Basically, the question started through a tiny argument I had with mine friend. My friend said he thinks it"s feasible to draw only 2 currently on the letter "W" and also make 6 triangles, and I played approximately with it, however I couldn"t really carry out it, so i told him i don"t think it"s possible, and also we require at least 3 lines.

We retained arguing, but I really couldn"t prove that we need much more than 2 currently on letter "W" to create 6 triangles except that ns couldn"t discover a method of law so. Is there any proof because that this kind of problem?

PS. Sorry. Ns didn"t clarify. Girlfriend CANNOT combine one or an ext triangles to kind a triangle and count that separately. Triangles must be within disjoint.For instance, if triangle ABC, BCD have the right to be an unified to form triangle ADC, then us don"t count ADC together a triangle.

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No one stated the W couldn"t be "cursive"...

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Perhaps not what the OP was searching for (and I"m conscious this article was brought ago from years AGO, however I just thought I"d share this) however this equipment does save $6$ triangles with just $2$ lines


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It just takes two lines to obtain six triangles? allow me shot to add a photo real quick.
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So much the max i can find with 2 lines is $5$. Ill save trying!


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Consider this 2 lines / but a bit more inclined and one intersecting each other. Currently the suggest where lock cross needs to be in among the main side in W. Sorry because that no having a picture.


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Three parallel currently one in ~ the top, one middle, one bottom, and, if tri inside tri is allowed, you have $6$. Of these, $3$ room true and $3$ an ext are built approximately their particular parent.


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