User:JMBell/Puzzles

Puzzles and Others:

"If someone is mad at you, step into his shoes, and walk a mile in the opposite direction. Then, no matter how angry he is, he cannot get mad at you anymore, because you have his shoes and are a mile away." - Confusedshius

1

For a barnstar:

Image 1; the line running horizontally through the lower middle is an error.

The three defined angles are each 45°; Determine the area of the shaded area in relation to the circle.

Solutions and proofs here (sign with ~~~~):

Solved by:

• User:Gerbrant (solution on my/his talk page)

2

For a barnstar:

The natural numbers 1, 2, ..., 11, n are to be written on the edges of an octahedron. On every vertex is to be written the sum of the four numbers, whose lines meet at that vertex (e.g. lines with numbers a, b, c, d meet at vertex with number a + b + c + d).

For which values of n will the sums on the vertices be equal?

Solutions and proofs here (sign with ~~~~:

$n\in \{0,3,6,9,12,15\}$

Proof: Each egde takes part in the sum of two vertexes. The sum of all vertexes is therefore 132 + 2n. Because there are six vertexes 132 + 2n must be divisable by 6, from which follows that n must be divisable by 3. As a crude upper-bound we have n + 6 < 38, or n < 32.

If we name the egdes A to L we must satisfy the following condition:

• A+B+C+D = 22 + n/3
• E+F+G+H = 22 + n/3
• A+E+I+J = 22 + n/3
• B+F+I+K = 22 + n/3
• C+G+K+L = 22 + n/3
• D+H+J+L = 22 + n/3

For $n\geq 18$ we find that 6 + 7 + ... + 11 + n < 2(22 + n/3). For n = 18 we have 6 + 7 + ... + 11 + 18 = 56. Let A = 18, it is now impossible to select three numbers x, y, z from {6, 7, 8, 9, 10, 11} so that A + x + y + z = 28.

• For n = 0 we have the solution (A, B, C, D, E, F, G, H, I, J, K, L) = (0,1,10,11,7,8,5,2,9,6,4,3);
• For n = 3 we have the solution (A, B, C, D, E, F, G, H, I, J, K, L) = (3,1,8,11,10,6,2,5,7,3,9,4);
• For n = 6 we have the solution ((A, B, C, D, E, F, G, H, I, J, K, L) = (6,1,6,11,9,10,3,2,5,4,8,7);
• For n = 9 we have the solution (A, B, C, D, E, F, G, H, I, J, K, L) = (9,1,4,11,2,7,10,6,9,5,8,3);
• For n = 12 we have the solution (A, B, C, D, E, F, G, H, I, J, K, L) = (12,1,2,11,3,8,9,6,7,4,10,5);
• For n = 15 we have the solution (A, B, C, D, E, F, G, H, I, J, K, L) = (15,1,2,11,3,8,9,6,7,4,10,5).

Q.E.D. --R.Koot 14:05, 13 July 2005 (UTC)