MathDB

4

Part of 1950 Poland - Second Round

Problems(1)

P interior of ABC with <PAB=<PBC =<PCA

Source: Polish MO Second Round 1950 p4

9/1/2024
Inside the triangle ABCABC there is a point PP such that ∠PAB=∠PBC=∠PCA=ϕ.\angle PAB=\angle PBC =\angle PCA = \phi. Prove that 1sin⁡2ϕ=1sin⁡2A+1sin⁡2B+1sin⁡2C\frac{1}{\sin^2 \phi}=\frac{1}{\sin^2 A} +\frac{1}{\sin^2 B} +\frac{1}{\sin^2 C}
geometrytrigonometryequal angles