Get Puzzled!

As a continuation we will consider two more puzzle which motivates one or more applications in our day to day life.

Puzzle #1: A school  math teacher wants to teach geometry to her class 8 students. She collects 2 hemisphere shells and a cuboid, such that the cuboid can be inscribed inside the shell. She also bought paints -red and blue. One hemisphere was painted red, while only 80% of the second shell could be finished with the red paint.. So she decided to paint the remaining by blue paint. The teacher put the cuboid inside the sphere, and sees that all vertices fall on the red colour (assume that the shell is semi-transparent). Will her observation be true irrespective of the way she had painted.

The video solution of the puzzle above : - 

Solution:  If we consider the two hemispherical shells as a sphere, we understand that 10% of the sphere is painted in blue. Let us now denote, the vertices of the cube as \(v1\), \(v2\), \(v3\), \(v4\), \(v5\), \(v6\), \(v7\)  and \(v8\). The chance that a given vertex is placed in the blue part of the sphere, is given by:

\[\mathbb{P}(\cup_i v_i \in Blue)  = 0.1 \times 8 = 0.8\]

Thus, the chance that all eight vertices fall on the red part of the sphere \(= 0.2\). This means that there is at least one way of arranging the vertices such that all will fall on the red paint. The assertion made by the teacher is right.

(Courtesy : NIT Lecture, IITB)

Another one from geometry

Puzzle #2: A statistician is interested in this particular geometrical issue to find a spectrum allotment. An equilateral triangle is inscribed in a circle. The altitude of the equilateral triangle is 3/2 r. Find the chance (probability) that a chord chosen at random will intersect with the interior of the equilateral triangle. 

Try to work this out using more than one method. Is the answer you get the same in all cases?

Solution: 

Method 1 :  Each chord of the circle is uniquely defined by its center. We can place the chord's center inside the equilateral triangle area, so that it always falls inside the interior of the triangle. Thus the chance of choosing a chord which is in the interior of the triangle is

\(\mathbb{P}(\text{chord} \in \Delta) = \frac{\text{Area of } \Delta}{\text{Area of Circle}} = \frac{3\sqrt{3}}{4 \pi}\)

Method 2 :  Another method to express the probability is as follows; Fix one of the end points of the chord at the vertex of the equilateral triangle.. The chords that can fall in the interior will fall in the angle range \(\frac{\pi}{3} \leq \theta \leq \frac{2 \pi}{3}\) . Thus the probability of the event of interest

\(\mathbb{P}(\text{chord} \in \Delta) =  \frac{\pi/3}{\pi} = \frac 13\)

Method 3 : The chord is defined by its distance from the center of the circle.Thus the chords with distance from the center less than \(r/2\) can be classified as the interior of  the triangle. Thus the chance that the chord chosen at random will fall in the interior of the triangle is given as:

\(\mathbb{P}(\text{chord} \in \Delta) = \frac{r/2}{r} = \frac 12 \)


Notice the three distinct answers we obtained. Can you think of a good reason to justify this.

If we look into the situation more clearly it is evident that the notion of uniformly random chord is not well defined. In that case all 3 methods are right . Unless a fixed notion(reference) has been foretold, there is no clear solution. Such class of problems (challenges!) are called ill-posed .

(Courtesy : Introduction to Probability Theory, IITB )

The Balancing Act

Lets us start with a puzzle somewhat common place. A shopkeeper gets a stock of dozen cricket balls out of which one is a cork ball (kookaburra) and the remaining are rubber balls.  He is asked not to touch balls with hands (as it has to be used for the match), but is given a "pan" balance which will show either equal, less than, or larger than for the weights on the pans.

Find the minimum number of uses of the balance to find out the heavy ball.

Hint: 1 >> Separate the balls

The extension to this puzzle is interesting, Let us say that we want to find the odd ball (which can be  heavy/light) from the lot of 12 cricket balls. 

Guess how?

The interesting fact is that the odd ball can be determined with its weight in three measures. The idea is very similar to the source coding - Huffman coding. We can form a coding tree and verify it easily.

Let Left pan be represented as L and right pan as R.

The stepwise solution is as follows,

1. Divide the 12 balls into groups of three.
2. Weight 2 groups (of three balls) --> 1st measure. There are three possibilities 
• L=R then discard both the group
• L>R , call case A
• L<R , call case B
3. (If case A or case B happened) In the cases L>R and L<R, replace the right set by any of the untouched group. --> 2nd measure Then, 
• L=R, then the balls removed from the right pan is the odd group and follows the case in the previous measure. (ie, if case A-- light, case B -- heavy)
• L!=R, the the case which happened in the previous measure follows(ie, if case A-- heavy, case B -- light)
4. From the odd group determined from 3, take any two balls and do the measure;  --> 3rd measure
• if L=R, the  odd ball is the unmeasured ball and follows the same weight as in point 3.
• if L!=R, then the odd ball is one which follows the case in point 3.

The reader now can try to generalize this for \(m\) measurements