[/quote]
,
and 
intersect at 
and 
.
be the circumcircle of the triangle 
;
and 
,
and 
intersects 
.
and 
meet at 
,
.
.
?
be a positive integer. A Japanese triangle consists of 
circles arranged in an equilateral triangular shape such that for each 
,
,
,
row contains exactly 
circles, exactly one of which is coloured red. A ninja path in a Japanese triangle is a sequence of 
,
such that in each Japanese triangle there is a ninja path containing at least 
,
and 
.
is prime, 
with 
and 
be a function such that for every regular 
-
we have 
.
for all reals 
.
)
are the altitudes of the triangle. The bisector of 
intersects 
at 
.
and 
.
passes through the midpoint of 
.
.
with 
and at most one of them being zero, show that![\[\frac{1}{\max(a^2,b)}+\frac{1}{\max(b^2,c)}+\frac{1}{\max(c^2,a)}\geq\frac{27}{4}\]](http://latex.artofproblemsolving.com/9/e/1/9e14a9ef2d1a0a98006bcf716fa33c3d9ea29bd5.png)
be an integer. Find all sequences 
satisfying the following conditions:![\[ \text{ (a) } a_i \in \left\{0,1\right\} \text{ for all } 1 \leq i \leq n^2 + n;
\]](http://latex.artofproblemsolving.com/3/c/5/3c509ec2e9013e8d3be492c8eb44a7c33841b74e.png)
![\[ \text{ (b) } a_{i + 1} + a_{i + 2} + \ldots + a_{i + n} < a_{i + n + 1} + a_{i + n + 2} + \ldots + a_{i + 2n} \text{ for all } 0 \leq i \leq n^2 - n.
\]](http://latex.artofproblemsolving.com/9/7/d/97d2a467d1c0dc8594ec024c3bb9b8c87ee85b19.png)
Y by baladin, Temp456, claserken, doxuanlong15052000, baopbc, aopser123, anantmudgal09, Akatsuki1010, trumpeter, Aryan-23, MathPassionForever, itslumi, v4913, HamstPan38825, csong, sotpidot, Lamboreghini, Lcz, Adventure10, Mango247, nguyenducmanh2705, balllightning37
Let 
be a triangle with incenter 
, and whose incircle is tangent to 
, 
, 
at 
, 
, 
, respectively. Let 
be the foot of the altitude from to 
. Suppose that the circumcircle of 
meets the incircle at two distinct points 
and 
, while the circumcircle of 
meets the incircle at two distinct points 
and 
. Prove that the radical axis of the circumcircles of 
and 
passes through the midpoint 
of 
.
Proposed by Danielle Wang
be a triangle with incenter 
, and whose incircle is tangent to 
, 
, 
at 
, 
, 
, respectively. Let 
be the foot of the altitude from to 
. Suppose that the circumcircle of 
meets the incircle at two distinct points 
and 
, while the circumcircle of 
meets the incircle at two distinct points 
and 
. Prove that the radical axis of the circumcircles of 
and 
passes through the midpoint 
of 
.Proposed by Danielle Wang
This post has been edited 1 time. Last edited by v_Enhance, Jun 29, 2016, 2:45 PM
,
,
be midpoints of 
,
,
,
be the Gregonne point. By radical axis on 
,
,
we see that 
and 
coincide. Similarly, lines 
and 
coincide. In particular lines 
,
passes through the midpoint of 
.
,
,
such that 
,
,
.
bisects 
and is isogonal to 
,
,
,
,
,
not concurrent, all six points 
are cyclic. Moreover, since 
,
,
.
similarly now and 
.
is in cyclic too, as 
.
again at 
.
is cyclic, and now ![\[ \measuredangle BVS' = \measuredangle PVS' = \measuredangle PQS = \measuredangle PTS = \measuredangle FED = \measuredangle XEZ = \measuredangle XVZ \]](http://latex.artofproblemsolving.com/4/e/5/4e59c855480e801d22a7e65f4a1d404c8a020182.png)
hence 
.
,
is cyclic too, so 
now; thus 
and 
,
.
.
,
is cyclic. But 
is cyclic, so we are done.
is called the lemon circle of 
.)
passes through 
,
and 
and a constant 
,
is a circle 
,
,
and 
is cyclic.
be the second intersection of the circumcircle of 
.
.
and 
.
.
.
,
after the Ravi Substitution and some cancellations. This in turn becomes 
,
.
with barycentric coordinates.
be the medial triangle of 
;
.
