AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
and the recursive relation
for all natural number 
. Find the general form of 
.
+1 w
.
be a triangle with 
let 
be the circumcircle of 
and let 
be its radius. Point 
is chosen on 
such taht 
and point 
is the foot of the perpendicular from to 
. Ray 
mets again at 
. Point 
is chosen on line 
such that 
and 
lie on a line in that order. Finally, let 
be a point satisfying 
and 
. Prove that lies on .
in the plane is assigned a real number 
It is known that 
whenever 
is the centroid of 
Prove that 
for all points 
such that 
for all 
be a positive integer. A board with a format 
is divided in equal squares.Determine all integers ≥3 such that the board can be covered in 
(or 
) pieces so that there is exactly one empty square in each row and each column.
be a positive integer. Given a sequence 
, 
, 
with 
or 
for each 
, , 
, the sequences 
, , and 
, , 
are constructed by the following rules: ![\[a_0 = b_0 = 1, \quad a_1 = b_1 = 7,\]](http://latex.artofproblemsolving.com/d/f/d/dfd49833078b6b676cd261cc358e8114b2c4b7f4.png)
![\[\begin{array}{lll}
a_{i+1} =
\begin{cases}
2a_{i-1} + 3a_i, \\
3a_{i-1} + a_i,
\end{cases} &
\begin{array}{l}
\text{if } \varepsilon_i = 0, \\
\text{if } \varepsilon_i = 1, \end{array}
& \text{for each } i = 1, \dots, n - 1, \\[15pt]
b_{i+1}=
\begin{cases}
2b_{i-1} + 3b_i, \\
3b_{i-1} + b_i,
\end{cases} &
\begin{array}{l}
\text{if } \varepsilon_{n-i} = 0, \\
\text{if } \varepsilon_{n-i} = 1, \end{array}
& \text{for each } i = 1, \dots, n - 1.
\end{array}\]](http://latex.artofproblemsolving.com/8/9/0/8908021c60992e39acdb4fcf6b587f96b88539e0.png)
Prove that 
.
and be natural numbers. Every one of the 
squares of the board is colored either black or white, so that no 2 neighbouring squares are the same color(the board is colored like in chess").In one step we can pick 2 neighbouring squares and change their colors like this: and can we ,in a finite sequence of these steps, switch the starting colors from white to black and vice versa.
be a real or complex inner product space. Suppose that 
and that 
Then
be an integer, and let 
be positive integers. Show that there exist positive integers 
satisfying the following three conditions:
for 
the remainders of on division by are pairwise different; and
denotes the integer part of real number 
, that is, the largest integer that does not exceed .)
, the angle bisector at 
meet the opposite sides at 
, respectively. Prove that if the quadrilateral 
is cyclic, then
table filled with natural numbers, we say it is a divisor table if:-th row are exactly all the divisors of some natural number 
,
-th column are exactly all the divisors of some natural number 
,
for every 
.
is given. Determine the smallest natural number , divisible by , such that there exists an divisor table, or prove that such does not exist.
be a right-angled triangle where ∠
=90°.Let 
be an altitude of that triangle and points and 
be the midpoints of and 
, respectively.If is the circumcenter of the triangle 
, prove that 
and are paralel.
, let be a point (
). Let 
be the anticomplement of , and let be the 
-anticomplementary conjugate of . Prove that the line 
is parallel to the real asymptote of the circular pivotal isocubic with pivot .
AC. Then ID = r, the inradius of 
ABC. Observe EF
BC and hence 
AEF = ABC = 
.AIF =.
= F D · DA. If a>c, then F A > DA and we have
− (s − a).
=
= (b + c − a)(a − c).
,
.
.Let E and F be respectively the midpoints of AB and AC.Suppose the incentre I of ABC lies on the circumcircle of triangle AEF,find the ratio BC/AB.
as incenter of 
then observe 
is isosceles right 
with 
and also 
then Pythagoras gives 
and also let 
now as 
now By angle bisector Theorem we get 
so by pythagoras we get 
on 
.by a bit of angle chasing Inradii=
.Prove that 
is tangent on the circumcircle of 
.Then use power of point 
.use 
.Solve the equation and we are done.
to 
and let the touch-point with 
,
,
,
and 
so we have 
substitute PGT, then do squaring to get 
.