Tag: solution

Solution to Devising Deactivation

We wanted to find a way for the invader to disable the radar undetected. Remember that Radar has a radius of 100 metres and rotates once every 12 seconds, and Invader runs at 7 m/s with their pack. Clearly, Invader waits until Radar has just passed it to begin. It takes Invader over 14 seconds to run straight to Radar. Maybe Invader could run a circle around Radar when it is about to catch up? For Invader yo circle Radar, they must be within 42/π, or 13.369, metres. That’s because the circle Invader runs must have radius 84 metres to be completed in 12 seconds. 7 m/s * 12 s = 84 m. This still leaves 86.631 metres left for Invader to run in 12 seconds, but they can only run 84 of them. So that’s a no. But what if Invader runs diagonally towards this circle? Invader then has 15 seconds to run √(1002+13.3692) = 100.89 metres. Invader can do 105! Invader does intersect the circle earlier on this path, so it’s not perfect, but it’s simple and works.

Solution to Absurd Algebra

The Absurd Algebraic graph.

We want to find an equation representing the equation on the left. We can begin with the operator modulo, represented by %. a%b gives the remainder when you divide a by b. For example, 7%3=1. x%1 – x%2 will give us -1 when x rounds down to an event number and 0 when it doesn’t. Add 0.5 and multiply by twice x%1 to get disconnected ups and downs centered at 0. If we add (x%1 – x%2) * -1 to this, we move up only the negatively sloped lines and get a zig-zag with peaks at 0 and 1. Add 2 to get our answer, the equation below.

y=2(x\%1-x\%2+0.5)(x\%1-0.5)+2.5

Solution to Crazy Quadrilateral

The Crazy Quadrilateral!

We wanted to find ∠FEG in the crazy quadrilateral. Since ΔABC ~ ΔADE ~ ΔHAG, we know that ∠ABC = ∠ADE = ∠HAG = 90°. This means ∠BAC = 30°. We know that ΔAGE is isosceles. It is obviously impossible to have ∠EAG equal ∠AEG. It is also impossible for ∠EAG to equal ∠AGE, because then EF would intersect AH. Thus, ∠AGE = ∠AEG, so they must both measure 75°. ∠FEG = ∠AEF – ∠AEG, so we have our answer. ∠FEG measures 15°.

Solution to Captivating Circles 6

Captivating Circle #6

We wanted to find the area of the largest piece of the small circle. The radius of the small circle is √2 / 2. To find the distance between 1 chord dividing the small circle and it’s centre, we use the equation below. This gives us 0.2649 units. The pythagorean theorem gives us the length of half the part of the chord in the small circle, in the second equation. This gives us 0.4298. asin(0.4298 / (√2 / 2) ) gives us half the section angle, making the section of the small circle encompassing the missing segment equal to 74.8653°. The triangle making the difference between the segment and sector has area 0.4298 * 0.2649 = 0.1139. The sector has area 74.8653π / 720 = 0.3267. π/2 – 0.3267 * 2 = 0.917. And there we have it: the area of the biggest piece in the smallest circle has area 0.9174 units squared

1-\int_{-\sqrt{h(2-h)}}^{\sqrt{h(2-h)}}\sqrt{1-x^2}\,dx – 2(1-h)\sqrt{h(2-h)}=\frac{\pi}{3}\\[16pt] x=\sqrt{(\sqrt{2}/2)^2-0.2649^2}=0.4298

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.

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.

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.

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.

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.

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.