The3Mathketeers

I think you can mathematically show that the 6-8-10 right triangle that PurpleWasp discovered is the only one. Consider the diagram below, which showcase the two possibilities for a non-congruent right triangle. Either the side length of 6 is the leg of the right triangle or it is the hypotenuse of the triangle.

Let's consider \(\triangle {\text{ABC}}\) first. We can use Pythagorean's Theorem to find the relationship between the lengths of the sides of the triangles, and simplify to a point where we can begin trying solutions.

\(h^2 = s_1^2 + 6^2 \\ h^2 = s_1^2 + 36 \\ 36 = h^2 - s_1^2 \\ 36 = (h + s_1)(h - s_1)\)

We have now changed the problem to factoring the number 36 where one factor is \(h + s_1\) and the other is \(h - s_1\). The number 36 has a lot of factors, but there are a lot of restrictions on these factors. For one, \(h \text{ and } s_1\) are both lengths of a triangle, so we need not waste time considering negative factors. We should also realize that \(h + s_1 > h - s_1\) since adding two positive numbers should always be greater than their corresponding difference. We can now list all the factors that meet this criteria.

\(36 = 36 \times 1 \\ 36 = 18 \times 2 \\ 36 = 12 \times 3 \\ 36 = 9 \times 4\)

Now, we have found a few candidates for possible factors of 36. We can solve the corresponding systems of equations and then find the mystery side lengths. I will start with the first set of factors.

\(\begin{cases} h + s_1 = 36 \\ h - s_1 = 1 \end{cases} \\\)

Probably simplest is to use the elimination method to solve this system of equations. This system would then simplify to \(2h = 37\). We can stop here and not go any further because we recall that \(h \in \mathbb{Z}\), and \(h = \frac{37}{2} \not\in \mathbb{Z}\). With this insight, we recognize that any two factors that have one even and one odd factor cannot have integer solutions. This leaves us with one possibility left, that is \(36 = 18 \times 2\). Let's solve this one with the elimination method!

\(\begin{cases} h + s_1 = 18 \\ h + s_2 = 2 \end{cases} \\ 2h = 20 \\ h = 10 \\ s_1 = 8\)

This answer is recognizable. This is the 6-8-10 right triangle that PurpleWasp discovered. We have no more possibilities for \(\triangle \text{ABC}\), so we move on to \(\triangle \text{EDF}\). 

Once again, we use the Pythagorean's Theorem to find a relationship that may give us insight on possible integer solutions.

\(6^2 = s_1^2 + s_2^2 \\ s_1^2 = 36 - s_2^2 \\ s_1^2 = (6 - s_2)(6 + s_2)\)

Since \(h = 6\) in this case, we know that \(s_2 < 6\) since the hypotenuse is the longest side of a right triangle. The question for this factor becomes when  \(6 - s_2 \text{ and } 6 + s_2\) are both perfect squares. Since \(0 < s_2 < 6\), \(6 - s_2\) will be a perfect square when \(6 - s_2 = 4 \text{ or } 6 - s_2 = 1\) because they are the only two perfect below 6. The only options are \(s_2 = 2 \text{ or } s_2 = 5\). Let's check both of those options.

\(s_1^2 = (6 - s_2)(6 + s_2) \\ \begin{align*} \text{Let } s_2 = 2:& 6 + s_2 = 8 \\ \text{Let } s_2 = 5:& 6 + s_2 = 11 \end{align*}\)

Notice how in both cases the \(6 + s_2\) term is not a perfect square. This means that \(s_1\) will be an irrational length, which is not acceptable given the parameters of the question. Therefore, no options exist for \(\triangle \text{EDF}\).

This leaves the 6-8-10 right triangle as the only option.

One strategy is to think of the "easiest" version of the problem and then applying the appropriate modifications to the true question at hand. I imagined a similar polynomial function \(g(x)\) with the same constraints except that \(g(p) = g(q) = g(r) = g(s) = 0\). I made this slight adjustment so that I can take advantage of the fact that now \(p, q, r, \text{ and } s\) are all now roots of \(g(x)\). Because of this, we can write \(g(x) = a(x - p)(x - q)(x - r)(x - s)\). Now, our goal is to minimize but ensure that \(g(t) > 0\).

As a result, we know that\(g(t) = a(t - p)(t - q)(t - r)(t - s) > 0\). Note that the original problem states that \(p, q, r, \text{ and } s\) are all distinct integers, so  \(t - p, t - q, t - r, \text{ and } t - s\) are all unique intergers involved in a positive product. We want to pick these integers strategically such that the product is minimized but still positive. By observation, I think it is clear that \(1 * 2 * 3 * 4\) is the smallest product of four unique integers that is also positive. The idea is to pick the smallest integers possible so that the product is the smallest possible. Now, we must pick a value for \(a\). Once again, we want the product to be as small as possible while still remaining positive and ensuring that the polynomial has integer coefficients, so picking \(a = 1\) is the best choice. 

