# Impossible triangles?

Which of these triangular jigsaws are impossible to finish?

You are given a limitless supply of triangular jigsaw pieces of type T1, T2, T3, T4 and T5

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Using these pieces, you need to try to make larger triangular shapes without any overlap.

Three triangle shapes are shown in the picture below. Is it possible to create two-coloured triangles of these shapes without moving the pieces already placed?

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Suppose next that you can move the pieces and choose two triangle types to work with. Which pairs of shapes can be used to make larger equilateral triangles? Which pairs of shapes can be used to make larger 30-60-90 triangles? See the video clip for a discussion on this part of the problem

NOTES AND BACKGROUND

This problem is all about either finding solutions or proving that there is no solution for any size of triangle shape.

For small triangle shapes, it is easy to check all possible configurations of pieces to check whether a solution exists. For larger triangle shapes the number of combinations of pieces gets larger extremely rapidly, and quickly reaches the point at which a check of all of the combinations is impossible, even on a supercomputer. Even if we have checked a large number of jigsaw sizes and found no solution, this does not necessarily mean that we cannot find a solution for a larger triangle shape. To find a solution, you often need to mine the depths of your cunning and ingenuity. To show that a solution does not exists you often have to use the concept of proof by contradiction .

Proving that a solution

**does not**exist is often much easier than proving that a solution does exist. Interestingly, if a solution can be found, then it is usually very quick to check that the solution is correct, although finding the solution in the first place might be exceptionally difficult.This behaviour underlies the notion of the mysteriously titled 'P vs NP' problem, the solution of which will earn the solver $1,000,000

Ideas concerning proof are discussed in the fascinating article Proof: A Brief Historical Survey

You might first like to try the slightly easier problem Equal Equilateral Triangles to get a feel for the possibilities.

Will the areas of the shapes be rational or irrational? What lengths of sides would be able to give you these areas?

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

$$\frac{a^2}{8}\sqrt{3}=\frac{n}{2}+\frac{m\sqrt{3}}{4}$$

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!

### Why do this problem

This problem will make students experiment, conjecture and prove. They will need to understand the interplay between rational and irrational numbers.### Possible Approach

This activity lends itself to hands-on experimentation with
cut-outs of triangles before algebraic proof is attempted. Without
cut-out shapes it is a good exercise in visualistion. Students
might need to be encouraged to pursue various lines of
enquiry.

Explaining a proof clearly to someone else is a very good way
to discover where the holes in the argument are. You could ask
students to explain their proofs to each other in pairs and then to
the whole class. Can the class point out areas where proofs or
explanations are unclear?

### Key Questions

How do we relate side-length and area of triangles?Which variables in the problem are rational and which are irrational?

Can we make use of the angle properties of the triangles?

### Possible Extensions

Can you suggest other regular shapes which you could prove to be impossible?### Possible Support

Try the easier question Equal
Equilateral Triangles first.

Struggling students could be asked to hear other students
proofs and comment on their clarity.