Atari Space

My friend Justin was reminiscing about some Atari-aged video games: Subspace/Continuum. The problem they inspired reminds me more of Asteroids.

Problem:

A turret in a video game can rotate but not change location. It shoots bullets that travel with a constant velocity V_B. An asteroid travels at a constant velocity V_A. Where should the turret aim to hit the asteroid?

Solution:

You'll never find where you're going if you don't know where you started. Let's put the turret at the center of a Cartesian plane. Add a point A for the asteroid at (x₀, y₀), and add a point for the eventual collision at (x₁, y₁). Label the distances traveled by the asteroid and the bullet D_A and D_B. Recall that little ditty: "Distance = Rate × Time." We can use this whenever we are dealing with constant velocities. I've already labeled arrows R_A and R_B to represent the rates in the diagram.(I'm listing the equation numbers (abbreviated "eqn") on the left. When I use those equations later, I'll show which equation number I used on the right.) Note: the bullet and the asteroid travel for the same amount of time once the bullet is fired, so T has the same value in each equation.

We have two equations and three unknowns, so we'll need to find some more relationships. I can find more information about D_A and D_B by breaking the distances into their horizontal and vertical components. I've added these components to the diagram below.Let β be the angle that the bullet's trajectory makes with the x-axis, and call the angle that the ship flies at with respect to the x-axis α.With all of these variables in play, let's take a moment to recap what is known and what we need to find.Now, we need to trade our variables down to get rid of some of those unknowns. I'll start with x₁. Now do the same with the sines to get rid of y₁. Now that we've gotten rid of x₁ and y₁, solve for T in equation 11 and substitute into equation 15... ...and simplify. Now we're down to one unknown: β. To find β, solve for sin(β), square both sides, and substitute 1 - cos²(β) for sin²(β). Then, use the quadratic equation to solve for the possible values of cos(β), and convert those values into the angles we want. Once we have β, we use equation 18 to find T and equations 4 and 5 to find x₁ and y₁.

I'm not going to attempt to finish this solution unless someone asks me to. I'm too likely to make a mistake somewhere in the tangled mass of algebra that is to follow, and Justin has access to math software that can do the manipulations for him. Comment if you have any questions.

Solution...

Hide solution.

S.A.T. Inequalities

This one comes right out of the S.A.T.s.

Problem:

Each of the following inequalities is true for some values of x EXCEPT

Solution:

The process of elimination is perfect for this problem. A quick look eliminates (A) as a possibility. Anyone could think of a value for x that would make (A) true; for instance, if x = 2, then x² = 4 and x³ = 8, so we can cross of choice (A).
What other kinds of numbers are there? What numbers are "tricky"? Fractions and negatives. If x = 1/2, then x² = 1/4 and x³ = 1/8. That fits with the last inequality, so we can cross off choice (E).
Letting x = -2 makes x² = 4 and x³ = - 8. Ordering those gives us x³ < x < x², so we can cross off choice (D).
Now we only have two possibilities left, and they both look kind of goofy. If you were running out of time on the S.A.T.s, you could just guess between the two earning an expected value of 3/8 of a point, but we're not running out of time. We've tried positive integers, negative integers and fractions, so what's left: negative fractions. If x = -1/2 then x² = 1/4 and x³ = -1/8. That lines up as x < x³ < x², just like choice (B).
Therefore, the only inequality that will not be true for a value of x is choice (C).

Solution...

Hide solution.

Squares of Triangles

Pythagoras was a Greek philosopher and mathematician who lived in the sixth century B.C.E. He acquired a following in his time, and there was a cult built up around him in which he was worshiped as a demi god¹. Today, he is most well known for his contribution to Euclidian geometry: the Pythagorean Theorem. Today's challenge is to prove the Pythagorean Theorem.

Problem:

Given any right triangle ABC, prove A² + B² = C² using only simple geometric formulas. (No trigonometric functions allowed).

Solution:

This problem seeped into my head as I was trying to fall asleep some nights ago, and I never could solve it lying in bed in the dark. It wasn't until I sat down with a pencil and paper and studied the problem from several different angles (during the second and third quarters of the Superbowl) that I came up with a solution.

The first thing I did was to redraw the diagram like this:
That way, I could visualize the quantities, A², B² and C² as the areas of the squares made from the sides of the triangles.

Next, I started to think about this:
Maybe we can find a relationship between a square with the side lengths of A + B and the grey square with lengths C.

The final thing you need to know is that the area of a triangle is equal to half the base times the height: The answer becomes clear when you draw in the rest of the triangles. Look at the red square.The lengths of each of its sides are equal to A + B, so it has the area (A + B)².

If you look at the triangles shaded in pink......you should see that each has a height of A and a base of B, so their areas are ½AB. Therefore, the area of the grey square is equal to the area of the red square from above minus the areas of the four triangles.

Solution...

Hide solution.

Weighing in on Icy Roads

Problem:

Does increasing your vehicle's mass increase your maneuverability on icy roads? Assume the vehicle and all contents are treated as a point mass.

Solution:

It all comes down to two things: momentum and friction. Momentum (p) = the mass of the car (m) times the car's velocity (v), which is a term for a directed speed. We can change momentum by applying a force for a duration of time. Change in momentum = force × time.

Δp = F⋅t
Δm⋅v = F⋅t

What kind of force are we talking about? When it comes to driving a car, it's almost entirely related to the force of friction. When you apply the breaks, you're using
the friction between the tires and the road to stop the car. When you turn the wheel, it's the friction between the tires and the car that will ultimately change your direction, and when you accelerate, without the friction between the road and the tires, you'd just be spinning your wheels.

The force of friction (F_f) = the normal force (F_n) times the coefficient of friction (μ). The normal force of any object is equal to the product of the objects mass, the acceleration due to gravity, and the cosine of the angle (θ) between the horizontal and the surface providing the friction.

F_f = F_n⋅μ
F_f = m⋅g⋅cos(θ)⋅μ

Now, let's look at the change in momentum due to the force of friction applied over time t:

Δp = F⋅t
Δm⋅v = F_f⋅t
Δm⋅v = F_n⋅μ⋅t
Δm⋅v = m⋅g⋅cos(θ)⋅μ⋅t

The m on each side of the equation represents the same thing: the mass of the vehicle plus contents. Dividing each side by m gives us:

Δv = g⋅cos(θ)⋅μ⋅t

Thus, taken as a point mass, the mass of your vehicle has no effect on your ability to change your direction or speed.

Solution...

Hide solution.

Welcome to J-Function Problem Solving

Here's the deal: you send me math or science problems, and I'll post solutions up to as often as one per day. I'll solve problems from any of the subject areas I tutor and also from logic or philosophy. Ask away.

Solution...

Hide solution.