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.
be reals such that 
and 
Prove that
Let be reals such that and 
Prove thatLet be reals such that and 
Prove that
be an acute-angled triangle 
with 
and incenter 
. Let 
be the mixtilinear circle at vertex 
, i.e. the circle internally tangent to the circumcircle of 
and also tangent to lines 
and 
. A circle 
passes through points 
and 
and is tangent to at point 
, with 
and being inside 
. Prove that:
be real numbers such that 
. Prove that the set of possible values of 
is the interval ![$\left[-\frac{2}{3}, \frac{2}{3}\right]$](http://latex.artofproblemsolving.com/4/f/2/4f22df2b391cf34b9b4f86634575e3e0ad23a386.png)
.
, 
, 
be positive integers such that 
. Alice and Bob play a game on an (initially uncoloured) 
grid as follows: cells green. other (i.e.uncoloured) cells blue., and , who has a winning strategy.
and a constant 
, it is given that 
and 
are also friendly polynomials. Prove that 
.
for which the sequence of real numbers with 
for all 
is periodic from the beginning.
be inscribed in the circumcircle 
and circumscribed about the incircle 
, with 
. The incircle touches the sides 
, 
, and 
at 
, 
, and 
, respectively. A line through 
, perpendicular to 
, intersects , , and at 
, 
, and 
, respectively. The line meets at 
(distinct from 
). The circumcircle of triangle 
intersects at 
(distinct from ). Let 
be the midpoint of the arc 
of . The line cuts segments 
and 
at 
and 
, respectively, and the tangents to the circle 
at and intersect at 
. Prove that 
.
be an acute scalene triangle inscribed in a circle . The angle bisector of 
intersects at 
and at 
. The point 
is the midpoint of 
. Let 
be the altitude in , and the circumcircle of 
intersects again at 
. Let 
be the midpoint of , and let 
be the reflection of 
with respect to . Prove that the triangles 
and 
are similar.
and determine the minimum value of 
as well as all values of 
which attain it.
such that 
(where 
), and a functional equation 
that satisfies 
for all , has a finite number of solutions. Find such .
is the function obtained by composing 
times, that is, 
)
" 
:
.
and 
, but because 
differs from 
, we have that 
, and denote 
.
, therefore 
,, and 
, so 
.
, we have that 
, and now just replacing it in 
, we obtain the desired result.
" 
:
and 
, again from the sin law we obtain that 
.
, therefore 
, so 
.
, we obtain 
, therefore replacing that in , we obtain that
, therefore 
, and thus the problem is solved.
is the reflection of 
through 
so 
and 
is not equal to so 
is a isosceles trapezium
meets 
at 
so 
are cyclic Thus 
is isosceles so 
be a convex quadrilateral with 
and let be the intersection point of it's diagonals.
. Remark. Easy and very nice problem !
. ( memorize it ! ):
. Thus, 
. 
.
. Thus, 
. 
a.s.o.
:
and 
are such that they have two equal sides and one eqaul angle, so 
or 
. In both cases, wanted equality follows almost directly.
doesn't work, because then 
, so the only case available is when the angles are suplementary.
and 
. We note that 
; hence it is sufficient to prove that the condition 
is equivalent to the condition 
.
if and only if 
. Since we know 
(or else triangles 
would be congruent and we would have 
) and 
, this later condition is equivalent to the condition 
, or 
, which gives us what we wanted.
, then we obtain 
, from the angle chasing, therefore this case doesn't work 
, because in the hypothesis we have that differs from .
and meet at P:
By the way, nice problem.
is equivalent to 
. Let be a point external to such that 
is a parallelogram. Since 
, 
is equivalent to 
bisects 
and meets the perpendicular bisector of 
at . This occurs if and only if 
is cyclic.
. Now note that since triangles and 
are isosceles,![\[\angle{PDB}=\angle{DAC}+\angle{ADB}=\angle{DBC}+\angle{ACB}=\frac{x}{2}\]](http://latex.artofproblemsolving.com/2/4/3/243bf0e4bc0a9f0e9ec83eaca95f376bec6ba779.png)
is cyclic if and only if 
or 
.
to such that 
. Angle chase to get the relationship between triangles 
and 
is SSA, so either 
or 
. The former gives , which is a contradiction. The latter gives the required. to such that 
. Again, angle chase to get that and 
which gives the required.
have same radii and hence if 
then either 
or their sum is 
.One gives contradiction to 
while other solves the problem 
So 
) and (
)
from here two cases exist.
which means Diagonals are equal contradiction.
Done!!!
and which means Diagonals are equal contradiction. 
or 
. From isosceles triangle 
and sine law:
and sine law:
and 
or 
and
and 
, 
, 
, 
. From Stewart's theorem in 
and 
, we have 
. Rearranging gives 
. Since 
by the condition of the problem, we obtain 
. be the reflection of across the midpoint of , and let be the reflection of across the midpoint of 
. Then, 
, 
, 
, and 
. Also, 
and 
. Therefore, 
and 
are equilateral triangles.
, we have 
. 
be the intersection of and 
, with 
. If we let 
, then 
, 
, and 
. on ![$[AE]$](http://latex.artofproblemsolving.com/0/e/e/0ee81ab9421e97824f7785dc2c44c0f5b04a03b1.png)
such that 
, we have as an equilateral triangle, and by similarity or symmetry, 
. Thus, 
. Similarly, taking on ![$[DE]$](http://latex.artofproblemsolving.com/4/f/5/4f55b2be1d3d9963afec61b4973bfecc6141b1ff.png)
gives us 
, we obtain the desired result.
.
.
and 
,
q.e.d.
is not equal with 60, we find # 