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Mathcounts Prep Site

February 2, 2007

Since my students (including my oldest son) are preparing for Mathcounts, I figured I’d post a link to a cool Mathcounts prep site: Elias Saab’s Mathcounts Drills.

This is a great source of practice problems for kids preparing for mathcounts, but also just a great source of problems in general.

So, I was trying a few for fun, and came across this one, which I thought was interesting and challenging: A triangle ABC has an area of 16 square centimeters. We extend the side AC to obtain a point M such that A is between M and C and AM=AC . We extend the side AB to obtain a point N such that B is between N and A and AB=BN . We extend the side BC to obtain a point P such that C is between P and B and BC=CP . How many square centimeters are in the area of the triangle MNP? (I don’t feel right about copying their image, so you’re going to have to draw your own diagram.)

With my rusty geometry and unsure how to approach the problem, I guessed that the area of the larger triangle might be four times that of the original triangle (just on a gut double the linears -> quadruple the area) but that isn’t right. So, let’s have some thoughts on how to solve this one. 🙂

6 Comments leave one →
  1. February 3, 2007 1:45 am

    Wow, what a challenging problem! I was completely stumped by this for over half an hour. I just thought of a solution, but do you have answer choices or know the correct answer? The answer I get is 7*original triangle or 112 cm2. Is that plausible? If so, I can explain or hint at how I thought about it.

  2. February 3, 2007 10:53 am

    Yes, the answer 7*original area is right, and I now know why.

    It uses a theorem I don’t think I previously knew, that the median of any triangle breaks that triangle into two regions with equal area.

    I didn’t see it at first, but on the site I posted, he does give full solutions to the problems after you submit your answers, if you scroll down far enough. Lots of great problems there — I highly recommend the site.

  3. February 4, 2007 3:08 am

    I didn’t know that theorem either, but I suppose I derived it while solving the problem. You had the original, interior triangle, right? Then when you extended the base, you created a new triangle adjacent to it. They have the same length base. And they have the same vertical height too, obviously, since they share that vertex. Therefore, they have the same area. (And therefore, bisecting the side of a triangle creates two equal-area triangles.)

    The site sounds intriguing, but given the level of my math skills, I fear that I would end up spending large amounts of time trying to figure the problems out!

  4. February 4, 2007 5:10 pm

    There’s a cool related theorem that says that the 3 medians of a triangle divide the triangle into 6 triangular regions of equal area. This also means that the 3 medians always meet at a single point (called the centroid), which itself seems like a non-trivial result to me.

    There’s a proof of this theorem here:

    Ask Dr. Math Proof of the 6 equal area triangles formed by the 3 medians

    A proof that all the medians intersect at a single point can be found here:

    Ask Dr. Math Proof that 3 medians of a triangle intersect in a single point

  5. February 5, 2007 2:05 am

    Ah, yes, now that you mention it, I do remember that formula. I’ll have to take a look at the proof (that all three intersect); I can’t intuitively explain it. (I mean it makes sense, but I can’t explain why it should always be true.)

  6. February 6, 2007 12:46 am

    What’s kind of nice about this one is that we don’t need a good description of the original or the resulting triangle.

    Follow ups:
    1. Is the resulting triangle similar to the original?
    2. Is the perimeter of the resulting triangle a constant multiple of the original?


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