How many lines are needed to create 6 triangles on W?

Question

Basically, the question started with a little argument I had with my friend. My friend said he thinks it's possible to draw only 2 lines on the letter "W" and make 6 triangles, and I played around with it, but I couldn't really do it, so I told him I don't think it's possible, and we need at least 3 lines.

We kept arguing, but I really couldn't prove that we need more than 2 lines on letter "W" to create 6 triangles except that I couldn't find a way of doing so. Is there any proof for this kind of problem?

PS. Sorry. I didn't clarify. You CANNOT combine one or more triangles to form a triangle and count that separately. Triangles must be INTERNALLY disjoint. For instance, if triangles ABC, BCD can be combined to form triangle ADC, then we don't count ADC as a triangle.

Can the triangles overlap? May their interiors contain a line segment? If so, you can actually create 8 triangles with two lines. — JosephSlote, Dec 05 '14 at 04:39
Alright I'm starting to believe it's impossible as well, though I'm not sure what math structure to hang the picture on. Graphs? A clever application of Euler's formula may help you... — JosephSlote, Dec 05 '14 at 05:27

score 1 · Answer 1 · answered Feb 17 '18 at 01:18

1

No one said the W couldn't be "cursive"...

Perhaps not what the OP was looking for (and I'm aware this post was brought back from YEARS AGO, but I just thought I'd share this) but this solution does contain $6$ triangles with only $2$ lines

answered Feb 17 '18 at 01:18

WaveX

5,440

Fmonkey2001 · Answer 2 · 2014-12-05T05:07:44.643

0

It only takes two lines to get six triangles? Let me try to add a picture real quick.

So far the max I can find with two lines is $5$. Ill keep trying!

edited Dec 05 '14 at 05:07

answered Dec 05 '14 at 04:41

Fmonkey2001

1,240

I don't think OP counts $BCF$ or $ACD$ – Ross Millikan Dec 05 '14 at 04:49
I'd assume the intent is for internally disjoint triangles: so I count only 4. – Nick Matteo Dec 05 '14 at 04:50
What if I moved $\overline {CD}$ and made it $\overline {CD'}$ where $D'$ was on $\overline {AD}$? Would that help? – Fmonkey2001 Dec 05 '14 at 04:51
@Fmonkey2001: I still count 4. – Nick Matteo Dec 05 '14 at 04:59
Why doesn't the intersection of $HC$ and $BE$ count as a vertex? I mean, with that one, we do have five triangles (if we call the point $I$, $IBF$ and $IFG$ are two triangles). – Avatrin Mar 31 '17 at 10:57
Move AF a bit up and we are done – J. P. C. Mar 31 '17 at 19:34

score 0 · Answer 3 · answered Mar 31 '17 at 19:32

0

Consider this two lines / but a bit more inclined and one intersecting each other. Now the point where they cross has to be in one of the central side in W. Sorry for no having a picture.

answered Mar 31 '17 at 19:32

J. P. C.

155

score -1 · Answer 4 · edited Dec 05 '14 at 05:27

-1

Three parallel lines one at the top, one middle, one bottom, and, if tri inside tri is allowed, you have $6$. Of these, $3$ are true and $3$ more are built around their respective parent.

edited Dec 05 '14 at 05:27

Vladhagen

4,878

answered Dec 05 '14 at 05:00

Robo

1

How many lines are needed to create 6 triangles on W?

4 Answers4