,
lies on the radical axis of both circles. By a well-known lemma, it suffices to show that 
,
and 
and 
respectively; we claim that pentagon 
is cyclic. To do this, we use barycentric coordinates with respect to 
and 
is of the form 
,
which is clear by subtracting the equations of both circles. Since 
,
.
;
and it is easy to verify that both sides are equal to 
,
,
,
and 
similarly with respect to 
.
,
,
are concyclic on the Lemoine circle of 
is concyclic by a homothety at 
,
.
,
,
concur at 
at 
and 
,
at 
at 
,
.
and two points 
Let 
be the cevian triangle of 
WRT 
cuts 
at 
respectively. Then 
are concurrent.
cuts 
at 
respectively. Clearly, 
is the radical center of 
and the pedal circle 
of 
of 
so if 
cuts 
again at 
lies on 
and 
lies on 
so the radical axis 
of 
hence from Lemma 
intersects the line passes through 
.
is the symmectric of 
are collinear.
.
are concyclic. Similarly, 
are concyclic implies 
.
be the averages of 
.
so 
and 
.
so 
and 
is the radial axis of 
is the radial axis of 
and the radial axis of 
then 
the radial axis of 
passes through the projection of 
i.e 
be the orthocenter of 
implies 
and 
.
are concyclic so 
implies 
.
implies 
lies on the radial axis of 
is another cevian triangle. Set 
,
,
.
are collinear (see 
;
is harmonic, and 
,
.
is the radical axis of 
is the radical axis of 
is the cevian triangle of 
so notice 
WRT 
we conclude that (by Cevian nest theorem) 
,
.
has equation 
On the other hand, the incircle has equation 
.
For some constant 
,
.
;
to find 
and after plugging in 
,![$j = \frac{(e+1)^2}{(e-1)^2}\left[\frac{8ef}{(e+f)(e+1)(f+1)}-1\right]$](http://latex.artofproblemsolving.com/b/0/4/b04ab15ea5de1ed4bdbe82fa0a4887647e731503.png)
.
where ![$l = \frac{(f+1)^2}{(f-1)^2}\left[\frac{8ef}{(e+f)(e+1)(f+1)}-1\right]$](http://latex.artofproblemsolving.com/a/5/0/a50bed7cf2b067058dfc715d9203772e9324f405.png)
.
,![$$\frac{2ef\overline{z}}{e+f}\left[\frac{(f+1)}{(f-1)^2} - \frac{(e+1)}{(e-1)^2}\right] + \frac{2z}{e+f}\left[\frac{f(f+1)}{(f-1)^2} - \frac{e(e+1)}{(e-1)^2}\right] =\frac{(f+1)^2}{(f-1)^2}-\frac{(e+1)^2}{(e-1)^2} $$](http://latex.artofproblemsolving.com/6/3/1/631c826c4d40ddc7d3c0d0b652da3720e5206256.png)
Taking 
,![$$\frac{ef(1+\overline{k})}{e+f}\left[\frac{(f+1)}{(f-1)^2} - \frac{(e+1)}{(e-1)^2}\right] + \frac{1+k}{e+f}\left[\frac{f(f+1)}{(f-1)^2} - \frac{e(e+1)}{(e-1)^2}\right] = \frac{(f+1)^2}{(f-1)^2}-\frac{(e+1)^2}{(e-1)^2} $$](http://latex.artofproblemsolving.com/c/2/4/c2465ca290f57a6f451d094cfeff4ba331066c7b.png)
Recalling that 
since 
,![$$\frac{ef}{e+f}\left[\frac{(f+1)}{(f-1)^2} - \frac{(e+1)}{(e-1)^2}\right] + \frac{1}{e+f}\left[\frac{f(f+1)}{(f-1)^2} - \frac{e(e+1)}{(e-1)^2}\right]$$](http://latex.artofproblemsolving.com/8/4/6/846177c05bdedff9647b3d4bfa02edc7710c3408.png)
![$$ + \frac{e+f-k}{e+f}\left[\frac{(f+1)}{(f-1)^2} - \frac{(e+1)}{(e-1)^2}\right] + \frac{k}{e+f}\left[\frac{f(f+1)}{(f-1)^2} - \frac{e(e+1)}{(e-1)^2}\right] = \frac{(f+1)^2}{(f-1)^2}-\frac{(e+1)^2}{(e-1)^2} $$](http://latex.artofproblemsolving.com/3/a/2/3a2d47f80c804f3e3b42feeb10e3eda3ed2d409e.png)
