Monday, December 25, 2017

Easy, yet interesting problem...

I think this is one of the rare problems where all the students solved it in the same way (whoever could solve it :)

While they were still working on this problem, I asked them about their opinion. Most of them guessed that the two areas will be equal. Couple of them thought the greener one might be bigger :)

Vaishnavi was the one who cracked it first...... She proved it beautifully that both the regions are of same size..... Though she could not explain and represent the solution clearly to me in the beginning, however, I was quite impressed with the way she pulled off this stint on the board to the class ....probably the after-effect of long interrogation session she had to go through :)

She said -

1) Radius of the smaller circles is half that of the bigger one. Hence the total area of four small circles is equal to the area of big circle... Note how she has explained this above!

2) Now, this bigger circle is made up of many smaller regions as shown above, labelled as a,b,c......j,k,l.

3) However, the 4 smaller circles include only 8 of these regions (overlapping parts considered twice) ...... ...Is she right? 

4) So, if we equate these two expressions, we get

a + b + c + d + e + f + g + h + i + j + k + l = (a + b + c + d) +  2( e + f + g + h)

This leads to   ( i + j + k + l )  =  ( + f + g + h )

thus proving that the yellow and green regions have same areas.


Jeetu explained this in similar yet little different way. He said -

1) Lets imagine that we have cut the four smaller circles out of paper.
2) Cut these circles further into pieces of shape and size as given in the diagram(12 pieces)
3) Place these pieces on their respective sections on the diagram. 
4) We will be able to place only 8 pieces on the diagram. The four petal shaped pieces will be left over. 

5) Now, because the total area of 4 circles = area of big circle, it implies that -

these spare petal-shaped pieces should be of the same size as that of the 4 sections between smaller & outer circles (i,j,k,l)


How did you solve this problem?
What was your guess?
What's your view about the solutions of students?
How about you trying this with your children/ students?
If so, we (I and my students) will be happy to know your experience and their solutions :)

Thanks and Regards

Rupesh Gesota

PS: Students belong to marathi medium government school (class-7 and 8) based at Navi-Mumbai. To know more about their Maths Enrichment program, check this link:

Sunday, December 24, 2017

Area of Flower

It seems I am getting addicted to listening to (& learning from) the different beautiful approaches of my students...  :-) 

And I am also delighted to see the growing interest of my students to solve more of the Geometry problems these days... 

So the above problem I saw on facebook and I was quite sure, they would like it - Area of Flower :)  The bounding shape is Square of side 2 units and the petals are formed using semi-circles. I had given this problem to them as home-work...

Three of them had solved it in the following way: (Method-1)

Consider one petal of the flower. Divide it into two halves as shown. Now, this half-petal is the circular segment whose area is the difference between the areas of circular sector with radius as 1 and right isosceles triangle with sides as 1 i.e. (pi / 4) - 1/2.  

Now the flower comprises of 8 such segments, thus leading to its area as [2(pi) - 4] 


One of the students, Vaishnavi visualized this in a different way. She sliced the upper semicircular section and just placed it below the lower section. It will look like this:

Now, the Area of flower (i.e. 4 petals) = Area of Square - Area of regions around the petals

Look at the figure above. She said that the Area of regions A and B are same. There are eight such identical regions, each with Area = (4 - pi) / 4. 

So, the total area of such 8 regions = (4 - pi)/4  x  8  = 2(4 - pi)

And while she did this, she also came up with one more method  - almost immediately -


If we remove the area of inscribed square from the area of circle, we get the area of 4 outer half-petals...... Doubling that will give the area of 4 full-petals i.e flower......  Side of the inner square can be found using Pythagoras theorem which is Sq.root of (2). Thus its area is 2.

So, the area of flower = (pi - 2) x 2 = same as above :)


Kanchan said that if we remove the two left and right semicircles from the square, we are left with two regions 'a' and 'd'

So the Total area of regions 'a' and 'd' = (4 - pi)

Now comes the beautiful part -- She saw these two regions 'a' and 'd' as part of the two upper and lower semicircles. So,if these two regions are removed from these 2 semicircles, then what will be left over is the 4 petals !

