A point P lies inside \usepackagegensymb∠ABC(<90\degree). Show that there exists a point Q inside ∠ABC satisfying the following condition: For any two points X and Y on the rays →BA and →BC respectively satisfying ∠XPY=∠ABC, it holds that \usepackagegensymb∠XQY=180\degree−2∠ABC.
Problem
Source: 2024 Korea winter program practice test P1
Tags: geometry
01.02.2024 21:03
Let K,L be on BA,BC s.t. BPKL is parallelogram. Take the point Q on isogonal of P WRT ∡ABC such that Q lies on perpendicular bisector of KL. It is easy to see that this works. Remark : Everything heavily motivated by second isogonality lemma
13.10.2024 09:04
math_comb01 wrote: Let K,L be on BA,BC s.t. BPKL is parallelogram. Take the point Q on isogonal of P WRT ∡ABC such that Q lies on perpendicular bisector of KL. It is easy to see that this works. Remark : Everything heavily motivated by second isogonality lemma I can't get why it works
21.10.2024 15:26
math_comb01 wrote: Let K,L be on BA,BC s.t. BPKL is parallelogram. Take the point Q on isogonal of P WRT ∡ABC such that Q lies on perpendicular bisector of KL. It is easy to see that this works. Remark : Everything heavily motivated by second isogonality lemma I can't understand the part "It is easy to see that this works." Can anyone elaborate a bit, please?
21.10.2024 16:51
invt wrote: I can't understand the part "It is easy to see that this works." Can anyone elaborate a bit, please? I also don't think the first sol in #2 is clear enough, I will post a solution I know, more intuitive i guess. We use the following fact, well known. Fact. In a quadrilateral ABCD, an isogonal conjugate of P wrt ABCD exists iff ∠APB+∠CPD=180∘. Now, since B,C has no meaning otherthan its angle, set B and C so that BP⊥AC and CP⊥AB. Now, let Q be the circumcenter of ABC, which is the isogonal conjugate of P in ABC. Now, if ∠XPY=∠BAC, it means that ∠XPY+∠BPC=180∘, which we can use our Fact which implies that Q is the isogonal conjugate of P in a quadrilateral XYBC. Hence, Q has an isogonal conjugate in XYBC, which is equivalent to ∠XQY=180∘−2∠BAC as ∠BQC=2∠BAC. Also, since ∠BAC is fixed, it is easy to prove this with moving points, where X is a variable point. (Guessing Q will be the main part)