[/quote]
,
and circumcenter 
.
and 
be points on the line segments 
and 
is a parallelogram. Prove that 
.
be an integer. Show that, if 
is an integer, then it is a perfect square.[/quote]
.
is at least rational, so that 
must be a perfect square then. Using Conway's topograph method, I have found out that the smallest non-negative pairs 
for which this happens are 
and 
,
,
The eigenvalues of that matrix are irrational, however, so that any calculation which uses powers of that matrix is a little cumbersome. There must be an easier way, but I cannot find it. Can you?
that satisfy 
for all 
and 
satisfies the following recursion formulas: 
,
.
interesting if 
of 
and 
is also a divisor of 
and 
be points on segment 
denote the circle passing through 
denote the circle passing through 
.
)
.
the feet of the altitudes lying on 
respectively. One of the intersection points of the line 
and the circumcircle is 
The lines 
and 
meet at point 
Prove that 
airline companies in a country. Between any two cities, there is exactly one 
and circumscribed about the incircle 
,
.
,
,
(distinct from 
intersects 
(distinct from 
be the midpoint of the arc 
of 
at 
and 
,
at 
.
.
but doesn't tell it to anyone. Bernaldo thens pick an arbitrary positive integer 
(not necessarily in 
.
chosen by Arnaldo.
be a set of 2025 positive real numbers. For a subset 
,
as the median of 
when all elements of 
.![\[
P(A) = \sum_{\substack{T \subseteq A \\ |T| \text{ odd}}} M_T, \quad Q(A) = \sum_{\substack{T \subseteq A \\ |T| \text{ even}}} M_T.
\]](http://latex.artofproblemsolving.com/1/d/f/1df2ea8487289111939a53dcc49576fbe6fdef37.png)
Find the smallest real number 
such that for any set ![\[
P(A) - Q(A) \leq C \cdot \max(A),
\]](http://latex.artofproblemsolving.com/c/d/e/cdea343d11c722a423e857aeb01202863405773e.png)
where 
denotes the largest element in 
in 
colors, there exists an arithmetic progression of size 
.
from this?
.
be foots of perpendicular lines from 
.
is cyclic and 
.
bisects 
Y by shinichiman, Akatsuki1010, anantmudgal09, e_plus_pi, Tumon2001, magicarrow, A-Thought-Of-God, HWenslawski, Adventure10, Mango247, Rounak_iitr
Let 
be a triangle with circumcircle 
and incentre 
. Let the line passing through and perpendicular to 
intersect the segment 
and the arc (not containing 
) of at points 
and 
, respectively. Let the line passing through and parallel to 
intersect 
at 
, and let the line passing through and parallel to intersect 
at 
. Let 
and 
be the midpoints of 
and , respectively. Prove that if the points 
and are collinear, then the points 
and are also collinear.
Proposed by David B. Rush, USA
be a triangle with circumcircle 
and incentre 
. Let the line passing through and perpendicular to 
intersect the segment 
and the arc (not containing 
) of at points 
and 
, respectively. Let the line passing through and parallel to 
intersect 
at 
, and let the line passing through and parallel to intersect 
at 
. Let 
and 
be the midpoints of 
and , respectively. Prove that if the points 
and are collinear, then the points 
and are also collinear.Proposed by David B. Rush, USA
This post has been edited 3 times. Last edited by hajimbrak, Jul 15, 2016, 9:20 AM
Reason: Added proposer
Reason: Added proposer
.
,
are concyclic ... 
are concyclic ... 
,
is the perpendicular bisector of 
,
combine 
is the tangency point of A-mixtilinear circle with 
,
is the isogonal conjugate of the A-Nagel line of 
WRT 
(well-known) . ... 
is parallel to the A-Nagel line of 
are collinear .
are collinear and 
so 
.
then 
.
.
so 
.
.
then since 
so 
.
