Try Triling ("Triangular-Tiling")

Question

You are given a grid. Some of the cells in the grid are labelled with positive numbers. You must partition the grid into triangles.

• There must be a triangle for each labelled cell
• Each cell must be completely contained within its triangle
• The vertices of each triangle must be intersection points of the grid
• Every triangle has one (or more) edges that are either horizontal or vertical
• The cell's label specifies the area of its triangle

Here is an example to illustrate the problem.

Observe that the restriction that every triangle must have a horizontal or vertical edge means that it is easy to work out its area. For example the triangle labelled 6 has base length 3 and height 4. So its area is $\frac{1}{2}(3 \times 4) = 6$ as required.

Here are two problems to solve, an easy one for practise and one that requires a little more effort. Both problems have a unique solution.

You can of course solve it any way you like, but I find it easiest to print out the pictures and then draw lines on them using a ruler. There is no need to use a computer - they're not that hard! It would be nice for an answer to specify some of the steps used to find the solution.

I came up this idea myself. I find the restriction on valid triangles gives a nice balance between having too many options and too few options when partitioning the grid. I'm not aware of anyone else having already used the idea, but it wouldn't surprise me if it were not original. But the instances of the problem used here must be original.

Answer 1

Here is the solved ‘hard problem’:

Long story short I started with the pair of 20s in the bottom and found that only these pair of triangles work as a configuration:

This then forces this since the 3 triangle has to be positioned this way to enable a 12 triangle to be formed…

Which then forces this

The 5 has only one way to be formed… Leading to this:

The three triangle has two options but one of which doesn’t allow the 5 triangle so it forces it to be like so: After that there is only one way to complete the grid.

Answer 2

Solution to the Practise problem:

And how to get there:

First, some observations:
Every triangle has (at least) one horizontal or vertical edge, and all the vertices must be on grid-points, so that edge has integer length.
If we call that edge the base of the triangle, its perpendicular height (to the vertex not on that edge) is also an integer.

So for each area there are a very limited number of options for base and height because they must be factors of 2*area.

No triangle can be 1-by-anything, because it would not contain any whole squares.
Prime area triangles can only be 2-by-P.
So area 2 triangles can only be 2-by-2, and must be a right triangle to include a whole square.

Adjacent numbers must have a horizontal/vertical edge between them.

Start by drawing edges between adjacent numbers, and completing all the forced area-2 triangles.

The 5 in the top-left corner must be 2-by-5, and nearby numbers on the edges mean there is only one possibility.

The other 5 on the left edge (R6C1) has a few possibilities, but must have a vertical 5-long edge on the left of the grid.

Area 15 triangles must be 5-by-6 or 3-by-10 (2-by-15 doesn't fit the grid). So the 15 on the bottom edge can only be 3-by-10, and must use all of the bottom edge. There are two ways it can fit without hitting other numbers, but only one of them avoids the left-edge 5. And that gives us two edges of that 5 triangle which can now be completed:

The 15 near the right (R7C9) must also be 3-by-10, and the only way to fit a horizontal/vertical edge is by using the entire right-hand edge of the grid. And the diagonally adjacent 6 gives us only one way to fit this triangle.

The remaining 2 is now forced. And with it, the nearby 4:

The 3 in R7C3 must be 2-by-3, and is sitting on a 2-long horizontal. There is only one way to complete it without hitting any numbers.
And that also forces the other 3's which must use up the space around it. That gives us two edges and a corner around the 5 in R5C7, forcing it to extend to the top-right corner:

Finally the last 15 must be 5-by-6, so must reach all the way into the apex on the left edge, and we are done.

Answer 3

Solution for the Practise Problem

I started

with the simpler triangles (smaller numbers) and from there worked my way up. 2 for example must be a 2x2 triangle, 3 must be a 2x3 triangle, 5 must be a 2x5 triangle etc.. And since the whole grid needs to be partioned, meaning each of the vertices (except the 5 in left top corner) must be a vertex of some other triangle/triangles, it didn't require too much work to work out the rest.