Lessons taught; Lessons learnt

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Archive for the ‘puzzle’ tag

An ‘average’ puzzle

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One of the teachers in my school regularly provides a ‘problem of the day’ to his students. This one, from last week, caught my eye, as containing some ‘hidden’ maths that would make a good lower-secondary investigation:

The average of a set of 64 numbers is 64.
The average of the first 36 numbers is 36.
What is the average of the last 28 numbers?

Solution

The total of all 64 numbers is 64 \times 64 = 4096.
The total of the first 36 numbers is 36 \times 36 = 1296.
The total of the last 28 is therefore 4096 - 1296 = 2800,
which makes their average \frac{2800}{28} = 100.

Comments

In its ‘raw’ form this is just a basic test that someone understands how to calculate means. It’s nice that the final number is an integer, though… and is it a coincidence that 64 + 36 = 100?

Let’s try the same problem with different numbers:

The average of a set of 51 numbers is 51.
The average of the first 17 numbers is 17.
What is the average of the last 34 numbers?

The average of the remainder is \frac{51^2 - 17^2}{51 - 17} = \frac{2312}{34} = 68.

Not only do we get a whole number, but the answer is 68 = 51 + 17.

Can we conjecture that the answer will always be the sum of the two set sizes? Well, it’s happened twice, so it must be true (!). Looking at the general problem:

The average of a set of n numbers is n.
The average of the first k numbers is k.
What is the average of the last n-k numbers?

The average of the remainder is \frac{n^2 - k^2}{n - k}.

Here we might get stuck, unless we remember (or discover, or are taught) at this point the difference of two squares identity: a^2 - b^2 = (a+b)(a-b).

This makes our average \frac{(n+k)(n-k)}{n-k} = n+k, and our conjecture is proven.

Extensions

Does a similarly nice thing happen if we specify extra conditions? For example, we could say that there are 5 numbers with an average of 5, and another 9 numbers with an average of 9.

Can we say anything about the median in these situations?

Written by Jon Ingram

February 28th, 2010 at 7:58 am

Posted in Maths, Puzzles

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Hinged square dissection

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In several recent posts I have referred to ‘linkages’ (which should more properly be called ‘hinged dissections‘. One good recent book on these is Hinged Dissections: Swinging and Twisting, by Greg Frederickson, but there are several classics out there that discuss hinged dissections, including Amusements in Mathematics by H. E. Dudeney.

Dudeney was responsible for one of the most well known hinged dissections, of which this is a simplified example. It converts a square into, well, something else.

The white circles are the hinges — move the coloured circles to move the corresponding parts of the square.

Can you predict what the end result of the transformation is? Can you prove it?


Geogebra applet (enable Java to see it).

(Source: dudeney-dissection.ggb.)

The Haberdasher’s Puzzle

Dudeney’s classic dissection, published in his ‘Canterbury Puzzles‘ in 1907, is a slightly altered version of this, which allows you to transform a square into an equilateral triangle. You can download a program which will allow you print out a template for this here.

Constructing Dudeney’s dissection takes a touch more effort than the dissection illustrated above, but the process is described incredibly well in this lesson plan, which demonstrates how to make a model of the dissection using foam rubber. I haven’t tried it yet, but it may make a tempting break from lesson planning next week!

Written by Jon Ingram

August 27th, 2008 at 4:31 pm

Posted in Maths, Puzzles

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A quickie on quadratics

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Fix two points in the plane, and consider all the quadratics which go through those two points.

Find the locus of the stationary points of these quadratics.

The following Geogebra worksheet might help. If you select “Show stationary point” and move the blue point, it will trace out the locus.


Geogebra applet (enable Java to see it).

(Source: quadratic_stationary_points.ggb.)

Thinking about this question kept me from going to bed until far too late on Thursday. My answer, and other thoughts, will appear tomorrow.

Written by Jon Ingram

August 23rd, 2008 at 8:20 pm

Posted in Maths, Puzzles

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