Hence Area of flower = ( pi - (4 - pi) )


Before simplifying the above expression, she further continued that there is one more method..... Regions 'c' and 'd' are same as 'a' and 'b'.... Hence if twice the total area of regions 'a' and 'b' is removed from the area of square, this too will yield the area of flower.

This is what she has shown adjacent to her first approach above... Now if you observe the two expressions obtained don't appear equal directly. So I teased her that the two answers are different. I was glad that she countered me saying that they are equivalent. She simplified her first solution to (2 pi - 4) orally :)


An Observation - Interpreting the answer.....

I was quite intrigued by the answer we got as the area of flower.... 2 (pi) - 4..... 

This meant if we remove a Square from 2 Circles (diameter = side), then we will get a Flower ! Really? ---  Oh !!

I shared this interpretation with wonder with my students and asked them if we could actually 'see' this...?  They stared at the diagrams for a while... I further thought aloud -

It implies that:-  1 Circle - Half Square = 2 petals !   (isn't it?)

Now, can this be 'seen' ?

And.... in less than half a minute while I was still pondering over this, I heard two YESes :) 

Would suggest you to think about this before reading further...


He said that Half square is the upper Rectangle. This upper rectangle comprises of upper semicircle and upper regions 'c' and 'b' as shown above.


Circle - Half Square = Circle - (upper semicircle + upper regions 'c' and 'b')
                                = lower semicircle - upper region 'c' - upper region 'b'
Now, we have two regions identical to upper regions 'b' and 'c' even in the lower semicircle if you observe the above diagram.... So, the above expression simplifies to:

lower semicircle - lower regions 'b' and 'c' = just two lower petals of the flower.

Now, isn't that just Waaaowww ??  :)

And yes, definitely this becomes super-exciting when this happens happens along with your students - all driven by them :)

  • Do let me know your thoughts about this lesson.....
  • How did you solve this problem?
  • Which approach did you like the most?
  • How about you trying this out with your students/ children?
  • If so, please do share your experiences and their approaches... I and even my students would love to know....  :)
Thanks and Regards
Rupesh Gesota
PS: Students belong to marathi medium government school (class-7 and 8) based at Navi-Mumbai. To know more about their Maths Enrichment program, check this link:

Saturday, December 2, 2017

Another Geometry Problem (Extension) : Part-2

I was pleasantly surprised to know about the amount of interest / attention drawn by my previous post on the geometric problem on various facebook groups... So many people had not just read and liked it, but had even left their comments with their approach of solving this problem. I would first like to thank all these people for sharing their methods. 

I did share some of the different methods with my students (construction of auxiliary lines in different ways) and they were quite intrigued by this fact that there can be so many ways of seeing and solving a geometric problem. 

For those who haven't read the previous post or are not aware of the original problem, you can find it here:

However, Special Thanks to following remark made by a professor/ researcher Michael de Villiers on this post:

"Good, but did you perhaps encourage your students to generalize further by exploring more 'kinks' as in this online, dynamic activity?"

This motivated me to try the extension of this problem with my students...

So then the next day, 

I asked them -- 

"The other day, in our original problem (fig (1)), when there were just two inclined lines between the two parallel lines, we saw that the measure of angle b was sum of the measures of angles a and c......, What will happen if there are three such lines between these parallel lines as in fig (2)...... or may be if there are four or more such lines between them as in fig 3 and fig 4? ........ I mean, will there still be any relation between these angles a, b, c and d in fig (2), the way we had  a + c = b in fig (1) ?"

They thought for a while and started working on their notebooks.. I stopped them...

"Wait. Can you first guess what could be the possible relation?"

These were their different guesses about the four angles in fig (2)

i) a + b = c + d

ii) a + d = b + c 

iii) a + b + c + d = 180

Then I asked them to continue with their work...

While some were engrossed in arriving at the solution, I could see that few were unable to find a break though... So I asked them to "actually construct this diagram using a scale and then measure the angles a,b, c and d to really find out if there is any relation..."

This idea was well received by this bunch... However I could see that not all were ready to do this construction / verification exercise....

