What is the volume of the biggest piece in this geometry tetrahedron problem?

[terryds]

Homework Statement

Volume of tetrahedron T.ABC = V
Point P is on the middle of TA, Q is on the expansion of AB making AQ = 2AB
A shape is made through PQ which is parallel to BC so that it cuts the tetahedron into 2 pieces.
What is the volume of the biggest piece?

The Attempt at a Solution

I sketch the problem

http://www.sumopaint.com/images/temp/xzohqlsrafmmqrck.png
Then, I have no idea how to continue

 

[BvU]

Hi Terry,

I am missing something ! :

Homework Equations

For this part of the template, you could think of this or this
​

Deleting (even just a part of) the template irritates the spirits that watch over us, so you really don't want to do that. A litte googling usually helps if your notes or textbook don't have anything useful. And in this case the second link might have given you an idea...

I notice you've drawn PM parallalel to AC instead of NM parallel to BC. Perhaps you could clarify ?

And, for the record, I don't have the answer, so we still need help.

--

 

[Ray Vickson]

Homework Statement

Volume of tetrahedron T.ABC = V
Point P is on the middle of TA, Q is on the expansion of AB making AQ = 2AB
A shape is made through PQ which is parallel to BC so that it cuts the tetahedron into 2 pieces.
What is the volume of the biggest piece?

The Attempt at a Solution

I sketch the problem

http://www.sumopaint.com/images/temp/xzohqlsrafmmqrck.png
Then, I have no idea how to continue

Nevertheless, you need to make an attempt; those are the PF rules.

 

[terryds]

Hi Terry,

I am missing something ! :

Homework Equations

For this part of the template, you could think of this or this
​

Deleting (even just a part of) the template irritates the spirits that watch over us, so you really don't want to do that. A litte googling usually helps if your notes or textbook don't have anything useful. And in this case the second link might have given you an idea...

I notice you've drawn PM parallalel to AC instead of NM parallel to BC. Perhaps you could clarify ?

And, for the record, I don't have the answer, so we still need help.

--

Nevertheless, you need to make an attempt; those are the PF rules.

I'm sorry. Now, I want to edit the thread but it's disabled now
Okay, I just add them in this post

- Relevant equations

V = Base Area * height

- My Attempts

I see the answer in the book, but it's very messy and I don't understand.

Please help

I don't understand why it is TN/NB = TM/MC (I think it should be TN/TB = TM/TC)
Then, TP/TA*TN/TB*TM/TC comes out which gets me more confused
I think the book has lots of typos, but I'm pretty sure the book's got the idea

(The book is written in Bahasa, this is the translation :
P is the center point of TA,
Q is on the expansion of AB => AQ = 2AB
The shape through PQ//BC that cuts the tetrahedron into two pieces is)

 

[haruspex]

I don't understand why it is TN/NB = TM/MC (I think it should be TN/TB = TM/TC)

TN/TB=TN/(TN+NB)=1/(1+NB/TN).
Likewise, TM/TC=1/(1+MC/TM).
So if TN/TB=TM/TC then TN/NB=TM/MC.

 

[terryds]

TN/TB=TN/(TN+NB)=1/(1+NB/TN).
Likewise, TM/TC=1/(1+MC/TM).
So if TN/TB=TM/TC then TN/NB=TM/MC.

Why is it 2/1 ? And, how does TP/TA*TN/TB*TM/TC come from? I really get nothing from the book answer

 

[Samy_A]

Why is it 2/1 ?

I don't know if it is the easiest way, but the theorem of Menelaus, applied to the triangle TAB and the line PNQ, shows that TN/NB = 2.

 

[haruspex]

Why is it 2/1 ?

Consider the centroid of triangle TAQ.

And, how does TP/TA*TN/TB*TM/TC come from?

The volume of a tetrahedron Is 1/6 of the triple scalar product of the vectors representing ithree sides adjacent to one vertex. So if you keep the angles fixed and vary the lengths, it is proportional to the product of the three lengths.

 

[terryds]

Consider the centroid of triangle TAQ.

The volume of a tetrahedron Is 1/6 of the triple scalar product of the vectors representing ithree sides adjacent to one vertex. So if you keep the angles fixed and vary the lengths, it is proportional to the product of the three lengths.

Isn't the volume of tetrahedron is a^3/(6√2) where a is the length of the edge??
But, how to know that the angle is fixed?? The shape is a bit different from the big initial tetrahedron I think.

 

[haruspex]

Isn't the volume of tetrahedron is a^3/(6√2) where a is the length of the edge??

For a regular tetrahedron, yes, but this is not.

But, how to know that the angle is fixed?? The shape is a bit different from the big initial tetrahedron I think.

We're keeping the three angles at T fixed.
Think of T as the origin. The volume of TABC is one sixth the triple scalar product of vectors TA, TB, TC. Likewise, the volume of TMNP is one sixth the triple scalar product of the vectors TM, TN, TP.
The triple scalar product of three vectors is the product of the magnitudes of the vectors, multiplied by a function of the angles between them.