and let 
and 
be its orthocenter and circumcenter, respectively. Let 
be the circle 
.
and the circle of radius 
cross 
and 
,
and circle 
are concurrent.
,
,
.
at 
,
be the midpoint of arc 
.
intersects 
,
and 
meet on 
.
.
be the second intersection point of the circumcircle circumscribed to 
with 
and 
and 
with 
,
and 
be the intersection points of lines 
and 
with lines 
line passes through the midpoint of 
are real numbers, find the smallest possible value of the expression:
and 
be prime numbers. Prove that if 
is a perfect square, then 
is also a perfect square.
is a perfect square.
be a positive integer. Aslı and Zehra are playing a game on an 
chessboard. Initially, 
stones are placed on the squares of the board. In each move, Aslı chooses a row or a column; Zehra chooses a row or a column. The number of stones in each square of the chosen row or column must change such that the difference between the number of stones in a square with the most stones and a square with the fewest stones in that same row or column is at most 1. For which values of 
reals, and 
.
?
be positive real numbers such that 
.
knowing that the denominators are positive real numbers.
such that: Among every 
equals to a positive integer power of 
.
Y by Davi-8191, artsolver, LoveIsABeautifulPain, me9hanics, AlastorMoody, MatBoy-123, Adventure10, farhad.fritl, ehuseyinyigit
Let 
be a triangle with 
. Let 
be the intersection point of the internal bisector of angle 
and the circumcircle of . Let 
be the intersection point of the perpendicular bisector of 
with the external bisector of angle 
. Prove that the midpoint of the segment 
lies on the circumcircle of triangle 
.
Olimpiada de Matemáticas, Nicaragua
be a triangle with 
. Let 
be the intersection point of the internal bisector of angle 
and the circumcircle of . Let 
be the intersection point of the perpendicular bisector of 
with the external bisector of angle 
. Prove that the midpoint of the segment 
lies on the circumcircle of triangle 
.Olimpiada de Matemáticas, Nicaragua
This post has been edited 1 time. Last edited by MellowMelon, May 17, 2017, 5:50 PM
Reason: add proposer
Reason: add proposer
and 
the perpendicular bisector of 
.
are symmetric in 
passes through 
,
is a rectangle and 
,
,
and 
,
.
are collinear.
and 
.
and 
so 
Finally, by Menelaus' Theorem in 
for transversal 
The last result is due to the Angle-Bisector Theorem. Evidently, 
so we are done! 
.By angle chasing we get triangles 
and 
are similar.So there exist a spiral similarity centered at Z that takes KD to NC .Then by lemma if 
is cyclic.Again angle chasing gives 
.
:
,
and 
the midpoint of 
,
is cyclic we get: 
,
we get 
and 
we get 
,
we get 
hence 
is cyclic.
.
to 
,
,
.
is tangent to 
,
to 
,
.
,
,
onto the external angle bisector of 
,
at 
,
bisects 
at a point 
of 
,
so 
,
,
,
,
.
,
and 
.
