Angles of tetrahedron

For discussing Olympiad level Geometry Problems
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samiul_samin
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Angles of tetrahedron

Unread post by samiul_samin » Wed Feb 14, 2018 8:24 am

Prove that,In any tetrahedron,ther is a verticle such that all of the angles connected with that verticle are less than right angle.

joydip
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Re: Angles of tetrahedron

Unread post by joydip » Wed Feb 14, 2018 11:01 am

yep, my counter example was wrong. Here's my solution :

The 4 points are not coplaner. There are 4 triangles, each can have atmost one non-acute angle. So there can be atmost 4 non-acute angles in total. So if we assume there is no such vertex as stated above, then each triangle and each vertex should have exactly one non-acute angle.
If a vertex has three angles $x,y,z$, then any 2 of them must be greater then the third (prove this yourself). Let $ABCD$ be the tetrahedron and let $AB$ be a edge with the largest length. Then $\angle ACB$ and $\angle ADB $ must be non-acute. $\angle CAB ,\angle BAD ,\angle ABD,\angle CBA$ are acute. So $ \angle DAC$ and $\angle DBC$ are both non-acute. But $\angle DCA+ \angle DCB >\angle ACB$ and $\angle CDA +\angle CDB> \angle ADB$. So,

$$\angle DCA+ \angle DCB+\angle CDA +\angle CDB>\angle ACB+\angle ADB\geq 90^\circ+90^\circ=180^\circ$$
Again,
$$\angle DCA+ \angle DCB+\angle CDA +\angle CDB= (180^\circ-\angle DAC)+(180^\circ-\angle DBC)\leq 90^\circ+90^\circ=180^\circ$$

This gives a contradiction. So there must be a vertex as stated above.
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Last edited by joydip on Wed Feb 14, 2018 1:15 pm, edited 4 times in total.
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samiul_samin
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Re: Angles of tetrahedron

Unread post by samiul_samin » Wed Feb 14, 2018 11:50 am

I don't know how to solve it but this problem is taken from 'Bigganchinta' and Nafis Tiham Vhai has given this problem.

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samiul_samin
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Re: Angles of tetrahedron

Unread post by samiul_samin » Thu Feb 15, 2018 7:06 pm

Thanks for the solution.I was thinking wrong about solving this problem.

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