*Consider an equilateral triangle in the*xy

*-plane, centered at the origin.*

*1) By suitably rotating the triangle about its center, is it possible for each quadrant to contain one fourth of the triangle's area?*

*2) By suitably rotating the triangle about its center, is it possible for each quadrant to contain one fourth of the triangle's perimeter?*

(See my previous post for animations.)

In short, the answer to problem (1) is "no." This can be proved from scratch or deduced from the information here.

As for problem (2), again the answer is no. This is not hard to see, if we begin with the premise that the only perimeter-bisecting lines passing through the center of the triangle are the three obvious ones that connect some vertex to the midpoint of its opposite side. More about this premise in a moment, but for now this fact is enough to solve the problem at hand. For suppose the triangle could be rotated so that each quadrant contains one fourth of its perimeter. In that case, the

*y*-axis would bisect the perimeter (since the quarter in

*x*> 0,

*y*> 0 and the quarter in

*x*> 0,

*y*< 0 would add to half), and likewise the

*x*-axis would also bisect the perimeter. There would exist perpendicular perimeter bisectors passing through the center. But none of the lines connecting vertices to midpoints of opposite sides are perpendicular to one another. In other words, there do not exist perpendicular perimeter bisectors passing through the center, and it follows that the triangle cannot be rotated in such a way that each quadrant contains one fourth of the triangle's perimeter.

Now as to that premise: I wrote up a couple of pages of discussion here about the fact that the only perimeter bisectors of an equilateral triangle are the obvious ones. (For amusement's sake, I approach the problem calculus-style.)

By the way, after a conversation with my colleague, I can describe the thought process reflected in these marginalia: what is being shown is that at any time when one quadrant has a quarter of the area, the quadrant immediately "behind" it will demonstrably not have a quarter of the area. Nice approach!

If you liked these problems, some related problems come to mind:

3) If possible, place an equilateral triangle in the

*xy*-plane in such a way that each quadrant has one fourth of the triangle's area.

4) If possible, place an equilateral triangle in the

*xy*-plane in such a way that each quadrant has one fourth of the triangle's perimeter.

5) If possible, place an equilateral triangle in the

*xy*-plane in such a way that each quadrant has one fourth of the triangle's area

*and*one fourth of the triangle's perimeter.

For further reading on these subjects, try "Lines Simultaneously Bisecting the Perimeter and Area of a Triangle," by Paul Yiu. Professor Yiu is the editor of

*Forum Geometricorum*, the journal that published my Pythagoras-free derivation of the identity sin

^{2}θ + cos

^{2}θ = 1.

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