the strat behind triangle problems in the TMUA

Added: 2026-05-14

[00:00:00]Today, I'm going to be solving a problem from the TMUA 2023 paper one. And it's a nice problem about triangles, and they like these types of problems about triangles. So, I'm going to be showing you the strategy behind them. Let's have a look. A triangle XYZ is called fun if it has the following properties: angle YXZ equals 30°, XY = √3 * A, and YZ = A, where A is a constant. For a given value of A, there are two distinct fun triangles S and T, where the area of S is greater than the area of T. We want to find the ratio of the area of S to the area of T.

[00:00:35]And we've got five options here. Now, they love these sorts of questions where they give you a little bit of information about the side lengths and the angles, and you've got to just be careful cuz here there are two different types of triangles which have these properties. Um okay, so the way we want to do this is think about the angle here. So, the angle here is at X. So, we're going to mark on X here.

[00:00:56]And what other things do we know about X in terms of lengths? Well, we know that XY is √3A. So, that's the first thing that we'll draw. So, X Y, and that length there will be √3 A. And now we Now we know that Z is such that YXZ has angle 30°. So, if I draw a very long line here, which has angle 30°, I know Z lies somewhere on that line.

[00:01:23]And the idea is there are two possible places that Z could be.

[00:01:27]And we use those two possible places to determine S and T. And they're determined by this fact here, that YZ = A. So, let me just change my pen color here. So, we just need to find two possible places for Z. And so, one is maybe there. So, I'll call that Z1.

[00:01:43]And one is maybe there, and I'll call that Z2. And notice that both of these red lines here have length A.

[00:01:51]And so, this triangle here would be Which one has smaller area? So, S is the greater, so this would be triangle T.

[00:01:59]And then this here would be triangle S.

[00:02:02]And so, we're going to find the area uh the ratio of the area of S to the area of T.

[00:02:07]Okay, cool. So, you all we need to do is work out these areas.

[00:02:10]So, in order to do this, there's a few things uh we could do.

[00:02:15]Perhaps we could work out these side lengths um first to to help us. Um and that's obviously going to be crucial.

[00:02:23]So, I'm going to call this length here uh let's call it uh B.

[00:02:28]I'll call this length here C.

[00:02:31]Like so.

[00:02:32]And one thing that we can notice is if we think about this here, this line this very long line I've drawn as the base of both the triangles, we can notice that we can think of this as our height as well. This is a maybe a bit more abstract thing, but this is a nice little trick here. If we call that our height here, then we know that this ratio here, the area of S to the area of T, so the area of S is going to be a half times H times B plus C.

[00:03:01]And then this area of T is going to be a half times B times H. But the idea is the half H will cancel on both sides.

[00:03:09]So, we're basically just looking at the ratio between B plus C to B.

[00:03:14]Okay, cool. How can we work this out?

[00:03:17]Well, probably what we want to do is use the cosine rule here to help us. So, let's look at B, but the exact same logic will be true for C. So, we can use the cosine rule on this triangle X Y Z1.

[00:03:28]And if we do that, we're going to get the A squared equals uh so, B squared plus root 3 A squared minus 2 B root 3 A cos of 30 degrees, like so.

[00:03:42]So, A squared equals B squared plus 3 A squared minus So, cos of 30 is root 3 over 2. And so, we're going to get minus 3 A B um like so. So, B squared plus two A squared minus three AB equals zero which factorizes very, very nicely.

[00:04:00]Um so, we get two A minus B uh times uh A minus B like so equals zero.

[00:04:11]Um and of course, this means that either we have uh B equals A or B equals two A like so.

[00:04:20]Um and essentially, this is the the two possible values of C here.

[00:04:27]Because if we think about this triangle and we think about this triangle, if we were to use a cosine rule on either of those triangles, we'd be giving the equation the same bits of information. This 30 degrees, this root three A, and the opposite side here being A in either case.

[00:04:47]And so therefore, these are going to be our two values. One is for this B and one is for C. And obviously, B is uh well, sorry, one is for B and one is for B plus C. So, this is actually B plus C.

[00:04:57]Uh and so, this will just be two A to A.

[00:05:00]And that is just two to one. And so, our answer here is B.

[00:05:06]Um a pretty nice question and the most important thing is how we set this up, which of course as we mentioned at the start as we focus on where the angle is, in this case X, and then we look at the lengths here um that we have to do with the res- necessary angles. So, in this case, the only thing we knew is XY was root three A. We draw that on and then the way I like to do it is draw a very long line at an angle of 30 degrees, and then we know that Z has to be somewhere on that line. And in this case, it's just using the cosine rule. I used a nice little trick here, which meant I didn't have to worry about what H was because I realized they have a common height, these two triangles. And since we're looking at the ratios, that will cancel out. Anyway, that's how we solved this problem. And there's a couple of ways uh other ways you can do this. So, let me know how you solved it in the comments down below.