Centroid - Center of GravityDate: 03/25/2002 at 13:52:54 From: Jack Grahl Subject: Unique centre of gravity of a triangle Dear Dr(s). Math, I would like you to help me understand a result I have been puzzling over: In a problem involving triangles (actually a problem involving cutting an irregular shape into two), I realised I had a solution if there was a point on a triangle through which any line divided the area of the triangle into two equal areas. My expectation was that there must be such a point, as in physics a long time ago I learned that everything has a centre of gravity. However, I tried to prove that any triangle has a (unique) centre point like this. I remembered (vaguely) from geometry that the triangle has a point called the centre of gravity, and realised that it had to lie on each of the lines between a vertex and the mid way point of the opposite side. These are easy to construct and all meet at a point. I tried to prove that any line through this point cut the triangle into two shapes of equal area, but wasn't able to. I also did some accurate constructions where it didn't look as if what I was trying to prove was right. The next day I tried another approach, and managed to prove that the point where these three lines meet is 2/3 of the way along each of them. This means that lines through this point, parallel to a side, divide the triangle into a similar triangle, with 4/9 of the area, and a trapezium, with 5/9. I'm pretty sure of my proof, and I've also found information about the centroid, or "centre of gravity" of a triangle, which confirmed it lies 2/3 of the way along each line. However, although this point is often described as a "Centre of Gravity", I can't find anything that shows it is a centre in the way I understand it, nor anything which explains that it isn't. My problem, in brief, is that I believed everything to have a single point, about which the area (or, in physics, mass), can be divided in 2 IN ANY DIRECTION, and that I seem to have shown that a triangle can't have one. I don't understand how, if this is so, physics can cope with an object like a flat, solid triangle, or how calculating a volume of a solid of rotation can work. Hopefully someone has time to deal with a question like this, but I currently don't study maths (except by myself) and so don't have a teacher I can ask. Thanks a lot, Jack Date: 03/25/2002 at 17:11:29 From: Doctor Rick Subject: Re: Unique centre of gravity of a triangle Hi, Jack. You have made many correct observations. The point you are considering is the centroid of a triangle: the intersection of the medians, the lines that join each vertex to the midpoint of the opposite side. The centroid is a unique point, and each median does divide the triangle into two triangles of equal area, but it is not true that ANY line through the centroid divides the triangle into regions of equal area. However, the centroid IS the center of gravity of the triangle. How can this be? It's simple: the center of gravity of a 2-dimensional figure is NOT defined as a point such that any straight line through the point divides the figure into two regions of equal area. Rather, if you rest the figure on any straight edge passing through the center of gravity, it will balance. That's not the same thing. The difference is that, when you balance the figure on an edge, parts of the figure farther from the fulcrum count more. If you've had physics, we are talking about "torque": the product of force times distance. The torque exerted by a small part of the figure is the mass (which we're assuming is proportional to its area) times the distance from the fulcrum (the edge on which we're balancing the figure). The figure balances if the sum of the torques from all the parts of the figure (counting parts on the left as positive and parts on the right as negative) is zero. If the figure is a triangle and the fulcrum lies along a median, then the parts on the two sides of the fulcrum are both triangles, and the point farthest from the fulcrum is the same distance on each side. You can convince yourself that, if you take thin strips parallel to the fulcrum, the same distance from the fulcrum on each side, they are of equal length; therefore both the area and the distance are equal, and the torques are equal and opposite. Thus each thin strip balances the strip on the other side. Altogether, the strips on the left side balance the strips on the right side, and the triangle balances. It's different when you set the fulcrum along a line through the centroid that is not a median. In your example where the line is parallel to one side of the triangle, that side has 5/9 of the mass - but it doesn't stretch out as far from the fulcrum, so its mass doesn't count as much as the smaller mass on the other side. It turns out that, indeed, the triangle will balance on this fulcrum. I've just been waving my hands in describing how it works, but it can be made quantitative and rigorous. A few years ago in response to a Geometry Problem of the Week, I wrote up a fairly rigorous proof concerning the center of gravity of triangles and quadrilaterals. If you're interested, I'll try to dig it up. It won't exactly answer your questions, but it will be sufficiently related that it should be of interest. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ Date: 03/27/2002 at 12:48:36 From: Jack Grahl Subject: Unique centre of gravity of a triangle Dear Dr. Rick, Thanks a lot for your answer. I think I now have a totally clear picture of what a centre of gravity is. Although if I had to define it, it would be something like this: A centre of gravity is a point on a figure through which any line has the following property: that the areas of each shape on either side of the line, multiplied by the respective distances between the centre of gravity of that shape and the line, are equal to one another. What you might call a recursive definition, but don't worry, it makes sense to me. I've been trying to get more information on working them out for other shapes, by building together triangles. Your answer suggested to me a method for finding the centre of a quadilateral, where, instead of dividing the shape into two pairs of triangles, you work with only one pair. After finding the centroids of the two triangles, and drawing a line through it, shouldn't you be able to use their areas (if you know them) to work out the ratio between the distances of each centroid to the centre of gravity? Then it's simply a case of dividing the line segment between them in this ratio to find the point. Anyway, what I wanted to say is that I'd love to see your proof, if you still have it. Thanks a lot for your help, Jack Grahl Date: 03/27/2002 at 13:32:21 From: Doctor Rick Subject: Re: Unique centre of gravity of a triangle Hi, Jack. Yes, that's a way to find the