From the people behind the
Alan Turing Cryptography Competition.

Ruth and Leonie have recently become interested in Pythagorean triples, which are whole number solutions to the equation $a^2+b^2=c^2$. Geometrically we can think of these as the side lengths of right angled triangles where each side length is a whole number.

Some Pythagorean triples are boring, for example $(6,8,10)$ satisfies $6^2+8^2=10^2$, but this is just the triple $(3,4,5)$ with each entry doubled. Call a Pythagorean triple 'primitive' if it is not a multiple of another Pythagorean triple.

There is a remarkable structure to the set of primitive Pythagorean triples. Given a triple $(a,b,c)$ we can imagine it having three children $$(a_1,b_1,c_1)=(a-2b+2c,2a-b+2c,2a-2b+3c)$$ $$(a_2,b_2,c_2)= (a+2b+2c,2a+b+2c,2a+2b+3c)$$ $$(a_3,b_3,c_3)= (2b-a+2c,b-2a+2c,2b-2a+3c)$$ Each of the children of a primitive Pythagorean triple are themselves primitive Pythagorean triples. Moreover, every single primitive Pythagorean triple is descended from $(3,4,5)$ in this way.

Ruth challenges Leonie to find lots of primitive Pythagorean triples where $\frac{c_1}{c}, \frac{c_2}{c}$ and $\frac{c_3}{c}$ are all larger than 2. After thinking about it for a bit, she realised this really only depends on $\theta$.

There exist angles $\theta_1$ and $\theta_2$ between $0$ and $90$ degrees such that a triple $(a,b,c)$ has $\frac{c_1}{c}, \frac{c_2}{c}$ and $\frac{c_3}{c}$ all larger than 2 if and only if the angle $\theta$ with $\tan(\theta)=\frac{a}{b}$ has $\theta_1\leq \theta\leq \theta_2$. What is the difference $\theta_2-\theta_1$? Write down your answer to the nearest half a degree.

This problem was first solved on Wed 24th February at 4:02:43pm

Mathsbombe Competition 2021 is organised by the The Department of Mathematics at The University of Manchester.

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