Author: zachary

Puzzling Probabilities

This challenge, though it appears simple, is far more difficult than you might expect. Knowledge of algebra, combinatorics, and graph theory, as well as ingenuity and effort, are necessary to successfully complete this challenge. Be warned.

Bob goes to Scam Casino to lose some money. He decides to play the game “Stupid Spin.” There are many equal pieces of the spinner, and only a few give Bob a point. Otherwise, the Casino gets one. The probability that Bob gets a point is a, and the first to b points wins. The challenge is to find the probability that Bob will win a given game of Stupid Spin.

Solution

Solution to Crazy Capture

2D capture example.

We wanted to find the minimum number of police officers to capture a target. Remember that all units and the target are points with constant speed, perfect logic, and instant reaction time. Consider the lines perpendicular to those through the target and officers. Clearly, the target cannot cross any of these lines, the officers would always block it off. Furthermore, the target cannot delay all the officers in this way at once. The inactive ones can close in. It follows that a closed shape can catch the target. The smallest 2D shape is a triangle. We realize that we can generalize this. In higher dimensions, we use perpendicular planes and such. But the smallest figure is always the simplex! The simplex has n+1 vertices. Therefore, our answer is that 3 constables are needed to catch the target in 2 dimensions.

Crazy Capture

All of your algebra, geometry, trigonometry, angle theorems, and pathmaking skills are required to have a chance of solving today’s challenge.

2D capture example.

In a mathematical capture, all police officers and the target are points that travel at the same constant speed in infinite space in n dimensions. They have infinite endurance and perfect reaction time. The target is instantly captured and cannot ecape when a police officer touches it. In 1 dimension, when n = 1, it obviously takes 2 officers surrounding the target. When n = 2, we can do it with three easily. We do this by dividing the plane into three, each officer can easily trap the target in a 120° slice. The challenge is to see how many officers it takes to capture the target when n equals 3, in three dimensions.

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.

Devising Deactivation

To solve this challenge, you will need to use geometry, trigonometry, and algebra, to do pathmaking.

Map of the radar that must be deactivated.

You are working for the government as a mathematician and they tell you of a problem. There is a plan afoot to compromise an enemy radar station. It is critical that the invading crew member is undedected, as the enemy must not know that their radar is sending them misinformation. However, the radar has a range of 100 metres and completes a rotation every 12 seconds. The fastest crew member can only sprint at 7 metres per second with their toolkit. Because of the radar’s length and rotation speed, it is nearly impossible to disable it undetected. None of the simple strategies work. The challenge is to find a path that the invader can take without being detected.

Solution

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

Absurd Algebra

Extreme algebra knowledge and skills are required to solve this challenge.

Absurd Algebraic Graph.

This time, we are faced with a strange graph. When x rounds down to an even number, the slope is 1. But when it rounds down to an odd number, the slope is -1. This makes an intriguing zig-zag pattern with peaks always at 2 or 3. But what equation could possibly give such a line? Well, that is the challenge.

Solution

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°.

Crazy Quadrilateral

You will need to use geometry, angle theorems, trigonometry, and your brain to solve this challenge.

Crazy Quadrilateral

We begin with a quadrilateral AEGH. We then draw a line from A to G to make ΔAEG and ΔAGH. ΔAEG is an isoscelese triangle. Next, we draw a line from point E to a new point, F, such that ∠AEF measures 60°. This makes a new point, D, where AG intersects FE. After that, we draw a line HC so that it is parallel to FE and intersects AG at point C. This makes ΔABC ~ ΔADE ~ ΔHAG. The challenge is to find ∠FEG if ∠HAG measures 90°.

Solution

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