[/quote]
,
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)
and 
with equal radii intersect at P and Q. Points B and C are located on the circle
The intersection point of the circumcircles of triangles XPC and YPB is called S. Prove that BC, XY and QS are concurrent.
and 
,
be the midpoint of 
.
on the extension of 
past 
on segment 
such that 
.
be on the opposite side of 
from 
such that 
and 
.
intersect the circumcircle of 
again at 
,
intersect the circumcircle of 
again at 
.
,
are collinear.
which satisfy ![\[f(x)\cdot f(y)\mid (1+2x)\cdot f(y)+(1+2y)\cdot f(x)\]](http://latex.artofproblemsolving.com/6/4/f/64f2720634f57b4d88db823b9931379386ac1c27.png)
for all non-negative integers 
and 
.
be natural numbers such that: 
and 
.
?
are taken on the legs 
and 
and 
.
and 
.
be the real polonomial such that all roots are real and distinct. Prove that there is a rational number 
that all roots of 
are real numbers
from the reals to the reals such that ![\[ \left(f(x)+f(z)\right)\left(f(y)+f(t)\right)=f(xy-zt)+f(xt+yz) \]](http://latex.artofproblemsolving.com/6/1/0/610e13a6f27341b5eed393923bb98e182537e254.png)
for all real 
, such that 
Prove that :
be nonnegative real numbers such that 
.
Where 
be the Fibonacci sequence: 
if 
.
so that there is a positive integer 
such that
and that 
such that 
.
Y by VulcanForge, itised, ETS1331, tree_3, centslordm, Smileyklaws, megarnie, tigerzhang, HrishiP, ike.chen, OliverA, ihatemath123, rayfish, samrocksnature, e_plus_pi, asimov, son7, ImSh95, crazyeyemoody907, rama1728, itslumi, pearlhaas, deplasmanyollari, pomodor_ap, Funcshun840
Let 
be a triangle with 
, incenter 
, and 
excenter 
. The incircle meets 
at 
. Define 
, 
. Show that the circumcircle of 
and 
are tangent to each other
be a triangle with 
, incenter 
, and 
excenter 
. The incircle meets 
at 
. Define 
, 
. Show that the circumcircle of 
and 
are tangent to each otherThis post has been edited 1 time. Last edited by Eyed, Jul 20, 2021, 8:51 PM
have orthocenter 
and orthic triangle 
,
be the foot from 
.
intersects 
at 
respectively, show 
and 
are tangent.
.
all lie on the circle with diameter 
.
,
and 
.
says 
cyclic, and finally
says that 
and 
are in fact tangent at 
.
denote the miquel point of 
.
,
,
and 
.
.
is cyclic
,
.
We can easily find, 
and 
.
.
:
So, 
cyclic.
and 
are tangent at 
.
Hence, proven
(normally 
)
,
be the feet of the altitudes from 
,
be the midpoint of 
be the inverse of 
be the midpoint of 
.
intersects 
and 
at 
is tangent to 
(at 
is the diameter of 
.
is the diameter of 
,
,
,
"
,
is tangent to 
is the midpoint of 
.![\[\angle PBD\stackrel{(1)}{=}\angle PDM_B\stackrel{(2)}{=}\angle TDM_B\stackrel{(3)}{=}\angle TM_AM_B\stackrel{(4)}{=}\angle TAB\stackrel{(5)}{=}\angle M_AAB\stackrel{(6)}{=}\angle B'BC\stackrel{(7)}{=}\angle M'BD,\]](http://latex.artofproblemsolving.com/4/6/c/46cacb9c3cd123cb8802e5fda9af04eb6d3f641a.png)
as desired.
and 
.
Finally 
,
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to 
.
,
,
,
.
Hence 
as well.
.
Therefore,
which implies 
is tangent to 
and 
be the 
and 
-
.
and 
intersect again at 
.
,
collinear, and also 
,
,
,
,
and 
meet 
again at 
and 
.
is cyclic, we have 
by Reim's. Analogously 
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so 
,
.
,
.
.
,
where 
is the midpoint arc of 
.
is cyclic.
.![\[ \measuredangle A'XI_A = \measuredangle IXI_A = 90^{\circ} = \measuredangle A'AM = \measuredangle A'AI_A \]](http://latex.artofproblemsolving.com/7/5/6/7566839e9b44d7e819c4468f0bf5ff6e42df4544.png)
since 
is the diameter of 
by definition. Hence, 
is cyclic.
