Category: Trigonometry

Kayoed Kangaroo

This challenge considers an upset kangaroo making a series of discrete jumps in order to retaliate, and thus involves trigonometry.

Kangaroo chasing Sheep

Red Kangaroos in Australia compete with sheep for food. Sheep is grazing when it bumps into Kangaroo. The kangaroo wishes to learn the sheep some respect by either pummelling it, or pushing it into a nearby pond. Sheep wants to escape Kangaroo, but is slower than Kangaroo. However, Kangaroo is only capable of moving in jumps. Ergo, if Sheep stays far enough away to avoid pummelling, but too close for Kangaroo to jump, it is safe, and thus can avoid retribution indefinitely. Precisely formulated:

  • Kangaroo can jump up to 8 meters
  • Kangaroo cannot jump fewer than 5 meters
  • Kangaroo can jump at most fource per second
  • Kangaroo can pummel the sheep they are within 1 meter
  • Sheep can flee at 10 meters per second
  • Sheep can flee for 1 second before before Kangaroo pursues
  • There is a nearby pond, a perfect circle and 100 meters in radius
  • Sheep does not like swimming or being pummelled;
  • Kangaroo wishes to punish Sheep for perceived slight

Therefore, the challenge is thus: what is the maximum distance from the pond from which Kangaroo can punish Sheep?

Solution

Seriocomic Set

This challenge requires some knowledge of calculus and trigonometry, but mainly algebra.

We begin this challenge with a set of 4 numbers which has the following properties. Can you find them? If you can, good for you. But do you understand the use of this week’s adjective?

a,\,b,\,c,\,d\in\mathbb Z\\[8pt] a\leqslant b\leqslant c\leqslant d\\[8pt] a+b+3=a+d=\cos'{\pi}\\[8pt] \int \frac{d-c}a+1\;\;\delta a\bigg\lt\cos'{(a+b)\pi}\bigg\lt\int\frac{1-a}b-2\;\;\delta b\\[8pt] 0\lt|bc|+1\leqslant a+b+c+d\lt5\\[8pt] bc\notin\{a^2|a\in\mathbb N\}

Note: when calculating integrals, assume the ‘+c’ that you would add at the end is equal to the variable c.

SOLUTION

Sophisticated Streetcar

Algebra, trigonometry, and calculus are used in today’s challenge.

You are taking a streetcar through a city. Your destination is in the middle of a large park, and you are late. You can jump of the streetcar at any time while it is on the side of the park. Your destination is 1 kilometre from the road. The streetcar moves at 15 m/s, and you move at 5 when you run, which you will to try to be on time. You calculate when you will need to jump off the streetcar to arrive as soon as possible, and jump of exactly then. For how long do you run?

Solution

Gangbuster Geometry

Geometry, trigonometry, and calculus are required for this challenge.

Diagram of pond after being hit by rock. waves are just for illustration purposes and are not part of the problem.

We left an unsolved puzzle at the end of Fascinating Frequencies 3. We want to find the time it takes for a stone to stop after hitting water. The stone has an initial velocity of 1 m/s, a mass of 1 kilogram, and radius 10cm. The force acting on it from gravity is 1kg*g down. The buoyant force acting on it is the weight of the water displaced by the stone. The water has density 0.9982 kg/m3. The friction from the water in any direction acts perpendicular to the surface, inwards, with a strength equal to the sine of the angle multiplied by the density of the water.

\vec F_G=1kg g\ down\\[8pt] \vec F_B=\frac{998.2kg gV_w}{m^3}\ up\\[8pt] \vec F_d=\frac{469.2kgA_f\vec V^2}{m^3}\ up
solution

Peculiar Products

Trigonometry, geometry, ad algebra are required to solve todays challenge.

Pi notation is to Sigma notation as plus is to minus. Sigma notation is the sum of a bunch of stuff, Pi is the product. Pi is also ≈3.1416, but we will note that as π. In Strange Sums, we evaluated an expression with 2 Sigmas. Here will will find the product of the outputs of an equation. Specifically, the equation for the area of a regular polygon. Better go find that one out right now if you don’t know it. We will express this function as A(n, s), where n is the number of sides and s the length of them. Now solve the expression below.

\prod_{k=1}^\infin\Bigg(A\bigg(k,\,\frac1{tan (90\degree)-\frac{360\degree}k}\bigg)\Bigg)
Solution

Recursive Renewals

You will need to do algebra, combinatorics, calculus, and trigonometry, and also use multiple matrices, to solve today’s challenge.

When you take a book out at the Leibniz Library, you must return it before the fine equation, f(x), rises above zero. If not, your fine is determined by the fine equation. The variable r represents the number of times you have renewed the book, beggining at zero. The variable d counts days and increases by 1 every day, beggining at zero. The variable t counts days and increases by 1 each day, beggining at 0. Each time you renew the book, the variable r in the fine equation, which begins at zero, increases by 1, and the variable t is reset to 0. You may not renew the book if someone else has it on hold or the fine equation is above 0. The probability that someone puts your book on hold on any given day is determined by the hold equation, h(x). How can you maximize the number of days you have with your book without paying anything?

f(x)=d^2+5d-7cos(r)+r^3-100\\[10pt] h(x)=\frac{t^2+3sin(t)+15}{100}
Solution

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

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

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