circle-square writeup
Problem statement
A square and a circle are situated on a line with length BR, where B is equal to the perimeter of the square and R is equal to the circumference of the circle. BR is equal to 40 centimeters total. As the position of the shapes changes left or right, their size changes at the same rate as the length of B or R, respectively. At which point on BR is the area of the shapes the same?
Official Problem Statement by Given an arbitrary point P on a line segment AB, let AP form the perimeter of a square and PB form the circumference of a circle. Find P such that the area of the square and circle are equal.
Official Problem Statement by Given an arbitrary point P on a line segment AB, let AP form the perimeter of a square and PB form the circumference of a circle. Find P such that the area of the square and circle are equal.
process
Although the work that I did with my team was based on guess-and-check and inputting different numbers to see if they would match up. While a lot of groups did this, we found that a couple tried to come up with formulas for each of the shapes' areas in respect to B or R. At first, I only used one variable. They used this equation: pi(L/2pi)^2 = (4/3L)^2
However, they set the length to only one variable (L), when it should have been one value for each shape. Also, they should have the second value be in terms of the first, because they needed to be able to solve the equation for the variable I chose (B) eventually. So, here's the real equation which was given to us by Gabe's team: pi((40-B)/2pi)^2 = (4/3B)^2
Now, if you solve the above equation for B, you should get something like 21.1 centimeters, which means that the shapes would have to be sitting on that point on line BR for their areas to be equal.
However, they set the length to only one variable (L), when it should have been one value for each shape. Also, they should have the second value be in terms of the first, because they needed to be able to solve the equation for the variable I chose (B) eventually. So, here's the real equation which was given to us by Gabe's team: pi((40-B)/2pi)^2 = (4/3B)^2
Now, if you solve the above equation for B, you should get something like 21.1 centimeters, which means that the shapes would have to be sitting on that point on line BR for their areas to be equal.
assessment and self-evaluation
Overall, I somewhat enjoyed this problem because it was very interesting to try to find a solution when you were not really given much information to begin with, but I feel like that is what made me a bit "stressed" at this problem. I felt like the information that we were given was very vague, and did not help too much when trying to come up with original ways to find a solution. That is why, throughout the problem, the main thing I learned was how to use creative thinking and apply it to mathematics when trying to find the answer to a certain problem. I, along with the rest of my group worked mostly on coming up with ways to connect what we were learning through the "guess and check" method to some kind of formula. We were not able to do so, but we worked well as a team trying to.
I think that I deserve an A on this problem, mostly because I never stopped trying to find a solution (even if it was through "guess and check"). I tried really hard to not get discouraged when it was getting more and more difficult to find solutions to the questions we were being presented. I also presented several times throughout the problem, and think that I did a good job at explaining my process and the methods I used.
I think that I deserve an A on this problem, mostly because I never stopped trying to find a solution (even if it was through "guess and check"). I tried really hard to not get discouraged when it was getting more and more difficult to find solutions to the questions we were being presented. I also presented several times throughout the problem, and think that I did a good job at explaining my process and the methods I used.