azjps using Evan's magic cleaner
https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py
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A few re-additions (otherwise throws error due to the missing xmin) are done using bubu-asy.py. This adds back the dps, real xmin coordinates, changes the linewidth, the fontsize and adds the directions to the labellings.
*/
pair A = (-37.57730,75.56340);
pair B = (-100.84251,-43.18470);
pair C = (39.83866,-44.14172);
pair D = (-83.79684,-11.19018);
pair M = (-101.25884,33.26400);
pair T = (-56.51806,-80.63562);
pair P = (-60.13324,10.70260);
pair Q = (7.98004,-26.66416);
import graph;
size(10cm);
pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps);
draw(arc(B,23.68739,-0.38976,8.63225)--B--cycle, linewidth(0.75) + blue);
draw(arc(B,23.68739,52.93070,61.95272)--B--cycle, linewidth(0.75) + blue);
draw(A--B, linewidth(0.5));
draw(B--C, linewidth(0.5));
draw(C--A, linewidth(0.5));
draw(circle((-30.23601,-4.57470), 80.47365), linewidth(0.5));
draw(circle((-40.49370,-34.26071), 49.06538), linewidth(0.5));
draw(M--T, linewidth(0.5) + blue);
draw(M--C, linewidth(0.5));
draw(B--M, linewidth(0.5) + blue);
draw(B--P, linewidth(0.5) + red);
draw(B--Q, linewidth(0.5) + red);
dot("$A$", A, N);
dot("$B$", B, W);
dot("$C$", C, dir(0));
dot("$D$", D, W);
dot("$M$", M, NW);
dot("$T$", T, SW);
dot("$P$", P, N);
dot("$Q$", Q, NE);
[/asy]](http://latex.artofproblemsolving.com/b/e/b/bebb2b1e5a02001dc44caa6d032bb893f4681966.png)
are collinear and that 
.
which means that 
.
are collinear.Let 
and
.
.
and 
.
it is not.
so that they switch places. Suppose 
intersect 
which implies 
.
is cyclic, this implies 
.
,
tangents 
at 
and we are done
is collinear and 
.
so 
.
and 
so the desired result follows.
.
so 
and:
so 
and we're done.
be the 
.
and from 
we get that 
