Note that X is the excenter of CDE by construction. So if Y is the foot of X to DE, and CD=a, CE=b,DE=c then YD=p--b, YE=p-a, we only need to prove now that YO is perpendicular to DE. OY and DE are perpendicular if and only if OD^2-OE^2=YD^2-YE^2, we know how much is YD and YE in terms of a,b,c, so we know the right side of the equation. If M,N are the midpoints of CA nad CB, then MD^2=[MA-DA]^2=[c-a]^2/4, analogously NE= [c-b]^2:4. OD^2=DM^2+a^2=EM^2+b^2, DM^2-EM^2=b^2-a^2, but OD^2-OE^2= [DM^2-EM^2]+[MD^2-NE^2], BUT WE ALSO KNOW THE VALUE OF THIS Parenthesis in terms of a,b,c, with a little bit computational we end the problem.

What a beautiful problem, so that just deserves a meh. Let: $a=BC$, $b=DE$, $c=AB$, $d=EB$, $e=DB$, $p=\dfrac{d+e+b}{2}$, $\alpha = \angle OAB$, $\gamma = \angle OCB$, $r=OA=OB=OC$. Comments:

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Hint 1/Lemma 1:

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Hint 2:

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Solution: One may reduce $(i)$ using $(ii)$ to $DO^2+R^2+(p-d)^2=EO^2+R^2+(p-e)^2 \Leftrightarrow DO^2-EO^2=p^2-2pe+e^2-p^2-2pd+d^2=e^2-d^2+2.e.\dfrac{b+d+e}{2}+2.d.\dfrac{d+e+b}{2}=bd-eb=b(d-e)$, and this simplifies everything! Now we just need $DO^2-EO^2=b(d-e)$ $(iii)$, and by the side condition we get $b(d-e)=b(A-C)$. After this I was unable to do something related to the circumcenter. Then, I had the (not so brilliant) idea of using cossine law on triangles $ODA,OEC$, getting: $DO^2=r^2+b^2-2br.cos\alpha,$ $EO^2=r^2+b^2-2br.cos\gamma,$ hence subtracting the relations we get that $DO^2-EO^2=2br(cos\gamma-cos\alpha)$, and then because of $(iii)$ we have to prove that $2br(cos \gamma - cos \alpha)=b(a-c)$ $\Leftrightarrow 2r=(a-c)(cos \gamma - cos \alpha)$, and if $P$ is the feet of the $B$ altitude, we easily get: $2r=\dfrac{a.c}{BP}$, and $cos\gamma=\dfrac{BP}{c}=\dfrac{a/2}{r}$, $cos\alpha=\dfrac{BP}{a}$, and then we just need to have: $a.c(\dfrac{1}{c}-\dfrac{1}{a})=a-c$, which is true. $\blacksquare$

I actually went to this contest years ago. At the time i was not to good in geometry so I didnt get it right. So the problem can go like this: Lets call F the foot of X to DE. lets say FD = b and FE = a. then we know that DA = a + b and DC = a + b. Because the excircle is tangent to BA and BC we can see that BD = Y - b. (where Y is the distance from B to the tangency point of BD with the excircle) similarly BE = Y - a (Where Y is the same distance but this time to the tangent that forms with BE and by tangency both Y we mentioned are equal). Lastly we extend DE to intersect the cricumcircle of ABC in points P and Q so that the points in DE are in the order P - D - E - Q. then we define PD = p and QE = e. Now, we use power of point on points D and E and we get to the following: p(a + b + q) = (Y - b)(a + b) q(a + b + p) = (Y - a)(a + b) As we see, there is a lot of common things so we substract both: (p - q)(a + b) = (Y - b - Y + a)(a+b) by simplifying and reareanging we get to p + b = q + a We now that PF = p + b and QF = q + e then, F is the midpoint of PQ and XF is perpendicular to PQ, then XF is the mediatrix of PQ and PQ are points on the circuncircle of ABC, then O must live on the segment XF, and we get to the result. OX is perpendicular to PQ which is just an extension of DE, the OX is perpendicular to DE. Although is a pretty solution its kind of random. You just plug the values on the power of point and you just get the right result. So no my favourite problem but its OK.

Let $AE\cap DC=Y$. $Y$ is the orthocenter of $XDE$. Note that $X$ is the $B-$excenter on $BDE$. We have $XA=XE$ and $XD=XC$ and $AD=EC$ which gives us that $XCE \cong XDA$. $\angle YAD=\angle DEY=\angle YXD$ so $A,X,Y,D$ are cyclic. $\angle ECD=\angle YDE=\angle EXY$ so $C,X,Y,E$ are cyclic. Also because of $XCE \cong XDA$, we know that radius of $(XCEY)=\omega_1$ which is $R_1$ and radius of $(XDAY)=\omega_2$ which is $R_2$ are equal. Let $O_1$ be the circumcenter of $\omega_1$ and $O_2$ be the circumcenter of $\omega_2$. $\angle ECO_1=90-\angle EAD-\angle ECD$ and $\angle BCO=90-\angle A$ so $\angle OCO_1=\angle A-\angle EAD-\angle ECD$ $\angle OAB=90-\angle C$ and $\angle O_2AD=90-\angle EAD-\angle ECD$ so $\angle OAO_2=\angle EAD+\angle ECD-\angle C$ $\angle OCO_1=\angle A-\angle EAD-\angle ECD=\angle EAD+\angle ECD-\angle C=\angle OAO_2$ and $OA=R=OC$ and $O_1C=R_1=O_2A$ so $OAO_2\cong OCO_1$. $Pow_{\omega_1}(O)=OO_1^2-R_1^2=OO_2^2-R_2^2=Pow_{\omega_2}(O)$ so $O$ lies on the radical axis of $\omega_1$ and $\omega_2$ which is $XY$. $X,O,Y$ are collinear and $XY \perp DE$ gives us that $XO \perp DE$ as desired.

The problem is equivalent to the following, viewed with reference triangle $ADE$. Restated wrote: Let $\triangle ABC$ have $A$-excenter $I_A$. Let $D$ and $E$ be points such that $A,B,D$ and $A,C,E$ are collinear in that order, and $BC=BD=CE$. Let $O$ be the circumcenter of $\triangle ADE$. Prove that $\overline{OI_A} \perp \overline{BC}$. We use lengths. Let $a,b,c,s,r_A$ be defined as usual, and let $R$ be the radius of $(ADE)$. We have $$I_AB^2-I_AC^2=(r_A^2+(s-c)^2)-(r_A^2+(s-b)^2)=(2s-b-c)(b-c)=ab-ac.$$Now we calculate $$OB^2=d(O, \overline{AD})^2+\left(\frac{a-c}{2}\right)^2.$$On the other hand, we have $$R^2=OA^2=d(O,\overline{AD})^2+\left(\frac{a+c}{2}\right)^2 \implies OB^2=R^2-ac.$$Since a similar relation holds for $OC^2$, it follows that $OB^2-OC^2=ab-ac$, and we're done by the perpendicularity lemma. $\blacksquare$