[/quote]
,
and some point 
.
intersects the circumcircle of 
at 
and the perpendicular from 
intersects the circumcircle of 
at 
.
does not depend on the choice of 
grid consisting of 2025 unit squares. Within each unit square is a car, facing either up, down, left, or right. If the square in front of a car in the direction it is facing is empty, it can choose to move forward. Each car wishes to exit the 
cars from the 
coins, all of which show 
on one side and 
on the opposite side. The coins are laid out in a row on the table. Peer and Solveig alternate taking turns. On his turn, Peer may turn over one or more coins, so long as he does not turn over two adjacent coins. On her turn, Solveig picks precisely two adjacent coins and turns them over. When the game begins, all the coins are showing 
for which Solveig can keep Peer from winning.
be an angle in the interval 
.
is irrational, and that 
and ![$ \cos[(k + 1)\theta ]$](http://latex.artofproblemsolving.com/b/3/a/b3a14c57a1a8b5a6585195c394181c5124f8c471.png)
are both rational for some positive integer 
,
.
be three positive reals such that 
.![\[ \frac { x^5-x^2 }{x^5+y^2+z^2} + \frac {y^5-y^2}{x^2+y^5+z^2} + \frac {z^5-z^2}{x^2+y^2+z^5} \geq 0 . \]](http://latex.artofproblemsolving.com/a/1/4/a14adf0f1e35df57c734c7df701d5a67da22dc3c.png)
be a triangle with altitude 
.
-
at 
,
and 
.
and 
on lines 
and 
such that quadrilateral 
is a rhombus.
be an integer. Two players, Alice and Bob, play the following game on the complete graph 
:
,![\[
\left( \frac{1}{4} - \varepsilon \right) n^2 + n.
\]](http://latex.artofproblemsolving.com/a/9/c/a9cca00e87c643af32eb75a2501b59060f3504fb.png)
such that the following inequality holds for all 
:![\[
ab + bc + ca \leq \frac{(a + b + c)^2}{3} + k \cdot \max \{ (a - b)^2, (b - c)^2, (c - a)^2 \} \leq a^2 + b^2 + c^2
\]](http://latex.artofproblemsolving.com/1/c/e/1ce66163068b499383adb4a4ecec86c5f4edaec3.png)
be positive real numbers such that : 
. Prove that : 
where 
are lengths of parallel sides and 
are other two sides.
,
,
,
,
in some order.
board that is entirely covered by dominoes, with no overlap. He jumps along the vertical lines of the board, always going from the midpoint of the vertical segment of a 
square to another midpoint of the vertical segment, according to the rules:
When the domino is horizontal, the cricket jumps to the opposite vertical segment (such as from 
to 
)
When the domino is vertical downwards in relation to its position, the cricket jumps diagonally downwards (such as from 
to 
When the domino is vertically upwards relative to its position, the cricket jumps diagonally upwards (such as from 
)
board.
Prove that
candies. A starts first. On a move, if in the box there are 
so that 
and 
,
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 
.
is a tangent to 
,
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 
.
