Investigation; Triangles

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) ½(m − n +1)(m − n + 2)

if n ≤ ½(m+1)

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: 2

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

Proof 1 (Combinatorial (counting) argument) The small triangles may be orientated as

(sitting) or

(hanging).

row 1 row 2 row 3 row 4 In row 1, In row 2, In row 3, In row 4, …

In row n,

there is there are there are there are

1 sitting triangle 2 sitting and 1 hanging 3 sitting and 2 hanging 4 sitting and 3 hanging … … 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. m

sm =

"(2n !1) n=1

m m # m(m +1) & = 2" n ! "1 = 2 % ( ! m = m(m +1) ! m = 2 $ ' n=1 n=1

m2

Proof 2 ( Proof by Induction) Initial step: s1 = 1 = 12 so the result holds for m = 1.

Inductive step: 2

Assume the result holds when m = n, so there are n small triangles when the edges are divided into n segments, and consider a triangle whose edges are divided into n + 1 segments.

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 2 n small triangles, by the assumption. Row (n + 1) contains 2(n + 1) − 1 = 2n + 1 small triangles. So the 2 2 total number of small triangles is sn +1 = n + (2n + 1) = (n + 1) , 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 1 b h factor . In particular, the base and height of a small triangle are and respectively, if the base and height m m m of the original triangle are b and h.

Suppose the original triangle has area A, i.e.

1 2

! b ! h = A . Then the area of a small triangle is

1 b h ! ! = 2 m m

1 1 ! !b!h = m2 2

1 !A. m2

Since the sum of the areas of all the small triangles must equal the total area (A) of the original triangles, there must 2 be m of them.

Proof 4 (Recursion) Let a size m triangle be a triangle whose edges have been divided into m segments, and let sm 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 sm = sm!1 + 2m !1 and s1 = 1.

Then we have:

sm = sm!1 + 2m !1 sm!1 = sm!2 + 2(m !1) !1 sm!2 = sm!3 + 2(m ! 2) !1 ! s2 = s1 + 2(2) !1 Adding the left and right sides of the equations, and eliminating common terms gives: sm

m # m(m +1) & "1( − (m − 1) = 1 + m(m + 1) − 2 − m + 1 = m2. = s1 + 2 ! n − (m − 1) = 1 + 2 ! % 2 $ ' n=2

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

Shirleen Stibbe

http://www.shirleenstibbe.co.uk