Challenges cover a variety of topics
You will need all your skills in graph theory, and pathmaking, algebra and combinatorics to solve today’s challenge. Good luck!
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.
In this captivating circles challenge, you will need to use some algebra, some combinatorics, and some serious brainpower.
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?
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.
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?
Today’s challenge requires matrices.
A mixture is made by mixing 3 powders: powders 1, 2, and 3. The powders are each made from different amounts of the same 3 ingredients. The ingredients are named A, B, and C. Powder ratio’s are A : B : C. Powder 1 is 1 : 1 : 2 and weighs 35 g/cm3, powder 2 is 5 : 2 : 3 and weighs 44 g/cm3, and powder 3 is 4 : 15 : 1 and weighs 43 g/cm3. The mystery mixture is 2 : 2 : 3. The challenge is: how much of each powder is in the mixture, and how much does the mixture weigh per cm3? Answer in 1 : 2 : 3 and g/cm3.
Today’s challenge requires the concept of Algebra.
5 glasses contain amounts of various liquids. The glasses are numbered 1 to 5. Non-prime glasses contain liquid A, and prime glasses contain liquid B. Odd glasses contain 750mL, and even ones contain 500mL. When you mix 2 glasses they then contain the same quantity of liquid in the same proportions. The challenge is to find the minimum number of mixes to ensure that each glass contains the same quantity of liquid A.
Graph theory and algebra are required to solve this challenge.
Our challenge begins with a set of points. There are points of red, orange, yellow, green, blue, and purple. These points are all interconnected as follows. Every red point connects to 1 red point, 2 orange points, and 3 green points. Each orange point connects to 1 red point, 2 orange points, 2 yellow points, and 1 blue point. All yellow points connects to 1 orange point, 1 other yellow point, 3 green points, and 1 purple point. Green points all connect to 1 red point, 4 yellow points, and 1 blue point. All of the blue points connect to 2 orange points, 3 green points, and 1 purple point. And finally, purple points connect to 4 yellow points, 1 blue point, and 1 purple point. The question is this: what is the minimum number of points for each colour?
This challenge challenges your algebra skills.
Joanne is doing a science experiment. She is trying to purify water so that there are no isotopes. This means that there will be only normal H₂O molecules. If Joanne has 9kg of water after all of this, how many kilograms of hydrogen atoms does she have?
Today’s challenge utilizes the concept of algebra.
let: a+5=2-b, ab+3=a-74This time, the challenge is to see if you find all solutions to the above system of equations.
You will need to understand calculus and linear algebra to solve this problem.
f(x) = g(x+1)-x,\space g(x) = f(2x)-2Today’s challenge is this: what is the derivative of f( g(x) )? Good luck!
You will need to understand algebra and angle theorems to do today’s challenge.
With today’s challenge, we’ll be trying to find the area of a square! All outer regions of this square have equal area, and the inner region, a square itself, has side length 1. Can you find the area of the big square? Good luck!