[/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.
with 
and 
,
be the midpoint of 
.
on the extension of 
past 
and point 
on segment 
such that 
.
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 lethan3, LLL2019, GoodMorning, Rounak_iitr
Let 
be a cyclic quadrilateral whose opposite sides are not parallel. Suppose points 
lie in the interiors of segments 
respectively, such that 
Let 
intersect 
at 
, and 
intersect 
at 
. Prove that lines 
and 
are either parallel or coincide.
Tilek Askerbekov
be a cyclic quadrilateral whose opposite sides are not parallel. Suppose points 
lie in the interiors of segments 
respectively, such that 
Let 
intersect 
at 
, and 
intersect 
at 
. Prove that lines 
and 
are either parallel or coincide.Tilek Askerbekov
This post has been edited 7 times. Last edited by ike.chen, Oct 24, 2022, 2:39 AM
.
.
,![\[\measuredangle EBS =\measuredangle EBA+\measuredangle ABS=\measuredangle CDE +\measuredangle SCD=\measuredangle CSE,\]](http://latex.artofproblemsolving.com/b/7/6/b76630a005860480ade6d75961db1de3a2cc172a.png)
hence done. 
,
,
;
.
bisects 
,![\[\measuredangle BSQ=\measuredangle BSE+\measuredangle ESQ\overset *= -\measuredangle SCE-\measuredangle EQS=-\measuredangle CSQ,\]](http://latex.artofproblemsolving.com/6/0/1/601ee3aa6f6c048bdd034bc76539f6a961b35ad2.png)
where * follows from similar triangles in claim 1, hence done. 
.
.
.
,
Law of sines on 
and using the ratios gives us 
.
.
.
gives us 
Substituting all of this into (1) and simplifying, we get 
Ratio Lemma on 
and doing some angle manipulations using the fact that 
gives us 
Substituting, we get 
.
.
.
decreases, the expression 
decreases. Then there is a maximum of one solution for 
.
works, so it must be the solution.
and 
.
.![[asy]
size(13cm);
pen rvwvcq = rgb(0.08235,0.39607,0.75294); pen sexdts = rgb(0.18039,0.49019,0.19607); pen dtsfsf = rgb(0.82745,0.18431,0.18431); pen wrwrwr = rgb(0.38039,0.38039,0.38039);
pair A = (-3.32,0.67); pair B = (-3.36110,-1.58058); pair C = (0.48650,-1.57101); pair D = (-2.24520,1.56709); pair P = (-3.33683,-0.25181); pair Q = (-1.86021,-1.57684); pair R = (-1.30536,0.48744); pair S = (-2.82745,1.08111); pair Xp = extension(B,R,C,P); pair Yp = extension(A,R,D,P); pair X = extension(A,Q,B,S); pair Y = extension(D,Q,C,S); pair U = extension(A,D,B,C); pair V = extension(A,B,C,D);
filldraw(circle((-1.44,-0.49), 2.20907), invisible, rvwvcq); draw(B--S--C, rvwvcq); draw(A--Q--D, rvwvcq); draw(X--Y, dtsfsf); draw(P--R, sexdts); draw(Q--S, sexdts); draw(A--B--C--D--cycle, rvwvcq); draw(A--V--D, rvwvcq); draw(A--U--B, rvwvcq);
draw((-6.02432,-1.58720)--(0.38772,0.74615), dashed + wrwrwr); draw((-3.28186,2.75799)--(-1.30944,-2.66221), dashed + wrwrwr);
dot("$A$", A, dir(170)); dot("$B$", B, dir(225)); dot("$C$", C, dir(315)); dot("$D$", D, dir(80)); dot("$V$", V, dir(90)); dot("$U$", U, dir(225)); dot("$P$", P, dir(200)); dot("$Q$", Q, dir(270)); dot("$R$", R, dir(30)); dot("$S$", S, 2*dir(130)); dot("$X$", X, dir(190)); dot("$Y$", Y, dir(20));
[/asy]](http://latex.artofproblemsolving.com/0/1/2/012817f60903356dc3718b2102a8d10c6770a090.png)
and 
.![\[ \measuredangle BSA = \measuredangle BAS+\measuredangle SBA= \measuredangle BCD + \measuredangle DCS= \measuredangle BCS \]](http://latex.artofproblemsolving.com/2/2/8/2284b46411e9b752f35fe3bb1876efe79d9168d7.png)
hence ![\[ US^2=UB\cdot UC. \]](http://latex.artofproblemsolving.com/1/1/e/11e4e28d51d6015074bd298eb4f7336f96565239.png)
Similarly, 
.