center of gravity of a quadrilateral. The GeoPoW problem was just this, and I stated that solution, but I found another method simpler geometrically. You can draw both diagonals of the quadrilateral, connect the centroids of the two triangles into which one diagonal divides the quadrilateral, and do the same for the other diagonal. The intersection of the two lines is the center of gravity of the quadrilateral. The tough part was to prove, to my satisfaction, that this really does the job. The approach I came up with is essentially the same as your recursive definition of the center of gravity. Annie Fetter, the moderator, thought I was being too picky. Maybe she's right, I don't know ... The problem is found here: The Center of Gravity of a Quadrilateral, July 14-18, 1997 http://mathforum.org/geopow/archive/19970718.geopow.html Here is what I wrote: PROBLEM: Find the center of gravity of a quadrilateral. Construction: Given a quadrilateral ABCD, find the midpoint of each side (the midpoints of AB, BC, CD, and DA are E, F, G, and H respectively). Take two opposite vertices (A and C) and draw lines connecting each to the midpoints of the two sides that meet at the opposite vertex: AF, AG, CE, and CH. Label the intersections of AF and CE, P, and of AG and CH, Q. Connect them. Repeat with the other two vertices (B and D); label the intersections of BH and DE, R, and of BG and DF, S. The intersection of PQ and RS is the centroid, O. Discussion: A tough problem, unless I'm missing something obvious! I found the construction fairly easily, but I had to use what I remember about centers of gravity from physics. Bringing what I know into the context of geometry is the hard part, starting with finding a useable geometrical definition of the center of gravity (centroid) of an arbritrary figure. First, the bonus: Prove that the six small triangles between the medians have the same area. PROOF: Consider triangle ABC in my figure. C /\ / \ \ / \ / \ \ / \ F / \ \ G / \ / \ / \ \ / \ / / \ \ / / \ \ \ / / \ \ /________________\________________\ A E B The areas of triangles CAE and CEB are equal because they have the same base (AE = EB) and height (distance of C from AB). Each is therefore half the area of triangle ABC. Likewise triangles ABF and AFC have the same area, again half the area of triangle ABC. Thus the areas of ABF and CEB are equal, and if we subtract the common area BFPE, we have the areas of PAE and PFC equal. Again, triangles PAE and PEB have the same area because their bases and height are equal, and likewise for PBF and PFC. So we have PEB = PAE = PFC = PBF, and in a similar fashion the remaining two small triangles can be shown to be equal to the others. The implication of the problem statement is that the definition of a centroid has something to do with this fact, that is, lines passing through the centroid dividing a figure into equal areas. But this is not the case. The centroid is the mean (average) location of all the points contained in a figure. Any line passing through the centroid can serve as a pivot and the figure will "balance" (supposing it is cut out of wood, for instance, of uniform thickness). In physics, this means the "moments" on both sides are equal (and opposite). It does not necessarily mean the areas on both sides are the same; a smaller area farther from the pivot can balance a larger area closer to the pivot. In the case of the medians of a triangle, the figures on both sides are triangles; therefore the area is distributed the same way, so equal moments imply equal areas. Other lines through the centroid give a triangle on one side and a quadrilateral on the other; the mass distribution is different - the quadrilateral has more mass farther out - so the triangle needs a larger area to balance it. Now I will try to put the centroid of a general figure in a classical- geometry context. I define the centroid by three postulates: 1. The centroid of a figure with a rotational symmetry is the rotation axis. 2. If two figures a and b are similar, their centroids A and B form similar figures with corresponding elements of a and b. 3. Given two figures a and b with centroids A and B, the centroid of the two figures taken together is the point C on the line AB, such that AC * Area(a) = CB * Area(b). I don't know if these are all necessary or sufficient, but they are consistent with the definition of a centroid as the mean location of the points in a figure, and they are sufficient for my purpose. I have one thing to prove before I can justify my construction: that this definition is consistent with the centroid we know. THEOREM: The centroid of a triangle is the intersection of its medians. PROOF: Given any triangle ABC, suppose its centroid is P. Construct the midpoints of AB, BC, and CA at D, E, and F respectively. Triangle ABC can be divided into parallelogram BDFE and triangles ADF and FEC. Call the centroids of these figures Q, R, and S respectively. Triangles ADF and FEC are congruent, so they have the same area, therefore by (1) above, the centroid T of the two triangles taken together is at the midpoint of RS. The centroid Q of parallelogram BDFE is on the diagonal BF by (1) above. Triangle ADF is similar to triangle ABC. Therefore by (2) above, triangle ARF is similar to APC; angle RAF = angle PAC, so R lies on line AP; and since F is the midpoint of AC, R is the midpoint of AP. Similarly, S is the midpoint of CP. Triangle RPS is similar to triangle APC. Therefore T, the midpoint of RS, lies on PF. Triangle ABC is the sum of parallelogram BDFE (with centroid Q) and triangles ADF and FEC (taken together, with centroid T), so by (3) above, the centroid P of ABC is on the line QT. Since Since P, T, and F are collinear by (4), and P, T, and Q are collinear by (5), then P, T, F, and Q are all collinear. Since Q, B, and F are collinear by (2), we can add B to the list, so in particular, P is on BF. But BF is a median of triangle ABC. By similar arguments, P is on the other two medians of ABC as well. In other words, the centroid of any triangle ABC is the intersection of its medians. Now I am finally ready to justify my construction of the centroid of a quadrilateral. A quadrilateral can be divided into two triangles in two ways: by cutting along either diagonal. If I make one such cut, I know the centroid lies along the line joining the centroids of the two triangles (P and Q), by definition part (3). I could at this point find the centroid by choosing O on PQ such that OP:OQ = Area(ACD):Area(ABC). But it is much easier with compass and straightedge to take advantage of the other pair of triangles and find another line (RS) on which the centroid lies. The centroid must of course lie at the intersection of the two lines. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ |
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