There is a triangle, ABC, with points Am, Bm, Cm at the centre on each side. Show..?

Question

..that OA+OB+OC = OAm + OBm + OCm
where OA,OB,OC,OAm,OBm,OCm are vectors.
and O is an arbitrary point.

How should you draw this, and then how to show it?

Answer 1

I'll assume Am is between A and B, Bm is between B and C, and Cm is between C and A--it doesn't matter which one's which but it'll give a starting point for showing.

If Am is the midpoint between A and B, then OA + OB = OAm + OAm. You can show this geometrically by drawing OA and OB, then drawing another copy of OA starting at B (let's call this OA'), and drawing another copy of OB starting at A (let's call this OB'). This forms a parallelogram (by definition--you've drawn two pairs of the same vectors), and if you draw the vector from O to where OB' and OA' meet up you'll get OA+OB, which is one of the diagonals.

Now note that if you draw AB, you'll have drawn the other diagonal. Since the diagonals of a parallelogram bisect one another, the point where they meet is also Am, and so you've shown that OA+OB is exactly OAm + OAm.

Now we can without loss of generality say this applies to the other points of the triangle as well--we haven't assumed anything except that Am was the midpoint of AB. So in addition to knowing OA+OB=OAm+OAm, we also know that OB+OC=OBm+OBm, and OC+OA=OCm+OCm. If we add up all three of these equations, we get

OA+OB+OB+OC+OC+OA = OAm+OAm+OBm+OBm+OCm+OCm

or 2(OA+OB+OC)=2(OAm+OBm+OCm)
or OA+OB+OC = OAm+OBm+OCm.

I hope that answers your question--I kind of mixed both the drawing and the showing together but that often makes a proof easier to understand :)

Answer 2

you cannot prove what is not always true. I give a counter example. Consider an eqilateral triangle with sides 2sm. and let O coincide with point A.
OA=0,OB=2,OC=2 giving OA+OB+OC=4
OAm=sqrt3=1.73.., OBm=1, OCm=1
hence OAm+OBm+OCm= 3.73... They are necessarily equal hence the proof cannot be done on something not true

Answer 3

Let Am, Bm and Cm be the midpoints of BC, CA and AB respectively.

=>
OAm = (1/2) (OB + OC),
OBm = (1/2) (OC + OA)
OCm = (1/2) (OA + OB)

Adding,
OAm + OBm + OCm
= (1/2) (2OA+2OB+2OC)
= OA+OB+OC

Answer 4

For convenience draw any scalene triangle, and place O inside the triangle.

OA + AA(m) = OA(m); OB + BB(m) = OB(m);
OC + CC(m) = OC(m).

Therefore:
OA + OB + OC + AA(m) + BB(m) + CC(m)
= OA(m) + OB(m) + OC(m)

But assuming A(m) is the midpoint of AB, etc.,

AA(m) + BB(m) + CC(m) = 1/2[AB + BC + CA]
Draw the appropriate conclusion.

Answer 5

ok, i'm going to take a crack at this one. the predicted value of the climate is barely one-a million/2 of the dimensions. in case you finished a small triangle via connecting the mid-factors of the three components, that is section is a million/4 of the part of the unique triangle and you will use Hero's formulation to locate it. as a fashion to try this you're employing the reality that the predicted value of a sum or difference of random variables is the sum or difference of their predicted values. you need to additionally calculate the predicted value of a made from random variables it somewhat is the made from the predicted values of those variables on condition that they are self sufficient. in this subject, it style of feels to me that this situation is happy. Hero's formulation, then, supplies the predicted value of the sq. of the section. The sq. root of it particularly is the respond.

Answer 6

You cannot prove it because it is not true.

Let A = (0,0), B= (0,10), C = (6,0)
Then Am = (3,5), Bm = (3,0), Cm = (0,5)
Let O = (1,1) be the arbitrary point.
Now use the distance formula to compute the distances and you will see that your conjecture is false.

Answer 7

Please note that all are vectors:
You should know (or easy to prove) that:
AmA+BmB+CmC = 0
So:
OA+OB+OC=OAm+AmA+OBm+BmB+OCm+CmC=
OAm+OBm+OCm+AmA+BmB+CmC=
OAm+OBm+OCm
What you want to prove