;
is a kite (with 
and 
)![\[ \measuredangle BSQ = \measuredangle USQ - \measuredangle USB =\measuredangle SQU - \measuredangle SCB = \measuredangle QSC \]](http://latex.artofproblemsolving.com/4/4/9/4490678845396e5bd91d78cb647bcd717ffd70e4.png)
so 
bisects 
;
.
and 
are perpendicular.
are perpendicular to the internal bisectors of 
.
is perpendicular to 
lies on ![[asy]
size(14cm); pen rvwvcq = rgb(0.08235,0.39607,0.75294); pen sexdts = rgb(0.18039,0.49019,0.19607); pen dtsfsf = rgb(0.82745,0.18431,0.18431); pen wrwrwr = rgb(0.38039,0.38039,0.38039);
pair A = (-3.32,0.67); pair B = (-3.36110,-1.58058); pair C = (0.48650,-1.57101); pair D = (-2.24520,1.56709); pair E = extension(A,C,B,D); pair P = (-3.33683,-0.25181); pair Q = (-1.86021,-1.57684); pair R = (-1.30536,0.48744); pair S = (-2.82745,1.08111);
filldraw(circle((-1.44,-0.49), 2.20907), invisible, rvwvcq); draw(A--B--C--D--cycle, grey); draw(A--C, grey); draw(B--D, grey); dot("$A$", A, dir(170)); dot("$B$", B, dir(225)); dot("$C$", C, dir(315)); dot("$D$", D, dir(80));
picture left = rotate(0)*currentpicture; picture right = rotate(0)*currentpicture; currentpicture.erase(); pair Xp = extension(B,R,C,P); pair Yp = extension(A,R,D,P); pair X = extension(A,Q,B,S); pair Y = extension(D,Q,C,S); draw(left, B--S--C, rvwvcq); draw(left, A--Q--D, rvwvcq); draw(left, X--Y, dtsfsf);
draw(right, C--P--D, rvwvcq); draw(right, A--R--B, rvwvcq); draw(right, Xp--Yp, dtsfsf); draw(right, P--R, sexdts);
dot(left, "$X$", X, dir(190)); dot(left, "$Y$", Y, dir(20)); dot(right, "$X'$", Xp, dir(290)); dot(right, "$Y'$", Yp, dir(110)); dot(left, "$Q$", Q, dir(270)); dot(left, "$S$", S, 2.4*dir(130)); dot(right, "$P$", P, dir(200)); dot(right,"$R$", R, dir(30)); dot(left, "$E$", E, dir(110)); dot(right, "$E$", E, dir(200));
add(shift(-2.5,0)*left); add(shift(2.5,0)*right);
[/asy]](http://latex.artofproblemsolving.com/7/0/5/7059a7108f6778ef3683b95a160277ba174b9ff2.png)
is the interior bisector of 
and 
.
and 
.
and 
.
and 
are collinear, we're done. ![\[ X' = \overline{BR} \cap \overline{CP} \qquad\text{and}\qquad Y' = \overline{AR} \cap \overline{DP}. \]](http://latex.artofproblemsolving.com/3/c/9/3c9dd8be2cecef23ad8999efcd3f27815e5775d7.png)
By the same argument as before, line 
is the internal angle bisector of angles 
and 
.
is a kite (with 
and 
)
and 
are corresponding points in 
and 
,![\[ \angle RX'Y' = 180^\circ - \angle BX'E = 180^\circ - \angle AY'E = \angle RY'X', \]](http://latex.artofproblemsolving.com/5/b/c/5bcf9f9b26ba63542c8bac143dcd371aa50dc771.png)
and so 
.
