IIT JEE hairy trig and algebra (part 1)

"Let ABC be a triangle, such that angle ABC is equal to pi over 6." Let me draw it. So that's A, B, and C. And they tell us that angle ACB, this angle right over here, is pi over 6. I'm assuming it's pi over 6 radians. Or we could view that as 30 degrees. And then they tell us, "Let a, b, and c denote the lengths of the sides opposite to capital A, B, C, respectively." So this is a, this is b, and this is c, right over there. The values of x, for which a is equal to this thing, b is equal to this thing, and c is equal to this thing is? So let's think about it. One, we need to find a relationship between A, B, and C in this angle. And then once we have that relationship, we can substitute them with these expressions. And then essentially try to solve for x. And the first thing you want to think about is, OK, how can I relate the length of b, this lowercase c, this lowercase a with this angle here.? And what should pop out into your brain is the law of cosines. Law of cosines. Law of cosines that tells us that c squared is equal to a squared plus b squared. It's essentially the Pythagorean theorem with an adjustment right here, because this isn't a right triangle. So a squared plus b squared minus 2ab times the cosine of C. Now in this case, angle C right here-- or I should say, in this case, it's angle ACB-- this over here is going to be 30 degrees. So what's the cosine of 30 degrees? And just in case you forgot it, although I assume if you're going to take the IIT exam, the JEE, you're going to need all the help you need in terms of time. So this is 30 degrees. That is 1. Then this is 1/2, the side opposite the 30-degree side. This will be 60 degrees. It's going to be a 30-60-90. And so this is square root of 3 over 2. Cosine is adjacent over hypotenuse, so square root of 3/2 over 1. So this, in this example, it's going to be c squared is equal to a squared plus b squared minus 2ab. And then this whole thing is square root of 3/2. So times square root of 3/2. And these 2's cancel out. And so we get c squared is equal to a squared plus b squared minus the square root of 3ab. Now the next thing I'd want to do is substitute for a, I'd substitute with this expression over here. For b, I want to substitute with this expression over here. And then for c, I want to substitute this expression right over here. But I want to be careful, I want to make sure that it's in a form that I can, in some way, simplify. Because if I just take this a over here-- this x squared plus x plus 1-- if I square it, I'm going to end up with something to the fourth power. This b, I'm going to end up with something to the fourth power. And then I have the square root of 3 business over here. So let's see if we can at least rearrange it, or at least maybe quarantine the square root of 3, little bit, to make it at least something that we might have some hopes of factoring, or solving in some way. And I want to make a little clarifying point here. In most Western exams, if you're taking the SAT or you're taking any type of-- even at a math competition-- if you find yourself grinding through math, multiplying huge polynomials times each other, you're probably doing something wrong. Most Western exams-- like the SAT-- you kind of have to see the trick, you kind of have to see the elegance in the problem. But once you see the elegance, the problem is usually pretty straightforward. In fact, if you're grinding through math, it's usually a giveaway that you're doing something wrong. On these JEE math problems-- and I've only recently gotten exposure to them. I obviously never took them, not having grown up in India. But on these, you have to kind of see the trick. And then you have to grind through math, and you have to use every trick at your disposal to make the grinding through math a little bit easier. So the test writers for the JEE are definitely a little, actually, a lot more cruel than the test writers for most Western exams that I've seen. So what I want to do is, let's just isolate the square root of 3ab. So let's add it to both sides. So the square root of 3ab plus-- so let's just add that-- square root of 3ab. And let's get the c squared onto this side of the equation. So let me subtract c squared from both sides of the equation. So minus c squared. And then minus c squared. And so I will have, on this side, those cancel out. And I just have the square root of 3ab is equal to a squared plus b squared-- these cancel out-- minus c squared-- I'm just rearranging the whole thing-- minus c squared. All of that, well right now, that's all it is. And now I can divide both sides of this by ab. So if I divide that by ab and then I divide the right-hand side by ab, that cancels out. And I've kind of quarantined that square root of 3. So I'll hopefully only have integer stuff on this side of the equation. So I have the square root of 3 is equal to a squared plus b squared minus c squared. So let's figure out what each of these things are. Let's expand them out, actually, before we rewrite it. So let's figure out what a squared is. So let's think about a squared. a squared is going to be equal to-- and this is the part where they clearly want to see whether you can multiply multiple polynomials. That's going to be x squared plus x plus 1 squared, or x squared plus x plus 1 times x squared plus x plus 1. And so this is going to be 1 times this whole thing, is x squared plus x plus 1. x times this whole thing, is x plus x squared plus x to the third. And you may or might not have noticed, I'm doing these a little faster than I traditionally do them, just in the hope of being able to finish this problem in a reasonable amount of time. And then finally, x squared times this whole thing, x squared times 1 is x squared. x squared times x is x to the third. And then x squared times x squared is x to the fourth. So a squared is going to be x to the fourth plus 2x to the third plus 3x squared plus 2x plus 1. So that's a squared. What's b squared? That looks a little bit more straightforward. b squared, so that's b. So b squared is going to be-- well, that squared is just x to the fourth minus 2x squared plus 1. And then c squared. That's more straightforward. So at least they weren't as painful as this. c squared is going to 4x squared plus 4x-- 2 times the product of these two, so plus 4x-- plus 1. So the numerator of our expression here. So let's rewrite this thing that we were simplifying. It now becomes the square root of 3 is equal to-- oh, and we don't want to-- well, let's just do one step at a time. So it's equal to a squared, which is this thing over here. So it is x to the fourth-- actually, let me scroll down a little bit. It's this thing plus this thing, minus this thing over here. So let me just do it. It's this thing plus the blue thing, plus x to the fourth minus 2x squared-- it'll be easier to add them up this way-- plus 1. And then it's minus this thing, minus c squared. So minus. So this is going to be a minus 4x squared minus 4x. And then you have a minus 1. And then when you add them all up-- so this will be the numerator of this expression. We have the square root of 3 is equal to-- so let me just do it down here. So we have 2x to the fourth-- let's scroll to the left a little bit. So we have 2x to the fourth plus 2x to the third. And then 3x squared minus 6x squared is minus 3x squared. 2 minus 4 is minus 2x, and then you have 2 minus 1 is-- then you have just plus 1. So the numerator, this thing over here, simplifies to this. It's equal to 2x to the fourth plus 2x to the third minus 3x squared minus 2x plus 1. And the denominator is ab. a is this thing over here. a is x squared plus x plus 1. And b, we saw it up here, b is-- actually, we have it written up over here. b is x squared minus 1. So b is x squared minus 1. Now at this point, we need to solve for x. If we just multiply this stuff times square root of 3 and went backwards, kind of unsolved it back to the form that we had way up here, we'd have a fourth degree polynomial with square roots of 3's in it. We still have a fourth degree polynomial. But at least this one has all integer coefficients, which at least gives me some hope that I might be able to factor it. I might be able to find some roots. And if I can find some roots that are common to the numerator and the denominator, maybe I can reduce this into a lower degree problem. So I'm going to leave you there. I'm pushing 10 minutes. In the next video, we're going to try to factor this beast up here and see if it has any common factors with this over here. And then try to solve for x.