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Impossible Triangles?

Age 16 to 18
Challenge Level

Luke from St Patrick's Bryndwr sent us in some good descriptions of how we could build triangles of the shapes given using different types of triangles to those in the picture, but looking at the angles in the corners. Good problem solving Luke!

This full solution came from a workshop on proof in which NRICH worked with several keen and enthusiastic 6th formers from many school

Board 1-- can't be completed
T2 triangle has lengths $1$, $1$ and $\sqrt{3}$. T1 triangle has sides of length $1$. Both triangles have areas of $\frac{\sqrt{3}}{4}$ . Since the board is made from a combination of T1 and T2 triangles, this means that the total area of the equilateral triangle must be a whole number multiple of $\frac{\sqrt{3}}{4}$. Now, since both of these types of triangle are found at a corner then one side of the triangle must contain a piece of length $\sqrt{3}$ and also a piece of length $1$. Thus the length of the sides of the equilateral triangle must be $A+B\sqrt{3}$ with neither of $A$ and $B$ zero.The area of a triangle with a side length of this form cannot be a whole multiple of $\frac{\sqrt{3}}{4}$, which is a contradiction. Thus the board cannot be tiled in this way.

Board 2 -- can't be completed
No. To see why, we make two points

1. If the board could be completed then the area of the board must be a whole number, because it is made from a sum of triangles with whole number areas.
2. If the board has a side of length $L$ then, because it is an equilateral triangle, the area would be $L\times L\times \frac{\sqrt{3}}{4}$. An equilateral triangle of area $1$ must have side length of $\frac{2}{3^{\frac{1}{4}}}$ and an equilateral triangle of area $2$ must have a side length of length $\frac{2\sqrt{2}}{3^{\frac{1}{4}}}$
Since there is at least one of each triangle touching each side, we know that the overall side length looks like ($n$ and $m$ are whole numbers)
$$L = \frac{2n\sqrt{2}}{3^{\frac{1}{4}}}+\frac{2m}{3^{\frac{1}{4}}}$$
The total area of the triangle will be
$$L\times L\times \frac{\sqrt{3}}{4}L= (n\sqrt{2}+m)\times(n\sqrt{2}+m)=2n^2+m^2+2nm\sqrt{2}$$
This can't be a whole number unless either $n$ or $m$ is zero! So the board is impossible!

Board 3 (This is the trickiest part, although each line is not too bad by itself)
Can we use these area tricks in this third case? We look again at the area from the formula for the area of a triangle and also the sum of its pieces

1.The areas of the two small triangles on the board are $\frac{1}{2}$ and $\frac{\sqrt{3}}{4}$. So, the total area will be $\frac{n}{2}+\frac{m\sqrt{3}}{4}$ if we use $n$ and $m$ of each triangle.
2. If the hypotenuse is of length $a$ then the area of the triangle is $\frac{a^2}{8}\sqrt{3}$

For the two expressions for the area to match, we would need

A bit of surd work turns this into

$$a^2 = 2m+\frac{4\sqrt{3}}{3}$$

Now, what could $a$ be? The side lengths of the triangles in question are $(1, 1, \sqrt{2})$ and $(1, 1, \sqrt{3})$, so for whole number $L, M, N$ we could have
$$a = L + M\sqrt{2}+N\sqrt{3}$$
Choosing $M=0$ gives a compatible area. So this board might still be possible based on a calculation of areas.We'll leave this part as an open challenge!