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 a triangle. Let 
be a circle passing through 
intersecting 
at 
and 
at 
. Let 
be the intersection of 
and . Further, let 
and 
be the intersections of 
and 
with the tangent to 
at 
. Now, let 
be the second intersection of 
and . Then, prove that , , , 
and are concyclic.
, 
is the midpoint of 
. 
is an arbitrary point on 
. Let 
be the intersection of 
and 
. The line 
intersects with the circumcircle of 
, for the second time at 
. 
is the circumcenter of triangle 
. Prove that 
.
where 
are integers such that 
for all 
.
a prime number with 
and 
when 
. then 
else 
.
be positive integers and 
be prime. We know that 
and 
. Prove that 
is a perfect square.
be a triangle with circumcentre 
. The points 
and 
are interior points of the sides 
and 
respectively. Let 
and 
be the midpoints of the segments 
and 
. respectively, and let 
be the circle passing through and . Suppose that the line is tangent to the circle . Prove that 
with 
such that ![\[ (a-1)(b-1)(c-1) \]](http://latex.artofproblemsolving.com/f/8/c/f8c33b59bdba1fdd12dec51f16a0b1b93e60754f.png)
is a divisor of 
is called good if there exists a permutation 
of the numbers 
, such that:
and 
have different parities for every 
is a quadratic residue modulo 
for every 
is considered a quadratic residue modulo if there exists an integer 
such that 
.
| 
with 
are positive integers and is prime (
), we get: 
be a real numbers such that 
Prove that
Where 
be non-negative real numbers. Prove that
Looking for an algebraic solution!
lie on the arc of the circumcircle of 
(this arc does not contain ) such that 
. Let denotes the midpoint of the chord 
. Show that 
.
be the circumcenter of the . Then 
. We draw a normal line from the point at 
and let it intersect the circumcircle in the point
. Since 
we have that is a line of symmetry of 
. According to this,
. Now, from the triangle inequality
.
, so
. That is why 
.
![\[
r_1 \ge r_a,\quad r_2 \ge r_b,\quad r_3 \ge r_c
\]](http://latex.artofproblemsolving.com/8/2/1/8214c052ab9ad2c979414f92b6707fa6de33664f.png)
because if any of the radii were smaller than these, the circles would not be able to reach and touch the incircle.![\[
\frac{r_1 + r_2 + r_3}{r} \ge \frac{r_a + r_b + r_c}{r}
\]](http://latex.artofproblemsolving.com/b/b/a/bbadb40e6aa197207abfa907981e0a4a88926969.png)
![\[
\frac{r_a + r_b + r_c}{r} = \frac{h_a - 2r}{h_a} + \frac{h_b - 2r}{h_b} + \frac{h_c - 2r}{h_c}
\]](http://latex.artofproblemsolving.com/5/3/2/5329fec1f02a728b5f34817e0ed43ded663516ae.png)
![\[
= \left(1 - \frac{2r}{h_a}\right) + \left(1 - \frac{2r}{h_b}\right) + \left(1 - \frac{2r}{h_c}\right)
= 3 - 2r \left( \frac{1}{h_a} + \frac{1}{h_b} + \frac{1}{h_c} \right)
\]](http://latex.artofproblemsolving.com/1/e/0/1e087ea87566fb4cf8cd2ae0cda3389247095eea.png)
![\[
\frac{1}{h_a} + \frac{1}{h_b} + \frac{1}{h_c} = \frac{a + b + c}{2S}
\]](http://latex.artofproblemsolving.com/a/4/a/a4ab49131584cf2028270e7248fcbb5d2c6a7adb.png)
![\[
3 - 2r \cdot \frac{a + b + c}{2S} = 3 - \frac{4S}{a + b + c} \cdot \frac{a + b + c}{2S} = 3 - 2 = 1
\]](http://latex.artofproblemsolving.com/0/f/3/0f3ada5ca22deeecc1644d79138942caf04d5134.png)
![\[
\frac{r_1 + r_2 + r_3}{r} \ge 1
\]](http://latex.artofproblemsolving.com/7/1/e/71ea14c8fb8e3860a1db427a0afcdb3b2fd6254b.png)
be real numbers such that 
.
.
is equivalent to 
,
.
and the triangle inequality imply that the left hand side of the desired inequality is at least 
.