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January 29, 2007
Recent math contest
The math contests at school have become a big thing. To prevent the kids from googling my blog, I'm showing the contest from December. Click to embiggie it.
So. Each one of the tiny triangles is an equilateral triangle. However, each of the smaller triangles can also build a larger equilateral triangle. How many triangles are in the figure in total?
Just for a hint, there are between 1000 and 2000. I'm just saying. Points to anyone who can express the solution algebraically. Maybe I'll even take someone out to dinner?
Posted by G at January 29, 2007 06:14 PM
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Comments
1458?
I got the area of the triangle, which is 162, 'n' cubed it twice, 'n' hoped that it was between 1000 'n' 2000, but I know I wouldn't of had guessed that if you didn't give out that hint b/c all I did was use trial 'n' error.
Posted by: Doug at January 29, 2007 08:50 PM
1665? I'm going to go wash my hands about a thousand times now.
Posted by: Long Story Longer at January 29, 2007 10:59 PM
I can do this algorithmically, but not sure how to express it algebraically.
To get the number of little triangles, take the sum of (2n - 1) where n goes from 18 to 0.
To get the sum of all possible triangles, add a "unit size" factor to the above, and take a sum of the sums, where n= the modulus of 18/u (unit size) where u goes from 1 to 18.
Not sure how to express this algebraically. Its been a while. :-)
Posted by: Brian at January 30, 2007 09:44 AM
I subscribe to the line from Peggy Sue Got Married: "in the future I will not have the slightest use for algebra, and I speak from experience. ...!"
Posted by: Blobby at January 30, 2007 10:07 AM
1665
Posted by: JP at January 30, 2007 02:06 PM
This questio is really uncovering a lot of school-aged math anxiety for me. Is there an extra credit project available?
Posted by: Bloghungry at January 30, 2007 09:44 PM
I can't do the math problem, but I could make a pyramid like that out of toothpicks to die for.
Posted by: MikeProv1 at January 30, 2007 10:50 PM
1665??
Posted by: David G at January 31, 2007 02:36 AM
Why do you do this? You know that math makes my brain hurt.
Posted by: mark at January 31, 2007 11:35 AM
1665 - will write it up when I get a chance.
Posted by: Dennis at February 2, 2007 12:42 PM
My mind = Boggled
Posted by: Todd HellsKitchen at February 4, 2007 10:53 AM
Havent had a chance to write up the formula so I thought I'd skect it out. Basic recursive development. If you let T(n) be the number of triangles for side of length n and look at T(n+1) - adding the next row of small triangles you see that T(n+1) is T(n) plus the number of triangles that must include at least one triangle in the last row. Start off on the left with the smaller one. the next largest with the smaller one in the left corner, etc. This leads to an odd/even break down - if its even the last largest triangle is the whole triangle, odd it isnt. Then slide all of them by one. Then handle the upside down triangles that must contain one in the last row. These can be added up, you get a recursive relationship and can solve for T(n).
Posted by: Dennis at February 4, 2007 04:01 PM
Meanwhile, several days later, the Galicious crowd waits for an answer from the hunky teacher.
Posted by: MikeProv1 at February 7, 2007 06:27 PM