# Triangle inscribed in a square

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?

• 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. – The Coding Wombat Mar 26 at 23:10
• 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. – 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.

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