We have fantasized about a mystery polynomial that has these properties, but can we make this dream a reality? Yes! All we have to do is craft a polynomial with these properties. There are infinitely many that would satisfy our dreams, but one of the simplest is to write \(g(x) = (x + 1)(x + 2)(x + 3)(x + 4)\). Now, \(g(0) = 1 * 2 * 3 * 4 = 24\). In this example, \(a = 1, p = -1, q = -2, r = -3, s = -4, \text{ and } t = 0\). This fits all the criteria and minimizes \(g(0)\).

While we were successful in generating \(g(x)\), we still have to apply this problem to \(f(x)\). Luckily, this is not a problem because we can take advantage of simple transformations to make the characteristics of \(g(x) \) apply to \(f(x)\). We can observe that if \(f(x) = g(x) + 1\), then everything still holds. This would make\(f(x) = (x + 1)(x + 2)(x + 3)(x + 4) + 1\), and \(f(t)\) is still minimized at \(t = 0\). Therefore, \(f(t) = f(0) = (0 + 1)(0 + 2)(0 + 3)(0 + 4) + 1= 1 * 2 * 3 * 4 + 1 = 25\).

We are done!

I know very little about the complex numbers realm of mathematics, but I suppose I know enough to find the answer to this particular problem although I doubt my method is the cleanest. or even the simplest because typically problems involving complex numbers can have intricate solutions. Regardless, I will present my thought process, and I hope you can follow the logic. I first started by completing the square of this complex quadratic, which gave me some insight about the real part of the product of the roots.

\(x^2 + 2x = i - 2x \\ x^2 + 4x = i \\ x^2 + 4x + 4 = i + 4 \\ (x + 2)^2 = i + 4 \\ |x + 2| = \sqrt{i + 4} \\ x_1 = -2 + \sqrt{i + 4} \text{ or } x_2 = -2 - \sqrt{i + 4}\)

Unfortunately, we are left with an undesirable term, \(\sqrt{i + 4}\). We will have to extract the real part from this expression to be able to find the product of the real parts of the complex solutions. For now, I will just assume I know what the real part is and simplify as much as possible while including the \(-2\) term since \(-2 \in \mathbb{R}\).

\(\Re(x_1x_2) = \left(-2 + \Re\left(\sqrt{i + 4}\right)\right)\left(-2 - \Re\left(\sqrt{i + 4}\right)\right) \\ \Re(x_1x_2) = 4 - \left(\Re\left(\sqrt{i + 4}\right)\right)^2\)

Once again, we have to extract the real part out of \(\sqrt{i + 4}\) somehow. The complex numbers are algebraically closed in almost all operations except arguably division because of issues arising from division by zero. As a result, we know that the square root of a complex number will yield another complex number. Because of the algebraic closure, we can write \(\sqrt{i + 4} = a + bi\) where \(a\) is the real part and \(b\) is the imaginary part and solve for each of its components.

\(\sqrt{i + 4} = a + bi \\ i + 4 = a^2 + 2abi + b^2i^2 \\ i + 4 = 2abi + (a^2 - b^2)\)

Now, we can write a system of equations and solve enough to extract the desired variable. In this case, we only care about \(a\) since that represents the real part of \(\sqrt{i + 4}\)

\(\begin{cases} a^2 - b^2 = 4 \\ 2ab = 1 \end{cases} \\ \begin{cases} b^2 = a^2 - 4 \\ 4a^2b^2 = 1 \end{cases} \)

Now, we have an easy substitution for \(a^2\), which will allow us to solve for \(a\) and finish this problem.

\(4a^2(a^2 - 4) = 1 \\ 4(a^4 - 4a^2) = 1\)

This is a quartic polynomial with only even-numbered exponents on the a-variable. This means we can treat this equation like a quadratic by letting \(y = a^2\) and then solving it again by another version of completing the square.