,
concur. Suppose they concur at 
is cyclic since 
.
as well. Therefore, 
is cyclic.
and 
are both cyclic.![\[ \measuredangle XDE = \measuredangle XDF = \measuredangle XDA = \measuredangle XIA = \measuredangle XII_A = \measuredangle XBI_A = \measuredangle XBE. \]](http://latex.artofproblemsolving.com/e/9/8/e98bcd18f2f731ae4904b4297aefae2f6e8b5442.png)
is cyclic.![\[ \measuredangle XI_A E = \measuredangle XI_A B = \measuredangle XCB = \measuredangle XCD = \measuredangle XFD = \measuredangle XFE \]](http://latex.artofproblemsolving.com/9/4/9/949b3b4e7398a99d7c9b8c666abf5e9c6a6fac0f.png)
from our previous claims. Now, draw a line 
tangent to 
at 
as well. It suffices to prove that 
,![\[ \measuredangle YAX = \measuredangle YDX = \measuredangle BDX = \measuredangle BEX = \measuredangle I_A EX = \measuredangle I_A FX \]](http://latex.artofproblemsolving.com/0/1/b/01bd39058d65d484f40c90be311c899874b21919.png)
Hence, 
be the midpoint of arc 
,
,
be the reflection of 
,
so 
,
.
is the 
is the orthic triangle of the excentral triangle, 
is the 
is cyclic.
and 
as desired. 
,
be the reflection of 
meets 
and 
are tangent at 
and 
,
is harmonic.
to 
.
and similarly 
,
and 
are indeed similar. I will prove that 
is the similarity ratio, which solves the problem because obviously 
as 
,
and 
.
as desired. 
be the center of 
be the center of 
and 
,
by 
respectively. Define 
.
intersect 
again at 
.
and 
,
.
,
.
intersect 
again at 
.
.
and 
,
,
.
,
are collinear. Moreover by the similarity, 
.
and 
are bases of the same angle, which implies that 
,
.
.
We use complex numbers with the assumption that 
have complex coordinates 
respectively. Since 
From 
we have 
By the definition of 
It remains to check 
Luckily, these differences have nice forms 
Hence 
is 
.
is cyclic. But by 
done. 
and 
are tangent.
observe easily that 
is cyclic. 
Consequently the desired tangency holds. 
as the Miquel point of 
Proof
It's well-known that 
is cyclic, 
are collinear, and 
is harmonic. Observe that 
so 
is cyclic and thus 
is the Miquel Point of 
![[asy]
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with 
denoting the image of a point 
Since![\[-1 = (T, I; B, C) = (\infty, I'; B', C'),\]](http://latex.artofproblemsolving.com/e/f/c/efc8605d0bd52e840ac8db50c954ba3c49ce87c5.png)
it follows that 
Thus, omitting the 
s
be a quadrilateral with 
and Miquel Point 
Suppose that 
if 
and 
and we're done. ![[asy]
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to hold true. Also, my original proof for 
so 
is cyclic.
is cyclic.
proving the claim. 
and 
as 
.
at ![\[\measuredangle(\ell, GI)=90^{\circ}-\measuredangle(I_AG,\ell)=90^{\circ}-\measuredangle I_AEG=90^{\circ} -\measuredangle BEG =90 ^{\circ} -\measuredangle BDG=\measuredangle GAI\]](http://latex.artofproblemsolving.com/e/1/b/e1bc301e0082980ca5cb2373f1e13c1fd1a62ebe.png)
finishing. 
.
.
Thus 
are concyclic. Now let 
.
,
is tangent to 
,
.
,
.
are tangent to 
,
are collinear, we have 
.
.
centered at 
.
,
.
.
we can easily check 
since in normalized barycentric coordinates they both have 
.
,
,
.
.
is cyclic. Now angle chasing gives 
so 
is tangent to 
,
is the other tangent.
is tangent to 
.
be the unit complex number corresponding to the direction of 
.
Both of these are equal to 
,
,
where 
.
is cyclic.
Let 
.
reveals that 
,
we have 
cyclic. But then since 
we also have 
is cyclic.
.
,![\[ \measuredangle TDE' = \measuredangle TBE' = \measuredangle TIM = \measuredangle TIA = \measuredangle TDA, \]](http://latex.artofproblemsolving.com/9/0/f/90f51c6e1e17f3372621352977969e82685990e9.png)
so 
lies on 
.
