be a quadrilateral inscribed in a circle 
.
and 
meet at 
and 
meet at 
and 
and the line through the incenters of 
and 
meet at a point lying on 
be an odd integer. Prove that there exists a prime number 
such that![\[
p \mid 2^{\varphi(n)} - 1 \quad \text{but} \quad p \nmid n.
\]](http://latex.artofproblemsolving.com/3/5/1/3514b64ecd5f7b2586d5bfeae59b7c9504d5b455.png)
positive integers not exceeding 
,
?
Y by Adventure10, mathematicsy, son7, Titusir, Mango247, kiyoras_2001, and 1 other user
Let 
be a triangle, 
its orthocenter, 
its circumcenter, and 
its circumradius. Let 
be the reflection of the point 
across the line 
, let 
be the reflection of the point 
across the line 
, and let 
be the reflection of the point 
across the line 
. Prove that the points , and are collinear if and only if 
.
be a triangle, 
its orthocenter, 
its circumcenter, and 
its circumradius. Let 
be the reflection of the point 
across the line 
, let 
be the reflection of the point 
across the line 
, and let 
be the reflection of the point 
across the line 
. Prove that the points , and are collinear if and only if 
.Attachments:
be the projections of the nine-point center on 
respectively. If 
are the midpoints of 
are the feet of the perpendiculars from 
,
.
.
are collinear iff 
.
is the centre of gravity of 
. 
is a triangle satisfying : 
are the midpoints of 
.
and 
with 
are the feet of the perpendiculars from O then 
. So 
are colinear 
be the midpoint of 
.
.
.
are collinear iff 
is the ninepoint centre. Then 
lies on 
.
is the reflection of 
is parallelogram, hence 
to 
.
which is equivalent to the condition 
,
.
in our calculations.I also don't think the original problem is solved,it is only solved in this particular case(when radius is 1).For solving the original problem one can approach by a similar manner but one needs to struggle a lot.Can one post a similar proof using the general 
)
with center 
and ratio 
.
be the image of 
under such transformation
on 
.
is the feet of perpendicular from 
and 
)
collinear 
is on the circumcircle of 
(easy to see that 
and now we have used the converse of the Pedal Line theorem).
are collinear if and only if reflections of 
,
,
,
;
to get
be the unit circle. Notice that 
and 
Hence, 
Similarly, 
We know 
and 
onto 
be 
respectively and let the reflections of 
over the opposite sides be 
respectively. Then we have that 
and 
,
,
.
where 
are the radii of the circumcircle, nine point circle of 
is the foot from 
are the midpoints of 
respectively. Observe that 
(other configurations are also very easily handled like this). Similar relations hold for 
.
respectively and 
.
be the altitudes of 
and 
respectively, the Nine-Point Center of 
and 
be 
and 
respectively. Because 
.
-
and 
.
is cyclic if and only if 
.
and 
are parallelograms, so 
and 
.
and 
are cyclic with diameters 
and 
and 
This means 
holds if and only if 
,
is cyclic, which finishes. 
.
Then, by collinearity criterion, we have 
which immediately implies the desired result. 
Similarly, 
and 
Then the collinear condition is equivalent to 
This is the same as 
The RHS is the same as 
so the collinear condition simplifies to 
Expanding, 
which rearranges to 
The second condition is saying that 
.
These two conditions are therefore equivalent, QED.
,
,
.
For the first quotient, we have
For the second quotient, we have
So, we have 
if and only if
which expands to give
On the other hand, the second condition of the problem is equivalent to 
or 
,
This matches our earlier condition, so we are done. 
are collinear. Note that if any two of 
are equal to each other, the collinearity is vacuously true. This will be important later. This is equivalent to 
Clearing the fractions, this is 
This then factors: 
Cancelling and then expanding, 
After expanding this and cancelling as many terms as possible, 
Note the equality case noted earlier. We haven't used an 
factor yet, so we suspect that this is divisible by 
.
Remove the 
Dividing by 
gives 
as the condition for collinearity. Now, since we assumed the circumradius is 1, 
is just 
so we are finally done.
