## Teaser 2992: Crystal cleavage calamity!

**From The Sunday Times, 26th January 2020** [link]

My “mystic” crystal stood on its triangular end face, with every side of the triangle being less than 10cm. It had three vertical sides and an identical triangular top end face, with the height being under 50cm. Unfortunately, during dusting, it fell and broke in two. The break was a perfect cleavage along an oblique plane that didn’t encroach on the end faces, so I then had two, pointed, pentahedral crystals.

Later, I found that the nine edge lengths, in cm, of one of these were all different whole numbers — the majority prime — and their total was also a prime number. Curiously, the total of the nine, whole-number, edge lengths, in cm, of the other piece was also a prime number.

What was the total for this latter piece?

[teaser2992]

## Jim Randell 9:44 am

on25 January 2020 Permalink |Here’s my first stab at a program. It’s a bit simplistic and slow, but it finds the required solution.

Run:[ @repl.it ]Solution:The total is 113 cm.The crystal is a triangular prism. The base of which forms a triangle with sides of 5cm, 7cm, 9cm.

There are two possible sizes for the original crystal – either 44 cm tall or 48 cm tall.

Each of these two sizes can break in two different ways, but in each scenario the sum of the edges of the “other” piece is always 113 cm.

If we think about the triangular prism as a cardboard tube then we can “unroll” it to get the following flat diagram:

The base of the prism consists of a triangle of sides

a, b, c, and the overall height of the prism ish.The fracture is formed by the red line, and the lengths of the fracture line across the vertical faces are

d, e, f.Note that

d, e, fare the hypotenuses of right-angled triangles with bases ofa, b, crespectively.And in order for the fracture to line up the lengths of two of the vertical sides of these triangles must sum to the length of the third.

Subject to the fact all the distances are integers such that:

a, b, c < 10; h < 50, we can find candidate solutions fora, b, c; d, e, f; x, y, z, such that each takes on a different value, the majority of them are prime, and their total sum is prime.We can then consider possible values for

hthat make the edges of the top piecea, b, c; d, e, f; h – x, h – y, h – zalso sum to a prime number.We find there is only one possible shape for the base of the prism (a (5, 7, 9) triangle), and for the corresponding lengths of the fracture lines (13, 25, 15).

There are two possible values for overall height of the prism (44 or 48), and in each case there are two possible sets of

x, y, zvalues, giving us 4 possible solutions.However in each one the sum of the lengths of the edges of the top piece is 113 (the lengths of the edges for the top piece are always: 5, 7, 9; 13, 25, 15; 13, 1, 25).

Here are the diagrams corresponding to the four possible solutions:

LikeLike

## Jim Randell 4:09 pm

on25 January 2020 Permalink |Here’s a faster program. It constructs the possible lengths of the fractures (which are the hypotenuses of right-angled triangles) up front, and then uses these to construct possible prism shapes.

It runs in a much more respectable 180ms.

Run:[ @repl.it ]LikeLike

## GeoffR 9:03 am

on30 January 2020 Permalink |I also found two possible solutions, with the same edge total of the second piece.

LikeLike