Showing posts with label circles. Show all posts
Showing posts with label circles. Show all posts

Monday, November 30, 2015

[PriOlym_20151130RTAC] Ratio with One Circle Overlapping Two

Question


Introduction
     This question is like this previous one, except it is of olympiad standard.  I illustrate the solution of this without algebra, by using Distinguised Ratio Units.  As before, I try to match parts to an equal number.  But here we have quite a mixture of different types of units.

Solution

Commentary
     Basically we make the triangle units to number 12 and do the same for the circle and square units.  It turns out that one triangle unit is the sum of one circle unit and square unit.  We deduce that 9 circle units (for the area of A) plus 6 circle units (for the area of B) is the same as 8 circle units and 8 circle units.  The reduction of circle units must be equally compensated by the increase in the circle units.  Thus one circle unit is the same as two square units.  From here, things become easy.


H02. Use a diagram / model
H04. Look for pattern(s)
H05. Work backwards
H09. Restate the problem in another way
H11. Solve part of the problem

Suitable Levels
Primary School Olympiad Mathematics
Primary School Mathematics (challenge)
* other syllabuses that involve areas and ratios
* anyone who is game for a challenge






Monday, May 25, 2015

[OlymPri_20150524MSCR] Round and Round, Twice or Thrice?

Question


Introduction
     If you think the answer is  2 cm (since the wheel rotates two rounds), you are wrong!  This tricky question is taken from a previous year Asia Pacific Mathematical Olympiad for Primary Schools competition.  In the original question, there was no colouring.  I added colours just to make the distinction between the wheel and the track a bit clearer.

Visualisation
     There is some subtlety in this question.  It is something we often do not notice unless we really think hard about it.  To get a handle on what is really happening, let us imagine we made a mark on the wheel in its original position, indicated by a blue dot at a twelve o’clock position.  Let us imagine rolling the wheel clockwise.  Note that the centre of the wheel will move in a bigger circle in an anti-clockwise (American: counter-clockwise) direction within the orange circular track.  Actually it does not matter which way you roll, the centre of the wheel goes round the centre of the track in a direction opposite to that of the turning wheel, which always remains in contact with the track.

Since the blue dot on the wheel turns two rounds, by the time the wheel reaches the bottom half, the blue dot must again be on top or at a twelve o’clock position.  Note however, the point of contact between the wheel and the circular track (indicated by a green dot whenever possible) is not the same as the blue dot!  The green dot is actually a dynamic dot (it is not always the same point on the wheel) whereas the blue dot is always the same dot on the wheel but the wheel is being rotated.

Note that at the halfway point, the green dot is at the bottom of the wheel while the blue dot is on top of the wheel.  Notice also that, relative to the wheel, the green dot goes in the opposite direction as the blue dot and they actually crossed over somewhere along the way!  Actually the green dot has made one-and-a-half turns with respect to the wheel.  Remember that the green dot is not a fixed point on the wheel, but it measures how much the wheel and the track have been in contact.


As the wheel continues to roll back up, the green dot makes another  1½  rounds.  So altogether the green dot moves through  3  rounds while the blue dot rotates only  2  rounds around the centre of the wheel!  The green dot is the one that matters.

Solution
     circumference of track (measured by green dot) : circumference of wheel = 3 : 1
     Since radius is proportional to circumference,  the radius of the track is  3  cm.

Remarks
     Yes, the solution is that short.  But the thinking behind it is profound.  But do we need to draw all the diagrams as in the visualisation above?  I did that to explain to you.  Actually I imagined it in my mind.  If imagination is difficult, you can act it out by drawing a big circle and then using a small coin to simulate the rotation around the track.  I actually drew a rough sketch by hand  to convince myself that my thinking was accurate.

     Notice that I used a proportionality argument.  If you know how to use the concept of proportion, you can make the working short and sweet.  There is nothing wrong in using the formula                     circumference = 2p ´ radius.
This formula just says that the circumference of a circle is proportional to its radius, and the constant of proportionality is  2p.  Your working would look like this
                                  circumference of track = 3 ´ circumference of wheel
                                      2p  ´ radius of track = 3 ´ 2p  ´ radius of wheel
After cancelling out the  2p,  you would get
                                       radius of track = 3 ´ radius of wheel = 3 cm
You get the same conclusion, but using the proportionality method, you do not need to bother about the  2p.

     By the way, the centre of wheel traces out the path of a circle of radius  2 cm.  See the red arrow in the first diagram on top.  The path traced out by the blue dot looks like an  epicycleIn the old days,people thought that the sun, moon and planets rotated around the earth in epicycles.  This also reminds me of Spirograph,which is a toy that allows you to use your coloured pencils to create very beautiful patterns.  [Click to search for images of Spirograph and patterns produced.]


H01. Act it out
H02. Use a diagram / model
H04. Look for pattern(s)
H05. Work backwards
H06. Use before-after concept
H09. Restate the problem in another way

Suitable Levels
Primary School Mathematics Olympiad
* anybody game for a challenge relating to imagination, circumference and lengths





Saturday, May 2, 2015

[Pri20150408YYS] Yin-Yang Semicircles?

Question


Introduction
     The diagram looks a bit like the Yin and Yang symbol, doesn’t it?  This problem can be solved easily using the correct insights.  I present two solutions: the first one is by direct calculation (in terms of p), and the second solution uses the powerful concept of ratio of similar figures.
     Whichever method is used, first we must make observations.  Can you see that there are three types of semicircles (small, medium and large)?  [H02, H04]

Let  S = area of small semicircle,  M = area of medium-sized semicircle,  and  L = area of large semicircle.

Note that  area of A  = area of C  = LM + S,  and  area of B = 2´(MS).  [H10, H11]

Solution 1 (by direct calculation)
     L = ½p(3)2 = 9p/2.   M = ½p(2)2 = 4p/2.   S = ½p(1)2 = p/2.
     area of A  = area of C  = 9p/24p/2 + p/2  =  3p
     area of B = 2´(4p/2p/2) =  3p
\ area of A : area of B : area of C  =  1 : 1 : 1.  (The areas are all the same)
Many pupils feel more comfortable using concrete approximations like  p  » 22/7  or  p  » 3.14,  but this tends to obscure relationships between entities, and makes the calculations messier.

Solution 2 (using similar shapes)
     An powerful idea is that the ratio of areas of similar shapes is the square of the ratios of their lengths.  When a figure is enlarged by a factor of (say) 5, we get a similar figure and the area becomes  52 = 25 times as large.  So  M = (2)2S = 4S  because the radius of the medium-sized semicircle is twice that of the small semicircle.  Likewise,  L = (3)2S = 9S.
     area of A  = area of C  = 9S – 4S + S  =  6S
     area of B = 2´(4SS) =  6S
\ area of A : area of B : area of C  =  1 : 1 : 1.  (The areas are all the same)

Commentary
     The second solution is neater because we do not need to deal with fractions or with pThe  ratio of areas of similar shapes is the square of the ratios of their lengths.  This concept is in fact required knowledge in secondary school mathematics, including GCE ‘O’ level “Elementary” Mathematics.

H02. Use a diagram / model
H04. Look for pattern(s)
H10. Simplify the problem
H11. Solve part of the problem


Suitable Levels
GCE ‘O’ Level “Elementary” Mathematics (“similar figures”)
* Primary School Mathematics (“areas”)
* other syllabuses that involve areas, ratios and or similar shapes.