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 positive real numbers such that 
Prove that we have![\[(a+b)(b+c)(c+a)\geq 8.\]](http://latex.artofproblemsolving.com/2/a/6/2a63f19de634fce70ff4ebc02ad13d442a20c378.png)
be the lengths of the sides of the triangle. Prove that:
such that 
Prove that 
with the property that ![\[f(x-f(y))=f(f(x))-f(y)-1\]](http://latex.artofproblemsolving.com/f/2/5/f25cc1e8ae0be1fd02b347fd94be4fab88af1d46.png)
holds for all 
.
be a function such that
for all 
. Find all such functions 
.
be a function such that
for all 
. Determine all such functions .
. Find all 
such that for all 
be the set of real numbers. Find all functions 
where there exist a real constant 
such that 
holds for all reals 
, 
, 
that satisfy 
.
with its center 
. There are two fixed points 
and one moving point 
on . The midpoint of the line segment 
is 
. 
is a fixed point on . Line 
intersects
at 
, and line 
intersects 
at 
. such that is always tangent to 
when varies.
such that
be a scalene triangle with orthocenter 
and circumcenter . Denote by , 
the midpoints of 
, 
. Suppose the circle 
with diameter meets the circumcircle of at 
, and meets line 
at a point 
. The tangent to at 
meets line 
at 
. Show that the circumcircles of 
and 
intersect at a point 
on 
.
such that
for all 
) for all their time and support. We received immense help from TestX01 for the proof of claim 2.19 and it's associated lemma.
" 
:
.
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.
such that: 
.
.
.
and 
,
q.e.d.
is not equal with 60, we find # 