![$$\frac{(e+1)(f+1)}{e+f}\left[\frac{f}{(f-1)^2}-\frac{e}{(e-1)^2}\right] + \frac{f+1}{(f-1)^2} -\frac{e+1}{(e-1)^2}$$](http://latex.artofproblemsolving.com/6/9/d/69d7b6f1fc4e1e45aeb7579f425cf477ac9751f9.png)
![$$+ \frac{1/2(1+e+f-ef)}{e+f}\left[\frac{f+1}{f-1} - \frac{e+1}{e-1}\right] = \frac{(f+1)^2}{(f-1)^2}-\frac{(e+1)^2}{(e-1)^2}$$](http://latex.artofproblemsolving.com/9/c/8/9c8b4bb5c89a2bd581276cb348fa6d40b6ffb121.png)
![$$(e+1)(f+1)[f(e-1)^2-e(f-1)^2] + 1/2(1+e+f-ef)(e-1)(f-1)[(f+1)(e-1)-(f-1)(e+1)] $$](http://latex.artofproblemsolving.com/d/5/5/d55d9af382a86d019a03c2177eb6396a1964a15b.png)
![$$= (e+f)[f(f+1)(e-1)^2-e(e+1)(f-1)^2]$$](http://latex.artofproblemsolving.com/a/5/c/a5c376f434b9eb7eda073981dd41402e37722047.png)
which is true after a bit of expanding! Thus 
in a similar fashion. Let 
be the midpoints of 
,
the incircle 
.
are collinear, etc. So we have that 
.
,
is the radical axis of 
.
,
is then the radical center of 
.
are collinear. Let 
be the A-excenter, and 
be the intersections of 
,
respectively. We easily see that 
,
lies on the circle 
.
yields that 
are collinear. Similarly 
are collinear. Since the sides of 
are parallel, 
are the corresponding feet of altitudes in the two triangles, we are done.
and 
.
denote the midpoints of 
respectively. Let 
,
.
is the perpendicular bisector of 
.
.
is the centre of 
,
,
.
,
concur. Thus, 
.
by Claim 1, yet 
since 
and 
is the 
and 
.
via linearity.
is the radical centre of 
.
and analogously,![\[P(N_c) = - N_cC \cdot IT_c.\]](http://latex.artofproblemsolving.com/e/9/7/e979a40e1bc09dcdf5a4f3420b65f66f701c3bb0.png)
.
Note that 
,
.
and 
.![\[DF \cdot IT_b = DE \cdot IT_c.\]](http://latex.artofproblemsolving.com/7/d/6/7d6ea33e652caa1634033a8414a7cb9e74835e29.png)
By symmetry, it now suffices to show:
However, note that 
,
cyclic and 
anti-parallel to 
.
,
cyclic and 
,
.
with 
,
lies on the ![$DE \cdot R_cN_c = 2 \cdot [DER_c] = [DEF] = 2 \cdot [DFR_b] = DF \cdot R_bN_b$](http://latex.artofproblemsolving.com/a/f/6/af65da69505f8539a441dc7f4288d455573fc5e2.png)
.
be the radical axis of 
of 
is cyclic, and thus we consider the radical center of the three circles 
,
.
,
,
.
of 
,
.
,
,
,
to realize that 
,
,
,
be 
,
,
(wrt the intouch triangle). Define similarly for 
.
by repeating the argument symmetrically for 
for some constant 
is on the circle. From here we can directly compute the power of 
and take the W by symmetry.
and 
respectively. Define 
and 
similarly but for 
and 
.
is the orthocenter of 
,![\[ \providecommand{\dang}{\measuredangle} \dang PX_1X_2 = \dang PBX_2 \dang EDC \qquad \dang QY_1Y_1 = \dang QCY_2 = \dang FDB, \]](http://latex.artofproblemsolving.com/2/7/e/27e3c4d4f3bff0926089c72d9ce92b002a9acafe.png)
and moreover 
and 
by symmetry. From this we observe that the quadrilateral formed by the lines 
,
,
,
must be a triangle and hence 
is a line. It's also not hard to angle chase that![\[ \providecommand{\dang}{\measuredangle} \dang Y_1X_1X_2 = \dang PX_1X_2 = \dang PBX_2 = \dang Y_1Y_2X_2 \]](http://latex.artofproblemsolving.com/9/3/1/9311fdae0f69ebd0f45ca2b1ac2637ab49398701.png)
which shows that 
lies on the radical axis.
