interesting if 
is composite and for every divisor 
of 
and 
is also a divisor of 
to 
: 
distinct points on a circle. Consider the set of distances between any two out of the 
Y by centslordm, Aritra12, HWenslawski, megarnie, deplasmanyollari, Rounak_iitr, cubres, Mathandski, ehuseyinyigit
Let 
be a circle with centre 
, and 
a convex quadrilateral such that each of the segments 
and 
is tangent to . Let 
be the circumcircle of the triangle 
. The extension of 
beyond 
meets at 
, and the extension of 
beyond 
meets at 
. The extensions of 
and 
beyond 
meet at 
and 
, respectively. Prove that ![\[A D+D T+T X+X A=C D+D Y+Y Z+Z C.\]](//latex.artofproblemsolving.com/8/2/6/8261276053071cc6cc06dda824a883426bf1d4ac.png)
Proposed by Dominik Burek, Poland and Tomasz Ciesla, Poland
be a circle with centre 
, and 
a convex quadrilateral such that each of the segments 
and 
is tangent to . Let 
be the circumcircle of the triangle 
. The extension of 
beyond 
meets at 
, and the extension of 
beyond 
meets at 
. The extensions of 
and 
beyond 
meet at 
and 
, respectively. Prove that ![\[A D+D T+T X+X A=C D+D Y+Y Z+Z C.\]](http://latex.artofproblemsolving.com/8/2/6/8261276053071cc6cc06dda824a883426bf1d4ac.png)
Proposed by Dominik Burek, Poland and Tomasz Ciesla, Poland
This post has been edited 1 time. Last edited by popcorn1, Jul 20, 2021, 9:56 PM
Which is equivalent to proving 
by using 
.
and 
Use the fact 
and 
,
,
and 
.
,
.
.
,
.
.
.
and that 
.
.
are all collinear. Let 
be the tangencies of 
.
;
;
;
.
Hence, 
,
.
Now, ![\[A D+D T+T X+X A=C D+D Y+Y Z+Z C\]](http://latex.artofproblemsolving.com/4/8/1/48154facf2671e8d8bad854a3ef5c7d2ed280d80.png)
is equivalent to![\[D V+A U+D T+T X+X A=CW+DS+D Y+Y Z+Z C\]](http://latex.artofproblemsolving.com/e/d/8/ed880f00c15da8eedfd501fe85916db5edd8ad1e.png)
or ![\[VT+TX+XU=WZ+ZY+YS\]](http://latex.artofproblemsolving.com/a/c/4/ac4ec0b3a65dc2ed4f19370cbac9bf3c5c0ff3dd.png)
. Which is true since 
,
,
.
is an isosceles trapezoid.
is the angle bisector of 
,
.
.
at 
,
(as 
and 
.
.
done.
,
.
.
.
,![\[ AD + DT + XA = CD + DY + ZC \]](http://latex.artofproblemsolving.com/1/f/8/1f8e76431a113edaf217bb8326eef35a8bf20697.png)
However, by Pitot, 
,![\[ BX + DT = DY + BZ \Leftrightarrow DT - DY = BZ - BX \ (\star) \]](http://latex.artofproblemsolving.com/0/f/7/0f76368681379530b9b1e346da65ecb7da3cc260.png)
By power of point, we have 
and 
.
,
by Pitot, it suffices to prove that![\[ \frac{DY}{DC} = \frac{BX}{BC} \]](http://latex.artofproblemsolving.com/0/c/7/0c774aa35719adc411afd653850d4c5ba0fb1e3f.png)
The easiest way to do this is by Law of Sine, which is![\[ \frac{DY}{DC} = \frac{\sin \angle DCY}{\sin \angle DYC} = \frac{\sin \angle TCY}{\sin \angle AYC} = \frac{\sin \angle XCB}{\sin \angle AXC} = \frac{\sin \angle BCX}{\sin \angle BXC} = \frac{BX}{BC} \]](http://latex.artofproblemsolving.com/d/6/f/d6f5dabce50355e40384e0e1b77443f25cb2b05f.png)
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to avoid confusion with the imaginary unit. Let 
at 
respectively, and let 
be the center of 
Assuming quadrilateral 
is positively oriented, we find
The other four lengths are found by transposing the coordinates 
,
,
.
like there is for the other three sides. So it remains to show that
Indeed, we compute
and so
This is indeed symmetric under the transposition 
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are the tangency points as shown above. Note that 
so 
.
so 
is a isosceles trapezoid. In particular 
,
are equal. But this follows by symmetry: the lengths of the tangents to 
be 
and the foot from 
be 
.
denote the radius of 
,
,
.
