Angles of tetrahedron
 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.
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 nonacute angle. So there can be atmost 4 nonacute 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 nonacute 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 nonacute. $\angle CAB ,\angle BAD ,\angle ABD,\angle CBA$ are acute. So $ \angle DAC$ and $\angle DBC$ are both nonacute. 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.
The 4 points are not coplaner. There are 4 triangles, each can have atmost one nonacute angle. So there can be atmost 4 nonacute 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 nonacute 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 nonacute. $\angle CAB ,\angle BAD ,\angle ABD,\angle CBA$ are acute. So $ \angle DAC$ and $\angle DBC$ are both nonacute. 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.