sticks

48, 52, 80, 84

Method: Some insight into how the sticks are likely to be arranged and a spreadsheet. I don't know if this solution is unique.

whim

Can you get an unique answer if you know the sticks are eating-sticks . 80 cm is a bit long for such an eating tool.



whim

oops double post


whim

You get close, but not exact, with a triangle that has sides 20, 19, and 16. The height (with 19 as the base) is 14.9462. There are four different triangles you can form with three of the chopsticks. Each of the triangles has three heights since there are three choices for a base. Here are the numbers for 15, 16, 19, 20:

Here's a useful formula for the area of a triangle given sides of length A, B, C:
S = (A+B+C)/2 (semi-perimeter)
Area = sqrt(S * (S-A) * (S-B) * (S-C))

To calculate the three heights of a triangle, calculate the area, and multiply it by 2 (area*2). Then, for each side (base), divide the area*2 by the length of the side. That is the height.

What values do work?

Whim,

Here are all the solutions for a relaxed set of conditions. Here, the only requirements are that all chopsticks are different lengths less than 50cm. It appears that 30, 28, 26, 24 is the smallest solution to your problem.

One issue: Does "but his two sticks are longer than yours" mean that both of his are longer than both of yours? If so, this is not a solution. You have to go to 60, 56, 52, 48 to satisfy that condition. I can send you a longer list if you wish.

3 24 25 26
4 12 13 15
6 20 25 29
7 12 15 20
8 9 10 17
8 10 17 21
8 21 29 35
8 24 26 30
9 15 28 41
11 12 13 20
11 24 25 30
12 13 14 15
12 13 20 21
12 13 30 37
12 13 37 40
12 15 17 25
12 15 20 25
12 15 26 37
12 15 37 44
12 19 20 37
12 36 39 45
14 24 30 40
15 16 25 39
15 17 25 28
15 17 28 39
15 17 39 44
16 18 20 34
16 20 34 42
17 24 25 26
20 25 29 36
21 29 35 48
22 24 26 40
24 25 39 40
24 26 28 30
24 26 40 42
36 39 42 45

Thanks mark



Whim