be positive real numbers such that : 
. Prove that : 
candies. A starts first. On a move, if in the box there are 
candies, the player chooses positive integer 
so that 
and 
,
.
such that 
and 
.
![$\ [\text{Blog Post 141}] $](http://latex.artofproblemsolving.com/e/2/e/e2e1563c16c5d526310409de610870dc615964d8.png)
Y by Davi-8191, HWenslawski, PRMOisTheHardestExam, Adventure10, Mango247, ItsBesi, Rounak_iitr
Let 
be a convex pentagon such that
![\[ \angle BAC = \angle CAD = \angle DAE\qquad \text{and}\qquad \angle ABC = \angle ACD = \angle ADE.
\]](//latex.artofproblemsolving.com/d/3/d/d3d9a82f4318190298a8f008d417a364e03f1fca.png)
The diagonals 
and 
meet at 
. Prove that the line 
bisects the side 
.
Proposed by Zuming Feng, USA
be a convex pentagon such that![\[ \angle BAC = \angle CAD = \angle DAE\qquad \text{and}\qquad \angle ABC = \angle ACD = \angle ADE.
\]](http://latex.artofproblemsolving.com/d/3/d/d3d9a82f4318190298a8f008d417a364e03f1fca.png)
The diagonals 
and 
meet at 
. Prove that the line 
bisects the side 
.Proposed by Zuming Feng, USA
This post has been edited 2 times. Last edited by djmathman, Jun 27, 2015, 12:15 AM
Reason: modified wording to reflect english version of ISL2006
Reason: modified wording to reflect english version of ISL2006
and
from which we obtain:
and
and hence
or otherwise
hence triangles 
and 
are similar. Now it's just angle chasing. Use this similarity to have some equal angles and see that 
and 
are cyclic and tangent to line 
and 
.
is intersection of AP and CD then 
so 
,
and factor 
that maps 
onto 
.
and 
are the intersection points of the diagonals of two quadrilaterals that are similar, we know that![\[\frac{\overline{FA}}{\overline{FC}}=\frac{\overline{GA}}{\overline{GC}}\]](http://latex.artofproblemsolving.com/5/c/a/5ca3c54f240e640bb0fa19d45faa56615afb90d4.png)
,![\[-1=\frac{\overline{FA}}{\overline{FC}}.\frac{\overline{SC}}{\overline{SD}}.\frac{\overline{GD}}{\overline{GA}}=\frac{\overline{SC}}{\overline{SD}}\]](http://latex.artofproblemsolving.com/2/5/a/25ac4d0e4d29ece28520a6f9407b8e04b8918050.png)
is the midpoint of 
are similar. Therefor, we can use the alternating segment theorem to prove that both 
and 
are tangent to the circumcircle of 
in points 
the intersection of 
is a symmedian of 
,
is cyclic and 
(1)
,
(2)
and 
we conclude that the quadrilateral 
is cyclic and so, 
we have that the quadrilateral 
is also cyclic.
we conclude that the segment line 
as a direct conclusion we have that 
and so, 
and 
.
.
we draw two lines parallel to 
respectively and we denote as 
their intersection point.
are isogonal conjugates with respect to the angle 
and similarly, the segment lines 
are isogonal conjugates wrt 
based on the 
we have that the segment line 
passes through the midpoint of the side segment 
is similar to triangle 
,
to 
,
for both circles, it is immediate that 
=
.
by AA similarity. Then a spiral similarity centered at 
,
and 
.
are cyclic, and line 
and 
.
implies 
,
,
.
and 
![\[ \frac{MD}{MC}=\frac{QD}{QA}\frac{RA}{RC}\]](http://latex.artofproblemsolving.com/f/b/a/fba9ae98729bef9fe77bf488a3ddefb0ee7ec9ff.png)
with line 
gives ![\[\frac{RA}{RC}=\frac{BC.DL}{BL.AD}\]](http://latex.artofproblemsolving.com/1/5/a/15a086d16d317dcd715929006592d45d289510ca.png)
with line 
gives ![\[\frac{QD}{QA}=\frac{CK.AE}{CD.KE}\]](http://latex.artofproblemsolving.com/e/7/f/e7fc5a1557f9a57c8de4e6326b7dd6791839e2ae.png)
and 
are similar so ![\[\frac{DL}{KE}=\frac{CD}{DE}\]](http://latex.artofproblemsolving.com/4/8/2/482031266c2cbf91fa187a5cfb763b81da3950af.png)
![\[\frac{BC}{DE}=\frac{AC}{AE}\]](http://latex.artofproblemsolving.com/3/8/b/38b48046d171c78b2cd7fba357ab88714f88596d.png)
and 
are similar, 
are respectively their angles bisector from 
so ![\[\frac{AC}{BL}=\frac{AD}{CK}\]](http://latex.artofproblemsolving.com/b/6/d/b6ddb9011386c54bbc0f60772f53edc6ca6d41d7.png)
![\[\frac{MD}{MC}=\frac{RA}{RC}\frac{QD}{QA}=\frac{BC.DL}{BL.AD}\frac{CK.AE}{CD.KE}=...=\frac{AC.CK}{BL.AD}=1\]](http://latex.artofproblemsolving.com/3/6/2/362379234ef58f9bd804f7819dd40376380280e9.png)
as required 
and let 
.
