MathLabs

Problem 1

Let ABCABC be a triangle with ∠A<60∘\angle A<60^\circ. Let XX and YY be the points on the sides ABAB and ACAC, respectively, such that CA+AX=CB+BXCA+AX=CB+BX and BA+AY=BC+CYBA+AY=BC+CY. Let PP be the point in the plane such that the lines PXPX and PYPY are perpendicular to ABAB and ACAC, respectively. Prove that ∠BPC<120∘\angle BPC<120^\circ.
Step 3 of 5: Identify O
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O=O△ABC,∠BOC=2∠A,∠BIC=90∘+∠AO=O_{\triangle ABC},\quad \angle BOC=2\angle A,\quad \angle BIC=90^\circ+\angle A
Detailed analysis

Since OM and ON are perpendicular to AB and AC at their midpoints, O is the circumcenter of ABC. Because angle A is less than 60 degrees, O and I are on the same side of BC, and the central angle BOC equals 2 angle A. The standard incenter angle formula gives angle BIC=90 degrees+angle A, so angle BOC<angle BIC.