Consider a square $ABCD$ of side length $1$. Have the intersection of the diagonals $AC, BD$ be $G$. Determine the value of largest (in terms of area) inscribed triangle, such that $G$ is the centroid of the triangle.

I made this, not sure if it is classic/generic.

Two questions:

$1.$ I'm not sure if the centroid is the best center, any feedback? If so, what is the solution?

$2.$ What about a rectangle? What difference would that make?

  • $\begingroup$ What do you mean by best center? I assume we're talking about an unweighted centroid, so the intersection of the three medians of the triangle, where a median is the line segment from a vertex to the midpoint of the opposite side. These three medians intersect in one point, G, the centroid. $\endgroup$ – The Coding Wombat Mar 26 at 23:10
  • $\begingroup$ Yes, we are talking about the intersection of the medians. By the best center, i mean, is there a center such that the difficulty remains, and such that the computation is nicer. $\endgroup$ – weareallin Mar 26 at 23:13

First, consider which sides of the square the vertices $P,Q,R$ of the triangle are on. If all three vertices are on two adjacent sides, then the triangle lies entirely on one side of a diagonal of the square, and thus so does the centroid. The centroid can't be at the square's center in this case.

The alternative is that some pair of vertices (WLOG $P$ and $Q$) are on opposite sides of the square. WLOG, this is the unit square $0\le x,y\le 1$ and the sides $P$ and $Q$ are on are the two segments $x=0$ and $x=1$. In order for the $x$ value of the centroid $G$ to be $\frac12$, the $x$-value of $R$ must be $\frac12$, putting $R$ at either $(\frac12,0)$ or $(\frac12,1)$, the midpoint of a side of the square.

WLOG, $P=(0,y_1)$, $Q=(1,y_2)$, $R=(\frac12,0)$. Then, from $G=(\frac12,\frac12)$, the midpoint $S$ of $PQ$ is at $S=(\frac12,\frac34)$. By the standard side-altitude formula, the area of $\triangle PRS$ is $\frac12\cdot\frac12\cdot\frac34=\frac{3}{16}$ and the area of $\triangle QRS$ is $\frac12\cdot\frac12\cdot\frac34=\frac{3}{16}$. For any triangle inscribed in the square with centroid $G$ at the square's center, the triangle's area is $\boxed{\frac38}$. There's nothing to maximize.

I don't know why you were concerned with the computation not being nice here; it all comes out very cleanly.

Would a rectangle change anything? No; this is a purely affine problem.

Now, a different center? That would require a very different approach, and would lead to a meaningful optimization problem. If you're interested in the difficulty, try it out yourself. Just at a glance, the circumcenter (sticking with the square) should be fairly straightforward.

  • $\begingroup$ Thank you! I will attempt other centers myself. $\endgroup$ – weareallin Mar 26 at 23:39

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