# Infinite descent method in geometry

What are the examples we can use to explain infinite descent as an efficient method of proofs in geometry?
I think one of the best may be proving medians of a triangle are concurrent by the infinite descent method when compared with other approaches. If we have a few more, it can be useful to make the topic much more interesting for students.

• I have come across another one that uses the same approach to prove that there cannot be an equilateral triangle formed by joining any three grid points of a two-dimensional square grid. Commented Jan 14 at 13:13

## 2 Answers

This is an example for what you are looking for. I didn't include the solution!

Four grasshoppers sit at four vertices of a square. Every second one of the grasshoppers jumps over another one and lands at the symmetric point (that is, if a grasshopper jumps from point $$A$$ over $$B$$ and lands at $$C$$, then vectors $$\vec{AB}$$ and $$\vec{BC}$$ are equal). Prove that the four of them will never form a square bigger than the original one.

• It's complicated how grasshopper can jump over B from A to C and how the mentioned two vectors are equal. Could you please explain bit more. Commented Jan 21 at 4:22
• When I have $A$ jumping over $B$, $B$ over $C$, $C$ over $D$ and $D$ over the new location of $A$, then I create a larger surface but it's not a square. Is this what you mean? That, if you manage to get a square, that the surface will be smaller? Commented Jan 22 at 14:52
• @Dominique If you consider the new locations of the grasshoppers as vertices of a quadrilateral, the question asserts that this quadrilateral will never be a square (neither bigger, nor smaller). To prove it, we can assume that originally the grasshoppers were sitting on the unit square with vertices at $(0,0), (0,1), (1,1), (1,0)$ in $\mathbb{R}^2$. We can quickly convince ourselves that when each grasshopper jumps over another grasshopper, it will again land at a point with integral coordinates. This shows that grasshoppers can never land on a smaller square, ... Commented Jan 22 at 18:51
• @Dominique since there is no square smaller than the unit square with vertices having integral coordinates. Now suppose the grasshoppers land on a bigger square. Then this process of jumping over another grasshopper is reversible and if grasshoppers undo their jumps, they can go back to the original unit square. But this means that if they start with the original unit square and perform the same jumps that take them from the bigger square to the unit square, they must land on a square smaller than the unit square, which is again not possible. Therefore they can't land on a bigger square. Commented Jan 22 at 18:55

If the vertices of a regular $$n$$-gon have integer coordinates, then $$n \leq 6$$.

Proof: Draw the vectors representing the sides of the $$n$$-gon so that their initial point is the origin. Then you have drawn a new regular $$n$$-gon whose vertices have integer coordinates and that is $$a$$ times the size of the previous one, where $$a$$ is a constant depending only on $$n$$. And if $$n > 6$$, then it is easily seen that $$a < 1$$.

• Is there any chance of drawing a regular hexagon with vertices of Integer coordinates. I think has connection with my previous comment to the post. Commented Jan 21 at 5:43
• @JanakaRodrigo No, it can't be done for $n = 3, 5, 6$. The proof by infinite descent doesn't seem to extend to these cases, however. It can be proved that the hypothesis implies that $\sin 2\pi/n$, $\cos 2\pi/n$ are both rational, and it's straight forward to prove that this isn't the case for $n = 3, 5, 6$, where the explicit values are well-known.
– Dave
Commented Jan 21 at 14:34