Let ABCABCABC be an acute triangle.
Let PPP be the foot of the altitude from AAA to BCBCBC.
Let OOO be the circumcenter of △ABC\bigtriangleup{ABC}△ABC.
Prove that if ∠BCA≥∠ABC+30∘\angle{BCA}\geq\angle{ABC}+30^{\circ}∠BCA≥∠ABC+30∘,
∠BAC+∠COP<90∘.\angle{BAC}+\angle{COP}<90^{\circ}.∠BAC+∠COP<90∘.
Note by Yuxuan Seah 6 years, 11 months ago
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Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
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to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
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