Proof that math exams have become easier over the last 169 years!

June 22nd, 2009 | Categories: general math, Problem of the week | Tags:

Solution to problem of the week #6

Way back in April, I posed the following problem.  ‘Consider a square pyramidal pile of identical cannonballs of radius r such that the bottom layer contains 16 cannonballs (such as the pile in the diagram above). Find the volume (in terms of r) of the pyramid that envelops and contains the whole pile’

Since then I have received several answers (check out the comments section of the original post for a few of them) and all but one of them were wrong.  In my opinion, this is nothing to be too ashamed about since I couldn’t solve the problem either and I am not about to berate my readers for failing to do something that I couldn’t do myself!

So if I couldn’t do it then how did I know that all of these answers were incorrect?  Well clearly I had cheated and had access to a worked solution.  ‘My’ problem was in fact problem number 15 from one the 1840 undergraduate final exams at Cambridge University and a worked solution is given in the (now fully digitized, thanks to google) text Solutions for the Cambridge Problems 1840,1841.

The solution is

\light v=\frac{2}{3}( 1+3 \sqrt{2} +sqrt{3})^3 r^3

You’ll find the official worked solution to this problem on page 20 of ‘Solutions for the Cambridge Problems 180,1841’ (sadly, the diagram is missing) but James Graham-Eagle of the University of Massachusetts Lowell sent me not one but two different solutions in this pdf file and they are much easier to follow in my humble opinion.  Thanks for that James.

I’m glad that this problem wasn’t in my final exam!

  1. June 22nd, 2009 at 13:37
    Reply | Quote | #1

    Hi Mike:

    You are completely right!

    I had an experience with this Purdue University Problem:

    “Find the minimum possible area of an ellipse which encloses a 3,4,5 right triangle.”

    Answer: A_{E}=((4π)/(3√3))×6=((8π)/(√3))

    The equation of the ellipse is


    I have derived it as a particular case of the equation proved in pp.123 (“To find the least ellipse which will circumscribe a given triangle”) of Examples of the Processes of the Differential and Integral Calculus by Duncan Farquharson Gregory, William Walton, Published by J. and J.J. Deighton, Harvard University, 1846 (Google book digitalized on Aug 6, 2007). Please note that this equation is not required to find the solution!

    The answer is given with more generality in the book I mentioned.

    Best wishes,

  2. June 22nd, 2009 at 13:41
    Reply | Quote | #2

    … and of course congratulations to James Graham-Eagle

  3. someone interested
    June 24th, 2009 at 16:50
    Reply | Quote | #3

    One thing to remember about looking at old exams was the expectation of how they were completed and what percentage of the answers would be solvable.

    An older exam (especially at university level) would have many questions, and not be solvable within a 3 to 6 hour time-frame. If a question had a difficulty from 1 to 4 and 4 was challenging everyone who looked at it, almost all questions would be a level 4, with some a level 3. The professors who set the exams found questions that were unique and challenging, often using questions that only became known within the previous 10 years (consider the classic exam with Stokes’ Theorem). Most (all) of the class would fail the exam, and those that passed the course were those who failed least.

    More recent exams – say within the last 30 years – started on the notion that they were “do-able” in the sitting time, but still questions were rarely anything less than a level 3. People would fail the exams, but it wasn’t a requirement by the prof for the entire class to fail. This was the beginning of the time when profs started to look at the notion of a balanced exam that tested different aspects of what was taught in class. Unfortunately, problem solving was also moved by the wayside, as the idea of problem solving under high-stress and time constraints did not become achievable in a regular exam period of 3 to 6 hours.

    Fast forward to today, and we see exams that only test what the students were taught. Many exams now also have level 1 and 2 questions on them, with minimal level 4. To use modern teaching practice, if a student is expected to achieve a level 3, then most of the exam would also be a level 3. Students will come out of an exam being able (in theory) to calculate their actual mark since bell curving doesn’t seem to exist as a practice very much. It also removes the competition with fellow classmates as the student is just being tested against his or her own knowledge instead of sucking less than his or her fellow classmates.

    Unfortunately, all of this leads to the notion that we expect students to do less as that might damage their fragile egos. Problem solving remains a thing of the past. Without practicing problem solving, the students can get entire degrees with nothing more than regurgitation of a professor’s notes and a little memorization.

  4. July 2nd, 2009 at 22:37
    Reply | Quote | #4

    The above comment by “someone interested” is very illuminating, I think.

    I agree with, for instance, “Without practicing problem solving, the students can get entire degrees with nothing more than regurgitation of a professor’s notes and a little memorization.”

  5. July 5th, 2009 at 17:38
    Reply | Quote | #5

    Your solution can’t be right; the units don’t match: cubic length on the left versus length on the right. Probably it is a typo: r => r^3. That would make the scaling correct, increasing r by a 2-fold would increase the volume an 8-fold as expected. The prefactor is just a geometrical constant for a ball-pyramid.

  6. Mike Croucher
    July 6th, 2009 at 04:16
    Reply | Quote | #6

    Hi Sander

    You are right – it was a typo and has been corrected now. Well spotted and thank you.

    Best Wishes,