[/quote]
,
be non-negative real numbers. Prove that
Looking for an algebraic solution!
such that 
.
has no integer solution.
,
,
,
be positive numbers. Prove that
![\[
r_1 \ge r_a,\quad r_2 \ge r_b,\quad r_3 \ge r_c
\]](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)
a prime number with 
and 
when 
.
else 
.
such that 
.
and 
is obtuse. Point 
is the reflection of 
in 
.
be an arbitrary point on line 
is in between 
.
cuts the circumcircle of 
.
and 
cut 
respectively. Prove that 
is tangent to the circumcircle of triangle 
.
from the non-zero reals to the non-zero reals, such that ![\[f(x+y) \left(f(x) + f(y)\right) = f(xy)\]](http://latex.artofproblemsolving.com/2/f/e/2fe8a09e628f60faedaa07ceb154891375763a83.png)
for all non-zero reals 
such that 
.
that satisfy the equation
for any 
.
.
and circumcenter 
.
be the incenter of triangle 
,
be a point on the arc 
.
be the circumcenter of triangle 
.
and 
meet on 
non-negative real numbers, then we have:
| 
with 
are positive integers and 
is prime 
)
such that 
and:
Y by
Let triangle 
be inscribed in the circle 
. A line through point 
intersects 
and 
at points 
and 
, respectively. Let 
be the reflection of across the midpoint of , and 
be the reflection of across the midpoint of . Prove that:
a) the reflection of the orthocenter
of triangle across line 
lies on the circle .
b) the orthocenters of triangles
and 
coincide.
Im looking for a solution used complex bashing
be inscribed in the circle 
. A line through point 
intersects 
and 
at points 
and 
, respectively. Let 
be the reflection of across the midpoint of , and 
be the reflection of across the midpoint of . Prove that:a) the reflection of the orthocenter
of triangle across line 
lies on the circle .b) the orthocenters of triangles
and 
coincide.Im looking for a solution used complex bashing
,
To do part (a), we find the vectors
Then if 
is the reflection of 
,
and then
and so 
.
has equation
![$$\frac{x^2(a+c) - z(ac+x^2)}{(a-b)(ac+x^2)} = \frac{ab\left[(a+c) - \overline{z}(ac+x^2)\right]}{(a-b)(ac+x^2)}$$](http://latex.artofproblemsolving.com/a/b/7/ab72d564430466676f50fba3c89b8e21000e3c70.png)
Similarly, the altitude from 
has equation
Intersecting these gives
This point 
is the orthocenter of 
.
-- then we will have 
as well by symmetry. Now
and so
which is indeed pure imaginary. 