.
so 
.
are concyclic. Therefore 
.
are concyclic. Therefore 
.
so 
.
and 
.
so 
.
or 
passes through the midpoint 
of arc 
so 
or 
.
and 
cyclic as others have shown, and it is clear that 
is harmonic, so 
is harmonic and 
.
,
-
so 
are collinear. But the fact that 
we get 
,
.
of triangle 
we get angle bisectors of 
and 
are perpendicular. Therefore 
.
,
.
.
.
and 
are perpendicular to 
or 
are collinear.
and 
one gets by straightforward angle chase: 
as desired.
are collinear
and the 
since 
,
is the 
and therefore 
are concyclic.
are collinear, the center of spiral similarity sending 
and 
is 
,
QED.
, since 
isn't invariant
)
.
![[asy]
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,
,
.
,
and let 
be the midpoint of side 
hence 
(because 
is parallel to 
and quadrilateral 
is cyclic. Since 
,
,
and 
are congruent, yielding 
.
,
;
.
,
.
are collinear, triangles 
and 
are similar, as well as triangles 
and 
.
or, equivalently, 
.
are parallel, triangles 
and 
are similar; hence 
is parallel to 
,
;
,
and 
.
,
.
implying that quadrilateral 
is cyclic. Since 
,
;
,
is the external angle bissector of the same angle, and thus 
,
.
,
,
is tangent to the circumcircle of 
.
is tangent to such circle as well. It follows from a well-known lemma that 
,
is perpendicular to 
.
to 
,
is parallel to 
;
is right.
is right, point 
.
,
.
,
.
.
,
be midpoint of arc 
is cyclic with 
so 
are collinear. Since 
and 
are tangents to circumcircle of 
,
is symmedian to triangle 
.
,
are collinear.
.
and 
are the tangency points of the 
.
so 
implies that 
is cyclic. Now by the given collinearity, 
and since 
we deduce 
are homothetic and thus 
.
be the reflection of 
we deduce that 
are reflections over 
.
so 
are the tangency points of the 
.
.
gives the well-known result that 
are collinear with the midpoint of 
.
.
so it suffices to show that 
.
be the reflection of 
be the reflection of 
,
.
are isogonal in 
so 
has angle bisector 
is 
,
is cyclic
cuts 
is cyclic
cuts 
quadrilateral 
is cyclic
then 
is the perpendicular bisector of 
then 
be the midpoint of ar
then we have 
passes through 
; it follows that 
is the bisector of 
then 
and it follows that : 
then 
, we come to a conclusion that 
:
by midpoint of 
.
are always concurrent.
.
and line 
,
with fixed point 
,
where 
is 
which tells 
.
we get 
are concurrent, hence \End Claim.
in the problem. \return;
with the three pairs of opposite sides of a complete quadrangle and a conic section circumscribed about the quadrangle form four pairs of an involution on 
to meet 
and so 
is cyclic.
,
is cyclic, and 
by Fact 5.
and since 
in 
,
and so this means 
.
are collinear and 
in isosceles 
,
,
are all 
.
,
be the ex-touch point. Its well known that 
and 
are isogonal.
and so 
are collinear, as required
,
and 
and 
.
,
is the perpendicular bisector of 
But it is well known (trivial angle chase) that 
is the mixtilinear intouch point.
,
.
is symmedian of 
.
,
the midpoint of the arcs 
on 
also let 
,
,
so 
so 
is cyclic, now since 
we have 
isosceles and now by Thales spam on those parallels we can get 
so indeed 
we have that 
so 
is isosceles, now by well known usa mixtilinear config we have 
colinear, 
colinear and 
colinear. So using these colinearities by angle chase 
so 
is cyclic and now since 
we have 
cyclic so now by the final angle chasing:
Thus we are done 
.
as cyclic quadraples.Also by Fact 5 
and as we have 
by similarity ratios
(
)
.
.
