Month: March 2019

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

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

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.

Captivating Circles 1

Today we begin a special series, captivating circles. This challenge requires trigonometry and geometry.

Captivating Circle #1

The first captivating circles challenge is to find the area of the shaded region in the circle in the diagram. All vertices of the white triangle are on the circumference of the big circle, and the small circle is tangent to all 3 sides of the triangle. You may find that this problem requires theorems and formula’s not normally required. Do not be afraid to research the tools you need to solve this problem or other captivating circles, it’s knowing which tools to use that’s the challenge.

Solution

Solution to Spring Forward

We want to know when Joanna’s muon left Vancouver. Since teleportation is instantaneous, it leaves at 2AM Toronto time as well as arriving. There are 2 things we must account for. 1, Vancouver time is normally 3 hours behind Toronto time, which brings us to 2, Vancouver would not yet have moved the clocks forward to daylight savings. This makes Vancouver 4 hours behind Toronto. Our answer is thus that Joanna’s muon teleported from Vancouver to Toronto at 10PM Vancouver time, just before daylight savings.

Solution to Mystery Mixture

We wanted to find the weight per cm3. In systems of equations with many variables and equations, we should start with a matrix. In the matrix below we have rows by A, B, C (%), W (g/cm3).We can reach the identity matrix with 3 row operations. First, R1 : (-43/584)R1 + (275/1168)R2 – (35/584)R3. Second, we do R2 : (7/43)R1 + (1/430)R2 – (3/215)R3. And R3 : -4R1 – 15R2 + (1/6)R3 finishes it off. Then, we can find that the mixture has 40 g/cm3. Finally, finding the 1:2:3 ratio is easy because in both powder 1 and the mystery mixture, A=B. Powder 2 : powder 3 must equal 11:3, making the ratio of A:B:C is 67:67:36. To get A:B:C to equivalent 2:2:3, we must add 129 1’s to the 1:2:3 ratio to get 129:11:3, we have A:B:C equals 196:196:294, or 2:2:3. Our answers are 129:11:3 and 35.86 g/cm3.

\left(\begin{matrix} 25 & 25 & 50 & 35 \\ 50 & 20 & 30 & 44 \\ 20 & 75 & 05 & 43 \end{matrix}\right) \overrightarrow{_{-\frac{43}{584}R_1+\frac{275}{1168}R_2-\frac{35}{584}R_3\rightarrow R_1}} \left(\begin{matrix} 1 & 0 & 0 & 0.6 \\ 50 & 20 & 20 & 44 \\ 20 & 75 & 05 & 43 \end{matrix}\right) \\[16pt] \left(\begin{matrix} 1 & 0 & 0 & 0.6 \\ 50 & 20 & 20 & 44 \\ 20 & 75 & 05 & 43 \end{matrix}\right) \overrightarrow{_{\frac{7}{43}R_1+\frac{1}{430}R_2-\frac{3}{215}R_3\rightarrow R_2}} \left(\begin{matrix} 1 & 0 & 0 & 0.6 \\ 0 & 1 & 0 & 0.4 \\ 20 & 75 & 05 & 43 \end{matrix}\right) \\[16pt] \left(\begin{matrix} 1 & 0 & 0 & 0.6 \\ 0 & 1 & 0 & 0.4 \\ 20 & 75 & 05 & 43 \end{matrix}\right) \overrightarrow{_{-4R_1-15R_2+\frac{1}{6}R_3\rightarrow R_3}} \left(\begin{matrix} 1 & 0 & 0 & 0.6 \\ 0 & 1 & 0 & 0.4 \\ 0 & 0 & 1 & 0.2 \end{matrix}\right)

Mystery Mixture

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.

Solution

Solution to Mixing Glasses

We wanted to find the minimum number of mixes to equally distribute liquid A. It is critical to note that liquid B’s positions do not matter. There are many ways to do this in the minimum mixes, but they are essentially the same. First, mix the glass with 500mL of liquid A with another. We now have 2 glasses with 50mL and 1 with 150mL. Mix one of each. Finally mix both glasses that now have 100mL of liquid A with the 2 that have none. And there we have it. It takes 4 mixes to get 50mL of liquid A in each of the 5 glasses.