.
and 
be the altitudes of triangle.Let 
and 
be reflections of points 
across the line 
,
and 
be the midpoints of 
and 
,
and 
.
and 
are collinear.
be an infinite set of translated copies (i.e., obtained by parallel translation) of a given ellipse 
in the plane, and let 
be a fixed straight line. It is known that every straight line parallel to 
fibonacci numbers 
)
and 
that 
'strong' and 
'strong'
![$f:[1,8]\to R$](http://latex.artofproblemsolving.com/c/e/c/cec97ea0372e37977e8a77e8e77518a2344c1c43.png)
that:
and the distribution function is 
then
and find the expression of 
be a continuous function . Suppose![\[f(x)=\frac{1}{t} \int^t_0 (f(x+y)-f(y))\,dy\]](http://latex.artofproblemsolving.com/c/0/c/c0c48c4259d4e98562c21a23cf4a6841704b8e82.png)
and all 
. Then show that there exists a constant 
such that 
, 
,
,
is differentiable and 
for all 
.
,
.
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 
.
:
the circumcenter of 
,
and 
,
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 
are the incenter and 
,
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 
with respect to the circumcircle of ![\[p\left(P,\Gamma\right)+\left(ux+vy+wz\right)\left(x+y+z\right)=0.\]](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 
since 
is on the perpendicular bisector of 
for complex numbers 
.
,
is the midpoint of arc 
not containing 
,
.
,
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)
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 
.
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 
.
be the midpoints of arcs 
,
.
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.
be the intersection of 
.
and 
as desired.
denote the midarc of 
,
,
,
,
,
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 
is a trapezoid. Let 
be the reflection of 
and 
,
and 
.
-
where 
and 
goes to feet of perpendicular from 
collinear, as it is equivalent to 
cyclic. Let 
,
,
.
so 
and 
so we are done. 
and 
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.
