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Science communication is important in today's technologically advanced society. A good part of the adult community is not science saavy and lacks the background to make sense of rapidly changing technology. My blog attempts to help by publishing articles of general interest in an easy to read and understand format without using mathematics. I also give free lectures in community events - you can arrange these by writing to me.

Sunday, 29 April 2018

A Couple of Challenging Geometry puzzles - Created by Martin Gardner - Recreational Mathmatics

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These Puzzles are mathematically challenging

I was inspired to look into Martin Gardner's (1914 - 2010) work by an article in the recent Physics World issue.  Gardner ran a column in Scientific American for 25 years and published books on recreational mathematics.  Some of his puzzles can take a life of their own and result in a few sleepless nights.

I have chosen two examples from the famous collection of his puzzles based on geometry.  I use these puzzles to show how trigonometry and geometry are equally efficient in solving problems.  Personally, I find trigonometry easier to use - it provides ready-made relations between the sides and angles of a triangle which might be more difficult to visualize in a geometrical context - particularly if you are using hand-drawn diagrams that may not be  constructed accurately.

The first problem that I have chosen is best illustrated by the following diagram (click on the figure to see a full page image)

I shall give the geometry based solution but let us first try the easier trigonometric solution of the above problem:

Geometrical Solution: If we look at the angles at the corner marked T in the diagram:

angle WTR = angle TRS = angle C
angle WTP = angle TPS = angle A

and angle WTR = angle WTP + angle PTR
or          C = A + angle PTR 

Essentially, to solve the puzzle, we need to prove that angle PTR = angle TQS 

One way to do this is to complete a right angled triangle that will include angle PTR.  This may be done by drawing the perpendicular PV from point P on an extended line TR - this gives us the following figure:

Line PV has been drawn parallel to the diagonal YR of the middle square to ensure that angle PVT is 90 degrees.  In the figure above, we can clearly see that right angled triangles QST and TVP are similar as all their sides are in the same ratio (1 :2), viz. 

QS  =  2,       ST  =  1     and   TQ  =  √5
TV  = 2√2,    VP = √2     and   PT = √10

This means that corresponding internal angles are equal.
 Therefore, angle facing side ST (angle B)  is equal to angle facing side PV  or  

         angle PTV  (or PTR)  =  angle B

Thus we can be sure that angle PTR = angle B, and 

                        angle C = angle A + angle B

Notice, we have used the properties of similar triangles to solve the above puzzle.  This method is used frequently in solving geometry problems and sometimes it can provide a quicker and more concise solution. To demonstrate this point,  I discuss another geometrical puzzle in the following:  


There are three different ways that one can solve this problem.  
Geometry, Trigonometry and Pythagoras Theorem. 
We shall try all three methods:

Solution using Geometry:  Here we have three similar right-angled triangles - their internal angles are the same.  It is straightforward to show that

Triangle ABC:  angle BAC = A, angle BCA = C and angle ABC = 90
Triangle BDA:  angle BAD = A, angle ABD = C and angle ABC = 90
Triangle BDC:  angle CBD = A, angle BCD = C and angle BDC = 90

In similar triangles, the ratios of corresponding sides are equal. This implies that

For angles 90 degrees and A ___  AC:BC = BC:DC = AB:BD
   or   25:a = a:16 = c:H
   or     a^2 = 25 x 16 = 400   or   a = 20 cm

For angles 90 degrees and C ___  AC:AB = BC:BD = AB:AD
  or   25:c = a:H = c:9 
  or     c^2 = 25 x 9 = 225    or   c = 15 cm

We also have a:H = c:9   or  20 x 9 = 15 x H   or  H = 12 cm

Therefore   H = 12 cm,  a = 20 cm and c = 15 cm

Solution using Trigonometry: 

In right angled triangle ABD,   tan A = BD/AD = H/9
In right angled triangle CDB,   tan C = BD/DC = H/16

But angle C = 90 - angle A;   therefore  tan C = tan (90-A) = 1/tan A

This gives us ---   tan A = H/9 and 1/tan A = H/16
On multiplying the two equations:  1 = H^2/(9x16)

            or   H^2 = 144   or H = 12 cm

Solution using Pythagorus Theorem:  

In the three triangles:

     a^2 + c^2 = 25^2     or   a^2 + c^2 = 625    eq.1 
     H^2 + 16^2 = a^2    or   a^2 - H^2 = 256     eq.2  
     H^2 +  9^2  = c^2    or   c^2 - H^2 =  81      eq. 3

Subtract the sum of equations 2 and 3 from equation 1 to obtain

2 x H^2 = 625 - (256 + 81) = 625 - 337 = 288

or H^2 = 144   or   H = 12 cm

Final Word:  Geometrical puzzles are really addictive.  I remember during my school days, my father had asked my younger brother and I to see if we can divide an angle in three equal parts just using a compass and ruler.  Of course, this is not possible but we spent weeks of our play time trying to work it out and used up a large amount of paper.  But it was lots of fun and we learnt many useful tricks and theorems in mathematics during this time.  It was one of the most productive period in term of return from our unsuccessful attempts. Struggle makes you wiser and more resourceful. 

Keep puzzling...

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