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.
so that the sum of its edges' lengths is minumum.
find, in terms of , the minimum value of 
for which there is a set of 
distinct positive integers that has sum greater than but every subset of size has sum at most 
with 
equilateral. Let 
be the area of the triangle of sides 
, 
and 
. Show that 
where 
is the circumradius and 
is the volume of the tetrahedron. be circumscribed about circle 
, and let 
be the orthocenter of 
. The circle touches line 
at 
. The tangent to the circle 
at meets at . Let 
be the midpoint of 
, and let the line 
meet again at 
. The tangent to parallel to meets the line 
at 
. Prove that 
is tangent to .
be an equilateral triangle, 
be a point inside it, and 
be the intersections of 
with the sides of . If 
are the midpoints of , 
, 
, show that there is a triangle with sides 
, 
and 
.
be two integers and 
be a positive integer such that 
. Show that ![\[(n-t)(d-t)>\dbinom{n-t+r}{r+1},\]](http://latex.artofproblemsolving.com/e/5/f/e5f9486f980870ff72858b57f7b50cd2aa9d142a.png)
for 
, 
, 
we have ![\[ 3(a+b+c) \ge 8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3+b^3+c^3}{3}}. \]](http://latex.artofproblemsolving.com/a/9/8/a98b6c29802355c491016bcce58228a020423526.png)
Prove that 
When does equality hold?
satisfy that: 
and 
. Find the maximum: 
be positive real numbers satisfying 
. Prove that![\[
3(x^{\frac{2}{3}} + y^{\frac{2}{3}}) \geq 4 + 2x^{\frac{1}{3}}y^{\frac{1}{3}}.
\]](http://latex.artofproblemsolving.com/f/3/c/f3cd114f909f388eacee142d3f90d07f6b149131.png)
be positive real numbers such that 
.![\[
\left( \frac{a + b}{c} \right) \left( \frac{b + c}{a} \right) \left( \frac{c + a}{b} \right) \geq 9.
\]](http://latex.artofproblemsolving.com/7/7/f/77f8bffb0d9b617655c7a8f4a8eb7245d60d8d9f.png)
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.
and 
Prove that
, such that 
Prove that :
equality cases : ?
to 
and let the touch-point with 
,
,
,
and 
so we have 
substitute PGT, then do squaring to get 
.