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I'm currently practicing some math tasks.
Here I found an task (121041) from the 4th round of the 12th Mathematics Olympics (years ago). The task is:

Show that there is a greatest number $m$ for which the following statement is true!
There is a convex polygon, under whose inner angles exactly $m$ are acute(or sharp?, idk. lets say smaller then 90° :D).

How would you proof this?

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    Hint: rephrase in terms of exterior angles. – 2017-01-04
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    You mean There is a convex polygon, under whose exterior angles exactly $m$ are unsharp(?) ? (I think, that the count of sharp inner angles and unsharp exterior angles has to be the same..) – 2017-01-04
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    Exactly. And what do you know about the sum of exterior angles in a (convex) polygon? – 2017-01-04
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    It has to be 360°? – 2017-01-04
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    And is there then a limit to how many external angles can be obtuse (unsharp)? – 2017-01-04
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    I would say the limit is 3? :D – 2017-01-05
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    It's $4$ if you count $90^\circ$ and make a rectangle, but otherwise, that's it. At any rate, it's conclusive proof that there _exists_ a largest $m$, no matter what exactly that largest $m$ _is_. Note that we're specifically asking about convex polygons because otherwise you could make the external angles negative (which corresponds to internal angles of more than $180^\circ$), thus counteracting all the large external angles. – 2017-01-05
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    and how would it be without an convex polygon (without overlaps)? – 2017-01-05
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    Then there is no maximal $m$. Think about a saw (or a star-shape, or anything spiky, really): the outline is a non-convex polygon, and can have as many sharp angles as you want. – 2017-01-05
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    But that is'nt true I think. Think about a octagon. How would u make 8 sharp angles there? I think, that there's also a rule.. – 2017-01-05
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    Note that you can't make all angles of an $n$-gon acute if $n$ is large enough. Asking "how many angles of an $n$-gon for a given $n$ can we make acute?" is a different question from the one you have. From looking at the saw shape, the answer is at least $\frac{n+3}2$ for odd $n$, and $\frac{n+2}2$ for $n$ even, but whether that's actually the best you can do, I don't know. – 2017-01-05

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