This one’s for all you so-called brainiacs out there. Think your smart huh? Try this one on for size. (Answer at the end of this node)

A right circular cone has a base of radius 1 and a height of 3. A cube is inscribed in the cone so that the one face of the cube is contained in the base of the cone. What is the length of an edge of the cube?

Say hello to a typical question that you might find at the annual event lovingly known as the William Lowell Putnam Mathematical Competition. Each year about 2,500 of the finest mathematical minds in the country gather in order to test their skills against each other in the competition. The test consists of only twelve questions and lasts for six hours. They must be some tough freakin’ questions. In last years competition, the median score was 1. This was out of a possible 120. Out of the 2,954 students who braved the test last year, only 50 got above 50%.

Originally devised in 1938 as a little friendly competition between schools, it is now deemed as one of the the toughest in the world. The 12 questions a devised by a committee of professors and increase in difficulty as one progresses through the test. (It only takes ‘em four or five months to come with the questions). The test is broken down into two three-hour sessions with a couple hour break for lunch. Forget about bringing in calculators or any notes you have. Not allowed, pencils and paper only..

There are both individual competitions and team competitions. Your top five finishers in the individual competition get a whopping $2,500 each. If your team comes in first you can expect to split $25,000. But its not about the money…

A decent finish in the competition (such as being in the first 500) gets your name on a list circulated throughout the mathematics field. From there, you can be expecting your phone to ring and offers to come flying in. Should your team win, your college’s math department just got a big shot in the arm.

Harvard has been on a roll in that department. Since 1985 their team has taken first place an amazing thirteen times. At one point, they had won 8 straight competitions. Past individual winners include two Nobel Prize and three Fields Medal recipients.

If you don’t do so well, don’t feel to bad because you're in some pretty good company. John Nash (A Beautiful Mind) tried the test twice and couldn’t crack the top five.

Answer to the problem? s=(9 X square root2-6)/7 – whatever the hell that means.

For whatever its worth - this was node #400

Source -December 23 2002 Time Magazine

Just because I am a smarty pants I'll answer the question from above:

A right circular cone has a base of radius 1 and a height of 3. A cube is inscribed in the cone so that the one face of the cube is contained in the base of the cone. What is the length of an edge of the cube?

To make this a little more informative I'll try to give an insight into the thought processes involved in answering the question.

First we draw a picture (always the first step!)

This is hard on E2 so just imagine a cone upside down on x-y-z axes cutting at 1 on the x and y axes and 3 on the z.

Now remember something you learnt in multivariate calculus or just think about it. The equation of this cone is z = 3(1 - sqrt(x^2+y^2)).

Now the sticking point (where the cube hits the cone and so can't get bigger) is going to be on the top face of the cube at the corners. Think about this for a while and make sure you agree (draw some pictures).

So now if we let the side length be 2s (2 times s just to make the maths easier) by rearranging our cone equation, the equation of the level surfaces on z = 2s

2s = 3(1 - sqrt(x^2+y^2))
x^2+y^2 = {(3-2s)/3}^2

so a circle with radius (3-2s)/3

but we want to fit a square inside this circle. By Pythagoras' Theorem the diagonal is sqrt(s^2+s^2) = sqrt(2)*s

so these must be equal (3-2s)/3 = sqrt(2)*s
3 = 2s + 3*sqrt(2)*s
s = 3/(2+3*sqrt(2))

now this is almost our answer but we need to express in the same terms as the answer above: so rationalise the denominator

s = 3(2-3*sqrt(2))/(2+3*sqrt(2))(2-3*sqrt(2))
s = (6-9*sqrt(2))/(4-9*2)
s = (9*sqrt(2)-6)/14 (multiply top and bottom by -1)

and we said side length = 2s = (9*sqrt(2)-6)/7 as required.

So there you are: these questions definitely aren't impossible but I'm sure most of them require more knowledge than this one.

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