Hello all. Welcome to the 8th of our Gifted World challenge series. Below the question, you will also find __the solution and featured responses__ we received for __Challenge number 7__.

This week's challenge is a mixture of economics, mathematics and general people skills. We don't have a particular 'right answer' in mind, so give it your best shot!

## Challenge #8

**Post your answer here - **__https://forms.gle/JbVZrdVgMVUuTUe68__

There is an apartment building with 12 floors (Ground + 12). The ground floor consists of parking space, and there is one apartment on each of the 12 floors. The electricity bill for the lift in the apartment comes to INR 6000 per month.Â

For many years, all the 12 apartments have just been splitting the bill equally between themselves, each paying INR 500 per month.

One day a lawyer moves into the first floor of the apartment. She finds this split of the bills unfair. Why should she, who only takes the lift up and down a single floor, pay the same amount for electricity as the person who lives on the top floor and goes up and down 12 floors each time?

__Can you help her devise a fair way of splitting up the electricity bill between the different floors. Please provide reasoning or justification for your solution.__

**BONUS:Â **

Consider a further complication. Often multiple people are taking the lift at the same time, such that the lift will stop at multiple floors on the way to the highest floor. In that situation, the lawyer living on floor 1 is nearly always going in the lift when there is someone anyway going to a higher floor. So the lawyer isn't causing additional usage of the lift. If that is the case, what is a fair way of splitting the bill?

**Post your answer here - **__https://forms.gle/JbVZrdVgMVUuTUe68__

Last day to submit answers is Saturday, July 27. The 3 best answers (as judged by the Gifted World team) will be featured on our blog. We don't have a particular 'right answer' in mind, so give it your best shot! Looking forward to some creative submissions.

This challenge has been given by Gifted World Mentor Siddharth Bharath.

## Solution to Challenge #7

__Challenge number 7__ involved doing geometry with limited tools. We received 16 responses to the questions. I will first share our solution, and then feedback on the answers we received.

__Our Solution__

__Our Solution__

Fold the paper twice to get a 90 degree angle (the corner of a rectangle). Now, there is a property of a circle (called __Thales Theorem__) which states that if you draw a triangle with one side being a diameter of a circle, and the third point being on the circle, the angle at that third point will be 90 degrees.

Put the vertex of the 90 degree angle created by your paper on any point of the circle, follow the angle to find the two points where it intersects the circle. Those two points are a diameter.

Repeat this again to find another diameter. The intersection of the two diameters is the centre of the circle.

__Best Answer we received__

__Best Answer we received__

This answer came from Aahana Das, 8, Greenwood High, Bengaluru. It involves multiple tries to get the length of the diameter right, but will work with the provided equipment.

1. Fold the page in half. The straight edge will act as a ruler. I shall refer to it as a paper ruler.

2. Place the paper ruler on the circle and mark a point on the paper where the circle's edge is there. Call the point on the circle "A".

3. Rotate one side of the paper along the circumference such that the marked point on the paper is on point A of the circle at all times.

4. At the place where the distance between the marked point on the paper and the opposite edge of the circle looks maximum, we mark a point on the paper at that edge of the circle. Call it point B.

5. Rotate the paper some more, making sure to keep paper ruler point at exactly point A on the circle.

6. If the distance from point A and the opposite edge of the circle increases further than point B, then we mark that point.

7. Repeat till the last point marked is at the same distance from point A as the diameter, i.e. the longest distance you can get from point A while still being within the diameter of the circle.

8. Place the paper ruler in such a way that point A and the last point marked are at the opposite edges of the circle and trace a straight line.

9. Repeat with another point C on the circumference.

10. Point of intersection is the center of the circle.

__Honourable Mention__

Honourable mention goes to Rishan Anoop Rao, Grade 8, The Future Kid's School, Hyderabad. We don't see how you can precisely construct medians of the triangles, but this was a different take on the geometry of the problem that we were not expecting.

"I first drew a circle using the glass. Then I drew two congruent circles such that they had one point in common with each of the other circles. I connected each of the common points with straight lines (I folded the paper until it became usable as a makeshift scale). Then I drew the three medians of the triangle using the curves of the circles as reference. This way I found out the centre of the triangle (it was equilateral- How, I don't know). I drew the same triangles on both sides of the circles and then extended the centre median into the main circle. The point where the three lines intersected was the centre.

I have no proof for this - I was just messing around with the question and this seemed to work!

Note: I think I could have solved it with only 9 lines, by leaving out the triangles and finding medians of imaginary triangles."

__Feedback on answers that didn't make it all the way __

Several of your answers involved drawing a circle with the mug on the paper, then tearing / cutting out that circle, and then making a semicircle with that. We feel that this cannot be done precisely with the tools you have, and therefore will not result in accurately finding the centre of the circle that is on the cardboard.

Some of you fold the paper to make a scale, then drew 2 lines on the cardboard circle, called those diameters and marked their intersection as the centre of the circle. Again, it isn't possible to know if that is an accurate diameter, and so the centre thus found is not accurate.

Two respondents suggested folding the paper once to create a ruler, then drawing a square that circumscribes the circle, and finally connecting the diagonals to find the centre. However, you can't precisely draw a square like this, and if the shape is not a square, then the intersection of the diagonals will not be the centre of the circle.

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