.
,
and 
.
.
.
,
is perpendicular bisector of 
and 
(the author's solution).
(I think this could fit as a 
)
again at 
and 
.
is parallel ot 
is tangent to 
,
then we have that 
,
is perpendicular to the angle bisector of 
.
,
.
in fact implies that 
is the Apollonius circle preserving the ratio 
.
.
,
was tangent to 
(and its cyclic variants); knowing this, we must have a lot of power of a point properties going on, which are best extracted by considering 
and 
.
. You can visit my blog post 
are 
to each other.
are 
to each other.
are collinear.
are collinear.
.
and 
.
.
.
so 
.
.
by Power of a Point. 
is a kite, so 
.
,
hence 
bisects 
.
,
.
,
.
.
symmetrically, so 
.
is tangent to 
meet at 
.
(and similarly 
)
so 
and 
.
Furthermore, 
Since we want to show that these are equal, we want to show that 
However, from quadrilateral 
,
and hence 
Similarly, 
,
,
Let 
Note that 
since quadrilaterals 
and 
both have 
.
so thus 
Hence, 
,
.
is tangent to 
,
is tangent to 
so 
,
is isosceles. Thus, we can angle chase to get
Similarly, we have 
,
,
.
which indeed implies that 
.![\[\angle ASB = 180-(\angle ABS+\angle BAS) = \angle BCD-\angle SCD = \angle BCS \]](http://latex.artofproblemsolving.com/5/e/0/5e0719c004dcf53e54a7d6c1cfe9cdc0ab241e74.png)
.
,
![\[\angle XSQ = \angle USQ- \angle USB = \angle USQ - \angle UCS = \angle CSQ.\]](http://latex.artofproblemsolving.com/1/3/5/13588c89319df3e3ac6477b12bbe068e5aed0e10.png)
as the angle bisector of 
as the angle bisector of 
.
.
and 
,
.
![\[\angle BRC=180-\angle RBC-\angle BCR=\angle DAB-\angle DAR=\angle RAB\]](http://latex.artofproblemsolving.com/1/0/1/10157f1ff81b82dc56fbe799c239c41b42b1057e.png)
so 
is tangent to 
at 
.
and 
,![\[WR^2=WA\cdot WB=WC\cdot WD=WP^2\]](http://latex.artofproblemsolving.com/6/e/0/6e068a2c72dae26472858ad28c948ca54c831f37.png)
so 
and similarly 
.
and similarly 
.![\[\angle(\overline{SQ},\overline{PR})=360-\angle A-\angle QSA-\angle APR=90\]](http://latex.artofproblemsolving.com/b/f/d/bfd128189e8147d0b81cbbb9ab09ec8b00ce8fff.png)
so 
.![\[\angle XSQ=\angle QSA-\angle BSA=\angle BQS-\angle BCS=\angle QSY\]](http://latex.artofproblemsolving.com/9/1/9/9199a92384a92ec42cfd23a6914f50cec310a854.png)
and similarly 
.
is tangent to ![\[\angle BSA = 180^{\circ} - \angle BAS - \angle ABS = \angle BCD - \angle SCD = \angle BCS,\]](http://latex.artofproblemsolving.com/e/6/9/e699aafbfea403a38cb570f3a19e26bdb36d595a.png)
implying the tangency.
is the intersection of lines ![\[TS = \sqrt{TB \cdot TC} = \sqrt{TA \cdot TD} = TQ,\]](http://latex.artofproblemsolving.com/6/4/c/64c94f7eee827d3d714d22c1a65f0c00b3197ed9.png)
implying that 
-
.
are opposites, as are lines 
.
is a kite, so 
being tangent to 
respectively.
and 
to get a cyclic quadrilateral, and then angle chase.
and 
where 
and 
.
.
.
.
.
and 
.
and 
.
.
.
,
by the SAS test.
and 
have side 
.