We have that 
is the Gergonne point of 
Let 
where 
.
be the reflection of ![\[ -1 = (B, X_1; F, F') \overset{DF \cap X_1X_2}{=} (B, X_1X_2 \cap BE, DF \cap BE, E) \overset{D}{=} (D, ?, F, E), \]](http://latex.artofproblemsolving.com/3/9/d/39d4c30582ed70475c1024962ce433f394f29be3.png)
where the 
is the harmonic conjugate of 
,
,
and 
,
,
,
.
and 
are coaxial.
and 
,![\[ \text{Pow}(X,t_B) = \frac{ (y+x)(y+z) \text{Pow}(X,\omega) - y^2 \text{Pow}(X, \omega_B) }{ (y+x)(y+z) - y^2 } = \frac{ (y+x)(y+z) \text{Pow}(X,\omega) - y^2 \text{Pow}(X, \omega_B) }{ \sum xy } \]](http://latex.artofproblemsolving.com/d/5/a/d5af0bbd9b50c0b5c9a89599d1dfbd62385466c0.png)
since it has leading coefficient 
when viewed as a quadratic function, and also vanishes when 
.![\[ f(X) = \text{Pow}(X,t_B) - \text{Pow}(X,t_C) = \frac{ (y^2 - z^2) \text{Pow}(X,\omega) - y^2 \text{Pow}(X, \omega_B) + z^2 \text{Pow}(X,\omega_C) }{ \sum xy } \]](http://latex.artofproblemsolving.com/7/6/c/76cee22b83974befb6b907a70dc6aa05dbe98f48.png)
Now we just need to compute 
.
.![\[ f(E) = \frac{ 0 - y^2 (-xz) + z^2 (xy) }{ \sum xy } = \frac{ xyz (y+z) }{ \sum xy }, f(F) = \frac{ xyz (-y-z) }{ \sum xy }\]](http://latex.artofproblemsolving.com/d/b/3/db302043354c2a971878323a56bb4a7ccb63b258.png)
so ![\[ f(K) = \frac{xyz}{\sum xy} \left( \frac{y(y+z)+z(-y-z)}{y+z} \right) = \frac{xyz}{\sum xy} (y - z) \]](http://latex.artofproblemsolving.com/8/a/2/8a27690c6636dc5bb2a6b222843d190d0cce0750.png)
and on the other hand,![\[ f(D) = \frac{ 0 - y^2 (xz) + z^2 (xy) }{\sum xy} = \frac{xyz}{\sum xy} (-y + z) \]](http://latex.artofproblemsolving.com/2/8/a/28ada5b5284fbcd2c9f444d6ff58be2b724a0d48.png)
so 
and we are done.
.
of 
also contains the midpoint of 
,
and 
concur at G, and that 
.
and 
so 
so 
is the radical axis of 
is the radical axis of 
concur at 
,
,
concur so by Cevian Nest 
and 
concur on the desired radical axis. Thus, both 
.
,
and 
and hence 
as the Gergonne point of 
.
Lets say now that 
Then their pairwise radical axes has equation,
But also 
for some constant 
Then we can actually compute this constant! Plugging in 
Thus we have the equation of 
Now to compute 
we will plug in 
where 
.
which is symmetric in 
is the radical axis of 
,
and, 
the midpoint of the 
has equation,
Then plugging in 
and thus we are done. 
which by definition is the Symmedian point of 
Let 
respectively. Thus, we need only show that 
is on the radical axis as 
is the 
and 
Then let 
is collinear.
we get that 
is on 
are concurrent, so 
and 
are perspective by Descargues.
and 
are perspective from the same point that 
is collinear, so 
is collinear.
As 
Note that by Incenter Excenter lemma, 
must have the center of 
and as 
we have that 
and similarly for 
Therefore as 
then 
so 
Same thing for 
Thus, as 
and 
it must be the radical center of 
so it must also be on the radical axis of 
which finishes
be the midpoints of 
and let 
be the midpoints of 
and 
are collinear by inversion about the incircle, so 
.
so 
and 
collinear. We get the same thing for other midpoints. 
so 
.
.
so 
cyclic. Analogously 
cyclic, hence 
cyclic. 
so 
bisects 
.
we get that there is an involution which swaps 
.
we get that 
are involutive pairs which is equivalent to 
are coaxial. 
is cyclic because it is 
,
and 
we get that 
.
so 
cyclic. Hence 
is cyclic because of coaxial circles. This means 
so 
.
so 
as desired.
