Solution is very clear but we should not assume that 2PM=AM because we want to find that whether 2PM=AM is true or not. In the solution we should have obtained the 2PM=AM equation without assume it is true.
He claims at first that P is the midpoint of AM, not all medians. He also claims point P is 2/3 along median AM. After solving, he determines and explains that, if he were to solve for point Q (2/3rds on median BN) the resulting equation would be equal to the original equal for point P. And if he were to do this for the final median (C to midpoint of AB) the result vector would also be that of point P. To summarize, he only assumed point P was 2/3rds of the median AM. Which was given.
this is a single variable calculus problem, using the weighted averages of x and y over its area, this is an old question so this answer is meant to be read for others.
The video is very well explained, but I fail to see how is this a coordinate-free proof. Even though it doesn't matter what the origin is, the fact is we do need _some_ origin to perform the calculations that will result in the symmetry mentioned at the end of the video, and _then_ we use that fact to conclude that where the origin is located is irrelevant. But that demonstration is merely an abstraction of calculations made using such an origin. I was thinking this would show a way to reason about the problem that wouldn't require a reference origin point at all, but I guess I am just approaching this from a wrong angle...
can u help me solve this: Using a graph, it can be seen that the curves y=x+ln(x) and y=x3−x intersect at the points (0.447141,−0.357742) and (1507397,1.917782) Find the coordinates (to three decimal places) of the centroid for the region bounded by these curves. x= y=
Not helpful at all. The problem asked IF the intersection is 2/3 of the way from any vertex. He did not prove that. He just wrote outright that P is at 2/3 of the way.
I think that he did. Maybe it would have been be more clear if he called the point P in his calculations to be P_A. And the point M to be M_A. So.. he defined P_A as a point that is 2/3 on the way from A to M_A (the intersection of the median from A with the BC side of the triangle). Then he derived the representation of this point in terms of vectors OA, OB, and OC. It turned out that it is 1/3 (OA + OB + OC). So again, P_A is equal to 1/3 (OA + OB + OC). The idea is then to note that all the calculations (and the result) will be analogous for P_B and for P_C, so P_A = P_B = P_C = 1/3 (OA + OB + OC). It means that the three points on the medians that are 2/3 from the each vertex are actually exactly the same one point. This is equivalent to the thing we had to prove. I hope it helps.
If you want to paint a visual in your mind, make O (0,0,0), A (1,0,0), B (0,1,0), and C (0,0,1). This makes an equilateral triangle in the first octant. Jut keep in mind that magnitudes don't actually matter for this problem.
Super cool.
It's amazing that this fact can be proved by simple algebra of vectors.
Very cool.
Solution is very clear but we should not assume that 2PM=AM because we want to find that whether 2PM=AM is true or not. In the solution we should have obtained the 2PM=AM equation without assume it is true.
He claims at first that P is the midpoint of AM, not all medians. He also claims point P is 2/3 along median AM. After solving, he determines and explains that, if he were to solve for point Q (2/3rds on median BN) the resulting equation would be equal to the original equal for point P. And if he were to do this for the final median (C to midpoint of AB) the result vector would also be that of point P. To summarize, he only assumed point P was 2/3rds of the median AM. Which was given.
this is a single variable calculus problem, using the weighted averages of x and y over its area, this is an old question so this answer is meant to be read for others.
Where is point Q? Why are you switching A and B? How is a position vector related to a triangle?...
You explained this so well thanks!!!
thank you so much for this video!!
I wish Iran's education was like this
Thanks, this cleared it up for me.
Are we just assume that OP = OA+ 2/3(AM)?
I think so.
It's vector addition
how we can use this thm in the problems?any example?
The video is very well explained, but I fail to see how is this a coordinate-free proof. Even though it doesn't matter what the origin is, the fact is we do need _some_ origin to perform the calculations that will result in the symmetry mentioned at the end of the video, and _then_ we use that fact to conclude that where the origin is located is irrelevant. But that demonstration is merely an abstraction of calculations made using such an origin. I was thinking this would show a way to reason about the problem that wouldn't require a reference origin point at all, but I guess I am just approaching this from a wrong angle...
can u help me solve this:
Using a graph, it can be seen that the curves y=x+ln(x) and y=x3−x intersect at the points (0.447141,−0.357742) and (1507397,1.917782) Find the coordinates (to three decimal places) of the centroid for the region bounded by these curves.
x=
y=
I don't think he can
Not helpful at all. The problem asked IF the intersection is 2/3 of the way from any vertex. He did not prove that. He just wrote outright that P is at 2/3 of the way.
agreed, it is a different problem if you need to show 2/3, see mathematics stack exchange
I think that he did. Maybe it would have been be more clear if he called the point P in his calculations to be P_A. And the point M to be M_A.
So.. he defined P_A as a point that is 2/3 on the way from A to M_A (the intersection of the median from A with the BC side of the triangle).
Then he derived the representation of this point in terms of vectors OA, OB, and OC. It turned out that it is 1/3 (OA + OB + OC).
So again, P_A is equal to 1/3 (OA + OB + OC).
The idea is then to note that all the calculations (and the result) will be analogous for P_B and for P_C, so
P_A = P_B = P_C = 1/3 (OA + OB + OC).
It means that the three points on the medians that are 2/3 from the each vertex are actually exactly the same one point.
This is equivalent to the thing we had to prove.
I hope it helps.
Try any number other than 2/3 and see what happens
where is O ??
it doesn''t matter
If you want to paint a visual in your mind, make O (0,0,0), A (1,0,0), B (0,1,0), and C (0,0,1). This makes an equilateral triangle in the first octant. Jut keep in mind that magnitudes don't actually matter for this problem.
lol, he's nervous
I know this guy might be smart with his material, but his voice makes me cringe sometimes 8:02
I know right, the little bastard!