In geometry class, it is usually first shown that the medians of a triangle intersect at a single point. Then is is explained that this point is called the centroid and that it is the balance point and center of mass of the triangle. Why is that the case?
This is the best explanation I could think of. I hope someone can come up with something better.
Choose one of the sides of the triangle. Construct a thin rectangle with one side coinciding with the side of the triangle and extending into it. The center of mass of this rectangle is near the midpoint of the side of the triangle. Continue constructing thin rectangles, with each one on top of the previous one and having having the lower side meet the two other sides of the triangle. In each case the centroid of the rectangle is near a point on the median. Making the rectangles thinner, in the limit all the centroids are on the median, and therefore the center of mass of the triangle must lie on the median. This follows because the center of mass of the combination of two regions lies on the segment joining the centroids of the two regions.
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$\begingroup$I think your approach is very simple and effective: you can substitute every chord of a triangle parallel to a fixed side with a mass at its midpoint, proportional to the length of the chord: the centre of mass of the triangle is the same as the centre of mass of those masses. But those masses are aligned along a median of the triangle, hence their centre of mass lies on that median.
For a different proof, one may divide every side of a triangle $ABC$ into $n$ equal parts and connect the dividing points to form $n^2$ equal triangles (see figure below for case $n=4$). Leaving aside $n$ triangles which have a side on $BC$ (yellow in the figure), the other $n^2-n$ triangles can be joined in pairs to form $(n^2-n)/2$ parallelograms (blue in the figure).
The centre of mass of every parallelogram, by symmetry, lies at its geometric centre. Hence the centre of mass of the blue region is the centre of mass of the $(n^2-n)/2$ centres (points $PQ\ldots U$ in the figure), which is the intersection point $V$ of the medians of triangle $UPR$ embedding them (see the Appendix for a proof). Notice that median $UQ$ of that triangle lies on median $AM$ of triangle $ABC$, and from $UV/QV=2$ one gets$$ {AV\over MV}={2n-1\over n+1}. $$It follows that for $n\to\infty$ point $V$ tends to the centroid of $ABC$.
The centre of mass $G$ of $ABC$ lies then on the line joining $V$ with the centre of mass $V'$ of the yellow region: as the mass of the blue region is $n-1$ times that of the yellow region we also have$$ {GV\over GV'}={1\over n-1}. $$When $n\to\infty$ we then have then $G\to V$ and that concludes the proof.
Appendix.
To prove that the centre of mass of equal mass points $PQ\ldots U$ lies at the intersection of the medians of triangle $UPR$, consider the points lying on rows $PR$, $ST$, and so on, parallel to $PR$. We can replace all the $k$ masses on a line with a single heavier point mass ($k$ times the small masses) lying at the midpoint of every line. The centre of mass of those large masses is the same as the centre of mass of all points $PQ\ldots U$.
But the midpoints of parallel lines $PR$, $ST$, and so on, lie on the median $UQ$ of triangle $UPR$: it follows that the centre of mass also lies on that median. And the reasoning can be repeated by considering rows $PU$, $TQ$, ..., parallel to $UP$, hence the centre of mass also lies on median $RS$. This completes the proof.
$\endgroup$ 7 $\begingroup$The centroid is the center of mass of the configuration where we have three point masses of mass 1 at each of the three vertices of the triangle.
Notice that we can replace two of those point masses by a mass of 2 at their midpoint (that is, at their center of mass). Therefore, the total center of mass is clearly on the line between a point and the midpoint of the opposite side, that is the median.
$\endgroup$ $\begingroup$Split a triangle into narrow stripes, parallel to one side.
A center of mass of each stripe is close to the middle of its length, which in turn is on the median. In a limit of the stripes' width approaching zero, centers of mass of all stripes get exactly on the median. Their weighted mean, i.e. the center of mass of the triangle, is on the same line.
$\endgroup$ 1 $\begingroup$I have a fairly intuitive proof but not very rigorous. Firstly, the median of any triangle divides it into two triangles of equal area. This is because the median goes from one corner and bisects the opposing side, resulting in two triangles which have the same base since each is one half of the original side. They also have the same height since they share the same top corner, being the one from which the median was originally drawn. This means that the centre of mass is along this line, since both sides are balanced along this line. Drawing a second median and making the same conclusion makes it obvious that the centre of mass must be at their intersection. It follows that all 3 medians intersect at one point since there is only one centre of mass.
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