,![\[D=\left(\frac{-a^2}{\left(a+b+c\right)\left(-a+b+c\right)},\frac{b\left(b+c\right)}{\left(a+b+c\right)\left(-a+b+c\right)},\frac{c\left(b+c\right)}{\left(a+b+c\right)\left(-a+b+c\right)}\right)=\left(-a^2:b\left(b+c\right):c\left(b+c\right)\right).\]](http://latex.artofproblemsolving.com/e/b/8/eb81ad93ca9f1544f50eb11119eef31bd67f0409.png)
Now, ![\[x+y+z=1\]](http://latex.artofproblemsolving.com/c/6/2/c621318c449929f0901592d85f0d545a3f543215.png)
![\[-b^2x+\left(c^2-a^2\right)y+b^2z=0\]](http://latex.artofproblemsolving.com/8/8/2/8823a20238cd31deff3aa79b81a87cdbde3b29c8.png)
![\[cy+bz=0.\]](http://latex.artofproblemsolving.com/5/b/9/5b9e6d6a1ddc1bf2d0ac1390cac17b596e2c053c.png)
This is ![\[Z=\left(\frac{a^2-c^2+bc}{\left(a-b+c\right)\left(a+b-c\right)},-\frac{b^2}{\left(a-b+c\right)\left(a+b-c\right)},\frac{bc}{\left(a-b+c\right)\left(a+b-c\right)}\right)=\left(a^2-c^2+bc:-b^2:bc\right).\]](http://latex.artofproblemsolving.com/0/e/9/0e95c34823d41f2d10ab8d10cbc03de435843b1b.png)
be represented by ![\[-a^2yz-b^2zx-c^2xy+\left(ux+vy+wz\right)\left(x+y+z\right)=0.\]](http://latex.artofproblemsolving.com/8/9/f/89f7c2485e0b983f4ca562a22ebcef2c40a6bb83.png)
If 
represents the power of a point ![\[p\left(P,\Gamma\right)+\left(ux+vy+wz\right)\left(x+y+z\right)=0.\]](http://latex.artofproblemsolving.com/a/3/1/a31e373ca2849fcc8d14ccdeb1d90e19055f2375.png)
Plugging in 
,![\[p\left(P,\Gamma\right)+\left(vy+wz\right)\left(x+y+z\right)=0.\]](http://latex.artofproblemsolving.com/2/3/1/231b72f1065d3d18a40e2307354d4cd05de2bafb.png)
Plugging in ![\[bv+cw=0.\]](http://latex.artofproblemsolving.com/3/0/5/305c6e61cce1e4e9d98825e6e6b6e1d39be1be06.png)
Plugging in ![\[a^2b^3c-b^3c\left(a^2-c^2+bc\right)+b^2c^2\left(a^2-c^2+bc\right)+\left(-b^2v+bcw\right)\left(a-b+c\right)\left(a+b-c\right)=0.\]](http://latex.artofproblemsolving.com/1/6/f/16f6a0c81f955994d9df7526cd8f08993343342b.png)
This is ![\[b^2c^2\left(a^2-\left(b-c\right)^2\right)-b\left(bv-cw\right)\left(a^2-\left(b-c\right)^2\right)=0,\]](http://latex.artofproblemsolving.com/4/e/b/4eb24da4983b5285d5178b69a94cd54a1fdf481d.png)
so ![\[bv-cw=bc^2.\]](http://latex.artofproblemsolving.com/c/0/5/c05222a782316a2d93728b85a9f888943832de6c.png)
Thus, 
and 
,![\[p\left(P,\Gamma\right)+\left(\frac{c^2}{2}y-\frac{bc}{2}z\right)\left(x+y+z\right)=0.\]](http://latex.artofproblemsolving.com/5/d/8/5d8009d4c64db0348a7decdfebdff3e2a46d8e11.png)
But 
,![\[p\left(M,\Gamma\right)+\left(\frac{c^2}{2}\cdot\frac{1}{2}-\frac{bc}{2}\cdot0\right)\left(\frac{1}{2}+\frac{1}{2}+0\right)=-\frac{c^2}{4}+\frac{c^2}{4}=0,\]](http://latex.artofproblemsolving.com/7/b/4/7b4034f9346b6d224f713ab1d6affb621b6a6ee8.png)
so 
be the reflection of 
be the reflection of 
across 
and 
are similar right triangles so 
,
as 
,
,
and 
all be on the complex unit circle. Using lowercase to represent the complex number of each point, we see 
for some real number 
is on the perpendicular bisector of 
for complex numbers 
.
,
is the midpoint of arc 
,
.
,
is right. Using the complex foot formula, the foot from 
,
,
and 
,
.
.
is a real number, which is to say 
is a real number, but this is 
,
are on the unit circle, 
and 
,
and 
are real. Hence, 
such that 
is an external bisector; hence 
.
