Math Is Fun Forum

Hi Mandy Jane,

My reply was :

Mandy Jane

If you post a fractions question you would like help with,  I'll show you how to do it.

Bob

Hi Jane Fairfax,

Ok, so I hadn't read the problem properly.  Now I have.

Bob

Hi Sameer,

That's a bit harsh.

Your method is shorter but does that make mine wrong?

Andrew Wiles took 7 years to prove Fermat's Last Theorem.  Fermat, himself,  claimed a shorter proof.  Does that make Wiles wrong?

Bob

OK got it.  But it's a bit of a chase of angles around the diagram so rather than add it to the above post, I thought it would be easier to show as a separate post.  My geometry software insists on showing A, B N and M on the same circle (because they are!) but that makes it hard to avoid making assumptions that haven't been justified.  So I started with a new diagram with A, B and M as before but with N deliberately shown not on the circle.  My aim is to show it is.  (see diagram below)

Let x and y be the angles shown on the diagram.

Then the following angles can be found by angle sum of a triangle, reflection properties, isosceles triangles, straight line etc:

MBC = y
BMC = y
MCB = 180 - 2y
CMA = 180 - 2x - y
MAC = 180 - 2x - y
ACM = 4x + 2y - 180
ACB = 4x

Now we do know that N lies on the perpendicular bisector of AB so

NCB = 2x
NCM = 2x + 2y - 180
CNM = 180 + 180 -2x -2y -x - 180 + 2x + y
        = 180 - x - y
NMC = 180 - 2x - y + x
        = 180 - x - y

So CNM = NMC

which fixes N on the circle as NCM is isosceles.

Bob

Ok Here's my answer to number 4.

It requires a ruler with width lower than the diameter of the smaller circle and smaller than the radius of the larger circle.

Lay the ruler across the circle so that it cuts both circles twice.

Draw parallel lines by using both edges of the ruler.

Label the points where the lines cut the large circle A, B, C, and D and the small circle points E, F, G and H. (My first and second diagrams have swopped over so see second diagram).

Join AC and BD and label the crossover J.

Join EG and FH and label the crossover I.

ABD = BDC ( parallels cut by transversal => alternate angles)
BAC = BDC (angles subtended by same chord)

So AJB is isosceles.

Similarly EIF is isosceles.

IJ is perpendicular to AB (and EF) because (oh drat I forgot to label the point where IJ cuts AB.  Let's call it X.)

AJ = JB (radii)  BAJ = ABJ (proved above) XJ is common to AXJ and BXJ.

So AXJ and BXJ are congruent so AXJ = BXJ = 90.

So IJ goes through the centre of the circle (perpendicular bisector of chord goes through centre)

Now refer to simplified first diagram.

Let this diameter cut outer circle at K and L

Turn ruler sideways and lay one edge along the diameter.

Draw the parallel line MN cutting the circle at M and N by using the other edge.

and another similar line OP as shown.

Join ML and KN to cut at R.

Join KP and OL to cut at S.

A similar proof to part one shows that the centre lies on SR.

Where SR crosses KL is the required centre.  See point T.

Bob

ps I still don't know what a 'Manhattan Plane' is.

hi Prakash,

Do you want a solution using differential calculus or, using quadratic graphs? (or both?)

Bob

Hi Nesic

The diagram below should help you reach a solution.

I started with 100 students (no loss of generality as you want percentages).

You can then work out the values 48 and 12 using the information given.

There are 40 students who don't do HW so you can work out 4 and 36.

Part A can be found by adding values to get the number in Pass.

And part B by calculating the percentage in the overlap as a percent of those in Pass.

Hope that helps.

Bob

Hi Elisa,

My own prefered way to explain fractions to students at any level I call my chocolate diagram method.  I haven't yet found a fractions problem that couldn't be done this way.  In the ideal mathematical world bars of chocolate would be sold in slabs of different length and width so you could actually make the fractions.  At the end of the lesson you eat the chocolate!:)

Have a look at my picture below.  You were asking to compare quarters and ninths so I did 4 x 9 = 36 and made my one bar of chocolate that size ( 4 x 9). 

[For a new problem you have to decide afresh what size of bar is needed for that problem.]

For your problem make two more the same (4 x 9).

Then split the first into quarters and shade 3.  That shows 3/4

The second is all shaded to show 9/9.

The third is split into 9 parts and has 2 shaded to show another 2/9.

The second and third taken together show 9/9 plus 2/9 = 11/9.

To decide which fraction is closer to 1, just look at how many squares of chocolate are needed to make the fraction up to one or how many need to be taken off to get down to one.

3/4 is 9 squares too small.

11/9 is 2/9 over which is 8 squares.

So 11/9 is closer by one square which is 1/36 as a fraction.

This method also lets you make equivalent fractions 

eg. 3/4 = 27/36

and to add fractions

eg. 3/4 + 2/9 = 27/36 + 8/36 = 35/36.

Hope that helps.

Bob

and finally ....

I've been coming back to part (iv) for a while now.  I cannot get it to yield to similar triangles (of which there are many) nor the cosine rule.

In desperation, I resorted to using 'Sketchpad' to try it out.  This program uses vector geometry and can be set to work to a fine degree of accuracy.  The diagram below shows why I'm giving up.

The two top angles can be shown analytically to be equal.  Sketchpad shows them differing by only 0.03 of a degree, which shows how accurately it can work.

The highlighted calculations show (iii), a result I stumbled upon connecting TM and RD and thirdly shows that (iv) does not seem to be correct. 

1a2b3c2212

If you spot this post, please check the problem as I don't think it can be done as formulated.

Bob

Maybe a diagram would help

Bob

hi Apothem,

Bobbym has already said how to calculate the slope.  [and as I post this, I see he has added another post but I'll leave mine so you have a choice of methods ...basically the same as  we are both using straight line graph formulas.]

Do that and then you can get an equation for all points on the line.

It's usual to write this as

'm' is the slope that you can calculate using bobbym's formula.

Now to find 'c'.

If (a,b) is one of your known points then c = b -ma will give you 'c'.

On any line (that is not parallel to either axis) you can always choose an 'x' that is on the line.

Find the corresponding 'y' by calculating y = mx + c using this 'x' and the calculated values of m, and c.

I hope that makes sense.  If you are having trouble post your coordinates and I'll work the numbers to make it easier.

Bob

hi leadfoot,

You are welcome.  As you can see, bobbym and I have raced to beat each other to  your solution. 

Gosh it's such fun posting to Maths Is Fun!

:)

bob