24

Here's a solution for 7 triangles:


17

First of all, by a simple geometry principle: $\triangle CED$ and $\triangle AED$ have the same base $|ED|$ and the area ratio between $\triangle CEF$ and $\triangle CFD$ has to be the same as the length ratio between $|EF|$ and $|FD|$ so, similarly, you can tell the area ratio between $\triangle AEF$ and $\triangle AFD$ as $2x$ and $3x$. Notice from ...


16

The area of ? is: Because: Working from there:


11

Solution:


11

Total number of triangles: w : white half parallelograms b : blue half parallelograms So... Image:


9

The answer is - Explanation-


9

It's possible if Professor Erasmus can reflect triangles when creating isosceles triangles. If we imagine them as being cut out of paper, this would be by flipping over the paper. This is an isosceles triangle if we set $x=\sqrt2-1$, making the bottom and right edges both have length $\sqrt2$. The two leftmost triangles form a symmetric isoceles triangles, ...


9

Using non-overlapping sticks I get 36. I'll give measurements and positions to make this easy to picture. Imagine we're drawing this on graph paper. Make one horizontal stick of width eight units. Pick a point somewhere above it, and attach sticks from it to each of 9 points on the line, all one unit apart. Now how many triangles are there? say the ...


9

Now that we have two correct answers, I figured I'd present my own approach. It's similar to Paul's but doesn't work with the ratios of the side lengths, but instead directly with the ratios of the areas. I'm splitting the solution into several spoiler paragraphs so that they can be used as individual hints towards the solution. As a corollary, note that ...


8

No, it's not possible. For any vertex, the three edges coming out of it must have distinct length, since otherwise the triangle must be isosceles as well as right, which forces an equilateral triangle for the opposite face. Therefore, the three angles at each vertex must correspond to the three angles of the right triangle; call them $\alpha$, $\beta$, and $...


8

Here is the solution to the puzzle (note the correction in "3,1,2,5" to "3,1,2,1,4", by comment here):


7

Here's an approach that I think is easier than the other approaches...


7

Here is a convex dodecagon made of $50$ of those triangles. Can it be done with fewer?


6

Here's one with six triangles (7 if you count triangles outside of triangles, which you don't):


6

I was trying to post this 5 minutes before the other answer, but got snookered by camp wifi


6

Assuming "transportation cost" means sum of distances to each of the three roads, and the side of the equilateral triangle has length $1$:


5

It is possible to do better than a hexagon, if an irregular polygon is acceptable. It is also possible to construct an equilateral triangle or a hexagon. On reflection (and thanks to @Hugh's comment) you probably can't make a square — but can get relatively close to a square. By taking $\sqrt{3}$ by $1$ squares, each made from two triangles, you could ...


5

I created 11 independent triangles with ...


5

Two 5-point stars should do the trick! This solution uses 10 'sticks' (each stick spans vertex-to-vertex) of exactly the same length in 2D. As someone commented briefly (sadly, it was deleted too quickly for me to reference), we get 10 triangles even with a single 5-point star: 5 large, 5 small, and the large triangles overlap the small triangles. Note: ...


5

Answer: size 1: 48 size 4 : 30 size 9 : 20 size 16 : 12 size 25 : 6 size 36 : 2 Total: 118


5

Here is one possible triangulation. Note that all triangulations (that do not introduce extra vertices) will have the same number of triangles, which depends only on the number of sides in the polygon. A sketch of the proof: A polygon with three sides is already a triangle, so the minimal number of triangles needed to triangulate it is exactly $1$. For a ...


5

My answer:


4

If the triangles must have a named(A to H) point at each corner then I think the answer is actually: Count: The triangles in one half only(left half in this case): (B,E,D) (B,E,G) (B,G,D) (A,D,E) (A,D,G) (A,G,E) (A,B,E) (A,D,B) so 16 for both halves plus the triangles which go over both halves (A,H,G) (A,C,B) (D,G,H) (D,E,F) (D,...


4

The professor has blundered. The triangle he claims to have constructed cannot exist in Euclidean space. (I apologize for the lack of pictures. I know they would make my explanation/proof easier to follow, but I do not currently have access to any tools to create or upload them.) Let's start by considering an isosceles triangle, $ABC$, that has been ...


4

Here are the solutions to both questions:


4

Here are the solutions to the five problems.


3

Here are at least two solutions (up to reflection and rotation)


3

with 7 lines could be same as the previous answer


3

I count 40 different triangles in the picture To check this I write them down in a sorted list of triples. Each triple is normalized in that way that the components of the triple are alphabecticaly ordered. 1. A B C 2. A B D 3. A B E 4. A B J 5. A C D 6. A C F 7. A C J 8. A D E 9. A D F 10. A D G 11. A ...


3

I think I managed to find


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