and 
.
Let 
and 
.
Let 
and 
and 
Prove that
Where 
such that:
x 
unit squares. Strawberries lie on some of the unit squares so that each row or column contains exactly one strawberry; call this arrangement 
.
be another such arrangement. Suppose that every grid rectangle with one vertex at the top left corner of the cake contains no fewer strawberries of arrangement 
by performing a number of switches, defined as follows:
Y by Davi-8191, Abderrahmane_Driouch, mathroyal, A-Thought-Of-God, sotpidot, chessgocube, megarnie, Adventure10, torch, lian_the_noob12, Rounak_iitr, and 2 other users
Let 
be the circumcircle of a triangle 
. Denote by 
and 
the midpoints of the sides 
and 
, respectively, and denote by 
the midpoint of the arc 
of not containing 
. The circumcircles of the triangles 
and 
intersect the perpendicular bisectors of and at points 
and 
, respectively; assume that and lie inside the triangle . The lines 
and 
intersect at 
. Prove that 
.
be the circumcircle of a triangle 
. Denote by 
and 
the midpoints of the sides 
and 
, respectively, and denote by 
the midpoint of the arc 
of not containing 
. The circumcircles of the triangles 
and 
intersect the perpendicular bisectors of and at points 
and 
, respectively; assume that and lie inside the triangle . The lines 
and 
intersect at 
. Prove that 
.
be the circumcenter of 
.
,
.
.
becomes an isosceles trapezoid, and hence we have 
.
and 
.
be the reflection of 
.
,
and similarly for 
,
are isosceles trapezoids. It follows that quadrilateral 
is an isosceles trapezoid, whence 
.
.
be the second point where the parallel line to 
and hence 
and 
are congruent. Since 
is a trapezoid inscribed in a circle, it is an isosceles trapezoid, and hence 
.
because 
(using again that 
,
and 
of 
.
.
.
,
,
is an isosceles trapezoid. We can prove in an analogous way that 
is an isosceles trapezoid. It follows that the segments 
,
all have the same perpendicular bisector. But then 
,
![[asy]
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[/asy]](http://latex.artofproblemsolving.com/6/2/9/629639b3a66085e7e4406f5ba9b5101df8b0df9d.png)
.
and we already know 
.
,
since 
.
.
.
.
so 
implying that 
so 
.
,
,
,
is an isoceles trapezoid; the perpendicular bisector of 
and 
lies on the perpendicular bisector, so 
at 
Since 
is cyclic and every cyclic trapezoid is isoceles, we must have that 
Now, some trivial angle chasing (too lazy to post it now; will complete it later) yields 
implying 
is cyclic, it must be isoceles. Similarly, so must be 
It follows that 
share the same perpendicular bisector, so 
be on circumcircle of 
such that 
and let
intersects with 
(if outside its same), intersects with 
.
is isosceles trapezoid, we have 
, 
and 
are isosceles which means 
, thus we have 
.
and 
is isosceles trapezoid, we conclude 
.
? 
such that 
is an isosceles trapezoid.
and 
be the perpendicular bisectors of 
,
lie on 
and let 
is an isoceles trapezoid, 
are collinear on 
Meanwhile, from 
,
is an isoceles trapezoid as well. Then from symmetry in 
,
is an isoceles trapezoid too. Hence, 
Therefore, 
,
as well. It follows that 
is an isoceles trapezoid, so by symmetry, 
Thus, 
and 
.
;
and 
,
are collinear. Thus 
.
,
.
.
.
.
.
.
and 
.
.
.
.
is symmetric wrt 
and 
are symmetric with respect to 
are collinear, which proves the claim, as we can use the same argument to show that 
are collinear.
by definition of ![\[ \frac{0-\frac{a+b}{2}}{0-\frac{2}{a+c}} = \overline{\left(\frac{0-\frac{a+b}{2}}{0-\frac{2}{a+c}}\right)} \]](http://latex.artofproblemsolving.com/2/4/a/24a085cf9f00adf5e5ea2b2b2081762a34a363d5.png)
This easily expands to 
,
.
as well as 
.
and 
.
,
cyclic 
,
.
be the circumcenter of 
,
.
,
,
.
.
,
,
.
,
,
.
.
are isosceles trapezoids, thus so as 
.
is on the perpendicular bisector of 
,
To see why, we phantom-point 
We prove that 
Next reflect 
which lies on 
since 
and 
is an isosceles trapezoid -> 
-> 
is an isosceles trapezoid -> 
-> 
proving our claim. We similarly see that 
and 
things:
and 
is cyclic, implying that 
This implies that 
so 
so 
is an isosceles trapezoid.
Since 
is isosceles, 
lies on this perpendicular bisector, and we are done.
.
.
ABTP is an isosceles trapezoid 
,
and AM = TX.
,
.
,
,
,
.
,
intersect the circumcircle of 
intersect the circumcircle of 
,
,
,
,
,
,
,
,
and the axis of symmetry between line 
and 
.
inversion.
is feet of perpendicular from 
.
and 
.
So 
cyclic.
cyclic, with 
.
we have 
.
is the perpendicular bisector of
,
,
and we have that 
which gives us that 
and thus 
proving the result. A similar argument proves 
.
.
![\[\angle TBM = \angle TBC + \angle B = \frac{\angle A}{2}+\angle B = \angle CTX'\]](http://latex.artofproblemsolving.com/1/4/3/143eb62b39fa4c84782cd2631442131670d72ac3.png)
since 
.
,
.
,
is isosceles. Thus, 
,
must also be perpendicular to 
is isosceles.
be the centers of 
.
are collinear and 
.
is an isosceles trapezoid, so 
,
and also 
.
.
is cyclic, we also have 
.
,
is cyclic, from which it follows that 
.
.
which is the perpendicular bisector of 
,
.
.
and 
are the perpendicular bisectors of 
and 
are symmetric about line 
.
and 
,
.
.
and 
are congruent.
is an isosceles trapezoid and line 
,
,
:
and 
are both isosceles trapezoids, they must be cyclic.
are isosceles trapezoids (we would have 
(which implies 
)
where 
(with 
)
.
.
,
.![\[AN' = AR - N'R = AR - AM = AN,\]](http://latex.artofproblemsolving.com/c/4/b/c4bb7d8d6d8c537c5079d9cbe2d598b9ef9058c6.png)
since 
is the average of 
be the midpoint of 
,
and 
Show that 
are collinear.
so 
is an isosceles trapezoid. Let 
Clearly, 
is cyclic. Thus, 
but 
since 
is cyclic.
& 
are isosceles trapezoids
is an isosceles trapezoid, it suffices to prove that the intersection of 
and 
is in fact the midpoint.
is isosceles,
is concyclic, forcing 
to be perpendicular to 
are collinear
,
are concyclic
is parallel to 
because 
and 
are congruent
criterion
is a cyclic quad as well
of 
are collinear.
lies on the circle of 
is perpendicular to 
![[asy]
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dot("$K$", K, dir(0));
[/asy]](http://latex.artofproblemsolving.com/e/e/b/eeb616fa9160b203f9175a7c4ec3b9229952a1de.png)
.![\[\measuredangle(\ell, \overline{OM}) = \measuredangle(\overline{AT}, \overline{AB}) = \measuredangle(\overline{AC}, \overline{AT}) = \measuredangle(\overline{ON}, \ell)\]](http://latex.artofproblemsolving.com/6/8/e/68ea7a32dd9794197c8014fba462ed93e61c95f2.png)
by 
rotations, so 
and 
.
lies on 
