and 
be positive integers with 
.
,
.
Y by awesomehuman, aansc1729, mathmax12, peelybonehead, GeoKing, Rounak_iitr, GoodMorning, Spiritpalm, Mogmog8, Tastymooncake2, centslordm
Isosceles triangle 
, with 
, is inscribed in circle 
. Let 
be an arbitrary point inside 
such that 
. Ray 
intersects again at 
(other than 
). Point 
(other than ) is chosen on such that 
. Line 
intersects rays 
and 
at points 
and 
, respectively. Prove that 
.
Proposed by Anton Trygub
, with 
, is inscribed in circle 
. Let 
be an arbitrary point inside 
such that 
. Ray 
intersects again at 
(other than 
). Point 
(other than ) is chosen on such that 
. Line 
intersects rays 
and 
at points 
and 
, respectively. Prove that 
.Proposed by Anton Trygub
This post has been edited 2 times. Last edited by brianzjk, Mar 23, 2023, 11:12 PM
a typo?
be projections of 
correspondingly. Clearly, points 
lie on a single circle with diameter 
.
,
lie on a single circle. Similarly, points 
lie on a single circle.
.
,
.
be the harmonic conjugate of 
be the point on 
such that 
.
with respect to 
if and only if there exists a point ![\[\measuredangle AEB=90^{\circ},\, \measuredangle CEB=\measuredangle BED.\]](http://latex.artofproblemsolving.com/c/8/1/c81264bcef52fc0df1ec43d896a23f6fe12c1eeb.png)
Since 
we find that 
.
through 
are collinear. Draw 
,
we only need to show that 
are collinear. But now![\[\measuredangle HFE=\measuredangle HDE=\measuredangle BDE=\measuredangle ACE=\measuredangle AFE,\]](http://latex.artofproblemsolving.com/7/3/5/735d11de2f7229873f6f7385650da1c93fbbb681.png)
done. 
and 
.
is tangent to 
,
.
passes through the minor arc midpoint 
of 
,![\[-1=(AM;BC) \stackrel D= (KE;BC) \stackrel A=(PE;XY).\]](http://latex.artofproblemsolving.com/2/6/7/26720c9856febdd47806ee22113779b6707c7e7d.png)
Bisector bundle lemma finishes.
?
and 
intersect 
at 
respectively not equal to 
,
.
,
are cyclic quadrilaterals. However 
therefore the radiuses of 
,
are equal and are thus reflections over 
so we can conclude.
and 
are congruent since 
by LoS. So it then follows that 
be 
,
.
so 
and similarly 
,
is perpendicular to 
.
.
.
we get 
and by PoP 
and by similar triangles again since 
,
is the projection of 
.
where 
is the 
as desired.
.
from right angles.
through 
at 
.
are collinear, as then 
and Appolonian circles finishes.
is divided by 
.
,![\[\frac{DQ\times \sin\theta}{AD\times\sin 90^\circ}\]](http://latex.artofproblemsolving.com/3/3/b/33bd3558c8918b852397012a2c88189cb305b7b8.png)
and the second ratio, by ratio lemma on 
,![\[\frac{EQ\times \sin\alpha }{EA\times \sin{90^\circ-\alpha}}\]](http://latex.artofproblemsolving.com/f/9/f/f9f075323fb2813d98301e097043fa97247b4683.png)
We equate the two, if equality holds working backwards we would be done.
This is true if 
,![\[\angle ADC=\angle EA'A=\angle ECA\]](http://latex.artofproblemsolving.com/2/2/1/221bc19bb04f170279747ca38597a8e72d2d2cf2.png)
But this is true clearly, by Shooting Lemma.
and 
intersect 
at 
,
,
are concurrent at 
is cyclic. Similarly, 
is as well. Now note that![\[ \text{radius}[(ADPB)] = \dfrac{AD}{2 \sin(\angle ABC)} = \dfrac{AD}{2 \sin(\angle ACB)} = \text{radius}[(ADQC)]\]](http://latex.artofproblemsolving.com/d/f/5/df5dd4612ed9d16da821a6d3bfca54ae700c0c26.png)
so 
and 
are symmetric about 
.
.![\[\angle XDE = \angle ABP = \angle ABC + \angle PBD = \angle ACB + \angle QCD = \angle ACQ = \angle EDY \]](http://latex.artofproblemsolving.com/a/1/1/a11296c059c29869dfe560a5c8f91cd1234cecac.png)
as desired.
.
.
. Notice that 
is tangent to the circle with diameter 
.
is right angled at 
so 
passes through midpoint of arc 
which is what we needed to show.
at 
,
.
.![\[(XY;KE) \overset{A}{=} (BC;EK') \overset{D}{=} (C,B;A,DK' \cap (ABC)),\]](http://latex.artofproblemsolving.com/b/4/2/b42e147a12738c40430edbdfd135de0e61963059.png)
to be the midpoint of arc 
.
,
,![\[KD^2 = KE \cdot KF = KA \cdot KK'. \quad \blacksquare\]](http://latex.artofproblemsolving.com/9/8/4/9840d4f37738a10a347c3f3f3bc255744e992a92.png)
.
and 
are cyclic, as
,
,
,
meet 
and 
meet 
in 
respectively.
.
.
and 
imply 
and 
.
.
gives that 
.
,
.
,
to 
(where 
.![\[\angle AES^\ast=\angle AEB=\angle AEC=\angle AET^\ast\]](http://latex.artofproblemsolving.com/b/1/a/b1aa82de5d28aae0de0ae8250f2606b39d49074f.png)
and 
,
and 
are reflections over 
meet 
.
,
.
So 
.
as the "main triangle". The new problem is:
not containing 
.
.
harmonic.
,
we have 
and 
,
and 
and 
are symmetric about 
so we are done.
so 
and 
are tangent.
we have 
passes through the antipode of 
which is the arc midpoint.
tangent to 
Do the same for 
.
are concurrent. Furthermore, this implies 
Simply project through 