So I could see the class divided into two groups - One who were working on the geometric solution without any construction, and the Other group who was doing the construction to 'actually see' the relation first....

Almost both the groups arrived at their respective results simultaneously i.e.

sum of the measures of angles a and c equals the sum of the measures of angles b & d.
a + c = b + d

And then each of these groups was asked to try with the other method - for verification of their result arrived using the first method. It was a beautiful exercise...

Finally they presented their approach to the class :

Kanchan's way

You can see that she has extended the two inclined lines to meet the parallel lines to form two triangles. Then she claimed that the angles 180-(180-c+d) and 180-(180-b+a) will be equal as they form a pair of alternate angles and hence, d-c = a-b  means 

a + c = b + d


While Vaishnavi had construction the auxiliary lines in a different way:

Vaishnavi's way
She has extended the upper inclined line beyond the upper horizontal line and then dropped a perpendicular from there to the lower horizontal line.

She now saw a pentagon and used the property of interior angles of pentagon to prove that a + c = b + d


Jitu instead drew two lines which were parallel to the two given parallel lines as follows: 

Jeetu's way
He then defined a new angle 'x' as interior alternate angles between these two newly constructed parallel lines.

Now if you focus on the upper two parallel lines, then x = b - a  while if you look at the two lower parallel lines, x = c - d , thus leading to 

b - a = c - d   which means a + c = b + d


It didn't take much time for them to arrive at the relation between angles in fig (3) and fig (4) i.e. more kinks between the two parallel lines. This is what one of them did immediately: 

She proved that 

a + c + e = b + d

and even others could infer the relation for more number of angles in fig (4),  

a + c + e = b + d + f


I would also like to share that before I gave these new set of (extended) problems to them, I also showed them couple of different approaches of solving the original problem that they had already solved the other day:

And if you observe closely, this will reveal to you the secret behind Vaishnavi's idea/ approach to solve the extended problems.

In fact, Its also interesting to note that the three methods of drawing the auxiliary lines used by the students in solving this extended problem are completely different than those used by them in solving the original problem (fig (1) few days back.


While we all could happily derive the relation between the angles that will be formed due to many kinks , I wanted my students to 'wonder' at this result..... I asked them if this result /relation was obvious / natural or does it 'surprise' them or 'makes them wonder' as to why this must happen so?

And this nasty question of mine made them stare at this problem again :-) :-) :-)


Meanwhile, what are your thoughts/ reactions about this result? Does that make you 'wonder'?   :-)


Rupesh Gesota

Monday, November 27, 2017

Another Geometry problem...

I knew that this problem can be solved in various ways.. And hence was curious to try it with my students...

They stayed up to my one expectation fully that they could solve this problem quickly. However I got only 3 different approaches (as against my expectation of 5 to 6 !) 

But when I realized that there are still 3 methods from just 7 students, then I was quite happy :-))

Almost all of them had extended the two oblique lines to meet the pair of parallel lines.
After this construction, there were two (little) different routes:

Most common method
In this method, first the Alternate angle property was used to find one of the angles of the Triangle. Then, using the property of angles of triangle, the third missing angle is calculated.

Finally, this third angle of the triangle forms a linear pair with x and hence it can be easily calculated..

However, one of them - Rohit - did a little twist...

He first calculated one of the angles of the Triangle (BGC) using the Alternate angles property like others. 

Then he directly used the Exterior Angle property to calculate x. 

Using Exterior Angle Property
And finally, my most favorite one ! 

This was done by Jitu...

This is how he had done it....

Instead of extending the two oblique lines like others, he dropped a perpendicular from one of the parallel lines to other.

He then focuses his attention on the newly formed Pentagon using this perpendicular line.

This pentagon has two of its angles as 90 while other two are given as 30 and 10. So he figured out the missing fifth angle (they had discovered the formula for "sum of the angles of a polygon" few months back)....

Now he used the property of sum of angles forming a circle to find the desired angle x.

Interesting way, isn't it?  :-)

In fact all the students were delighted by this construction/ approach....