.
,
lies on ![[asy] pair A = dir(110); pair B = dir(200); pair C = dir(340); draw(unitcircle, lightred);
draw(A--B--C--cycle, lightred);
pair D = dir(270); pair M = midpoint(A--B); pair Z = extension(origin, A+C, A, dir(90)); pair N = midpoint(A--C);
draw(circumcircle(A, D, Z), dashed+heavycyan); draw(A--Z, lightred); draw(A--D, lightred); draw(D--M--Z--cycle, heavycyan);
pair T = extension(M, N, D, C); draw(M--T--D, dashed+cyan); pair L = dir(90); draw(circumcircle(Z, T, C), dotted);
dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$M$", M, dir(M)); dot("$Z$", Z, dir(Z)); dot("$N$", N, dir(225)); dot("$T$", T, dir(T)); dot("$L$", L, dir(L));
[/asy]](http://latex.artofproblemsolving.com/a/7/c/a7cecc5d7fb86f45c4e004ec7a73dc48790f459a.png)
be the 
also by spiral similarity, so ![\[ \frac{ZM}{ZD} = \frac{LB}{LD} = \cos A/2 = \frac{MN}{DC} \]](http://latex.artofproblemsolving.com/2/3/6/2362132557880a051e2bdd8f561028fee5fa0122.png)
plus 
.
,
,
.
.
,![\[\dfrac{DE}{EM}=\dfrac{DE}{CN}=\dfrac{DC}{CZ}.\]](http://latex.artofproblemsolving.com/f/8/5/f85d3974e4fbcdb12175baa12ca8de2111157ddb.png)
Furthermore, ![\[\angle DEM=90^\circ+\angle MEB = 90^\circ + \angle ACB = (\angle ZCN+\angle DCE)+\angle ACB=\angle ZCD.\]](http://latex.artofproblemsolving.com/f/0/8/f084653f76a3424f960476b3c58ccd5747ef05db.png)
Hence 
,![\[\triangle DEC\sim\triangle DMZ\quad\implies\quad \angle DMZ=\angle DEC = 90^\circ.\]](http://latex.artofproblemsolving.com/0/2/7/027818d0f6ff208fb7d397c9c6f4f05ae7d28c06.png)
Combining this with 
establishes the desired cyclicity.![[asy]
size(250);
defaultpen(linewidth(0.8));
pair A = dir(120), B = dir(215), C = dir(325), D = dir(270), M = (A+B)/2, N = (A+C)/2, E = (B+C)/2, P = dir(90), Z = extension(A,P,origin,N);
draw(A--B--C--A--Z--C^^Z--N^^B--D--C^^unitcircle);
draw(circumcircle(D,A,Z),linetype("4 4"));
draw(D--M--E--cycle,linetype("8 8"));
label("$A$",A,dir(origin--A));
label("$B$",B,dir(origin--B));
label("$C$",C,dir(origin--C));
label("$D$",D,dir(origin--D));
dot(M^^N^^E);
label("$M$",M,dir(origin--M));
label("$Z$",Z,dir(N--Z));
label("$E$",E,NE);
label("$N$",N,dir(Z--N));
draw(rightanglemark(Z,N,C,3)^^rightanglemark(D,E,C,3));
[/asy]](http://latex.artofproblemsolving.com/d/d/f/ddf843b3473946918faa725a01fa3337c7a9b515.png)
and 
.
therefore 
.
.
.
.
for some 
.
.
.
Which is equal to.
as desired.
and 
are spirally similar with an angle of 
.
through 
and 
,
,
is the reflection of 
with 
.
then since 
is a parallelogram, 
and 
so this follows from angle bisector theorem.
with line 
to 
to 
,
with 
and 
,
,
and 
are both isosceles, there must a spiral similarity at 
to 
at 
to 
.
is just the midpoint of 
is the center of 
,
.
and 
(so lines 
and 
"correspond" to each other), we find that 
with 
and hence lies on the perpendicular bisector of 
.
