35 of 100: Most People Don't Find Them All

There are $6$ squares in the original picture (four small, one medium and one large). If we take out each square like the picture above, then each square contains $8$ triangles (four small and four large, where each large one consists of two small ones). But $4$ of small triangles of the medium square are counted twice this way (once as smaller triangles of itself and once as larger triangles of small squares).

Thus in total we have $6\times 8-4=\color{#D61F06}44$ triangles.

8 triangles that are only a given color. 4 colors. 32 single color triangles.

2 triangles that are red and orange. 4 neighboring color pairs. 8 two color triangles.

4 triangles that use three colors.

Total 44 triangles.

16 single triangles

16 triangles made of two of those, there are 4 in each colored area.

8 triangles made of four of the simple ones, these take in two colored areas. The ones in the two areas which are side by side will have the longest side horizontal, either at the top or bottom, with the third vertex then down or up from there. Another four will be where the two areas are above each other, with the longest side of the triangle vertical, on the left or right.

4 triangles each made of 8 simple triangles. These will have a diagonal as their longest side and two sides of the overall triangle as legs. Either diagonal works and the third vertex can be either up or down from it.

I think there are two stages to solving this:

1) Find all the types of triangle

2) Count how many triangles there are of each type

To help with part (1), note that all of the triangles are right-angled isosceles triangles. [The smallest angle in the diagram is $45^{\circ}$ and all of the other angles are multiples of $45^{\circ}$ . The angles in a triangle add up to $180^{\circ}$ , so the only way to do this is with $45^{\circ}$ + $90^{\circ}$ + $45^{\circ}$ .]

A second observation is that the number of small triangles that makes up a bigger triangle is always a power of 2. [The side-lengths are either multiples of the side of the small triangle, or of the hypotenuse of the small triangle, which involves $\sqrt{2}$ ]

In the diagram below the smallest triangle is shown in yellow. There are 16 of these.

Putting 2 of the small triangles together forms this yellow triangle. The right angle of this triangle can appear in any of the four corners of the four coloured squares. This means there are 16 of these:

This yellow triangle is formed of 4 of the small ones. This time I have marked the places where the right angle of this triangle can go. There are 8 places, so there are 8 of these triangles:

Finally there are triangles that are 8 of the small ones. There are 4 of these:

In total there are $16 + 16 + 8 + 4 = 44$ triangles of different sizes.

Et voilà.

I solve these puzzles in the most boring possible way using a graph based algorithm. The edges of the graph are pairs of nodes that connect in a straight line regardless of intervening intersections.

With this method there is no dependency on the overall shape, triangular, rectangular grid, etc. found in these puzzles.

[Aside: I was not impressed with the presentation in the style of click bait "most people ...", "only x% of people .." Suggested title "Hot Cross Boxes" ]

2017 July 9 Computer implementation. There is one subtlety, the orientation of each edge must be noted in order to detect when two edges connect into a single line not two independent sides of a triangle. No change in orientation, no triangle. Apart from that the edge to straight line mapping and ordered search for circuits in increasing node index method in code gets the correct count.

This is my solution. The pictures are self-explanatory.

Count the squares. In each square there is 8 triangles, plus 12 combinations.

16 of 16 of 4 of 4 of 4 of

Total: 44

Solved it by SHEER LUCK.

By counting, l obtained this:

$16+ (8\times 2) + (4\times 2) + (1\times2) + (1\times 2) = 44$