This is problem 30 on the grade 11 and 12 2025 math kangaroo test. Last question. The sides AB and CD of the convex quadrilateral ABCD are each divided into three parts by points E, F, P, and Q so that AE equals EF equals FB and DP equals PQ equals QC. The diagonals of AEPD and FBCQ intersect at M and N respectively. The areas of triangle AMD, triangle EMP, and triangle FNQ are 154, 112, and 99 respectively. What is the area of triangle BCN? Our options are A57, B70, C72, D86, and E141. Okay, so what stands out to me first is that calculating the area of these things directly might be kind of difficult because we don't really have bases, we don't really have heights, and we can't really connect them to the trisections that we've established of AB and CD. So what can we do instead? Well, what we can do is incorporate these triangles here, AME and FNB. And by doing that, we actually get to link the whole triangle QFB and CFB and DAE and PAE to the trisected areas. And that makes life a whole lot easier. What do I mean by that? Well, let's do a little exercise and let's drop the heights down. So I have here a little diagram that should be kind of useful. And it should be kind of illustrative that we have height 1 for our first triangle ADE, and then we have height 2, and then height 3, and then height 4. What it's saying here is that as you move along with respect to this top line, as you move parallel to it, you go down the exact same amount, then you go down the exact same amount, and you go right and go down the exact same amount. And actually, we get that h1, h2, h3, h4 are in an arithmetic sequence of one another, which is really quite nice because that allows us to make a lot more powerful statements about the areas attached to these things. And then sort of by virtue of what an arithmetic sequence is, we get that h2 minus h1 equals h4 minus h3. So then from there, what we can do again is actually get some areas this time because if you notice, all of the triangles have the exact same base of this trisected length. So then now we know that their areas are directly proportional to these arithmetic sequenced heights. And now we can start making the very powerful statements. Okay, so we get that triangle ADE minus triangle, like the area of triangle PAE, is in fact equivalent to the area of triangle QFB minus the area of triangle CFB. Okay, this is quite useful, but we're still dealing with these shaded triangles here. But actually, if we sort of write out what ADE, PAE, QFB, and CFB mean, they can be just decomposed into the triangle we care about, like ADM, so triangle ADM. And then also this sort of junk triangle they both have in common, so AEM, so plus triangle AEM. And then this one can be written as PME, triangle PME, plus triangle AEM. And if you notice something, these actually cancel. So we get that these two statements are equivalent to one another, and then we also, by the same logic, get that QFB minus CFB is in fact equivalent to QNF, like triangle QNF, minus triangle CNB. And we know how to calculate exactly one of these terms. We know how to calculate this term over here. We know that ADM has area 154, and we know that PME has area 112. So 154 minus 112 is equal to 42, is equal to QNF, or yeah, is equal to QNF minus CNB. And we know the area of QNF is 99. So writing this out, we get 99 minus area of CNB is equal to 42. With a little bit of algebra, we get that area of CNB is equal to 57. So our final answer choice on the final question is A. Thank you so much for watching, and I hope you gained something from this. Good luck on the next Math Kangaroo.

convex quadrilateral

triangle area

diagonals intersection

arithmetic sequences

geometry problem