A Nice Introduction to TMUA Geometry

Added: 2026-06-01

[00:00:00]Today, I'm going to be solving a nice geometry problem from the American mathematics competition. This is a really nice problem set or set of problems resources that you can use if you're preparing for an Oxford or Cambridge mathematics application. Square ABCD has side length two.

[00:00:15]A semicircle with diameter AB is constructed inside the square and tangent to the semicircle from C intersects Sorry, this and the tangent to the semicircle from C intersects side AD at E as shown here. What is the length of CE? So, this line here. And we've got a bunch here. We probably won't need to use these. Let's solve this. So, we know the side length of this square here is two. Very nice. And here, what's something that's generally quite useful to do when we are drawing dealing with a semicircle or circle and tangents is to draw the line from the center of the circle to the point of tangency like so. And maybe I'll call this point F for the time being. And we can say that that's a right angle there.

[00:00:57]And we're trying to work out the length CE. And it's actually quite nice that we can work out this length here, CF.

[00:01:03]There's a few ways we can do this. One way is to maybe draw this line here and notice since this is a right angle, we're thinking hmm maybe Pythagoras' theorem might be useful. Or since the side length of the square is two, we know the radius of the circle is one.

[00:01:15]So, that will have length one. And we can work out this length here from C to let's call that G um by doing Pythagoras on this right angle triangle. This is one, two. So, this length will be root five and then use that to work out CF.

[00:01:30]Another way we can work out the length CF which is maybe slightly slicker, doesn't really require us to think about G is to notice that this is a circle or semicircle and we've got two tangents here. This guy and this guy. And a nice little circle theorem we can apply is if you have two tangents that meet outside uh two tangents of the same circle that meet outside the circle at the same point, in this case C, then their lengths will be the same. And so, since this guy here has length two, this length here will also be two.

[00:01:59]Very nice. And since we're trying to work out CE, all we need to do is work out this little extra bit here, which I'll call X for the time being.

[00:02:06]And again, we can apply the same theorem. E to F and E and A E to A are both tangents of the same circle that meet outside the circle at E, and thus list this length here must also be X.

[00:02:17]And if that guy there is X, that means this length here is 2 - X, like so.

[00:02:23]And generally speaking, these sorts of problems we do eventually want to use Pythagoras' theorem, and now we can quite nicely on this right-angled triangle here. I kind of messed it up there, but the hypotenuse is CE, which has length 2 + X. This length here is 2 - X, and this length here is 2. We can just use that um to work out X. So, 2 - X squared + 2 squared = 2 + X all squared. These guys are very, very similar. One's 2 - X, one's 2 + X. So, when I expand them, I'm going to get similar terms. The only different ones are going to be - 4 X, and on this side I'll get a 4 X, and then I've got + 4.

[00:02:58]And so, this gives me that X must be a half, and so therefore the length CE is 2 + X, so 5 over 2. And that's our answer.