divide a plane fig into two equal areas with a st. line through its centroid?

[bulai]

Can we say that we can divide a plane fig into two non-symmetric but equal areas with straight lines that cut across the fig through its centroid?

 

[blamocur]

I believe we can. Here is a hint: when the straight line is rotated around the centroid the areas on both sides change continuously, at least for bounded figures.

 

[bulai]

Thanks blamocur. I arrived at this conclusion from the physics experiment of finding centre of gravity using a plumb line. However, the mathematical proof still remains vague to me as I could see that every individual plane figures need separate proof to arrive at the conclusion. For example, I could prove it easily in the case of a parallelogram by drawing a straight line connecting two opposite sides through the centroid of the figure (which is the point of intersection of the two diagonals). But I am still looking for general mathematical proof. Can you help?

 

[Dr.Peterson]

Can we say that we can divide a plane fig into two non-symmetric but equal areas with straight lines that cut across the fig through its centroid?

Does the wording of your question (the plural) imply that there will always be more than one such line, or that every line through the centroid will divide it into equal areas? Or, as blamocur implies, merely that there is at least one such line? (The latter doesn't require the point to be the centroid.)

 

[Dr.Peterson]

Thanks blamocur. I arrived at this conclusion from the physics experiment of finding centre of gravity using a plumb line. However, the mathematical proof still remains vague to me as I could see that every individual plane figures need separate proof to arrive at the conclusion. For example, I could prove it easily in the case of a parallelogram by drawing a straight line connecting two opposite sides through the centroid of the figure (which is the point of intersection of the two diagonals). But I am still looking for general mathematical proof. Can you help?

This is NOT true of the centroid. Balancing is not the same as forming equal areas. (It requires equal moments.) So there is no proof that it is true for all figures, and for all lines.

Here, for example, is a triangle, showing the centroid O and one of the many lines that divide it into equal areas:

​

The line does not pass through the centroid, and a parallel line that does would not divide the triangle into equal parts.

This is a common misconception.

 

[bulai]

This is NOT true of the centroid. Balancing is not the same as forming equal areas. (It requires equal moments.) So there is no proof that it is true for all figures, and for all lines.

Here, for example, is a triangle, showing the centroid O and one of the many lines that divide it into equal areas:

View attachment 31827​

The line does not pass through the centroid, and a parallel line that does would not divide the triangle into equal parts.

This is a common misconception.

Dr. Peterson, good morning. Thanks for your insights but my argument still stands. Please allow me to elaborate... In the case of a plane figure of uniform mass distribution, the centroid and center of gravity (or center of mass) are the same (assuming that we consider uniform gravitational field). Now, a centroid is a geometrical center of any plane figure and that would imply that when we draw a line (line of division) through that point and extends the line across the opposite sides of the figure (we mean closed plane figure) it does divide the figure into two equal areas which are predominantly non-symmetric (exception being in the cases of equilateral, isosceles triangles and parallelograms where the area splits are symmetrical).
Now equal areas would mean an equal number of point masses that are distributed in a non-symmetric manner with respect to the line of division of the plane figure but are present in pairs of equal moment arms with respect to the centroid (pivot) on either side of the line of division. We could thus arrive at equal measures of clockwise and anti-clockwise moments to create the balance at the centroid &/or at the center of gravity.
Thus in cases of plane figures of uniform mass distribution, we can have the situation where we can balance the plane at centroid (this point being the center of gravity as well) while the line of division that passes through this centroid and which cuts through the plane on opposite boundaries does split the plane into two equal halves (same area).
Am I missing something here? Kindly explain and oblige.

 

[bulai]

This is NOT true of the centroid. Balancing is not the same as forming equal areas. (It requires equal moments.) So there is no proof that it is true for all figures, and for all lines.

Here, for example, is a triangle, showing the centroid O and one of the many lines that divide it into equal areas:

View attachment 31827​

The line does not pass through the centroid, and a parallel line that does would not divide the triangle into equal parts.