.
similarly because I like to define the shape from 
m
.
,
lies on the radical axis of 
gives 
.
passes through radical axis 
and 
so 
cyclic. Similarly 
and project it to 
and 
we get 
passes through midpoint of 
.
so 
so 
and since 
cyclic 
with 
and 
.
so 
are homothetic and 
.
so 
.
all pass through 
of 
over 
and 
and we have ![\[\frac{\frac{PF^2}{PA\cdot PC'}}{\frac{BF^2}{BA\cdot BC'}}=\frac{\frac{RD^2}{RA'\cdot RC}}{\frac{BD^2}{BA'\cdot BC}}=\frac{\frac{QE\cdot UE}{AQ\cdot CU}}{\frac{BD\cdot BF}{BA\cdot BC}}=\frac{\frac{GD}{AG}\cdot\frac{GF}{CG}}{\frac{[FBD]}{[ABC]}}=\frac{\frac{[DGF]}{[AGC]}}{\frac{[FBD]}{[ABC]}}=\frac{[DGF]}{[FBD]}\cdot\frac{[ABC]}{[AGC]}=\frac{LG}{BL}\cdot\frac{BE}{GE}=-(G,B;L,E)\stackrel D=-(X,D;F,E)=1,\]](http://latex.artofproblemsolving.com/0/7/6/076aa9632c9802f9a9c12e448659aec49cbac87e.png)
so forgotten coaxiality lemma gives 
is coaxial to the incircle and 
is the radical axis of 
,
lies on the radical axis of 
are collinear, so 
and 
.
be the midpoints of 
,
.
is on the radical axis of ![\[\text{Pow}_\omega(T_B) - \text{Pow}_{\omega_B}(T_B) = 0.\]](http://latex.artofproblemsolving.com/b/d/8/bd8e5b5d35b5ae631f7838dd7fe11d3fa2248395.png)
,
,
,
.
be the radius of ![\begin{align*}
\frac{\text{Pow}_\omega(M) - \text{Pow}_{\omega_B}(M)}{\text{Pow}_\omega(B) - \text{Pow}_{\omega_B}(B)} &= \frac{[MB_1B_2]}{[BB_1B_2]}\\
&= \frac{MC'\sin C'}{BT}\\
&= \frac{r\sin^2 C'\cos B'}{r\frac{\sin^2 B'}{\cos B'} + r\sin A'\sin C'}\\
&= \frac{\sin^2 C'\cos^2 B'}{\sin^2 B' + \sin A'\sin C'\cos B'}\\
&= \frac{\sin^2 C'\cos^2 B'}{\sin^2 B' + (\sin A'\sin C'-\cos A'\cos C')\cos B'+\cos A'\cos B'\cos C'}\\
&= \frac{\sin^2 C'\cos^2 B'}{\sin^2 B' - \cos(A'+C')\cos B'+\cos A'\cos B'\cos C'}\\
&= \frac{\sin^2 C'\cos^2 B'}{1+\cos A'\cos B'\cos C'}\\
\frac{\text{Pow}_\omega(M) - \text{Pow}_{\omega_B}(M)}{\text{Pow}_\omega(B) - \text{Pow}_{\omega_B}(B)} &= \frac{\sin^2 C'\cos^2 B'}{1+\cos A'\cos B'\cos C'}\\
\text{Pow}_\omega(M) - \text{Pow}_{\omega_B}(M) &= r^2 \frac{\sin^2 B'}{\cos^2 B'}\frac{\sin^2 C'\cos^2 B'}{1+\cos A'\cos B'\cos C'}\\
\text{Pow}_\omega(M) - \text{Pow}_{\omega_B}(M) &= r^2 \frac{\sin^2 C'\sin^2 B'}{1+\cos A'\cos B'\cos C'}.
\end{align*}](http://latex.artofproblemsolving.com/8/b/c/8bcbafedd2b82d59596bfb666b3f2787e4509675.png)
So,![\[\text{Pow}_{\omega_B}(M) = \text{Pow}_\omega(M) - r^2 \frac{\sin^2 C'\sin^2 B'}{1+\cos A'\cos B'\cos C'} = \text{Pow}_{\omega_C}(M).\]](http://latex.artofproblemsolving.com/0/7/1/071b5983e78fcc1fa59346e7e52e054a41dfa6e5.png)
