lol IMO 1985 solution works

Well, I'm amazed this made a contest and wasn't a total flop. I wouldn't have solved it in contest though, am I a hypocrite? Solution: Call the center of the semi-circle as $O$. We obviously have that $\angle DOC$ is constant since $\overline{CD}$ is given to be constant. We will actually show later in the solution $\angle CED = \angle COD$. Introduce $X \coloneqq CA \cap DB$ and call its foot on $AB$ as $P$. [asy][asy] import olympiad; size(200); pair O = (0,0); pair A = (-2,0); pair B = (2,0); //yeah, C and D are randomly taken from geogebra; pair C = (-1.61, 1.19); pair D = (0.44, 1.95); pair X = extension(A,C,B,D); pair E = foot(X,A,B); // so, A is center, and place B,A in a directed angle way; path circ = arc(O,B,A); draw(A--X, red); draw(B--X, red); draw(A--B, red); draw(C--E, red); draw(D--E, red); draw(circumcircle(C,E,O), magenta+dashed); draw(circ, blue); markscalefactor = 0.03; draw(anglemark(B,E,D), deepgreen); draw(anglemark(C,E,A), deepgreen); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$O$", O, S); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$X$", X, dir(X)); dot("$E$", E, dir(S)); [/asy][/asy] It is not very hard to prove that $E$ is unique on the line $AB$. But observe that $P$ satisfies the same angle condition(recall the orthocenteric configuration), so we can actually conclude that $E = P$. From here, we get that $EODP$ is cyclic since its the nine-point circle of $\triangle XAB$. This is sufficient to conclude the problem now. Observe that \[\angle CED = \angle COD\]which is enough(comes from cyclic quadrilateral $CEOD$) to prove the first part. The second part is immediate, since $O$ always lies on $\odot(CED)$ and is independent of the position of $CD$ on $\omega$. $\blacksquare$

Complete the semicircle, reflect $C,D$ over $AB$ and introduce the center, after easy chase you get $\angle CEA = \angle DEB = 90^\circ - \frac{\angle COD}{2}$.

MrOreoJuice wrote: Complete the semicircle, reflect $C,D$ over $AB$ and introduce the center, after easy chase you get $\angle CEA = \angle DEB = 90^\circ - \frac{\angle COD}{2}$. Bro why so much reflection and stuff xDD, also hi again sir Define phantom point $E'=(COD) \cap AB$ where $O$ is the center of $\omega$ Easy to see that $\angle CEA=\angle OCD=\angle ODC=\angle DEB$ which means $E=E'$ This means that $\angle CED=\angle COD$ which must be fixed as the length of $CD$ is constant $\blacksquare$ Done!

MrOreoJuice wrote: Complete the semicircle, reflect $C,D$ over $AB$ and introduce the center, after easy chase you get $\angle CEA = \angle DEB = 90^\circ - \frac{\angle COD}{2}$. And then do p4 lolmaxx xx Also the part b should be given to prove before part a

Let $P = AC\cap BD$. Then $E$ is the foot of the altitude from $P$ to $AB$, so $(CED)$ is the NPC of $\triangle PAB$ which always passes through the midpoint $O$ of $AB$, and we're given $\angle COD$ is constant.

This question is so easy yet so beautiful, I am sure there exist at least 20 different solutions to it Reflect D about AB, call it T. Observe that $A,B,C,D,T$ are concyclic and $C,E,T$ are collinear $\angle COD$ is fixed as the length of $CD$ is fixed. $\angle CTD = \frac{\angle COD}{2}$ and $\angle CED = 2\angle CTD = \angle COD \implies O,E,C,D$ are concyclic

Even simpler: Let $O$ be the center. Note that $OC = OD$ and $EO$ is external angle bisector. So $(CED)$ goes through fixed point $O$ and $\angle CED = \angle COD$ is constant.

Sol:- Part B:- Let $O$ be the center of the circle.Equal inclination implies $EO$ is external angle bisector of $CED$ and since $OC=OD$ implies $O$ lies on perpendicular bisector of $CD$. This implies $O$ is midpoint of arc $CED$ and hence $ O \in (CED)$ which is the fixed point. Part A Consider the variable chord $CD$ at $2$ locations say $C_1D_1, C_2D_2$ and consider the diagram below. $C_1D_1=C_2D_2$ and $OC_1=OC_2=OD_1=OD_2 \implies \Delta C_1OD_1 \cong \Delta C_2OD_2$ by $SSS$ congruence criterion . So we have $\angle C_1E_1D_1=\angle C_1OD_1=\angle C_2OD_2=\angle C_2E_2D_2$. Hence $\angle CED$ is constant.

Looking all the reflection and intersection solutions here I had the urdge to post this T_T how people think so complex? For part B, Define $O \equiv (CED) \cap AB$ angle chasing gives $\angle OCD=\angle ODC \implies OC=OD$.So $O$ lies on the $\perp$ bisector of $CD$ again $O$ lies on the diameter as well . Since the centre lies on the perpendicular bisector of any chord and on the diameter and 2 straight lines can intersect at only one point, we have $O$ is the circumcentre. For part A,see that $\angle CED=\angle COD $ since $C,D$ lie on the same semicircle $\angle COD \not \geq 180$.Cosine rule on $\triangle COD$ gives $\cos \angle COD=\frac{CD^2}{2R^2}-1$ so $\cos \angle COD$ is constant,since $\angle COD \leq 180,\angle COD$ is constant as well.So is $\angle CED$ $\blacksquare$ L567 wrote: Let $O$ be the center. Note that $OC = OD$ and $EO$ is external angle bisector. So $(CED)$ goes through fixed point $O$ and $\angle CED = \angle COD$ is constant Yeah exactly , the moment I saw this I knew I had to fact 5 this out.Dk how people are motivating reflections and other things.

Firstly, let $\angle CEA= \angle DEB= \theta$. Now, we reflect point $C$ about line $AB$. Then, by properties of reflection, $C'$ lies on the circle and $\angle AEC'=\theta$. As the vertically opposite angles are equal, i.e, $\angle DEO= \angle AEC'$, we get that $C', E$ and $D$ are collinear. Let $CC'$ meet $AB$ at $F$. Then, $\angle EFC'=90^{\circ}$. Then, by angle sum property in $\Delta FEC'$, we get that $\angle FC'E= \angle AC'D= 90^{\circ}-\theta \implies$ angle subtended by chord $CD$ on the circumference $=90^{\circ}-\theta$. Therefore, $\angle COD=2(90^{\circ}-\theta)=180^{\circ}-2\theta$. Again, $\angle CEA+ \angle DEB +\angle CED=180^{\circ}$ (angles on a straight line). $\angle CED=180^{\circ}-2\theta$. But, then $\angle CED= \angle COD\implies$ the circumcircle of $\Delta CED$ passes through the centre of the circle, which is a fixed point. Also, as $CD$ is a chord of constant length, the value of $90^{\circ}-\theta$ is constant $\implies$ the measure of $\angle CED$ is also constant, and we're done!

I have discussed this question in my RMO 2023 video on my channel "little fermat". Here is the Video