and so 
.
counter-clockwise, defining the rest, clearly, counter-clockwise too.
,
and 
(well, it's given that 
and 
,
the second intersection of 
are collinear (similar for 
)
and 
and look: 
.
,
and 
,
intersect 
and 
intersect 
at 
,
bisects 
.
be the intersections of 
.
is the center of the spiral similarity 
Hence 
lies on the circumcircles of 
and 
that is 
is tangent to 
and 
.
and it follows that 
Further, we have 
and this leads to 
by side-angle-side similarity. Angle chasing thus gives 
Since it is given that 
,
,
so 
by side-angle-side similarity, and 
.
,
.
,
In the same way we can prove 
colinear therefore 
and the claim is true.
have 
colinear by radical axis, thus we are done 
and 
are cyclic.By the easy angle chasing we can show that 
is common tangent of 
and 
.
is common tangent and 
so 
so this implies that 
and 
are cyclic quadrilaterals due to the well known spiral similarity lemma. Now some angle chasing gives us that 
and 
are similar.![\[\measuredangle DCB = \measuredangle DCA + \measuredangle ACB = \measuredangle EDA + \measuredangle ADC = \measuredangle EDC \]](http://latex.artofproblemsolving.com/1/d/d/1dd3d156bcfffff5c35c6b942b0c1ed308f0f7ef.png)
Further since ![\[\frac{BC}{CD}=\frac{AC}{AD}=\frac{CD}{DE}\]](http://latex.artofproblemsolving.com/5/5/4/55441a4aa2e994601dd70ce2fb9cf5523bdd18d7.png)
which implies that 
as desired.![\[\measuredangle DPE = \measuredangle DEP + \measuredangle PDE = \measuredangle DEC + \measuredangle BDE = \measuredangle DEC + \measuredangle CDE + \measuredangle BDC = \measuredangle CDE = \measuredangle DAE \]](http://latex.artofproblemsolving.com/8/c/6/8c666e663d361677a56b131c6869255ffaab20f5.png)
which indeed implies that ![\[\measuredangle PDC = \measuredangle BDC = \measuredangle CED = \measuredangle PED\]](http://latex.artofproblemsolving.com/e/4/c/e4c146b32020e2b79a8d45f795da140473efb0d5.png)
so 
at 
at 
it follows that line 
.
is the pole of 
since 
and 
Hence 
by DDIT on 
at 
which will immediately prove the desired result by PoP.
implies that 
then 
and 
,
implies 
.
,
.
afterwards.![\[
[APM]=\frac{i}{4}
\begin{pmatrix}
a & \overline{a} & 1\\
p & \overline{p} & 1\\
m & \overline{m} & 1
\end{pmatrix}
=0.
\]](http://latex.artofproblemsolving.com/0/3/3/033fe09037538ef11f847e287132514026dfce3b.png)
it implies that 
are collinear.
It suffices to show that 
are concurrent. Let 
Then by ratio of areas ![$$\frac{AX}{XC} = \frac{[\triangle ABD]}{[\triangle BCD]}, \frac{AY}{YD} = \frac{[\triangle ACE]}{[\triangle CDE]}.$$](http://latex.artofproblemsolving.com/e/3/e/e3e3c5aa99cb6b64257536a79567a04b4daecceb.png)
Now, observe that 
say with ratio 
Thus ![$$\frac{CX}{XA} \cdot \frac{AY}{YD} \cdot \frac{DM}{MC} = \frac{[\triangle BCD]}{[\triangle ABD]} \frac{[\triangle ACE]}{[\triangle CDE]} = \frac{r^2}{r^2} = 1,$$](http://latex.artofproblemsolving.com/1/a/4/1a487ceb079e9c7b04b77291c29add0569237377.png)
and we are done by Ceva's Theorem. QED