,
so 
.
,
.
The perpendicular bisector of 
.
and 
are symmetric about the perpendicular bisector of 
is isosceles, 
,
Pentagon 
is cyclic.
.
is a rectangle, so 
.![\[\angle AMY = \angle AMO = 90,\]](http://latex.artofproblemsolving.com/2/4/b/24bb220c921fa01b27e8da4babdeb1be1716604d.png)
.![[asy]
size(250);
pair A, B, C, O, M, X, D, E, F, Z, Y;
A = dir(120); B = dir(210); C = dir(330); O = (0, 0); M = .5A + .5B; X = dir(90); D = dir(270); E = dir(45); F = dir(165); Z = extension(A, X, O, E); Y = D + Z - A;
draw(A--B--C--cycle^^circumcircle(A, B, C)); draw(Y--F^^O--Z); draw(D--A--Z--Y--cycle^^E--F); draw(dir(195)--foot(O, Y, Z), red); draw(circumcircle(A, Y, Z), blue+dashed);
dot("$A$", A, NW); dot("$B$", B, SW); dot("$C$", C, SE); dot("$D$", D, S); dot("$E$", E, 2dir(0)); dot("$F$", F, W); dot("$M$", M, dir(210)); dot("$Y$", Y, SE); dot("$Z$", Z, NE); dot("$O$", O, S);
[/asy]](http://latex.artofproblemsolving.com/e/b/b/ebb70d1d6a5105e2829327e9c8a736e8a2afe679.png)
inversion at 
-
.
collinear.
,
.
,
,
.
.
.
So we have 
.
.
Plugging in we have,
Thus we have 
.
and 
by,
Plugging in the coordinates of 
we find that,
Then plugging in the coordinates of 
so we find 
.
It is easy to verify 
.
are collinear.
.
and 
as desired.
,
,
,
,
,
denote 
,
is the perpendicular bisector of 
,
so 
,
,
,
,
which suffices.
let 
and let 
Observe 
and 
Next,![\[\angle CZN=\angle ZAN=\angle CAD=\angle MAD=\angle MZD.\]](http://latex.artofproblemsolving.com/b/9/e/b9e9933f59e8017a1a2e71afe9d8bab3ae59fa6a.png)
Thus 
Then there exists a unique point 
that is on 
and 
by the fundamental lemma of spiral similarity. Now,![\[\angle DTM=\angle DAM=\angle DAB=\angle DCB\]](http://latex.artofproblemsolving.com/a/c/5/ac5a47c918ece76c483a33db74d5b9121500e7bc.png)
so 
and 
be the foot from 
is a trapezoid. Let 
be the reflection of 
and 
,
and 
.
-
where 
and 
collinear, as it is equivalent to 
cyclic. Let 
,
,
.
so 
and 
so we are done. 
meet 
at 
,
which implies points 
implying 
as desired.
.
should be cyclic, so we let 
and prove 
.
Thus 
is collinear and 
is cyclic since 
,
is cyclic. This is evident since![\[ \angle ZKM = \angle ZCN = \angle ZAN = 180 - \angle MAZ. \]](http://latex.artofproblemsolving.com/c/9/b/c9bdb258911b9bc142365ba71014fa8b39760e68.png)
be the midpoints of 
.
,
so it remains to prove that 
by 
,
and 
hence we should also have
as required.
Let 
Let 
denote the reflection of 
Let 
with 
are collinear
Furthermore, 
since 
However, 
implying that 
![\[\angle NN'D = 90 - \angle ADE = 90 - \angle ADN - \angle NDE = \angle DNE - \angle CAE = \angle DAE = \angle MAD\]](http://latex.artofproblemsolving.com/3/1/8/318e9e27ac6e07b678cee1fc6cc27d1e91eaa62e.png)
as desired.
which implies our problem.
