Author: zachary

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.

Mixing Glasses

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.

Solution

Solution to Crazy Connections

We wanted to find the minimum number of points of each colour for a set. The most important thing to notice is that if one type of point connects to another type at a different frequency than vice versa, than the ratio of their sizes is the inverse of that difference. Put simply, if red points connect to 2 orange points and each orange point connects to 1 red point, there are twice as many orange points as red points. Repeat this to find a collective ratio of 1:2:4:3:1:1. Since red points connect to 1 other red point, we can thus find our answer. There is a minimum of 2 red points, 4 orange points, 8 yellow points, 6 green points, 2 blue points, and lastly, 2 purple points.

Crazy Connections

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?

Solution

Solution to Opposite Lines

let: f(x)=-g(x),h(x)=1/g(x)

In this challenge, we wanted to find the antiderivative of f(x), given the above, and that the derivative of h(x) is 0.5. First, lets find f(x) in terms of h(x).

f(x)=-1/h(x)

Then we can find the derivative of f(x) to be -2. The antiderivative of this, f(x), is -2x+c. The antiderivative of f(x), therefore, is -x^2+cx+c

-x^2+cx+c