MathDB

6

Part of 1986 Poland - Second Round

Problems(1)

[A'B'C' ]<= 1/4 [ABC]

Source: Polish MO Recond Round 1986 p6

9/9/2024
In the triangle ABC ABC , the point A′ A' on the side BC BC , the point B′ B' on the side AC AC , the point C′ C' on the side AB AB are chosen so that the straight lines AA′ AA' , CC′ CC' intersect at one point, i.e. equivalently ∣BA′∣⋅∣CB′∣⋅∣AC′∣=∣CA′∣⋅∣AB′∣⋅∣BC′∣ |BA'| \cdot |CB'| \cdot |AC'| = |CA'| \cdot |AB'| \cdot |BC'| . Prove that the area of triangle A′B′C′ A'B'C' is not greater than 1/4 1/4 of the area of triangle ABC ABC .
geometryGeometric Inequalitiesareas