.![\[YD+DC+CZ = XA+AD+DT.\]](http://latex.artofproblemsolving.com/5/c/1/5c1dc842fb9a7040025b1e49d72de4568713f460.png)
Let the foot from 
be 
and the foot from 
be 
.
is equal to![\[YQ - QD + DC + ZS - SC = YQ+ZS.\]](http://latex.artofproblemsolving.com/3/4/c/34c9c2dbafbec83815d156645fa1f4f14cc472f8.png)
We analogously have 
.
,
at 
.
Similarly 
.
,
.
gives 
as desired.
.
Note that 
so 
.
at 
.
and 
are equidistant from 
and 
.
which is true.
and similarly 
By pitot's theorem we know that 
Now using PoP and Pitot's theorem again we get that this condition is equivalent to 
as 
and 
which finishes the proof 
But 
then 
.
hence 
then we have 
Similarly, we have 
.
or 
.
Therefore, 
or 
therefore 
we reduce the desired equality to proving 
,
,
,
,
.
.
and 
are reflections across the diameter from 
,
.
.
and 
.
is a supplement to 
,
.
We have
as desired.
be the antipode of 
is perpendicular to 
,
passes through 
both share a midpoint of ![\[ XA + AD + DT = ZC + CD + DY.\]](http://latex.artofproblemsolving.com/6/8/f/68f62439acc00fcdfc729d2e206630ab7a9d6e2f.png)
Letting 
denote the length of the tangent from ![\[ f(X) + f(T) = f(Y) + f(Z).\]](http://latex.artofproblemsolving.com/c/6/d/c6d8494d9b296574f395a254ee62b393d2c6a583.png)
Because 
and 
,
.
share a midpoint, so 
,
.
to both sides of this equation, it suffices to show that 
.
,
,
,
.
,
,
,
at 
,
,
,
respectively.
by symmetry.
as desired.
,
,
,
,
,
,
,
and 
are congruent and 
via angle chasing, and 
![\[\measuredangle XAT = \measuredangle XAP + \measuredangle PAT = \measuredangle BAI + \measuredangle ICT = \measuredangle IAD + \measuredangle BCI = \measuredangle IAY + \measuredangle ZAI = \measuredangle ZAY\]](http://latex.artofproblemsolving.com/6/d/6/6d6d5f07e49cf502163789e7a985cb949257f6a7.png)
Thus, arcs 
and 
subtend equal angles on 
we have 
it suffices to show that,![\[BX+DT=DY+BZ\]](http://latex.artofproblemsolving.com/0/e/b/0eb57c2ffdaa11abcabb066aa7bc428ff7f42139.png)
However note that by Power of a Point at points ![\[BX = \frac{BC}{BA}\cdot BZ \text{ and }DT = \frac{DA}{DC}\cdot DY\]](http://latex.artofproblemsolving.com/0/9/b/09bb9715ba3e38a4a401610b686cf97c0f366865.png)
Thus the desired rewrites to,
But this is clear since 
and![\[\measuredangle BAZ = \measuredangle XAZ = \measuredangle XCY + \measuredangle YCZ = \measuredangle XCY + \measuredangle TCX = \measuredangle TCY = \measuredangle DCY\]](http://latex.artofproblemsolving.com/b/4/6/b461c320578f21f647cbd9ca77834053f45c3b45.png)
![\[\frac{BZ}{BA}= \frac{\sin \angle BAZ }{\sin \angle BZA} = \frac{\sin \angle TAD}{\sin \angle CYD}=\frac{DY}{DC}\]](http://latex.artofproblemsolving.com/3/6/1/3619ed73dbd485a310895681dcd93bfc7c6afc7f.png)
which finishes the proof.
be the midpoint of the arc 
(not containing 
.
,
.
and 
.
.
.
(since 
.
,
.
and 
are congruent. Therefore, 
.
.
and thus is an external angle bisector of 
Thus, 
Similarly, 
which implies that 
Add 
to opposite sides, and we get 
or 
while
where the last step is by Pitot’s.
Note that 
by POP.
while 
where the last step follows from the fact that 
Simple LOS finishes.
