### Random Post

### Recent Posts

### Recent Comments

### Archives

### Categories

- article (11)
- enigma (1,176)
- misc (2)
- project euler (2)
- puzzle (44)
- site news (46)
- tantalizer (48)
- teaser (3)

### Site Stats

- 183,172 hits

Advertisements

Programming Enigma Puzzles

3 October 2012

Posted by on **From New Scientist #2885, 6th October 2012** [link]

Joe thinks he has found, by trial and error, a new property of triangles. He asked Penny to mark the points A along the sides of a triangle of her choice, each point to be a percentage P of the length of the side from the nearest corner. Then she had to join the points to form a second triangle and repeat the process to form a third triangle with points B. Penny found that the lengths of the sides of this third triangle were 52 per cent of the corresponding sides of her original.

What was the value of P?

[enigma1718]

Advertisements

%d bloggers like this:

Another problem that’s solved with a bit of analysis and algebra. I’m not quite sure what the general property of triangles Joe thinks he has discovered, as it seems to me that Penny has to choose her particular triangle with care (or luck) in order for the situation described in the puzzle to come about. Anyway, here’s a SymPy solution.

Solution:P = 20%.In order for the B triangle to be similar to (in the mathematical sense) the initial triangle under the transformation given in the puzzle, it is a necessary condition that the triangle is equilateral.

The equation simplifies to:

or:

which factors as:

hence p = 1/5 = 20% as we want the root less than 50% (the new points are a certain percentage length from the

nearestcorner).Although the text of the puzzle doesn’t make it clear, the accompanying diagram shows the smaller triangles being twisted in the same direction. I think the variation of this puzzle where the B triangle is constructed using the same technique as the A triangle, but twisted in the opposite direction (as shown in the diagram below) makes for a more pleasing puzzle as the scenario works for any triangle. And the solution is the same.

Here’s a more complete solution that doesn’t start by assuming the triangle is equilateral (although it deduces it), but instead makes use of repeated application of the cosine rule. Again it uses SymPy.

It is interesting to note that for any triangle (not only equilateral), the ratio of circumference of the 3rd triangle and the original triangle is a quadratic function of P: r = (a3+b3+c3)/(a+b+c) = 3*P^2 – 3P + 1. So, if the ratio of the circumferences is 52%, then solving this quadratic equation r=3*P^2 – 3P + 1=0.52, one gets P=0.2 (20%) and P=0.8 (80%).