So 
so if 
we have 
which implies our claim 
is the midpoint of 
,
is right angled,
Again by angle chasing,
Again as 
is cyclic we get,
Finally 
Which means 
,
.
parallel, so 
so 
.
be 
respectively, and let 
and 
so 
is cyclic. Since 
,
,
.
is a diameter of 
,
so 
.
so 
is tangent to 
implies 
so 
.
,
so 
.
.
.
,
cyclic, so 
,
.
so 
sets 
are concyclic.
we conclude 
Now observe that 
point 
is parallel to isogonal of 
which is 
thus we are done.
Did I mention I HATE BARY?
gives 
gives 
is equivalent to 
which follows from substitution, so we are done. Thank goodness.
be the 
e
![\[\angle{BYV}=\angle{BAI}=\angle{BIV}\]](http://latex.artofproblemsolving.com/e/0/5/e05d0a90d96bdd79ae932910f24ebbcea26c64eb.png)
resulting in ![\[\angle{BNU}=\angle{BYV}=\angle{BIU}\]](http://latex.artofproblemsolving.com/1/0/e/10e176c01b7c9d85a586be7ee9b46c78032b68b6.png)
so 
is cyclic as well.
and 
.
![\[\angle{BVY}=\angle{BI_aI}=\angle{BCI}=\angle{BCT}\]](http://latex.artofproblemsolving.com/5/e/6/5e67b7dcbaf74ff358d933a8bebb20712455242b.png)
![\[\angle{TVC}=\angle{TBC}=\angle{ABC}+\frac{\angle{ACB}}{2}\Rightarrow \angle{YVA}=\frac{\angle{ACB}}{2}=\angle{VI_aA}\]](http://latex.artofproblemsolving.com/3/8/0/38003433206e31270ac9d6d5b612d4bfe41f55fa.png)
Since 
,
.
,
and 
and 
,
is a rectangle centred at 
and the claim would be proved. But just use another angle chase, THIRD TIME:![\[\angle{MVI}=180-\angle{MIV}-\angle{VMI}=180-\angle{VCA}-\frac{\angle{ABC}}{2}=\angle{ABV}-\frac{\angle{ABC}}{2}=90\]](http://latex.artofproblemsolving.com/d/4/c/d4c011a8eff65098b7ec7d09569688b1dd0e675e.png)
and 
is a parallelogram. This means nothing else than 
.
,
,
and ![[asy]
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.
Let 
at 
bisects 
,
,
,
.![\[ \providecommand{\dang}{\measuredangle} \dang ZIX = \dang ZIB + \dang BIX = 90^{\circ} + \dang BVY = 90^{\circ} + \dang WIA = \dang WIX,\]](http://latex.artofproblemsolving.com/2/4/f/24f2e939ba56c71a1f489805cac33fd033d4a8e0.png)
as desired.
I think the whole reason why this took me so long was that I was trying to find ways to prove that 
.![\[ \angle BPU = \angle BAI = \angle A/2 = \angle BIC - 90^{\circ} = \angle BIU, \]](http://latex.artofproblemsolving.com/5/6/1/561cda5ef75ce8eb228b3cd9c9e1878fe371dc02.png)
as desired.
,
,
,
,
,
.
.
is an angle bisector in 
.![\[ \angle YBI = \angle PBI = \angle PUI = \angle XUI = \angle YVI, \]](http://latex.artofproblemsolving.com/e/4/d/e4de659c51b70fb5b7800d958411bd0a3928bcaa.png)
which implies the claimed concyclicity. The proof of the claim is easy from here: write ![\[ \angle BVI = \angle AYI = 90^{\circ} - \angle A/2 = \frac{1}{2}\cdot (180^{\circ}-\angle A) = \frac{1}{2} \angle BVC, \]](http://latex.artofproblemsolving.com/6/7/6/676c0c5d27fd2eb41de7ce6896e3fca5fff25819.png)
as desired.
