Let ABCD be a cyclic quadrilateral with circumcenter O. The internal angle bisectors of ∠DAB, ∠ABC, ∠BCD, ∠CDA create a convex quadrilateral Q1. The external bisectors of the same angles create another convex quadrilateral Q2. Prove Q1, Q2 are cyclic, and that O is the midpoint of their circumcenters.
Problem
Source: Israel TST 2 2025 p3
Tags: geometry, cyclic quadrilateral, circumcircle, angle bisector
20.12.2024 05:55
It is well known that Q1,Q2 are cyclic. (Follows directly from easy angle chase) Let O1,O2 be the circumcenter of Q1,Q2, respectively. When M is the midpoint of O1O2, we will prove that M is the circumcenter of ABCD. Assume that the angle bisecters of ∠DAB, ∠CDA meets at X,X1, and the angle bisectors of ∠ABC, ∠BCD meets at Y,Y1. (X1,Y1 are the intersection of external bisectors) Defiine Z,W similarly with X,Y so that Q1 is XZYW. When X2 and Y2 are the midpoints of XX1 and YY1, X2 and Y2 are each circumcenters of AXDX1 and BYCY1. Also, since Q1 and Q2 are similar quadrilaterals, it is easy to see that O1X∥O2X1, and O1Y∥O2Y1. This implies that MX2∥O1X and MY2∥O1Y. It suffices to prove that O1X⊥AD, as this gives that M=O since X2 lies on the perpendicular bisector of segment AD, and MX2⊥AD means that M lies on the perpendicular bisector of AD. Doing this all four sides give the desired result. Let ∠O1XZ=α, ∠O1ZY=β, ∠O1YW=γ, and ∠O1WX=δ. We can see that ∠XZY=α+β, and it goes the same with other angles. Using the relations like ∠XZY=π−12(∠CDA+∠BCD), we can derive to a conclusion that gives a linear equation about α,β,γ,δ where the constants are only the four angles of ABCD. Solving this, we can easily get α=π2−∠CDZ, which implies that O1X⊥AD, as desired.
04.01.2025 07:50
mb i am on the train on the phone so.i will sketch solution..i also dont have diagram on paper so if my memory is bad oops i will change latwr at home lemma: if a, b are lines that pass through two distinct fixed point at infinity and moving at a constant speed each then their intersection moves at a constant speed. proof: let lines be l1 and l2. consider a fixed line m perpndicular to l1 and n perpendicular to l2. note that m intersect l1 and n intersect l2 also move at fixed speed because l1 at time t is parallel to l1 at time t' and use addendo in parallel lines with transversal m and the line/vector that l1 is moving in, same for l2. now an affine transform such l1 l2 are perpendicular means we can set m,n as directions of x,y axis such m intersect n is origin. conclusion is now obvious (say bc tan^-1 speed l1/speedl2 is fixed) now we angle chase akin to one of my previous post on an rmm 3 or look at above. it is easy to see that the diagonals of Q1 and Q2 coincide and are perpendicular. now take with reference to Q2. well-known Q1 negatively similar to Q2. thus we define map from Q2 to.Q1 by reflecting across one.diagonal of Q2 then dilating at intersection of diagonals. now on for convenience let PQRS be Q2 and transformation involves reflect in PR wlog. then suppose P goes to P'. etc. now move P' constant speed i.e on PR easy to see that A,B.are feet of S' on PQ, QR and C,D feet of Q' onto RS,SP. now clearly CDSQ' cyclic. hence check the slope note direction of Q'D which is fixed. because rightangle to PS. now CDQ' = CSQ by bowtie whcih is fixed. hence CD is fixed slope. similarly AB fixed slope. now by our lemma, perp bisector of AB, CD are fixed slope hence intersection, which is centre of (ABCD), moves on a line. now by similarity, Q1 has P' moves just expands in size so centre moves on a line through PR intersect QS now centre of Q2 is just fixed. now this means the points are deg 1,1,0 hence dilating from centre of Q2 by 2 the centrw of ABCD we get our phantom point which is deg 1. this means we check (1+1)+1=3 cases by combi null lol. use P' as PR intersect QS, P and P reflect over PR intersect QS. ill update soon why rhis wokrs im home slay, now we deal with the special cases. P'=P: direct symmetry of Q1 and Q2 over PR, so the midpoint is true because (ABCD) has centre on PR by symmetry. P' = P reflect over PR intersect QS: same as P=P', symmetry over QS P=PR intersect QS. Let this point be X. because PXQ and RXS are right, they add to 180 so X is on the isoptixook cubic of PQRS hence skibdii isognonal conjugable. the conjugate is centre of (PQRS) because of altitude and cirucmcentre isogonals. yet as Q1 is now just 0 radius so centre is X, we use SIX POINT CIRCLE (xonk) lemma so midpoint of circumcentres of Q1 Q2 is centre of PEDAL CIRCLE xoink!!!! now we are done
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04.01.2025 11:15
as a sidenote because this problem generalizes the six point circle configuration, #2's synth solve shows shows an elementary proof of midpoint of isogonal conjugates is centre of pedal circle