[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  = L – M + S,  and  area of B = 2´(M – S).  [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/₂ – ^4p/₂ + ^p/₂  =  3p

     area of B = 2´(^4p/₂ – ^p/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´(4S – S) =  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 p.  The  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.

Heuristics Used

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.