[/quote]
,
that satisfy the equation
.
.
and 
be points on the side BC such that 
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 
.
.
be reals such that 
and 
Prove that
be a cyclic quadrilateral. The point 
is the reflection of 
the intersection of 
is the reflection of 
.
be the reflection of 
.
concyclic.
. Prove that: 
at 
at 
intersects the circumcircle 
of triangle 
and 
intersect 
and 
respectively.
such that 
for all real number 
and 
is an equilateral . Let 
and 
at 
and 
at 
,
on 
satisfies 
and 
.
be the point where the perpendicular from 
to line 
meets the angle bisector of 
.
holds.
be a positive integer such that 
.
.
Y by Adventure10
In the acute triangle 
, the midpoint of side 
is called 
. Point 
lies on the angle bisector of 
such that 
. Point 
lies on the angle bisector of 
such that 
. Line segments 
and 
intersect in point 
. Prove that is the midpoint of .
![[asy]
import geometry;
unitsize (1.2 cm);
pair A, B, C, M, X, Y, Z;
A = (0,0);
B = (4,1.5);
C = (0.5,3);
M = (B + C)/2;
X = extension(M, incenter(A,B,M), B, B + rotate(90)*(incenter(A,B,M) - M));
Y = extension(M, incenter(A,C,M), C, C + rotate(90)*(incenter(A,C,M) - M));
Z = extension(A,M,X,Y);
draw(A--B--C--cycle);
draw(A--M);
draw(M--interp(M,X,2));
draw(M--interp(M,Y,2));
draw(B--X, dotted);
draw(C--Y, dotted);
draw(X--Y);
dot("$A$", A, SW);
dot("$B$", B, E);
dot("$C$", C, N);
dot("$M$", M, NE);
dot("$X$", X, NW);
dot("$Y$", Y, NE);
dot("$Z$", Z, S);
[/asy]](//latex.artofproblemsolving.com/5/4/f/54f32dbc0eef6f62486c7cea9fc71e53af39afe8.png)
, the midpoint of side 
is called 
. Point 
lies on the angle bisector of 
such that 
. Point 
lies on the angle bisector of 
such that 
. Line segments 
and 
intersect in point 
. Prove that is the midpoint of .![[asy]
import geometry;
unitsize (1.2 cm);
pair A, B, C, M, X, Y, Z;
A = (0,0);
B = (4,1.5);
C = (0.5,3);
M = (B + C)/2;
X = extension(M, incenter(A,B,M), B, B + rotate(90)*(incenter(A,B,M) - M));
Y = extension(M, incenter(A,C,M), C, C + rotate(90)*(incenter(A,C,M) - M));
Z = extension(A,M,X,Y);
draw(A--B--C--cycle);
draw(A--M);
draw(M--interp(M,X,2));
draw(M--interp(M,Y,2));
draw(B--X, dotted);
draw(C--Y, dotted);
draw(X--Y);
dot("$A$", A, SW);
dot("$B$", B, E);
dot("$C$", C, N);
dot("$M$", M, NE);
dot("$X$", X, NW);
dot("$Y$", Y, NE);
dot("$Z$", Z, S);
[/asy]](http://latex.artofproblemsolving.com/5/4/f/54f32dbc0eef6f62486c7cea9fc71e53af39afe8.png)
This post has been edited 2 times. Last edited by nsato, Feb 13, 2023, 9:58 PM
.
- rectangle, 
.
lay on middle lines of 
parallel to 
,