.
is the symmedian of 
so 
![[asy]
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[/asy]](http://latex.artofproblemsolving.com/c/d/5/cd5c778cd979449715087a799bfbb3bd67be277c.png)
and 
.
and 
as,
from the parallel condition.
has center 
we have,
Then to show that 
note that 
as 
and 
bisects 
.
and equivalently 
be the intersection of 
with 
.
and 
.
Note then that 
has center 
with 
.
so we can rewrite 
as the perpendicular bisector of 
and delete 
where the last step follows from noting that under the aforementioned spiral similarities we have 
in isosceles 
.
.
as desired. Then we find that 
is the perpendicular bisector of 
.
as desired. 
.
and 
are isogonal with respect to 
.
at 
collinear. This is easy as,
but we also have,
so we are done. 
are concyclic.
,
.
,
is cyclic. Since 
,
.
are parallel, thus 
,
.
,
.
and let 
.
.
,
and 
is similar to 
.
,
and 
,
are collinear, as desired. 
be the midpoint of arc ![\[\frac{TI}{TV}=\frac{AI}{YV}=\frac{AX}{XV}=\frac{UI}{UV}\]](http://latex.artofproblemsolving.com/8/6/5/8658cbb62e916e55ec0f70d768cafa708fc05a40.png)
so 
.
.![\[\angle BIV=\angle BIC-90^\circ=\frac12 \angle BAC=\angle BAI=\angle BYV\]](http://latex.artofproblemsolving.com/4/0/6/4063ffc8be49f951df0c1a1db0a7dd5dcaddf05a.png)
so 
.
,
.
,
.
.
.
are collinear. Furthermore, 
.![\[\angle YVI=\angle YBI=\angle ABM_B=\angle AVM_B=\angle XVY'\]](http://latex.artofproblemsolving.com/c/f/c/cfc57c34877fd2c801a9ca75e0ca492ad77171c9.png)
so 
is symmedian in 
.
tangent to 
.
inversion, we can see that 
.
so we are done.
,
and suppose 
.
,
(because the angle + above equality).
!
(Ez to prove with trig).
cyclic.
Line 
;
.
.
.![\[\frac{KV}{KI}=\frac{VY}{IA}=\frac{XV}{XA}=\frac{UV}{UI}\implies -1=(K,U;V,I)\overset{B}{=}(A,C;V,BI\cap (ABC))\]](http://latex.artofproblemsolving.com/3/8/7/38722b59470bc8fab6863fd4a268b9a1ec92505e.png)
Since 
is the midpoint of the arc 
are collinear.![\[\measuredangle MIV=\measuredangle ADC-\measuredangle IUC=\frac{\measuredangle A}{2}+\measuredangle B-\frac{\measuredangle A+\measuredangle B}{2}=\frac{\measuredangle B}{2}\]](http://latex.artofproblemsolving.com/b/3/7/b379c5b2cee9f90fcb586c60a1a0d1c226e06d97.png)
![\[\measuredangle VMA=\measuredangle VCA=\frac{\measuredangle A+\measuredangle C}{2}\]](http://latex.artofproblemsolving.com/7/0/f/70f8f48f5870b04f47c3fbf0902d1491b92ba1d7.png)
Hence 
yields 
.
we conclude that 
are concyclic. 
gives 
.![\[\frac{TU}{TD}=\frac{MI}{MD}.\frac{VU}{VI}=\frac{AI}{ID}.\frac{XU}{AI}=\frac{XU}{ID}\]](http://latex.artofproblemsolving.com/7/b/1/7b15450847a03dd836cb9e1f8c9aabf5c5cdd3e8.png)
Thus, 
are collinear. 
,
yields 
.
we have 
.![\[\measuredangle WAI=\measuredangle IAW=\frac{\measuredangle C}{2}=\measuredangle ZIM\]](http://latex.artofproblemsolving.com/a/2/c/a2c99c88f40238965088289ef3a72b811880a2e3.png)
Thus, 
are collinear as desired
