Shirleen Stibbe http://www.shirleenstibbe.co.uk

Investigation: Triangles Proofs Workshop

This was a 45 minute session, where students were invited to work in small groups in the breakout rooms, supervised by a tutor. We gave them the m = 10 picture to help them come up with a conjecture.

Problem

The edges of a triangle are divided into m equal segments by inserting m − 1 points. Lines are drawn through each of these points parallel to each of the three edges, forming a set of small triangles. How many of the small triangles are there? Justify your answer.

Example

In the example below, m = 10.

Shirleen Stibbe http://www.shirleenstibbe.co.uk

Triangle investigation: Notes for tutors Proofs Workshop

Purpose

To let them come up with a result themselves, and then find a way of proving it

To demonstrate that there are different ways of proving things

The problem can be extended to challenge the more experienced students

Problem

The edges of a triangle are divided into m equal segments by inserting m − 1 points. Lines are drawn through each of these points parallel to each of the three edges, forming a set of small triangles. How many of the small triangles are there? Justify your answer.

Tutor notes

• There is a handout for the start of the session, stating the problem and giving them an example for m = 10 - it's quite difficult to draw large examples.

• This is very much their own investigation, so they shouldn't be led to the solution. Some may need some gentle prodding, by asking them to summarise their thoughts, or being a sounding board for their ideas

• Initial investigation Let them investigate the number of triangles with sides of length 1/m (size 1) to start with.

They may need reminding of 'specialise', 'conjecture' etc., but they should all manage to come up with the solution (m2) quite quickly.

• If the previous session has overrun, this is probably as far as they'll get.

• Please encourage (force!) them to write out full proofs of their results before they move on, so that I can summarise them in the plenary session afterwards. The handout offers 5 different proofs - they may come up with even more.

• If they complete the size 1 task, they could go on to look at size 2, 3, ..., n There isn't a handout for these. The general result for the number of size n triangles is

m2 − (3n − 3)m + ½(5n2 − 9n + 4) if n ≤ ½(m+1)

½(m − n +1)(m − n + 2) if n > ½(m+1)

The result for the total number of triangles is

½ n(n + 1)(4n + 1) if m = 2n

½ (n + 1)(4n2 + 7n + 1) if m = 2n + 1

NB: please don't give them the handout at the end of the session - it would ruin the drama of the plenary session. Tell them that there is one, and they'll get it at the next Workshop session (after John's presentation).

Shirleen Stibbe http://www.shirleenstibbe.co.uk

Triangle investigation results (handout for students) Proofs Workshop

Investigate:

The edges of a triangle are divided into m equal segments by inserting m −1 points. Lines are drawn through each of these points parallel to each of the three edges, forming a set of small triangles. How many of the small triangles are there?

Specialise:

Try a few examples. Let sm be the number of small triangles for m divisions.

m = 2, sm = 4 m = 3, sm = 9 m = 4, sm = 16 Conjecture:

The number sm of small triangles formed by dividing the edges into m segments is m2.

Proof 1 (Combinatorial (counting) argument)

The small triangles may be orientated as

(sitting) or (hanging).

In row 1, there is 1 sitting triangle In row 2, there are 2 sitting and 1 hanging In row 3, there are 3 sitting and 2 hanging In row 4, there are 4 sitting and 3 hanging … … … In row n, there are n sitting and n − 1 hanging

So the number of small triangles in row n is n + (n −1) = 2n − 1.

The total number of small triangles is the sum, over m rows, of the number of small triangles in each row.

sm = (2n!1)n=1

m

" = 2 nn=1

m

" ! 1n=1

m

" = 2 m(m+1)2

#

$%

&

'(!m = m(m+1)!m = m2

Proof 2 ( Proof by Induction)

Initial step: s1 = 1 = 12 so the result holds for m = 1.

Inductive step: Assume the result holds when m = n, so there are n2 small triangles when the edges are divided into n

segments, and consider a triangle whose edges are divided into n + 1 segments.

row 1

row 2

row 3

row 4

Shirleen Stibbe http://www.shirleenstibbe.co.uk

The first n rows form a triangle whose edges have been divided into n segments, and these n rows contain n2 small triangles, by the assumption. Row (n + 1) contains 2(n + 1) − 1 = 2n + 1 small triangles. So the total number of small triangles is sn +1 = n2 + (2n + 1) = (n + 1)2, and the result holds for m = n + 1.

This completes the inductive step, and the result holds for all m ≥ 1 by the principle of induction.

Proof 3 (Geometric proof)

The small triangles are all similar to the original triangle, since the lines are all parallel, and they are scaled by the

factor 1m

. In particular, the base and height of a small triangle are bm

and hm

respectively, if the base and height

of the original triangle are b and h.

Suppose the original triangle has area A, i.e. 12 !b !h = A . Then the area of a small triangle is

12!bm!hm

=1m2

!12!b !h =

1m2

! A .

Since the sum of the areas of all the small triangles must equal the total area (A) of the original triangles, there must be m2 of them.

Proof 4 (Recursion)

Let a size m triangle be a triangle whose edges have been divided into m segments, and let be the total number of small triangles in a size m triangle. A size m triangle may be considered to be a size (m − 1) triangle, with an additional row containing 2m − 1 small triangles. So and s1 = 1.

Then we have:

Adding the left and right sides of the equations, and eliminating common terms gives:

sm = s1 + 2 nn=2

m

! − (m − 1) = 1 + 2 ! m(m+1)2

"1#

$%

&

'( − (m − 1) = 1 + m(m + 1) − 2 − m + 1 = m2.

Proof 5 (Proof by pictures - supply your own words and generalisation)

sm

sm = sm!1 + 2m!1

sm = sm!1 + 2m!1sm!1 = sm!2 + 2(m!1)!1sm!2 = sm!3 + 2(m! 2)!1!

s2 = s1 + 2(2)!1