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During the IMC 2019 contest i ended up with the following question in elementary 2D Euclidean geometry: Let $\mathrm{CDE}$ be any nondegenerate triangle inside a circle, consider the regions with areas labelled $S_1, S_2, S_3$ as on the picture. Prove that: $$S_1 + S_2 + S_3 > S(CDE)$$ It must be known and i would be happy if anyone knows the elementary solution (it seems that i can d it by rough 5-pages calculations with cases).

It is also interesting to find the best possible constant $\lambda$ such that: $$S_1 + S_2 + S_3 \geq \lambda \cdot S(CDE) $$ My guess is $\lambda = \frac{4\pi}{3\sqrt{3}} - 1\approx 1.418399$, that comes from equilateral inscribed triangle.

enter image description here

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  • $\begingroup$ CDE "worst" case is when the points CDE are on the circle. $\endgroup$ – Moti Aug 11 '19 at 15:22
  • $\begingroup$ @Moti if you know the proof of that, then by a known fact that area is maximal for inscribed equilateral, the problem follows. But note that there are examples when difference between LHS and RHS arbitarly small. $\endgroup$ – Rybin Dmitry Aug 11 '19 at 19:14
  • $\begingroup$ What are LHS and RHS? $\endgroup$ – Moti Aug 12 '19 at 1:32
  • $\begingroup$ @Moti Left Hand Side, Right Hand Side, feel free to ask $\endgroup$ – Rybin Dmitry Aug 12 '19 at 8:26
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My proof is valid only when the circum-circle of CDE lies completely inside the original circle.

enter image description here

Locate H, the ortho-center of CDE.

EH extended will cut CDE at E’ and the circum-circle at E’’

By properties of the ortho-center, HE’ = E’E’’. Hence, [CHE’] = [CE’E’’]

The other parts of CDE can be off-set similarly. Result follows.

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  • $\begingroup$ Cute construction, due to convexity we can require only points $E'$ and others to lie inside original circle. Unfortunately i don't know a short way to reduce to such case. $\endgroup$ – Rybin Dmitry Aug 12 '19 at 8:30
  • $\begingroup$ @RybinDmitry I think the key point is how the "non-degenerate triangle" is defined. I mean the original question could have been phrased differently. $\endgroup$ – Mick Aug 12 '19 at 8:36

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