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March 3

Maths competition (again)

I recently did a maths competition and there was a problem I did not get. There was an equilateral triangle split into 9 smaller equilateral triangles (just like this without the bottom row). The question is: When a small triangle is surrounded by 3 other small triangles, the result is called a four-triangle. If the integers 1 to 9 are put in the 9 small triangles (1 per triangle) so the sum of the 4 integers in each of the four four-triangles is the same, what is the least possible value of this sum?" How would I solve this? I have literally no ideas beside guess/check. 72.128.95.0 (talk) 02:27, 3 March 2011 (UTC)[reply]

There are only 3 such triangles that I can see. If you add the three 4-triangles together you get an overlap on three triangles. This gives a minimum that may be possible of sum 1 to 9 plus minimum of 3 whch is 1+2+3 for a total of 45+6=51. This gives 17 for each of the 3 4-triangles. If you manage that you're done. It may not be possible, I haven't tried, but I think it probably is possible. Dmcq (talk) 03:59, 3 March 2011 (UTC)[reply]

right, but how do you find these three triangles? 72.128.95.0 (talk) 04:23, 3 March 2011 (UTC)[reply]

If you look at where the three 4-triangles go you'll see there are three small triangles where a pair of the 4-triangles overlap. Dmcq (talk) 04:35, 3 March 2011 (UTC)[reply]

9
1 6 2
5 8 3 4 7
gives four triangles 1269, 1358 and 2347 which total 18, 17 and 16, which has grand total 51. I couldn't make each four-triangle the same value at this minimal grand-total; is that a requirement? -- SGBailey (talk) 10:27, 3 March 2011 (UTC)[reply]

Yes, it is a requirement - each four-triangle must sum to 17. There are two distinct ways of achieving this (modulo permuting numbers within a four-triangle and permuting the arrangement of four-triangles within the big triangle). Gandalf61 (talk) 11:03, 3 March 2011 (UTC)[reply]

In case you're wondering (1,2,6,8), (2,3,5,7), (1,3,4,9) is one solution. Taking 1,2 and 3 to be the overlaps leads us to formulate three equations: a+b=14, c+d=13, and e+f=12, where a,b,c,d,e and f are the numbers 4 to 9 inclusive. Neither e nor f can be 6; neither a nor b can be 7. Setting c and d to 6 and 7 gives the other possibility, I think. This is just a note about how one might show that 17 is a realistic possibility (in a general case). Grandiose (me, talk, contribs) 22:06, 3 March 2011 (UTC)[reply]

Mathematician "Sura man chi" (phonetic)

A mathematician "Sura man chi" has/had a set of mathematical relationships/ expansions, although expanded incorrectly, led to the correct expansion result

Hojw do I find these relationships98.25.109.175 (talk) 17:03, 3 March 2011 (UTC)[reply]

How can I find values that give the least bad values for these expressions?

Say I have these two expressions:

   1.35x -y
   3.75y -x

where x and y are both > 0 and I want to find values for w and d which maximize the least of the two expressions. How can I do this? What about if there are three expressions and three unknowns like this:

   1.35x -y -z
   3.75y -x -z
   4.21z -x -y

Any help appreciated! —Preceding unsigned comment added by 193.61.28.76 (talk) 20:01, 3 March 2011 (UTC)[reply]

This kind of problem is known as a "linear programming" problem. Our article will tell you more about it and the various solution methods than you probably wanted to know. –Henning Makholm (talk) 20:49, 3 March 2011 (UTC)[reply]

Basic combinatorics problem

This is a seemingly-easy combinatorics problem that I unfortunately could not get (I was never very good at combinatorics). It says: How many four digit numbers not beginning with 0 can you form with the numbers {0,1,2,3,4,5} if the digits cannot repeat AND the number must be even? My first instinct was to multiply 5*5*4*3, but the last choice is 3 which means that it's possible I already used up all the even numbers. How do I make sure that I do not? thanks. —Preceding unsigned comment added by 72.128.95.0 (talk) 23:44, 3 March 2011 (UTC)[reply]

Start by selecting the first and last digits. There are 5 possible choices for the first one and 3 for the last one, but 2 of the 3×5 combinations are forbidden because they would use the same digit twice. After that, selecting the middle digits is easy. –Henning Makholm (talk) 23:55, 3 March 2011 (UTC)[reply]

My guess: 5*5*4*3 gives you all the numbers both odd and even. Half of them will be even, so the answer is 5*5*4*3/2 or 150. Am I wrong? 92.15.6.227 (talk) 20:05, 5 March 2011 (UTC)[reply]

Yes, you are wrong. Because the first digit cannot be 0, the zero will be available for the last digit position slightly more often than either of the other digits will. Therefore there will be slightly more even numbers than odd before you exclude the odd ones. –Henning Makholm (talk) 20:32, 5 March 2011 (UTC)[reply]