MathDB

3

Part of 1992 Poland - Second Round

Problems(1)

ABC is isosceles if |BP| x|CQ| x|AR| = |PC| x |QA| x|RB|

Source: Polish MO Recond Round 1992 p3

9/9/2024
Through the center of gravity of the acute-angled triangle ABC ABC , lines are drawn perpendicular to the sides BC BC , CA CA , AB AB , intersecting them at the points P P , Q Q , R R , respectively. Prove that if ∣BP∣⋅∣CQ∣⋅∣AR∣=∣PC∣⋅∣QA∣⋅∣RB∣ |BP|\cdot |CQ| \cdot |AR| = |PC| \cdot |QA| \cdot |RB| , then the triangle ABC ABC is isosceles.
Note: According to Ceva's theorem, the assumed equality of products is equivalent to the fact that the lines AP AP , BQ BQ , CR CR have a common point.
geometryisosceles