Angles of tetrahedron

For discussing Olympiad level Geometry Problems
samiul_samin
Posts: 999
Joined: Sat Dec 09, 2017 1:32 pm

Angles of tetrahedron

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
Posts: 48
Joined: Tue May 17, 2016 11:52 am

Re: Angles of tetrahedron

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.
Attachments fo3.png (16.58 KiB) Viewed 418 times
Last edited by joydip on Wed Feb 14, 2018 1:15 pm, edited 4 times in total.
The first principle is that you must not fool yourself and you are the easiest person to fool.

samiul_samin
Posts: 999
Joined: Sat Dec 09, 2017 1:32 pm

Re: Angles of tetrahedron

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

samiul_samin
Posts: 999
Joined: Sat Dec 09, 2017 1:32 pm

Re: Angles of tetrahedron

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