be an acute triangle with 
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 
and 
,
be the midpoint of arc 
at 
.
intersects 
,
and 
meet on 
.
and 
respectively such that 
.
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 
A circle passes through 
and crosses the segments 
respectively. The circumcircles of 
and 
The segments 
and 
intersect at 
Let 
and 
be the reflections of 
and 
respectively. Prove that the points 
and 
are concyclic.
,
prove that the triangle must have a right angle.
be 
be arbitrary points on sides 
Prove that 
.
![\[\frac{\sin^2A+\sin^2B+\sin^2C}{\cos^2A+\cos^2B+\cos^2C} = 2\]](http://latex.artofproblemsolving.com/9/2/a/92a524bbdca92e9dd4091275714bf346bad96cd7.png)
is a pentagon whose vertices lie on circle 
.
and 
meet 
at 
holds, then what is the value of 
?
Y by anantmudgal09, rmtf1111, dooo203, Adventure10, Mango247
et 
be a convex hexagon which has an inscribed circle and a circumcribed. Denote by 
and 
the inscribed circles of the triangles 
and 
, respecitively. Let 
, be the external of 
and 
; lines 
, 
, 
, 
, 
are analoguosly defined. Let 
be the intersection point of the lines and , 
are analogously defined.
Prove that
are concurrent.
be a convex hexagon which has an inscribed circle and a circumcribed. Denote by 
and 
the inscribed circles of the triangles 
and 
, respecitively. Let 
, be the external of 
and 
; lines 
, 
, 
, 
, 
are analoguosly defined. Let 
be the intersection point of the lines and , 
are analogously defined.Prove that
are concurrent.This post has been edited 1 time. Last edited by aleksam, Sep 6, 2017, 4:26 PM
;
,
,
,
,
similarly. Let 
be the external common tangent of 
,
,
,
,
similarly. Let 
,
,
,
similarly.
,
,
are concurrent.![[asy]
unitsize(150);
pen n_purple = rgb(0.7,0.4,1),
n_blue = rgb(0,0.6,1),
n_green = rgb(0,0.4,0),
n_orange = rgb(1,0.4,0.1),
n_red = rgb(1,0.2,0.4);
pair f(pair P, pair O, real r) {real d = distance(O, P), c = aCos(r / d); pair Q = O + dir(P - O) * r; return rotate(c, O) * Q;}
pair O = (-0.1, 0.1), A[] = {}, B[] = {}, I[] = {}; string s = "ABCDEF";
A.push(dir(0));
real x = distance(O, origin), r, a = 16 * x^2, b = 4 * (1 + x^2) * (1 - x^2)^2, c = -3 * (1 - x^2)^4;
if (x != 0) r = sqrt((-b + sqrt(b^2 - 4 * a * c)) / (2 * a)); else r = sqrt(3) / 2;
for (int i = 0; i < 5; ++i) {pair X = f(A[i], O, r); A.push(2 * foot(origin, A[i], X) - A[i]);}
for (int i = 0; i < 12; ++i) {A.push(A[i % 6]);}
for (int i = 0; i < 18; ++i) {I.push(incenter(A[i % 6 + 5], A[i % 6 + 6], A[i % 6 + 7]));}
for (int i = 0; i < 6; ++i) {pair X1, X2, Y1, Y2; X1 = reflect(I[i + 6], I[i + 5]) * A[i + 6]; X2 = reflect(I[i + 6], I[i + 5]) * A[i + 5]; Y1 = reflect(I[i + 6], I[i + 7]) * A[i + 6]; Y2 = reflect(I[i + 6], I[i + 7]) * A[i + 7]; B.push(extension(X1, X2, Y1, Y2));}
for (int i = 0; i < 12; ++i) {B.push(B[i % 6]);}
for (int i = 0; i < 6; ++i) {pair X1, X2; X1 = reflect(I[i], I[i + 1]) * foot(I[i], A[i], A[i + 1]); X2 = reflect(I[i], I[i + 1]) * foot(I[i + 1], A[i], A[i + 1]); draw(X1--X2, gray(0.6));}
draw(unitcircle);
draw(circle(O, r), gray(0.6));
for (int i = 0; i < 6; ++i) {draw(incircle(A[i], A[i + 1], A[i + 2]), gray(0.6)); draw(I[i]--I[i + 1], gray(0.6));}
draw(circumcircle(I[0], I[1], I[2]), gray(0.6) + dashed);
for (int i = 0; i < 6; ++i) {draw(O--A[i], n_purple);}
for (int i = 0; i < 3; ++i) {draw(I[i]--I[i + 3], n_blue);}
for (int i = 0; i < 6; ++i) {draw(A[i]--A[i + 1]^^A[i]--A[i + 2]);}
for (int i = 0; i < 6; ++i) {dot(A[i]^^I[i]^^B[i]);}
dot("$I$", O, origin);
for (int i = 0; i < 6; ++i) {label("$"+substr(s, i, 1)+"$", A[i], A[i]); label("$"+substr(s, i, 1)+"_1$", B[i + 6], dir(dir(B[i + 5] - B[i + 6]) + dir(B[i + 7] - B[i + 6]))); label("$I_"+substr(s, i, 1)+"$", I[i]);}
[/asy]](http://latex.artofproblemsolving.com/d/5/b/d5bc67f344b49e52be05bc0f7f12d9dcd22e455d.png)
;
,
,
,
,
similarly.