This is a common misconception.

In the above example, how will you prove the area of quadrilateral ACFD = area of triangle DBF ??

 

[bulai]

This is NOT true of the centroid. Balancing is not the same as forming equal areas. (It requires equal moments.) So there is no proof that it is true for all figures, and for all lines.

Here, for example, is a triangle, showing the centroid O and one of the many lines that divide it into equal areas:

View attachment 31827​

The line does not pass through the centroid, and a parallel line that does would not divide the triangle into equal parts.

This is a common misconception.

It can be proved that if we consider a straight line, say (D^F^), parallel to the base AC, but which also passes through the centroid (O), then the resultant splits of a trapezoid (ACF^D^) and the upper triangle (D^F^B) are of equal areas. That is to say that the straight line (D^F^) passing through the centroid (O) and cutting across the opposite sides of the given triangle (ABC) (which are sides AB & CB respectively) divides the triangle ABC into two equal areas which are non-symmetric

 

[bulai]

This is NOT true of the centroid. Balancing is not the same as forming equal areas. (It requires equal moments.) So there is no proof that it is true for all figures, and for all lines.

Here, for example, is a triangle, showing the centroid O and one of the many lines that divide it into equal areas:

View attachment 31827​

The line does not pass through the centroid, and a parallel line that does would not divide the triangle into equal parts.

This is a common misconception.

Dr. Peterson, thank you for drawing my attention to this problem. I tried finding the areas of the possible trapezoid (bottom half of the triangle) and the possible triangle (at the upper part of the given triangle) by drawing a line segment, say D^F^ parallel to the base AC and passing through the centroid O. The result is, area of lower trapezoid ACF^D^ :: area of upper triangle DFB = 5::4 which implies that the resultant splits are of unequal areas. That proves conclusively that not every line segment, drawn through the centroid could split a plane figure into two equal halves. The exception is the three medians as in any triangle.

 

[blamocur]

Thanks blamocur. I arrived at this conclusion from the physics experiment of finding centre of gravity using a plumb line. However, the mathematical proof still remains vague to me as I could see that every individual plane figures need separate proof to arrive at the conclusion. For example, I could prove it easily in the case of a parallelogram by drawing a straight line connecting two opposite sides through the centroid of the figure (which is the point of intersection of the two diagonals). But I am still looking for general mathematical proof. Can you help?

Draw an arbitrary line through your fixed point (as @Dr.Peterson pointed out it does not have to be the centroid). Let A and B be the areas on each side of the line. If you rotate the line 180 degrees the areas on the side of the line will change continuously and at some point will have to pass through the position where the areas are (A+B)/2 each (https://en.wikipedia.org/wiki/Intermediate_value_theorem).

 

[Dr.Peterson]

I tried finding the areas of the possible trapezoid (bottom half of the triangle) and the possible triangle (at the upper part of the given triangle) by drawing a line segment, say D^F^ parallel to the base AC and passing through the centroid O. The result is, area of lower trapezoid ACF^D^ :: area of upper triangle DFB = 5::4 which implies that the resultant splits are of unequal areas. That proves conclusively that not every line segment, drawn through the centroid could split a plane figure into two equal halves. The exception is the three medians as in any triangle.

I'm glad you didn't just stick to your false assumptions (e.g. "equal areas would mean an equal number of point masses that are distributed in a non-symmetric manner with respect to the line of division of the plane figure but are present in pairs of equal moment arms with respect to the centroid"), and tested reality. Proof is important, isn't it?

And your approach of calculating a specific counterexample is good.

Incidentally, I keep having to rediscover this fact for myself; with each median of a triangle dividing it into equal areas, it is very easy to generalize this and imagine it is true of all lines through the centroid. I made my GeoGebra drawing in part to observe this in action, and make sure I was right. I was inspired by this page, and a picture near the bottom showing bisectors of a triangle:

In my drawing, I can move D and F is constructed to make a bisector, using the method on that page.