.
is cyclic. 
just simply note that![\[ \measuredangle SDT = \measuredangle BDT = \measuredangle BET = \measuredangle I_aET \]](http://latex.artofproblemsolving.com/4/2/3/423c9190b260a36ebd7064a83973b876eee9ac2f.png)
which shows that the arcs have the same measure, hence the tangency is proven.
meets 
again at 
,
is cyclic. Then by the Radical Center theorem, we know that 
concur at some point 
all lie on the circle with diameter 
,
is cyclic. Observe that 
.
,
.
is cyclic, and so is 
,
to 
,
meet 
again at 
and 
are tangent is equivalent to showing 
.
,
,
,
.
and 
.
has equation 
and line 
has equation 
.![\[\left|\begin{matrix}z&\overline z&1\\a&\overline a&1\\d&\overline d&1\end{matrix}\right|=0\]](http://latex.artofproblemsolving.com/e/6/c/e6c1866674bbd65c84c9dcc4022ed7e0de196f15.png)
,
,
.
would be a factor of each numerator because 
or 
implies 
.![\[\ell_3(z)=j(bj+cj+2bc)z-bcj^2(b+c+2j)\overline z+(j^2+bc)(-j^2+bc)=0\]](http://latex.artofproblemsolving.com/9/e/c/9ec3c22a990f5232f6a9443985119369d0a034ee.png)
is![\[C(z)=k_1\ell_2(z)\ell_3(z)+k_2\ell_3(z)\ell_1(z)+k_3\ell_1(z)\ell_2(z)=0\]](http://latex.artofproblemsolving.com/b/c/8/bc84c32234c919e49fa40c6816154fa3c4bf1ffc.png)
where 
,
,
are chosen such that the 
and 
terms vanish.
.
.
,
.
and 
,
.
,
.
,
So the other intersection is 
.
which is true by inspection.
at 
.
at 
,
.
which is true by inspection.
,
,
intersect at one point 
is a diameter of 
is perpendicular to 
is perpendicular to 
,
is perpendicular to 
,
is perpendicular to 
are collinear
,
is perpendicular to 
,
are concyclic
,
and 
are parallel and homothety yields that the circumcircles of 
are tangent to each other
.![\[\angle BTC = \angle BTD + \angle DTC = \angle BED + \angle DFC = \angle FEI_A + \angle EFI_A = \angle BI_AC\]](http://latex.artofproblemsolving.com/d/d/e/dde7e8bff79dd4217562da0d66fa463545d9a7c9.png)
![\[\angle ITD = \angle BTD - \angle BTI = \angle BEA - \angle BI_AI = \angle IAD\]](http://latex.artofproblemsolving.com/5/d/0/5d0b1b43444a3f48a32d15f8693595ce28d36b1d.png)
First we notice,![\[\angle TAI = 180 - \angle IDT = 180 - (90 + \angle BDT) = 90 - \angle BDT = 90 - \angle TFC.\]](http://latex.artofproblemsolving.com/d/8/2/d829ed186c3bed26a87a60f336c0066c3d1b5eb8.png)
we get that 
is an orthogonal system. Hence 
are collinear.
lye on the circle with diameter 
.![\[\angle BPD=\angle BPI+\angle IPD=\frac{\angle C}{2}+\angle IAD=\angle EI_AI+\angle I_AAE=\angle BEA=\angle BED\]](http://latex.artofproblemsolving.com/4/7/f/47f70099d48c250a1caadf9d40dd6e6bd8862500.png)
Thus, 
.
and 
it is the miquel point of 
.
.![\[\angle PI_AE+\angle DKP=\angle DBE=\angle DPE\]](http://latex.artofproblemsolving.com/7/b/5/7b5db6e6b3ff09931ceb1bc4af6633563e04f438.png)
If 
is tangent to 
at 
then![\[\angle DPE-\angle (PD,l)=\angle (l,PE)=\angle PI_AE=\angle DPE-\angle DKP\]](http://latex.artofproblemsolving.com/1/f/5/1f5f96332ad6ea92146ab826d6d15c0b923632d0.png)
Hence 
.
and 
are tangent to each other at 
meets line 
.
,
is cyclic, since then it follows that 
.![\[\angle PBE = \angle PII_A = 180^{\circ} - \angle PIA = 180^{\circ} - \angle PDA = \angle PDE.\]](http://latex.artofproblemsolving.com/b/e/3/be39e44bb4561878bf84871af078808cc4e3f4ef.png)
![\[ \angle KDP = \angle BEP = \angle PFI_A.\]](http://latex.artofproblemsolving.com/1/d/1/1d1e38652f52a60d5b0d7465806d0c760515c93d.png)
and has lots of nice properties (such as its use in 
of 
and 
)
,
:
L
,
with 
,
)
,
,
we have that :
and 
,
.