I asked them if any of them could think of more ways... They thought for a while but couldn't think of any... So I gave them a hint - I began from what Jitu had done...

"Jitu created two new points on the two lines.... I can already see two points on these lines.."

"yes sir, we can even join E and F (refer diagram above)"

So I did this construction as suggested by them...

"So what next?" I asked them.

They thought for a while.... and soon realized there is nothing that can help them forge ahead...

One of them exclaimed - Sir we can drop a perpendicular from the point E on the upper line to the lower line...

"Okay... So how will this situation be better than the present one?"

"Here, we get a Quadrilateral with three known angles, so we can find the fourth unknown i.e. X"

I looked at the class for their views... And all agreed with this.... 

Immediately the other said that we can even draw a perpendicular here similar to what Jitu did on the right hand side,... We will directly get x in one step itself... 

I could see a smile on Jitu's face :)

"How about drawing a parallel line somewhere , instead of perpendicular?" I threw this idea to them...

It didn't take much time for one of them to say--

Yes we can draw a line from the vertex of angle x, parallel to the given two parallel lines...

"Will that help?" I asked the class...

Yes... Use alternate angles property twice and we are done ! - shouted one of them.

."Good.. Any other way?"

I could see them quite surprised now :)

"Fine... So let this be homework.. If you find any other way, then share with the class tomorrow... Meanwhile, also write down your solutions systematically with reasoning at home... I will see them in our next class.. "


1) How did you solve this problem ?

2) Your views about the approaches of students..?

3) your opinion about the way this class was taken ?

Will be happy to hear your responses.. 

Thanks and Regards

Rupesh Gesota

PS: These students are from Marathi medium Government school , studying in class 7 and 8... We play with Maths after their school hours.. To know more, check the website

Tuesday, November 21, 2017

My students solved it better than me :)

This is an interesting problem that was given to about 30 Maths Teachers in one of the PD workshops that I attended recently.  I must confess that almost all of us, baring very few, struggled quite a lot and for quite a long to find its solution.. In fact, many of us could not even arrive at the desired solution :) However, I had a gut feeling that my students 'will' be able to reach the destination, and that too in a 'proper' way....

Yes, I was quite sure that my students would do this problem-solving much 'better than me' !  ..... Why?

Because I have been taking care and effort since past 2 years, to see to it that I do not pollute their minds or damage their mathematical thinking by feeding them 'my' methods or by enforcing upon them some 'standard' methods directly....

Makes you reflect?  :-)

So here is the problem:

So, I told this incident to my students while giving this problem... (that it was given to teachers and we all took lot of time, etc).. Naturally, the first thing they asked me was-

"Sir, could you solve it?"

"Yes, I could..."

I could see one of them saying to the other, "'Sir' will 'obviously' solve it....!"

"Wait ! It took me lot of time and effort.... and I was not so happy with my work.... However, I feel you will do better than me....and hence I am giving this to you.... "

Some of them smiled at me, while others had already dived into the problem.. :)

They sought clarification about some requirements mentioned in the problem... I clarified those with examples... and then there was silence for about 10 minutes...

I told them mid-way that they can even work in pairs or groups if they wish.... But none of them did.... It seems they wanted to give their own shot for some time..

Would suggest you to give a try before reading their solutions :)

Two of them - Sahil and Rohit - told me that they are on the verge of completion... have eliminated many possibilities out of the total 120 and are working on the final six ones ....

I understood what they were saying / doing... Did you?  :-)

Soon, they declared that ---- "There is No solution."

I felt Waaaow.... but contained my excitement and rather probed them if they have a proof to support their claim...

Yes sir, we can prove it...

I asked them to pen down their solution systematically while others were still solving... 

In some time, Vaishnavi too declared her accomplishment and this was followed by the rest of the class in next 10-15 minutes....

Every student who was done, was directed to 'write down' their solution / approach in a comprehensible form...

Only when everyone claimed that they have solved it, we began with the Whole-Class Discussion...

Rohit pushed Sahil for sharing.... I had seen both of them working together in the end....

Sahil's work:

He says that there is only one way to make 14 and 13. For 14 - 2,3,4,5 should be together while for 13 - 1,3,4,5 should be together. Hence he concluded that 1 and 2 should be on the two extremes.

