A point $P$ lies inside $\usepackage{gensymb} \angle ABC(<90 \degree)$. Show that there exists a point $Q$ inside $\angle ABC$ satisfying the following condition: For any two points $X$ and $Y$ on the rays $\overrightarrow{BA}$ and $\overrightarrow{BC}$ respectively satisfying $\angle XPY = \angle ABC$, it holds that $\usepackage{gensymb} \angle XQY = 180 \degree - 2 \angle ABC.$
Problem
Source: 2024 Korea winter program practice test P1
Tags: geometry
math_comb01
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 $\measuredangle ABC$ 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
invt
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 $\measuredangle ABC$ 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
invt
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 $\measuredangle ABC$ 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?
Seungjun_Lee
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 $\angle APB + \angle CPD = 180^{\circ}$. Now, since $B,C$ has no meaning otherthan its angle, set $B$ and $C$ so that $BP \perp AC$ and $CP \perp AB$. Now, let $Q$ be the circumcenter of $ABC$, which is the isogonal conjugate of $P$ in $ABC$. Now, if $\angle XPY = \angle BAC$, it means that $\angle XPY + \angle BPC = 180^{\circ}$, 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 $\angle XQY = 180^{\circ} - 2\angle BAC$ as $\angle BQC = 2\angle BAC$. Also, since $\angle BAC$ is fixed, it is easy to prove this with moving points, where $X$ is a variable point. (Guessing $Q$ will be the main part)