about the complex points 
and 
is ![\[\frac{(a - b) \overline z + \overline a b - a \overline b}{\overline a - \overline b}.\]](http://latex.artofproblemsolving.com/f/6/7/f679fc29a2d534c82a031603eff4f33d47aab7ba.png)
When ![\[\frac{(a - b) \frac1z + \frac ba - \frac ab}{\frac 1a - \frac1b} = b + a - \frac{ba}z.\]](http://latex.artofproblemsolving.com/9/5/0/9505addaa74e47ce7df4a09fa5005bc314cce53c.png)
![\[\begin{vmatrix}
a & \overline{a} & 1 \\
b & \overline{b} & 1 \\
c & \overline{c} & 1 \\
\end{vmatrix} = 0.\]](http://latex.artofproblemsolving.com/9/8/3/983e48403ff08297e5c6d89d28f1a2c288eba2ac.png)
So, we just need to verify that![\[\begin{vmatrix}
b + c - \frac{bc}{a} & \frac1 b + \frac1 c - \frac{a}{bc} & 1 \\
c + a - \frac{ca}{b} & \frac1 c + \frac1 a - \frac{b}{ca} & 1 \\
a + b - \frac{ab}{c} & \frac1 a + \frac1 b - \frac{c}{ab} & 1 \\
\end{vmatrix} = 0\]](http://latex.artofproblemsolving.com/b/5/d/b5da6625564ea54e8ebe7951d53585ee44dc6b9b.png)
is equivalent to ![\[|a + b + c| = 2.\]](http://latex.artofproblemsolving.com/2/c/5/2c584ae8d73379ff25ebf7f2e54782bc2b403721.png)
The determinant simplifies to![\[\begin{vmatrix}
a + \frac{bc}{a} & \frac1 a + \frac{a}{bc} & 1 \\
b + \frac{ca}{b} & \frac1 b + \frac{b}{ca} & 1 \\
c + \frac{ab}{c} & \frac1 c + \frac{c}{ab} & 1 \\
\end{vmatrix} = 0\]](http://latex.artofproblemsolving.com/d/b/a/dba4c5687adb31a8f0a83461087f37556211c1de.png)
So, ![\[\sum_{cyc} \left(\left(b + \frac{ca}{b}\right)\left(\frac1 c + \frac{c}{ab}\right) - \left(c + \frac{ab}{c}\right)\left(\frac1 b + \frac{b}{ca}\right)\right) = 0.\]](http://latex.artofproblemsolving.com/9/9/e/99e156b75f325ebcbfce4d97a9bc0f0bc3a871d6.png)
This becomes ![\[3\sum_{cyc}\left(\frac bc - \frac cb\right) + \sum_{cyc}\left(\frac{c^2}{b^2} - \frac{b^2}{c^2}\right).\]](http://latex.artofproblemsolving.com/c/8/0/c803232ec27df22c895cd28662547c7009c56fe5.png)
This can factor to ![\[\frac{(a - b)(b - c)(c - a)}{abc}\left(3 - \frac{(a + b)(b + c)(c + a)}{abc}\right) = 0.\]](http://latex.artofproblemsolving.com/3/0/a/30aff2577772dbe4df4f8a78932937abb9e1576a.png)
Since none of the points are equal to each other, this tells us that ![\[\sum_{sym} \frac ab = 1.\]](http://latex.artofproblemsolving.com/0/c/a/0ca3261ebaed45b039bb9c679d0f5f7e96ee3990.png)
But the condition for 
is equivalent to ![\[(a + b + c)(\overline{a + b + c}) = 3 + \sum_{sym} \frac{a}{b} = 4\]](http://latex.artofproblemsolving.com/1/1/5/115b9e390876f761e37e9178d51c639f9b4ebb9c.png)
, so we're done.
and 
be the foot of the altitude and perpendicular from 
of 
is collinear with 
and 
on the Simson Line of 
and 
are homothetic at 
.
are collinear with 
.
and 
.
,
,
be the feet of the altitudes from 
.
is a homothety of 
centered at 
.
is 
,
is 
,
is 
.
We claim that 
.
.
,
and 
,
,
as desired.
.
Similarly, we obtain that
Now, if 
,
.
and we must have
Thus,
and thus 
et cetera, and 
are collinear iff 
,
,
are collinear. Now
,
as the unit circle. 
have the complex numbers 
respectively and also 
.
are collinear.
.
.
Upon conjugation,
Now, 
thus for 
The above 
magically factorizes as,
Obviously, 
since 
as desired. 
)
and cyclic variants, appropriate translation and reflection gives the simplified 
and cyclic variants. The collinearity criteria is then the following being self conjugating: 
,
.
.
.
.
,
,
gives 
.
,
.
,
,
For 
we must have 
which is equivalent to proving
and thus 
and we are done! 
![[asy]
import graph; size(10cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -7.694081287150866, xmax = 0.4768291617537498, ymin = -2.0950501157930743, ymax = 5.236581455833256; /* image dimensions */
pen zzttqq = rgb(0.6,0.2,0);
draw((-3,4)--(-3.844974712047039,0.7797212109761972)--(-3.33,0.2)--cycle, linewidth(0.6) + zzttqq);
/* draw figures */
draw((-3,4)--(-3.844974712047039,0.7797212109761972), linewidth(0.6));
draw((-3.844974712047039,0.7797212109761972)--(-3.33,0.2), linewidth(0.6));
draw((-3.33,0.2)--(-3,4), linewidth(0.6));
draw(circle((-1.898703000807973,1.9900321027017454), 2.2919044545091602), linewidth(0.6));
draw((-7.142406349521041,0.32024070081159844)--(-2.7227998066557872,0.682269179718536), linewidth(0.6));
draw((-6.377568710431095,0.9996570055727055)--(-1.8987030008079708,1.9900321027017447), linewidth(0.6));
/* dots and labels */
dot((-3,4),linewidth(3pt) + dotstyle);
label("$A$", (-3.160045973118789,4.088602301854922), NE * labelscalefactor);
dot((-3.844974712047039,0.7797212109761972),linewidth(3pt) + dotstyle);
label("$B$", (-4.134381221453342,0.7218399090949367), NE * labelscalefactor);
dot((-3.33,0.2),linewidth(3pt) + dotstyle);
label("$C$", (-3.4205118315844616,-0.020970131714573053), NE * labelscalefactor);
dot((-6.377568710431095,0.9996570055727055),linewidth(3pt) + dotstyle);
label("$H$", (-6.555749016819409,1.0884214876762532), NE * labelscalefactor);
dot((-1.8987030008079708,1.9900321027017447),linewidth(3pt) + dotstyle);
label("$O$", (-1.8577166807904262,2.0434629687170514), NE * labelscalefactor);
dot((-7.142406349521041,0.32024070081159844),linewidth(3pt) + dotstyle);
label("$D$", (-7.105621384691384,0.3745520978073737), NE * labelscalefactor);
dot((-2.7227998066557872,0.682269179718536),linewidth(3pt) + dotstyle);
label("$E$", (-2.59087983795306,0.6253710726261692), NE * labelscalefactor);
dot((-4.578237152753622,0.5275271794198101),linewidth(3pt) + dotstyle);
label("$F$", (-4.64566605473781,0.6060773053324157), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/3/4/5/34593f2a939bc67b82dcd6e1ad436d52a5429653.png)
