Advanced Geometric Problem & Solution (Twice a week update) Now 4921 problems posted.
<Tri-2026-2816 >, hint
Since M is the center of circle ABPQ ,by Brocard theorem M is the orthocenter 0f triangle CHT and CMㅗTH.
<Tri-2026-2816> , general, orthocenter
<Taiwan MO 2000>
<Solid-2026-275>, hint
<Solid-2026-275> tetrahedron, area
<Taiwan MO 1997>
<Tri-2026-2815>, hint
Consider equilateral triangle ABC and a right triangle 30-60-90.