Prove in this case that one can always find among the triangles two that are congruent.

Question

Four congruent right triangles are given. Adriana can cut one of them along the altitude and repeat the operation several times with the newly obtained triangles. Prove that no matter how Adriana perform the cuts, she can always find among the triangles two that are congruent.

The source of the question and answer is given below for reference. http://tech.mit.edu/V135/N10/technical/solutions.pdf

Basically,

1. I can't understand why the triangles are similar in that ratio $\left (\frac ah\right)^m\left (\frac bh\right)^n $.
2. What is the meaning of "weight" assigned to each such triangle? Is it the invariant?

A simpler solution will be extremely appreciated.

Answer 1

Why are the triangles similar in that ratio?

When you draw the altitude of a right triangle with sides $a,b$ and hypotenuse $h$, you get two right triangles that are similar to the first one (because they share an angle in addition to the right angle). Their hypotenuses are the sides of the original triangle: $a$ and $b$. Therefore they have been scaled by factors $\frac ah$ and $\frac bh$ respectively.

The ratios $\frac ah$ and $\frac bh$ are the same for smaller similar triangles. So whenever we cut any right triangle in the problem, we get a copy of it scaled by $\frac ah$, and another copy of it scaled by $\frac bh$. So for example if we cut the original triangle, and then cut each of the smaller triangles we get, the result is

• one triangle scaled by $\frac ah$ and then by $\frac ah$ again, for a total scaling factor of $(\frac ah)^2$;
• two triangles scaled by $(\frac ah)(\frac bh)$: one that was scaled by $\frac ah$ then by $\frac bh$, and one that was scaled by $\frac bh$ and then by $\frac ah$;
• one triangle scaled by $(\frac bh)^2$.

We don't actually care about the exact sizes that much. The important takeaway here is that the operations "cut along the altitude, and take the smaller piece" and "cut along the altitude, and take the larger piece" are commutative: whichever order we do them in, we get a triangle of the same size.

We can keep going with these triangles. If we take a triangle with hypotenuse $h(\frac ah)^2$ and cut it in half, we get a piece with hypotenuse $h(\frac ah)^3$ and a piece with hypotenuse $h(\frac ah)^2 (\frac bh)$.

A triangle that's the result of $m+n$ cuts, $m$ of which have taken the $\frac ah$-scaled piece and $n$ of which have taken the $\frac bh$-scaled piece, has hypotenuse $h(\frac ah)^m(\frac bh)^n$.

What's up with the weight?

A generic right triangle constructed in this problem has been scaled from the original by a factor of $(\frac ah)^m (\frac bh)^n$. Knowing the pair $(m,n)$ uniquely determines this triangle, so we will represent it by that ordered pair.

If we take an $(m,n)$ triangle and cut it along the altitude, we get an $(m,n+1)$ triangle and an $(m+1,n)$ triangle. This motivates the weight function: it is just a function such that the total weight is always conserved. If the weight of an $(m,n)$ triangle is $\frac1{2^{m+n}}$, then we've turned a triangle of weight $\frac1{2^{m+n}}$ into two triangles of weight $\frac1{2^{m+(n+1)}} = \frac1{2^{(m+1)+n}}$: two triangles of half the weight.

The total weight is the invariant in this problem.

Is there a simpler solution?

"Simpler" is a matter of taste, but here is another approach.

We start with four congruent triangles with hypotenuse $h$. If we want to eliminate congruent triangles, at some point we'll have to cut at least three of these in half, so let's go ahead and do that. This gives us three congruent triangles with hypotenuse $a$, and three congruent triangles with hypotenuse $b$.

Of the three congruent triangles with hypotenuse $a$, we'll have to cut at least two in half at some point, so let's go ahead and do that now. This gives us two congruent triangles with hypotenuse $\frac{a^2}{h}$ and two congruent triangles with hypotenuse $\frac{ab}h$. We get two more congruent triangles with hypotenuse $\frac{ab}h$ (and two with hypotenuse $\frac{b^2}{h}$) when we are forced to cut two of the three congruent triangles with hypotenuse $b$.

But now, we're back in the same situation where we started: we have four congruent right triangles (with hypotenuse $\frac{ab}{h}$). This forces us to keep going forever without eliminating all the congruent triangles at any point.

Slightly more formally, suppose that there is a solution, and the shortest solution takes $n$ cuts. Well, the above argument outlines $7$ cuts that have to be done at some point, so the resulting situation after we do them has to be solved in $n-7$ cuts. But the resulting situation contains four congruent right triangle, which we assumed take $n$ cuts to solve: contradiction!