Playing around with math puzzles I came across this:
www.city.waltham.ma.us/phy2web/PUzzles/PU_Tuka_Tile.htm
Intuitively, I know the answer is “Z” but I’ll be damned if I can find all the polygons.
Ok, I think I may have it. In figure “Z” the two smaller triangles meet at a point and form a hexagon of sorts, although it is not really an open figure like all the other easy ones. It’s the only way I can make it work though. Think that is valid?
I make it as being figure Y. 4 hexagons and 9 rectangles.
Its y there are 32 million obvious.
whole rectangle 4
left rect 4
right rect 4
upper rect 4
lower rect 4
upper left rect 4
upper right rect 4
lower left rect 4
lower right rect 4
36
now for the hard part think of the center point as a spoke.
now mentally remove the bottom spoke and the right spoke
you have a hexagon that is the whole rect minus the upper left rectangle.
that hex is 6
(now do the some on the other sets of spokes)
remove right, and upper spoke = 6
remove upper and left spoke = 6
remove left and bottom spoke = 6
24
for a total of
I don’t think so, or else figure Y would have at least 2 Octogons.
You’ve totally lost me here.
i didnt spend too long but im almost positive that the answer is “Y”- i count 9 quads and 4 hexagons; 9 quads = 36 million and 4 hexagons = 24 million and adds to 60 million…
so the 9 quads are: (think of them as quadtrants on a graph)… one in each quadrant= 4; + combine the adjacent quadrants (+4 more) and then the entire tile (+1) = 9 quadralaterals
now for the hexagons… if you take out one quadrant in the “graph”, you are left with a six sided figure, you can do this 4 times so thats where the 4 hexagons come from.
I had ruled out the answer “Z” because i count at least 4 traingles, 5 quads, 2 pents, and 4 hexagons; adds up to 66 million… i would explain this but its too hard to put answer “Z” into words. For X, i came up with 4 triagles, 3 quads, 2 pents and 1 hexagon (this is also hard to explain; shapes dont translate well into words :- ) )
Do these hexagons have sides of equal length?
No, why?
I have been under the misconception that a ----gon has sides of equal length, obviously not, it only requires ----sides and ----angles. If I’d known this I’d have found the tile puzzle much easier 
What I meant was draw the figure, but don’t put in the line segment from the center to the bottom, and don’t put in the line segment from the center to the right.
leaving a figure like
XXXXXXXXX000000000
XXXXXXXXX000000000
XXXXXXXXX000000000
XXXXXXXXX000000000
000000000000000000
000000000000000000
000000000000000000
000000000000000000
the ‘0’ are the hexagon left without the spokes.
then there are hexagons for each other three quadrants.
sorry about the capitalization problem.
i think my keyboard needs a new clutch. cause i can’t seem to shift.
No, a regular or equilateral polygon has sides of equal length. An irregular polygon has sides of unequal length. Most people do tend to think of a regular polygon when they hear the term polygon, though.
Wow, I’m impressed. I figured that each of the named figures had to be used at least once, although that doesn’t explain the inclusion of “y” which immediately doesnt have any triangles. Octagons aren’t mentioned so I ignored them. I will have to re-look at “z” because I only found 3 hexagons and ended up with sixty million. Can you describe where the four hex’s you found are? I got two by removing a triange and then the double triange one as well. Did not discover a fouth.
Take out top left triangle. What’s left is one hexagon. Do the same for the lower right triangle, that’s the second. Take out the lower left quadrilateal. That’s the third. Do the same for the top right quadrilateral. There’s the fourth.
Ack. Ignore that. That makes the four pentagons, not hexagons
No, that’s two pents and two hexes. I need sleep. :smack:
here’s the explanation of why the answer is “y”:
http://home.earthlink.net/~biggiefries/y.jpg
Ignore the superflous little lines, I was too tired to take them out after I merged my layers.
So, what do I win by being the first one to post the correct answer?
Yeah, thanks, I get it… now.
QED: You get to impress me, which is really hard to do. I am pretty sure Y is the right answer although I managed to squeeze all 6 types of figures out of Z AND get it to equal 60. What do you think, do I have a viable alternative or is a hexagon composed of two triangles touching at a point too much of a stretch?