[/quote]
,
be positive integers and 
be an odd prime such that:![\[
a^p \equiv 1 \pmod{p^n}.
\]](http://latex.artofproblemsolving.com/3/4/8/34815b8211ded2f37b0653525dc4b436356e6eee.png)
Prove that:![\[
a \equiv 1 \pmod{p^{n-1}}.
\]](http://latex.artofproblemsolving.com/0/7/a/07a4ed2ab3c94f41028080f7789cfb6e2973955c.png)
passes through 
and 
and meets lines 
and 
again at points 
and 
(distinct from 
)
denote the orthocenter of triangle 
Prove that if lines 
are concurrent, then triangle 
and 
are similar.
satisfying 
satisfying![\[\begin{aligned}
\begin{cases}
a+b+c+d=0,\\
a^2+b^2+c^2+d^2=12,\\
abcd=-3.\\
\end{cases}
\end{aligned}\]](http://latex.artofproblemsolving.com/8/b/1/8b1923e0303c0aa79332ee4540ee1885a9cf3f08.png)
crosses are placed. A set of 2025 cells we will call balanced if no two of them are in the same row or column. It is known that any balanced set has at least 
crosses.
,
onto 
past 
.
be on the perpendicular bisector of 
such that 
are on the same side of ![\[\frac12\angle ALE=1.4\angle ABE+3.4\angle ACE-558^\circ\]](http://latex.artofproblemsolving.com/c/6/3/c63eeb053f1fa8a29bef93516814ae121382219e.png)
Let the reflection of 
and 
be 
and 
and 
such that 
.
and 
intersect the circumcircles of 
and 
at 
and 
,
and 
intersect 
and 
at 
and 
.
intersect 
.
such that 
![\[
C(p-1, k-1) \equiv (-1)^{k-1} \pmod{p}
\]](http://latex.artofproblemsolving.com/4/0/7/4072235afbaa1f2d8d4c9ca044615a113a4306ce.png)
for 
,
is the binomial coefficient 
choose 
.
satisfy that: 
and 
.
intersect at points 
and 
.
be a function such that for every regular 
-
we have 
.
for all reals 
.
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 ![[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 
.
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 
.
. 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 
.
is a tangent to 
,
hence 
bisects 
.
,
.
,
.
.
symmetrically, so 
.
is tangent to 
(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.
tangent to 
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 
.