,
be the intersection of lines 
,
,
,
.
,
,
,
and 
)
,
(Cuz 
,
![[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 = -9, xmax = 4, ymin = -5, ymax = 9.8; /* image dimensions */
draw((-3,6)--(-4,3)--(1,3)--cycle, linewidth(0.7));
draw((-3,6)--(-4,3), linewidth(0.7));
draw((-4,3)--(1,3), linewidth(0.7));
draw((1,3)--(-3,6), linewidth(0.7));
draw((-3,6)--(-1.162277660168379,-3.4868329805051377), linewidth(0.7));
draw((1,3)--(-1.162277660168379,-3.4868329805051377), linewidth(0.7));
draw((-4,3)--(-0.5811388300841897,-1.7434164902525695), linewidth(0.7));
draw(circle((-7.511328635098222,3.5698101949859686), 5.1242471195534), linewidth(0.7));
draw(circle((-2.7565835097474296,-1.9868329805051381), 2.189020589670635), linewidth(0.7));
draw(circle((-1.5,1.1981019498596834), 3.08169378477153), linewidth(0.7));
draw(circle((-3.209430584957906,1.2905694150420945), 1.883388681284223), linewidth(0.7)+dashed);
draw((-3,6)--(-0.5811388300841897,-1.7434164902525695), linewidth(0.7));
draw((-2.4188611699158105,4.139620389971937)--(-4.179782467020843,-0.3236084406375195), linewidth(0.7));
draw((-4.179782467020843,-0.3236084406375195)--(-0.5811388300841897,-1.7434164902525695), linewidth(0.7));
draw((-2.4188611699158105,3)--(-4.179782467020843,-0.3236084406375195), linewidth(0.7));
draw((-4.179782467020843,-0.3236084406375195)--(-1.837722339831621,0), linewidth(0.7));
draw((-3,6)--(-4.179782467020843,-0.3236084406375195), linewidth(0.7));
dot((-3,6),linewidth(3pt) + dotstyle);
label("$A$", (-2.9202501646632117,6.219693079558107), NE * labelscalefactor);
dot((-4,3),linewidth(3pt) + dotstyle);
label("$B$", (-4.450345658129821,3.09574811373045), NE * labelscalefactor);
dot((1,3),linewidth(3pt) + dotstyle);
label("$C$", (1.2875124423699638,2.9682401559415657), NE * labelscalefactor);
dot((-2.4188611699158105,4.139620389971937),linewidth(3pt) + dotstyle);
label("$I$", (-2.325213028315086,4.264571060128553), NE * labelscalefactor);
dot((-0.5811388300841897,-1.7434164902525695),linewidth(3pt) + dotstyle);
label("$I_A$", (-0.3488396825873824,-2.2595861134026767), NE * labelscalefactor);
dot((-2.4188611699158105,3),linewidth(3pt) + dotstyle);
label("$D$", (-2.325213028315086,2.4369569984878825), NE * labelscalefactor);
dot((-1.837722339831621,0),linewidth(3pt) + dotstyle);
label("$E$", (-1.5814166078799288,-0.07069950469350209), NE * labelscalefactor);
dot((-1.162277660168379,-3.4868329805051377),linewidth(3pt) + dotstyle);
label("$F$", (-1.1776414082151292,-4.087200175043347), NE * labelscalefactor);
dot((-4.179782467020843,-0.3236084406375195),linewidth(3pt) + dotstyle);
label("$R$", (-4.917874836689063,-0.4744747043583013), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy]](http://latex.artofproblemsolving.com/f/b/2/fb216a478c0cf7cc398838a78267002ed8b9d182.png)
.
and 
.
all intersect at a point .(Just angle chase for this )
Observe that 
. Let 
.
and self-intersecting 
coincide at a point 
.
,
.
so 
.![\[\angle DPI = \angle CPI - \angle CPD = \frac B2 - \angle CFD = \frac A2 - \theta = \angle IAD. \ \blacksquare\]](http://latex.artofproblemsolving.com/2/c/d/2cdcd5626acbc3799ada95d444b047afc38ca20a.png)
Thus done.
we only need
Clearly all segments are easy to calculate
.
this is true because 
and putting the pairs to line 
.
cyclic is trivial since 
and then just angle chase. The point which swaps these is 
which also swaps 
so by ddit it also swaps 
so we have 
cyclic. And the rest follows like everyone.
