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Finding a formula for Sticky Triangles
By Tom Neill (P4062) on Tuesday, April 24, 2001 - 06:01 pm:

I've been working on April's Bernard's Bag - sticky triangles.

I've worked out that if r is the number of rows, and t is the number of triangles and m is the number of matches.
then t = r2
but
I've worked out that for each row in turn
m = t + 2 (for the first row)
m = t + 2 + 3 (for the second row)
m = t + 2 + 3 + 4 (for the third row)
m = t + 2 + 3 + 4 + 5 (for the fourth row)
m = t + 2 + 3 + 4 + 5 + 6 (for the fifth row)
and so on.
How do I write this as an equation?

I am 6 but I like maths a lot and am very good at it.

Tom Neill

By Sean Hartnoll (Sah40) on Tuesday, April 24, 2001 - 08:08 pm:

Hi Tom,

I don't know what the sticky triangles problem is, but I can tell you what the equation is. It is

m = t + r×(r+3)÷2

Do you know what this means?

We can check that it works.

For the first row, r=1, so we have
m = t + 1×4÷2 = t + 2 ! which is what you gave above.
For the second row, r=2, so we have
m = t + 2×5÷2 = t + 5 = t + 2 + 3 ! Also like above.

Perhaps you can substitute in values or r for the other cases to see that it works.

It would be quite hard for me to explain in a simple way how you get the equation above, but I hope you can see that it works.

Sean

By Michael Doré (Michael) on Wednesday, April 25, 2001 - 01:43 pm:

One way to get it is as follows.

You say that if the row number is k then:

m = t + 2 + 3 + 4 + ... + (k-2) + (k-1) + k

where t = r2 (where r is the number of rows).

What we mean by the ... is that the sum is continued up to k. For example, 1 + 2 + ... + 10 is shorthand for 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10.

Let's write the sum for m down twice. You probably know that it doesn't matter what order you add the numbers in, so we'll change them around a bit.

m= t +2 +3 +4 +... +(k-2) +(k-1) +k
m= t +k +(k-1) +(k-2) +... +4 +3 +2
Now let's add those together, to find a sum for m+m. We'll add up the terms in pairs, which is why I've lined them up above each other.
t + t = 2t
2 + k = k + 2
3 + k-1 = k + 2
4 + k-2 = k + 2
(Are you noticing a pattern?)
...
k-2 + 4 = k + 2
k-1 + 3 = k + 2
k + 2 = k + 2

So:
m + m = 2t + (k+2) + (k+2) + ... + (k+2)
where the (k+2) appears (k+1) times.

So we get:
2m = 2t + (k+1)×(k+2)

But we didn't want twice m, we wanted m, so we halve the sum:
m = t + (k+1)×(k+2)÷2

If you like you can now replace the t with r2 to get the formula as:

m = r2 + (k+1)×(k+2)/2

So that should give the number of matches on the kth row (and there are r rows in total).

(Note: I haven't checked the original problem - I'm just using the expression you've already worked out and assuming it's correct.)
By The Editor:

I have edited Michael's message to make it a little clearer for younger readers.

Tom did very well with this problem, and you can see his published solution here.