Captivating Circles 6

This is the final Captivating Circles challenge, this challenge requires the concepts of geometry and trigonometry.

Captivating Circle #6

We begin this challenge with 2 concentric circles. The smaller circle divides the larger 1 into 2 equal pieces. We then draw 2 parallel chords in the large circle, which divide it evenly into three. They do not divide the small circle into three equal pieces. The challenge is then to find the area of the biggest piece in the smallest circle if the large circle has radius 1.

Solution to Captivating Circles 5

We can simulate the decrease in size for Jack and Jill with the javascript below:

  1. let jack = 1π;
  2. let jill = 2.25π;
  3. while( jack > 0 ) {
  4. jack -= jill * 0.000001;
  5. jill -= jack * 0.000001;
  6. }
  7. console.log(jill);

We will find Jill’s size area to be 6.333 in the console. Divide by π and square root to get Jill’s radius after facing Jack: 1.42.

Captivating Circles 5

You will need to use algebra to solve this challenge. Some programming would be extremely useful as well.

In the video-game “Battle of the Blobs”*, a blob is a circular creature that deals damage over time proportional to its current area. A blob’s area will shrink proportionally to damage dealt to it. The damage over time is not dealt at short intervals, rather is always being dealt, but with the same rate. Our challenge begins when 2 blobs, Jack and Jill, battle each other. The radius of Jack is 1, and the radius of Jill is 1.5. Clearly, Jill will kill Jack. The question is what will be Jill’s radius after fighting Jack, assuming the only change in a blob’s size is caused by damage from the other.

*”Battle of the Blobs” is not a real video-game.

Solution to Captivating Circles 4

Captivating Circle #4

This challenge is extremely long. The reason for this is that there was a slight error in the drawing of the diagram. 1 line was misplaced. Apologies for the inconvenience. The diagram, when simplified to a graph, should have been rotationally symmetric. As it is, it is just barely not. because of this, the number of shapes is many times greater than it should have been. The solution to the challenge as it is will be posted nevertheless. We can label each section with letters A to L. We can then list the shapes by the sections they contain, like AB, CDG, and so on. All in all, there are 2037 shapes.

Captivating Circles 4

You will need all your skills in graph theory, and pathmaking, algebra and combinatorics to solve today’s challenge. Good luck!

Captivating Circle #4

This captivating circle has been divided in to many sections. By combining these sections, you can make many different composite shapes. The challenge is to see how many shapes you can make. This includes the entire circle.

Solution to Captivating Circles 3

Captivating Circle #3

We wanted to find the maximum number of slices for our circle. This can be done by finding an equation for the count you have an need to see where they intercept, which requires comlex factorials, making it difficult to solve. The solution is 5. We can also use trial and error to obtain this. The equations will be posted seperately.

Captivating Circles 3

In this captivating circles challenge, you will need to use some algebra, some combinatorics, and some serious brainpower.

Captivating Circle #3

For our newest captivating circle, we are dealing with splitting oranges. The orange starts with N slices in a circle. It is then divided into 2 rings with A slices and B slices. In the diagram, N=9, A=3, and B=6. We then count one 3 and one 6. One of these rings is splitted to make 3 rings in all. We then count these rings. This continues until we can no longer split any rings as they all contain 1 slice. It happens that at the end, the number 1 is counted at least 3 more times than 2, which is counted at least 3 more times than 3, etcetera until the largest number counted. The challenge is this: what is the maximum value for N?

Solution to Captivating Circles 2

Captivating Circle #2

We wanted to find the ways to complete the circle, and the maximum distance along connections for 2 points. There are only 2 ways to fill the circle. If we notate them by the number of nodes travelled clockwise with each line, they are 3-3-3-3-3-3-3, and 3-3-3-2-4-4-2. The maximum distance for 2 points would be 2 points 3 circle segments apart. Of course, these would not be connected but would have to have 2 points connected within the 3 segments between them. The length of this is sin(360/7)/2 for the line and π/7 for the arc. This makes our answers 2 and ≈0.9078.

Captivating Circles 2

Today, we have a new captivating circle, and it is the first challenge to require the use of combinatorics. It also requires some trig and pathmaking.

Captivating Circle #2

Our second captivating circle is a circle with 7 points on its circumference. The points are equally spaced. Each point connects to 2 others through the circle and 2 others through a straight line. No points can connect to another through both a line and arc. The circle has circumference π. How many possible different sets of connections can be made? And of these, what is the highest possible distance that must be traversed to get between 2 points?

Solution to Captivating Circles 1

Captivating Circle #1

In this challenge, we wanted to find the area of the shaded region. First, we must note that the white triangle is a right triangle. the side that is 1.01dm is a diameter of the large black circle. With that information, we can find that the area of the big circle is 2550.25πmm2, and the area of the triangle is 990mm2. We also know that the semiperimeter of the triangle is 110mm. We can use the formula A=rs to find the radius of the small circle to be 9 mm. We then find it’s area to be 81πmm2. Putting this all together we get an anser of 2631.25πmm2 – 495mm2.