Let $ABCD$ be an isosceles trapezium inscribed in circle $\omega$, such that $AB||CD$. Let $P$ be a point on the circle $\omega$. Let $H_1$ and $H_2$ be the orthocenters of triangles $PAD$ and $PBC$ respectively. Prove that the length of $H_1H_2$ remains constant, when $P$ varies on the circle.
Problem
Source: BdMO 2023 Higher Secondary National P4
Tags: geometry, orthocenter
TheMathBob
13.02.2023 20:30
Am I wrong or it is trivial? We don't even need the isosceles trapezium. $A,B,C,D$ lying on $\omega$ are enough. Because putting in the complex, where $\omega$ is the unit circle, we get $H_1 = p + a + d$ and $H_2 = p + b + c$, so the length$|H_1H_2| = |a+d - b - c|$ which does not depend on $p$
v4913
14.02.2023 22:55
By angle chasing $(ADH_1), (BCH_2)$ are reflections of $(ABCD)$ over $AD, BC$ and $\widehat{AH_1} + \widehat{BH_2} = \widehat{DABC}$ so there is a point $M$ on $\widehat{AB}$ such that $AMDH_1, BMCH_2$ are parallelograms, thus by homothety at $M$, $H_1H_2 = 2 \cdot M_1M_2$ where $M_1, M_2$ are midpoints of $AD, BC$.
rudransh61
29.11.2024 13:42
TheMathBob wrote: Am I wrong or it is trivial? We don't even need the isosceles trapezium. $A,B,C,D$ lying on $\omega$ are enough. Because putting in the complex, where $\omega$ is the unit circle, we get $H_1 = p + a + d$ and $H_2 = p + b + c$, so the length$|H_1H_2| = |a+d - b - c|$ which does not depend on $p$ I also thought the same .... it was i guess an easy one on the test