\(4(y^2 - 4y) = 1 \\ 4(y^2 - 4y + 4) = 1 + 4*4 \\ 4(y - 2)^2 = 17 \\ (y - 2)^2 = \frac{17}{4} \\ y - 2 = \pm \frac{\sqrt{17}}{2} \\ a_1^2 = 2 + \frac{\sqrt{17}}{2} \text{ or } a_2^2 = 2 - \frac{\sqrt{17}}{2}\)

There is no need to go any further than this since we only need the square of the real part. However, we now need to choose between these two viable solutions. To choose, first recall that the a-variable represents \(\Re\left(\sqrt{i + 4}\right)\), so \(a \in \mathbb{R}\). Since \(2 + \frac{\sqrt{17}}{2} > 0, a_1 \in \mathbb{R}\). However, since \(2 - \frac{\sqrt{17}}{2} < 0, a_2 \not\in \mathbb{R}\). This means that \(a^2 = 2 + \frac{\sqrt{17}}{2}\) is the only correct solution for this particular situation. We finally have the information we seeked from the beginning of the problem.

\(\begin{align*} \Re(x_1x_2) &= 4 - \left(\Re\left(\sqrt{i + 4}\right)\right)^2 \\ &= 4 - \left(2 + \frac{\sqrt{17}}{2} \right) \\ &= 2 - \frac{\sqrt{17}}{2} \end{align*}\)

We are done!

If you could envision rotating \(\triangle \text{ABC}\) around \(\overline{\text{AB}}\), then the diagram would look like the following diagram. I realize the diagram is not drawn to scale. As of now, I have not yet worked out how to draw 3-dimensional figures efficiently.

The surface area of right cones are \(\text{SA}_\text{right cone} = \pi r^2 + \pi rL\) where r is the radius of the right cone and the L is the slant height of the right cone. We already know the radius of this cone, but we can find L with the use of Pythagorean's Theorem.

\(L^2 = 1^2 + 3^2 \\ L^2 = 1 + 9 \\ L^2 = 10 \\ L = \sqrt{10} \text{ or } L = -\sqrt{10}\)

Since we are dealing with lengths, reject \(L = -\sqrt{10}\). Now that we have found all the necessary lengths, we can find the surface area of this right cone.

\(\begin{align*} \text{SA}_\text{right cone} &= \pi r^2 + \pi rL \\ &= \pi * 3^2 + \pi * 3 * \sqrt{10} \\ &= 9\pi + 3\sqrt{10}\pi \\ &= (9 + 3\sqrt{10}) \pi \end{align*} \)

The question asks what the surface area is pi times what number? Well, that number would be \(9 + 3\sqrt{10}\).

I did not forget about y = x in my answer. In fact, here is a direct quote from my answer contradicting your claim:

"... [W]hen m = 1 and b = 0, the linear function is its own inverse. This means that \(y = x\) is its own inverse!"

Also, the question specifically concerns itself about "how many linear functions" there are with a particular property. Even if I had conveniently missed y = x, the answer would still be infinitely many, as I later concluded in my answer. 

Despite your insinuation, I am not AI. I am answering these questions genuinely, and I continue doing my best to generate high-quality answers geared towards the mathematical ability of the asker. If you have spotted a mistake in any of my previous answers, you can tell me. I will fix the mistake and give you credit accordingly for spotting the error.

We can find the inverse of an arbitrary linear function and find its inverse and determine how many instances there are when the linear function is an inverse of itself.

\(y = mx + b\) is the typical form of a linear function. Let's find its inverse.

1) Swap the variables x and y.

\(x = my + b\)

2) Solve for y-variable.

\(x = my + b \\ my = x - b \\ y = \frac{1}{m}x - \frac{b}{m}\)

Now we know that for an arbitrary linear equation \(y = mx + b\), \(y = \frac{1}{m}x - \frac{b}{m}\) is the inverse. When are these equations equal? These equations are equal when the coefficient of x and the constant term are the same! This leads to a system of equations.

\(m = \frac{1}{m}; b = -\frac{b}{m} \\ m = \frac{1}{m} \\ m^2 = 1 \\ m = 1 \text{ or } m = -1\)

There are two possibilities for m. Now, let's find the corresponding values for b that make linear functions their own inverse.

\(\text{Let } m = 1: \\ b = -\frac{b}{m} \\ b = -b \\ 2b = 0 \\ b = 0\)

This mean that when m = 1 and b = 0, the linear function is its own inverse. This means that \(y = x\) is its own inverse!

Now, let's try find the other value of b when m = -1.

\(\text{Let } m = -1 \\ b = -\frac{b}{m} \\ b = -\frac{b}{-1} \\ b = b\)

The left-hand side and the right-hand side of the equation are the same equation. This means that the value of b can be anything, and the equation will hold true. In other words, \(y = -x + b\) where b is any real number has the property that the linear function is its own inverse. This means that y = -x and y = -x + 1, y = -x + 3, y = -x + 1000000000000 are all inverses of themselves.

This means there are infinitely many linear functions that have inverses of themselves.