# Tag Info

25

Here's a solution for 7 triangles:

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First, observe Next, let's These events are

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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 ...

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

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The proof is in two parts, corresponding to the two linkages which are joined to each other at a single point. For each part, I'll try to both explain in words and illustrate on the picture you've provided.

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Solution:

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

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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 ...

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 ...

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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, ...

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A generalization to arbitrary centrally symmetric regions A planar region is called centrally symmetric with respect to the origin, if for every point $P$ in the region also its reflection with respect to the origin is in the region. Examples for centrally symmetric regions are for instance circles, squares, regular hexagons, regular polygons with an even ...

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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 $... 8 Imagine putting the three points within the circle one-by-one. We are only interested in the angular coordinates of these points. We use the first point to define the zero point for this angular coordinate. The second and third point have angular coordinates running from$-\pi$to$+\pi$. The combinations of both angular coordinates for which the triangle ... 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): 8 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? 7 Assuming "transportation cost" means sum of distances to each of the three roads, and the side of the equilateral triangle has length$1$: 7 Here is a geometric proof: The angle p is therefore 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 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 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 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 My answer: 5 The answer is because, for example, if T2 has side-lengths then a triangle T1 such as More generally, we can consider T2 with side-lengths and T1 with side-lengths 4 Thanks everyone for their answers. Here's mine which is close in spirit to a couple of the existing ones. Consider a circle centered at the origin. Without loss of generality, let one of the points lie on the positive$x$-axis, call it$p_1\$. We can do this just by rotating the coordinate system and the fact that the probability the origin is one of the ...