# Touching triangles at their vertices

What is the minimum number of non overlapping congruent triangles arranged in the plane, such that each vertex of the triangles coincide with exactly three triangles?

• Can someone give me an example of this? – risky mysteries Jul 28 '20 at 1:21
• @risky In case you haven't seen it yet, check out Jaap's answer. – Daniel Mathias Jul 28 '20 at 11:42

## 3 Answers

The number of triangles in my best solution is

20

but I don't know if this is optimal.

If you take kites formed from four 30-30-120 triangles, you can place a number of them in a ring, like this: This picture shows six such kites because that was easiest for me to draw, but you can remove one and make a loop with 5 kites, for a total of 20 triangles. @humn kindly made a picture of that: If you use four kites then they start touching each other, so you'll get too many triangles at a vertex. @humn made a picture of that too: Addendum:

I previously had an incorrect solution, as I used similar triangles instead of congruent ones (i.e. I used some triangles of a different size but the same shape). As requested by @humn, I'll keep that incorrect solution with 12 triangles available below:

The triangles are 20-80-80 isosceles triangles so that three can go together to form the corner of an equilateral triangle. There are then 3 smaller copies needed in the middle to fill things up and make sure there are three triangles at every vertex. • These triangles are not congruent. They are only similar (same angles, different side lengths) – happystar Jul 28 '20 at 7:43
• @happystar I have changed my solution, so now it should be valid. – Jaap Scherphuis Jul 28 '20 at 8:31
• I think the four kite (16 triangles) solution is valid if the condition is taken to mean each vertex is a vertex of exactly three triangles. – Daniel Mathias Jul 28 '20 at 11:40

A liberal interpretation of “coincides” in the puzzle statement ...

“each vertex of the triangles coincides with exactly three triangles”

... allows for a vertex to touch a side, and not always another vertex, of another triangle as in these failed attempts ... ... that led to this pair of ...

... 16-triangle liberally-interpreted solutions. • Very nice solution, and in my opinion a completely valid one that improves on mine. – Jaap Scherphuis Aug 1 '20 at 21:06

suppose plan area is A and no of triangles are n,

A = n * 1/2 * h * b

so if we want n,

n = 2A /(hb), that is the minimum no of triangle