[/quote]
,
such that the following equation holds:![\[
7^a + 1 = 2b^2
\]](http://latex.artofproblemsolving.com/e/0/8/e08d3fc7feefd7b4acc4a0f9c7e8bcfa328b5791.png)
have the property that 
is a perfect square for infinitely many positive integers 
?
and 
respectively be the incentre, circumcentre and circumcircle of triangle 
are chosen on 
,
is the foot of the altitude from 
to 
.
are similarly defined, prove that lines 
and 
concurr on 
.
and let 
and 
be tangency points of the incircle of 
,
and 
respectively. Let the circle with centre 
and 
.
,
and 
,
.
and 
be feet of perpendiculars from 
and 
respectively. Prove that 
,
and 
are concurrent.
that is not isosceles at 
,
be its incircle, which is tangent to 
at 
,
and 
intersect the line passing through 
at 
and 
,
and perpendicular to 
intersect 
and 
at 
and 
,
are collinear and that 
are concurrent.
positive integer numbers such that the least common multiple (LCM) of all these numbers is not a perfect square. Mykhailo consecutively hides one of these numbers and writes down the LCM of the remaining 
numbers that are not hidden. What is the maximum number of the 
satisfying: 
.
min
is a prime number and 
are positive integers. Prove that there exist infinitely 
are divisible by 
if 
if 
primes, and let 
be the sum of the divisors of ![\[\left|\sum_{d|n}\frac{\mu(d)\sigma(d)}{d}\right| \geq \frac{1}{n}, \]](http://latex.artofproblemsolving.com/8/e/b/8eb88de7cd7ea8e2853dbc80cae24bdc6b77f9be.png)
be a sequance of natural integers. Proof that, any number in the sequance can be written only as 
for any 
and not necessarily distinct primes 
for 
.
are same in some case.
![\[2^{xy}f(xy-1)+2^{x+y+1}f(x)f(y)=4xy-2,\forall x,y\in\mathbb{R}\]](http://latex.artofproblemsolving.com/d/1/d/d1de7196fd3ba60732c14fa2b156ec880b4829d0.png)
has no integer solutions
play a game with a stack of 
coins from the stack. The player who takes the last coin wins. Prove that there are infinitely many integers 
Y by Rounak_iitr
Let 
be an acute triangle, 
its circumcircle and 
its circumcenter. The altitude from 
intersects in a point 
and the segment 
intersects the circumcircle of 
in a point 
. Finally, let 
be the midpoint of 
. Show that 
is parallel to 
.
be an acute triangle, 
its circumcircle and 
its circumcenter. The altitude from 
intersects in a point 
and the segment 
intersects the circumcircle of 
in a point 
. Finally, let 
be the midpoint of 
. Show that 
is parallel to 
.
be the antipode of 
are collinear which is clear since simple angle chase leads to 
.
.
are parallel. Also, lines 
and ![\[2\measuredangle OEA = 2\measuredangle ODC = \measuredangle DOC = 2\measuredangle DAC = 2(90+\measuredangle BCA)= 2\measuredangle BCA\]](http://latex.artofproblemsolving.com/5/3/2/532ded7694cac26301b1822876d8d03caca9938c.png)
from which it is clear that 
and thus, 
as desired.![\[\measuredangle OFC = \measuredangle ODC = \measuredangle DCO = \measuredangle DEO\]](http://latex.artofproblemsolving.com/6/2/7/6272992e1397377e9e78c4c3cebf19b08d9440ac.png)
from which using the previous result it follows that 
,
and 
are isosceles trapezoids. Now, we also note the following.
is isosceles.![\[\measuredangle CFD = \measuredangle COD = 2\measuredangle CBD = 2\measuredangle FBD\]](http://latex.artofproblemsolving.com/9/d/e/9def22fa7ffb19ff62bbda273d2784f355631c6a.png)
from which the claim is clear.
.
due to the previously proved parallel lines, we have that![\[\frac{BM'}{M'E}=\frac{BR}{OE}=\frac{BR}{RD}=1\]](http://latex.artofproblemsolving.com/c/0/1/c01eb6709198b971a6cfc3e5bc9087767e2467df.png)
from which it is clear that 
must be the midpoint of 
and we conclude that 
and 
.![[asy] import olympiad; unitsize(12); pair A, B, C, O, D, E, M, P, X; C = origin; B = (13,0); A = (5,11); draw(A--B--C--cycle); path w = circumcircle(A, B, C); draw(w, red+dotted); O = circumcenter(A, B, C); P = foot(A, B, C); D = IP(w, A--(P-5*(A-P)), 1); draw(A--D); path w1 = circumcircle(O, C, D); draw(w1, blue+dotted); E = IP(w1, A--C); M = (B+E)/2; draw(D--E--B--cycle); draw(M--O--E); X = IP(B..C,D..E); dot(A); dot(B); dot(C); dot(O); dot(P); dot(D); dot(E); dot(M); dot(X); label("$A$", A, N); label("$B$", B, SE); label("$C$", C, SW); label("$O$", O, NW); label("$P$", P, SW); label("$D$", D, SW); label("$E$", E, NW); label("$M$", M, NE); label("$X$", X, SW); [/asy]](http://latex.artofproblemsolving.com/a/e/8/ae8cf382a0f5a4c91d214b8a5f73acd051c950b4.png)
to show that 
.
is cyclic, 
.
.
is cyclic, 
and also since 
,
.
.
be the reflection of 
through 
.
,
and 
are collinear so by homothety 
.
.
and since 
.
,
.
.
.
.
(crucial). Now since 
so 
is an isosceles trapezoid. Therefore 
so 
.
