This is problem 25 on the grade 11 and 12 2025 math kangaroo contest. On a semicircle with diameter AD, points B and C lie on the diameter, and points E, F, G, and H lie on the arc of the semicircle. How many triangles can be formed with their vertices at three of these eight points? Our answers are 15 triangles, B, 50 triangles, C, 51 triangles, D, 52 triangles, or E, 54 triangles. Take your time, try to solve the problem yourself. Okay, so there are eight points in total, you know, A, B, C, D, E, F, G, H, that's eight points. And we need three of them to make a triangle. So like naively, we could say like maybe A choose three. And that's definitely a really good starting point. But as we sort of like draw in triangles to maybe confirm that we're right, we get some hiccups. So like, let's start with like a triangle with zero points on the diameter. That seems okay to me. And if we like draw out more of them, it should be clear that no three of these points are collinear. And therefore, like, if a thing has, if a triangle has zero points on the diameter, then it is, it's valid, we like that, that's totally okay. Okay, what if it has maybe one point on the diameter? So like B here. Well, actually, we should also get that that's okay. Because again, no three are collinear. That's also valid. Okay, how about two points on the diameter? Again, also valid. And that's because, yeah, no three are collinear. Because if they were, then this F would have to lie on the diameter, which sort of contradicts our definition. So like, if we have a triangle with zero points on the diameter, then we should get that that's okay. And therefore, like, if a thing has maybe one point on the diameter, then it is, it's valid, we like that, that's totally okay. So that's valid. And then finally, let's go over maybe the one problematic case. If there are three points on the diameter. If there are three points on the diameter, then fundamentally, because the diameters align, all three of them are collinear. And this is a degenerate triangle. And we don't like that. So if there are three points on the diameter, that is a problem. Okay, well, we've counted the number of ways to pick three points. But now, we just want to, like, and we sort of confirmed that most of the combinations are probably okay, except for the ones with three points on the diameter. So then we just need to count the number of, like, combination, or yeah, the number of combinations that have three points on the diameter. Or basically, the number of ways we can pick three points from the four point set A, B, C, D. And this basically just turns into four, choose three, because there are four points on the diameter, on A, D, like A, which are A, B, C, and D. And we need to pick three of them. So then we get that there are four, choose three invalid points out of the eight, choose three total configurations. And we get that this is 56 minus four equals 52 valid triangles. Which points us to answer choice D.

triangles

semicircle

vertices

combinatorics

collinear