[/quote]
,
the point of intersection of the tangents at points 
and 
of the circumscribed circle of 
is the midpoint of the segment 
.
meets the circumcircle 
.
.
:
be reals such that 
and 
Prove that
Where 
Prove that: 
is the sum of 
square numbers
Let 
and 
is the sum of 
is the sum of 
,![\[\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\geq 4+\frac{8(a-c)^2}{(a+b+c+d)^2}.\]](http://latex.artofproblemsolving.com/7/b/6/7b6e32d32d89ca058a06974e7276f0a5874d90a7.png)
such that 
that satisfy the equation
.
and 
,
and 
be the incenter of triangles 
and 
,
is the Euler line of triangle 
.
,
,
,
are odd primes and![\[2^ap^b=(p+2)^c-1.\]](http://latex.artofproblemsolving.com/5/8/5/58518a8ebed14836b72af743ff100eae2efba5e2.png)
we have 
.
.
Y by
In triangle 
we have 
. The point 
is the midpoint of 
. The line through parallel to 
intersects 
in 
. The midpoint of line segment 
is 
. The lines 
and 
are perpendicular.
(a) Prove that triangles
and 
are similar.
(b) Prove that
and are perpendicular.
![[asy]
unitsize(1 cm);
pair A, B, C, D, E, M;
A = (0,0);
B = (4,0);
C = (2.6,2);
M = (A + B)/2;
D = (A + C)/2;
E = (C + D)/2;
draw(A--B--C--cycle);
draw(C--M--D--B);
dot("$A$", A, SW);
dot("$B$", B, SE);
dot("$C$", C, N);
dot("$D$", D, NW);
dot("$E$", E, NW);
dot("$M$", M, S);
[/asy]](//latex.artofproblemsolving.com/2/b/e/2becd6322b16aaf932d335309f266b04c29080bc.png)
Be aware: the figure is not drawn to scale.
we have 
. The point 
is the midpoint of 
. The line through parallel to 
intersects 
in 
. The midpoint of line segment 
is 
. The lines 
and 
are perpendicular.(a) Prove that triangles
and 
are similar.(b) Prove that
and are perpendicular.![[asy]
unitsize(1 cm);
pair A, B, C, D, E, M;
A = (0,0);
B = (4,0);
C = (2.6,2);
M = (A + B)/2;
D = (A + C)/2;
E = (C + D)/2;
draw(A--B--C--cycle);
draw(C--M--D--B);
dot("$A$", A, SW);
dot("$B$", B, SE);
dot("$C$", C, N);
dot("$D$", D, NW);
dot("$E$", E, NW);
dot("$M$", M, S);
[/asy]](http://latex.artofproblemsolving.com/2/b/e/2becd6322b16aaf932d335309f266b04c29080bc.png)
Be aware: the figure is not drawn to scale.
This post has been edited 6 times. Last edited by nsato, Feb 13, 2023, 8:10 PM
.
.
11 minutes ago 
.
be the reflection of 
We can see that 
collinear. (Because 
)
is a parallelogram 
,
)
.
is the reflection of 
.
= 
= 
,
= 
= 
.
similar 
Since 