cyclic 
,
cyclic and 
.
with 
,
is cyclic.
consisting of reflections in lines 
,
,
,
in that order. It consists of four reflections in lines through 
,
![\[\ell_{AB} \stackrel{\overline{I_AI_B}}{\longrightarrow} \overline{AB} \stackrel{\overline{I_AA}}{\longrightarrow} \overline{AF} \stackrel{\overline{I_AI_F}}{\longrightarrow} \ell_{FA} \stackrel{\overline{I_AI_D}}{\longrightarrow} \ell_{AB}\]](http://latex.artofproblemsolving.com/9/6/6/9660c8cad7fdef1e0623c3eb3609c0c4fb49bad0.png)
where 
because 
:
,
,
are concurrent iff 
.
:![\[\frac{\sin \angle CAD}{\sin \angle DAE} = \frac{\sin \angle CED}{\sin \angle DCE} = \frac{CD}{DE}\]](http://latex.artofproblemsolving.com/5/4/3/543e91e330b05637e4e94b9ac968a5fcf396eaa1.png)
by LoS. Multiplying symmetric equalities we deduce 
by inversion distance formula. Thus 
,
concur; i.e. 
,
,
concur.
Let given hexagon 
,
,
and given that 
.
are concurrent.
of segments 
,
and 
.
,
and 
.
and 
.
and similarly 
and 
.
.
which sends 
.
are coaxial and so 
are concurrent.
be the midpoint of the biggest arc 
of 
and 
be the midpoint of the biggest arc 
of 
to external angle bisector between lines 
.
and 
.
and 
.
.
.
be the respective centres of these six incircles. Note that 
since 
are concyclic, so 
is cyclic. Likewise, we obtain 
which together with 
at 
and along 
.
to see 
so 
.
we have 
as desired. 
.
.
as desired. 
and the incircle 
of 
then draw the circumcevian triangle 
of 
WRT 
where 
by a well-known fact --- For a cyclic quadrilateral ABCD, the line connecting the incenter of 
and 
is parallel to the bisector of (AD,BC) , so I guessed 
have same midpoint which led me to use the condition of the existence of incircle.
and 
and 
: Points 
,
and 
,
lie on a circle with center 
are collinear since lines 
and 
both bisect 
.
Now, we see that 
,
Now inversion about a circle 
inverts cyclic hexagon 
.
and so on, which ultimately gives you that 
which implies that the diagonals 
are concurrent at a point 
due to Trigonometric Ceva (and perhaps Phantom Points, depending on your argument)
: Points 
lie on line 
lie on line 
lie on line 
concur at the point 
and of 
which will prove our claim.
: 
.
is the tangent from 
which means that 
and the others follow analogously.
and 
are the midpoints of arc 
not containing 
and so it implies that 
is parallel to angle bisectors of 
,
and 
from an external point 
with center 
)
are collinear, similarly this means that points 
and points 
are collinear, implying that lines 
and 
