Let ω be a semicircle with AB as the bounding diameter and let CD be a variable chord of the semicircle of constant length such that C,D lie in the interior of the arc AB. Let E be a point on the diameter AB such that CE and DE are equally inclined to the line AB. Prove that (a) the measure of ∠CED is a constant; (b) the circumcircle of triangle CED passes through a fixed point.
Problem
Source:
Tags: geometry
29.10.2023 13:57
lol IMO 1985 solution works
29.10.2023 14:45
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 ∠DOC is constant since ¯CD is given to be constant. We will actually show later in the solution ∠CED=∠COD. Introduce X\coloneqqCA∩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 △XAB. This is sufficient to conclude the problem now. Observe that ∠CED=∠CODwhich is enough(comes from cyclic quadrilateral CEOD) to prove the first part. The second part is immediate, since O always lies on ⊙(CED) and is independent of the position of CD on ω. ◼
29.10.2023 15:33
Complete the semicircle, reflect C,D over AB and introduce the center, after easy chase you get ∠CEA=∠DEB=90∘−∠COD2.
29.10.2023 15:45
MrOreoJuice wrote: Complete the semicircle, reflect C,D over AB and introduce the center, after easy chase you get ∠CEA=∠DEB=90∘−∠COD2. Bro why so much reflection and stuff xDD, also hi again sir Define phantom point E′=(COD)∩AB where O is the center of ω Easy to see that ∠CEA=∠OCD=∠ODC=∠DEB which means E=E′ This means that ∠CED=∠COD which must be fixed as the length of CD is constant ◼ Done!
29.10.2023 16:53
MrOreoJuice wrote: Complete the semicircle, reflect C,D over AB and introduce the center, after easy chase you get ∠CEA=∠DEB=90∘−∠COD2. And then do p4 lolmaxx xx Also the part b should be given to prove before part a
29.10.2023 17:18
Let P=AC∩BD. Then E is the foot of the altitude from P to AB, so (CED) is the NPC of △PAB which always passes through the midpoint O of AB, and we're given ∠COD is constant.
29.10.2023 17:30
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 ∠COD is fixed as the length of CD is fixed. ∠CTD=∠COD2 and ∠CED=2∠CTD=∠COD⟹O,E,C,D are concyclic
29.10.2023 20:05
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 ∠CED=∠COD is constant.
31.10.2023 14:37
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∈(CED) which is the fixed point. Part A Consider the variable chord CD at 2 locations say C1D1,C2D2 and consider the diagram below. C1D1=C2D2 and OC1=OC2=OD1=OD2⟹ΔC1OD1≅ΔC2OD2 by SSS congruence criterion . So we have ∠C1E1D1=∠C1OD1=∠C2OD2=∠C2E2D2. Hence ∠CED is constant.
05.11.2023 07:19
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≡(CED)∩AB angle chasing gives ∠OCD=∠ODC⟹OC=OD.So O lies on the ⊥ 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 ∠CED=∠COD since C,D lie on the same semicircle ∠COD≱.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.
15.03.2024 15:17
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!
18.10.2024 06:30
I have discussed this question in my RMO 2023 video on my channel "little fermat". Here is the Video