Professor Erasmus has constructed a special isosceles triangle that he modestly calls the "Professor-Erasmus-triangle". The professor claims that he can cut his triangle into three smaller triangles, so that any pair of smaller triangles can be put together to form a new isosceles triangle.
Has the professor once again made a mathematical blunder, or do such triangles indeed exist?
It's possible if Professor Erasmus can reflect triangles when creating isosceles triangles. If we imagine them as being cut out of paper, this would be by flipping over the paper.
This is an isosceles triangle if we set $x=\sqrt2-1$, making the bottom and right edges both have length $\sqrt2$.
The two leftmost triangles form a symmetric isoceles triangles, and the rightmost two form a right isosceles triangle. For the outer two, the left triangle can be flipped into the position of the middle one to form the same right isosceles triangle.
The professor has blundered. The triangle he claims to have constructed cannot exist in Euclidean space.
(I apologize for the lack of pictures. I know they would make my explanation/proof easier to follow, but I do not currently have access to any tools to create or upload them.)
Let's start by considering an isosceles triangle, $ABC$, that has been constructed from two of the component triangles. WLOG, let $AB$ be the base. Notice that if this is an equilateral triangle, there is no way to "add" a third triangle to get a bigger isosceles triangle. So it can't be an equilateral triangle.
The "Professor-Erasmus-Triangle" (PET) must have the two equal sides equal in length to the longest edge of $ABC$, since we can't add something to $ABC$ and make it smaller. (Not exactly rigorous, I know, but this isn't Math.SE, so I'll be a little more flexible.)
If $ABC$ is a "short" isosceles triangle, that is, $AB$ is the longest edge, we can create the PET by extending $AC$ to some point $D$ where $BD=AB$, creating third component triangle $BCD$. (Sorry I don't have pictures.) Now, since we assumed that $AB=BD$ is longer than $BC$, we can say that either $BD$ or $CD$ is the longest edge of $BCD$. Suppose that we have a copy of one of the first two components of $ABC$, that we can add to $BCD$ to make a different isosceles triangle. So, this triangle shares an edge with $BCD$. It can't share $BC$, because it takes both component triangles to make an isosceles triangle that way. However, if it shares either of the others, it either shares the longest edge or has an edge of equal length (by virtue of making an isosceles triangle). This means that it is too big to "fit" in $ABC$, which means that isn't one of the component triangles. All of this means that our first triangle, $ABC$ can't be "short".
So, what if it's tall? So then $AC=BC>AB$. In this case, we can create the third compenent and the PET by extending $CB$ to a point $D$ where $AC=AD$. We can also find the first two component triangles by choosing a point $E$ on $CB$ were $ED=AD$. So now we have the Professor-Erasmus-Triangle $ACD$, with smaller triangles $ABD$, $ABE$, and $ACE$. Also, just by its construction, we know we can combine $ABE$ with either of the other smaller triangles to make an isosceles triangle. So, the only question that remains is, (how) can we combine $ABD$ and $ACD$ to make an isosceles triangle?
By their construction as parts of an isosceles triangle, we know that $AC=AD$ and $\angle ACE=\angle ADB$. Also, by combining with the other component triangle we get isosceles triangles, so $BC=AC$ and $DE=AD$. Then, by transitivity, $BC=DE$. Removing the segment $BE$ gives us $BD=CE$. This means that the two triangles have side-angle-side congruence.
In order for the two triangles to be combined into an isosceles triangle, they must have angles that are supplementary to each other (they make $180^{\circ}$ together). Since the triangles are congruent, these angles have to be right angles. However, if $\angle ACE=\angle ADB=90^{\circ}$, that leaves the third angle of the PET, $\angle CAD$, at $0^{\circ}$, which can't be. If $\angle CAE=\angle BAD=90^{\circ}$, that means that $\angle CAD>180^{\circ}$, which is also impossible. Finally, if $\angle AEC=\angle ABD=90^{\circ}$, then the third component triangle, which is between those two, has two zero-degree angles, $\angle AEB$ and $\angle ABE$. Since none of these are possible, that means that these two triangles, as constructed, cannot make an isosceles triangle. This, in turn, means that the PET cannot be "tall".
Since the Professor-Erasmus-Triangle is not "short," is not "tall," and is not equilateral, we are forced to conclude that it does not exist.
Note that this result does not necessarily hold for non-Euclidean spaces. For example, in elliptical geometry, where interior angles of a triangle add up to more than $180^{\circ}$, you could have an equilateral triangle with three $90^{\circ}$ angles, with the smaller triangles formed by trisecting one of the angles.
The only way to divvy up an isosceles triangle into three smaller triangles without flipping is to start with an isosceles right triangle and halve it to make two smaller isosceles right triangles. The third triangle is degenerate and coincident with the hypotenuse of one of the smaller right triangles.
If you alter the sectioning, you won't be able to create an isosceles triangle out of at least one pair of triangles.
If you alter any of the angles of the starting triangle, you'll have a scalene triangle and/or you won't be able to create an isosceles triangle out of at least one pair of triangles.
The degenerate triangle can't be created by a "cut", so Professor Erasmus's claim "that he can cut his triangle into three smaller triangles" is false, which means the Professor-Erasmus-triangle can't exist as described. However, he hasn't necessarily made a mathematical blunder because the geometry works, albeit with degeneracy.
