Alright, back in seventh grade a sadistic math teacher presented us with a geometry problem.
First, draw a large ‘H’. Then, using three straight lines, create seven triangles.
I filled notebooks trying to figure this one out and for the life of me, I can’t get more than six three-sided figures. What do y’all think?
-Ozzy:confused:
Draw the H. The points at the ends of the sides are A, B, C, and D. The points where the bar in the middle intersects are E and F. Let G be the midpoint of EF. Draw segments AG, BE, and BG. Let the intersection of BE and AG be H. The seven triangles are AHE, EGH, AGE, BEF, BFG, BEG, and BGH.
Are “triangles inside of triangles” allowed to be counted? For example, connect the top right of the H with the bottom left, and connect the top left of the H with the bottom right. Do it so that these two diagonals, along with the horizontal bar across the H, all intersect in a single point. That in itself gives you 6 triangles if you count all of the triangles. From there, it’s easy to draw a third line and get seven (and more) triangles.
:smack: D’oh, sorry, I forgot that detail. No, there are no triangles within triangles. You should be able to fill them in with seven different colors and end up with seven full colored triangles. Thanks.
-Oz
This link purports to solve it though I really can’t follow the ASCII line drawing instructions.
Interesting. The solution I gave has exactly seven triangles, but some of them are inside each other.
Astro -
Good call. It took a few seconds to decipher but that’s the right answer. At last, I can sleep easy at night.
-Oz
Neat.