So this leaves 3,4,5 in the middle zone which gives rise to 6 possibilities, as listed by him.

He then ruled out each of these possibilities with reasoning. For example: 

Option 1 ----  1,3,4,5,2 does not yield 6
Option 2 ---- 1,3,5,4,2 does not yield 7

and so on.....

Since none of the options work in the favor of given conditions, so the problem does not have any solution.

I asked others if they want to ask anything to Sahil.... All agreed with his work...

Jeetu said that his approach was also very similar to this....

Tanvi came forward. This is her work:..

She too first found out that 1 and 2 will be on the two extremes.

Later, she wrote down all the possible ways of making every number from 6 to 15 (though she later said there was no need to write for 13 to 15 again :)  

She then cancelled out all the options which used both 1 and 2 to make a number.

Then she started with 6-sized bogie. She chose to test one of the options of 6 i.e. 5+1
This would mean placing 5 besides 1. This forces 7 to be made using 4,3. However the option (5,3) for making 8 fixes the position of 3 at the center and hence 5 next to 1.

Thus the train looks like this:  1,5,3,4,2

This structure cannot however make a small train of size-10. 

So this means the initial assumption of 6 to be made using 5 and 1 would be wrong, thus leaving only (4,2) for 6. 

Again, sizes 7 and 8  will fix the positions of 3 and 5 respectively, eventually and surprisingly leading to the same arrangement 1,5,3,4,2 which does not help us for giving 10.

Thus any of the options of 6 leads to an impossibility for the size 10 and hence it is not possible to make such a train....

Again, I looked at the class for their agreement/ disagreement. .. All agreed...

Vaishnavi came up with her approach:

Her initial approach is same as that of Tanvi.... Listing down and ruling out the options having 1 and 2 together. However then she chose to work from the other end (bigger numbers), unlike Tanvi who worked from 6.

Done with 15,14 and 13, she said 12 will also be possible with 3,4,5 anywhere in the center.....  1  ? ? ? 2

Analyzing the only remaining option for 11 i.e. 5,4,2, its clear that 3 will have to be next to 1. This rules out two more options (5,1) for 6 and (5,4,1) for 10.....    1 3 ? ? 2

Analyzing the only remaining option for 10 i.e. (5,3,2) its clear that its impossible to make 10 because 3 and 2 are already 2 spaces apart.

Thus its not possible to complete this order.

All agreed to this.

Kanchan said that her approach is also somewhat similar to Vaishnavi's (starting from the bigger numbers).. This is her work:

She first fixed up 1 and 2 in the two extremes like others. However, she did not list all the possibilities for 6 to 14 like her two peers.

After finishing off  the work of 12 to 15, she listed the two options for 11:
(5,4,2) and (1,2,3,5)

She ruled out the latter (since 1 and 2 are apart) and hence the former got fixed. This meant either 4 or 5 should be next to 2.  .... (note that Vaishnavi had directly concluded that 3 will be next to 1 using this case)

She then listed down the three options for making 10 and one-by-one ruled out those too with appropriate reasoning
(5,4,1) -- Not possible as 5 and 4 will be next to 2.
(1,2,3,4) -- Not possible as 1 and 2 are 3 spaces apart
(2,3,5) -- Not possible as (2,4,5) have to be together to make 11 as discussed above.

Thus it is not possible to make 10 and hence this problem has no solutions.

Remember, I told you in the beginning - about my confidence that my students would be probably solving this better than me... And I am so happy and even proud that they did so !!

You want to see my work?  This is it :))

How about you trying this problem with your students / children?

Could you arrive at the correct solution on your own? What/ How was your approach?
I will be happy if you can share your or your students' approach with me, if its different than any of the above....

What are your views about the thought processes of students above? 

Waiting to hear from you, 

Thanks and Regards

Rupesh Gesota

PS: These students are studying in class-7 and 8 in a navi-mumbai based marathi medium government school.. I am working with them on their Maths Enrichment since past couple of years after their school hours. To know more, check the link: