For math problems and their solutions.

82 responses total.

TIME magazine of 23 December 2002 has an article describing the William
Powell Putnam Mathmematical Competition. This is an annual intercollegiate
competition with just 12 problems to solve in six hours. The median score
on the 2001 test was 1 point out of 120 possible. 

The following problem was given as "an easy one":

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

The solution is given in the article so, if you haven't already done it,
consider it as a practice problem. Here is a revised version of the
problem:

  "A right circular cone has a base of radius 1 and a height of H. A
  rectangular parallelpiped is inscribed in the cone so that one face
  of the parallelpiped is contained in the base of the cone. What
  are the dimensions of the parallelpiped that has the largest possible
  volume?"

(Show all work....)

Dumb question:  does a cube inscribed in a cone in this way have all of 
its volume inside the volume of the cone?   And what is a parallelpiped?

Yes, it would have all its volume in the cone.  A "parallelpiped" is a solid
whose opposite edges are parallel.  Includes rhombus, as well as square.

So, in the isnatnce of this problem, the parallelpiped in question would 
be a cube because that would satisfy the maximum volume requirement 
whereas a rhomboid solid would not, correct?

(Or is there such a thing as a rhombus which exceeds in area a square if 
both can be inscribed in the same circle?)

Re #1:  The equation is fairly easy to solve.  When the corner of
the cube exactly touches the surface of the cone, the edge length
l of the cube obeys the following relation:

l/sqrt(2) = 1 - l/3.  Solving for l,

(1/sqrt(2) + 1/3) l = 1    ===> l = 1/(1/sqrt(2)+1/3)

I'll leave the general case to someone else.

In the restatement as a general case in #1, I *think* the angles would
all be right angles, but the height would not necessarily equal the width
and length, as they would in the first statement, which specified a "cube".

I don't remember the formula for the area of a rhombus and don't feel like
trying to derive it, so I cannot compare the area of a square of side
x with that of a rhombus of side x.  I *think* they would be the same,
but I've been wrong before. :)

Rectangular parallelpiped = right prism = cuboid. See
http://mathworld.wolfram.com/Cuboid.html

russ failed to "show all work" wrt "cube" problem, so would get no credit.

other also flunks. 

How could I flunk if I haven't even gotten to the point of attempting a 
solution yet?

I'm sorry - I thought your statement in #4 was your solution but without
showing all your work. Here is a fresh exam sheet...... 

I find it amusing that Rane doesn't think that showing all the
work I actually did wasn't "sufficient".  I find it even more
amusing that he's presumptuous enough to assign grades here.

I've been playing "exam administrator". But I'll be glad to drop that
role, and just consider this an interesting math problem that I cobbled
up based on the one published in TIME. The solution for the new problem
has some interesting properties.

What I was driving at, russ, is that I would like you to explain the
logic by which you derived your departure equation for the cube case.
I did it differently, and don't recognize your logic. 

Taking a half cross-section through the cone and cube on a diagonal of
the face which lies on the base of the cone, you get a 2-D figure where
a rectangle of height l and width l / sqrt(2) (one edge and half a
diagonal of two faces of the cube) is inscribed in a right triangle of
height 3 and width 1 (the cross-section of the cone).  The equation
describing the width y of the triangle at a given height x is y = 1 - x/3,
while the equation describing the location of the corner of the cube
which lies on the surface of the cone is x = l, y = l / sqrt(2).  Ergo,
l / sqrt(2) = 1 - l/3, from which the rest follows.

I thought the cross-section step was so painfully obvious that it did
not need to be stated; calculating where the corner meets the cone *is*
the problem.

Good. My solution recognizes that the height 3 triangle and the height
3-l triangle (above the cube) are similar, and therefore

                   3/1  =  (3-l)/(l/sqrt(2))

which of course gives the same result.

Now...on to the volume maximization problem?

While y'all are working on maximizing the rectangular parallelpiped's volume,
I will observe that while both russ and I arrived at the solution he
shows in #5, although by slightly different paths, the solution given
in TIME was (9sqrt(2)-6)/7. For a moment I worried about my solution,
but of course they are equal. If anyone new to algebra is following
this, you might attempt the conversion. There is a common identity
that can be used.

Mulling further over the cube inscribed in a right circular cone: consider
that the cone is of height H and base radius R, and call the side of
the inscribed cube h. Then, by the same procedure already described

                 h = (Rsqrt(2))H/(Rsqrt(2) + H)

This is the same relation for the resistance of a circuit consisting of
two resistances of Rsqrt(2) and H in parallel. There is therefore in one
sense an analogy between the geometric construction and an electric
circuit. Is there a physical interpretation of this analogy? (The above
relation also describes an *analogue computer* for calculating the size of
the cube from the height and base radius of the cone.) 

It's been over a week with no "offers" for the rectangular parallelpiped
volume maximization in a cone, so I'll give a quick outline of a solution,
and the item can move on.

Call h the height, a the length, and b the width of the inscribed
right parallelpiped. The volume is then

                         V = abh                                  (1)

The dimensions are constrained by considering the similarity of the
right triangles defined on a plane placed to contain two opposite
lateral edges of the parallelpiped (same procedure as in cube
problem):

             (H-h)/(0.5*sqrt(a^2+b^2) = H/1                       (2)

The simplest way to proceed is to recognize that *given h*, a and b can be
chosen to maximize A = ab, the base area, and hence the volume. The
relation for that from eqn (2) is

                   A^2 = a^2*[4*(1-h/H)^2 - a^2]                  (3)

Differentiating A^2 w.r.t. a^2 and setting equal to 0 determines the
stationary point for A^2. Doing this yield

                     a = b = (1-h/H)*sqrt(2)                      (4)

It is not surprising at all that the base is a square.

To find the height, find the stationary point of the volume w.r.t. h,
i.e.

                   V = a^2*h  = 2*h*(1-h/H)^2                     (5)

Differentiate this w.r.t. h and set equal to zero, to obtain

                         h = (1/3)*H                              (6)

and from (4) and (6)

                         a = (2/3)*sqrt(2)

I found it interesting that the base dimensions are constant, independent
of H, while the volume of the parallelpiped is simply

                       V = (2/3)^3*H                              (7)



                          

That is intersting.  You can drop the requirement that the parallelepiped be
rectangular, and while the solution is no longer unique, the one you found
is still maximal.

Here's one from the most recent Putnam competition:

Shanille O'Keal shoots free throws on a basketball court.  She hits the
first and misses the second, and thereafter the probability that she hits
the next shot is equal to the proportion of shots she has hit so far.  What
is the probability she hits exactly 50 of her first 100 shots?

1/2 * 1/3 * ... * 1/50

Interesting observation in #18, which had not occurred to me. Indeed, since
the volume of a parallelpiped is just the base area x the height, the
base in the above solution could slide anywhere in the base of the cone
with the volume of the parallelpiped remaining constant. Therefore the problem
could be made to *appear* more difficult (by removing the "rectangular"
specification), although it would still be the same problem. One would
have to recognize the volume property of the parallelpiped before one
could proceed with confidence. 

This response has been erased.

Re #19: Nope.

Re new problem in #18: much to my surprise, I conclude that all outcomes
in n throws are equally likely, except for 0 and n hits. Hence, the
probability of hitting exactly 50 (or any other number not 0 or 100) on
her first 100 shots is 1/99. I concluded this from working the probability
distributions out up to n = 5 - and then took the (mathematical) "leap of
faith". I do not, however, have an "elegant" proof for all n. 

Do you have an inelegant one?

hmmmm.....

I was wondering if anyone here knows of a FREE program that
can do XYZ calculations. My TI-89 takes forever to re-calc
and build a wireframe....and anyways the display is too 
tiny and doesn't have a good resolution..

I figured there has to be a program to run on you computer
that does all the same things?

What do you mean by an XYZ calculation?

3-d graphing

I'm sure there must be free programs out there.  Most mathematicians use
either Mathematica, Maple, or Matlab.

  I've never used it, but I believe some of the GNU folks were working on
  something called "octave" which was supposed to fill a matlab/maple-like
  niche in the free software world..

oh...I will try that out, thanks

  I *think* that's what Octave is for, anyway..  Also, I have no idea how
  far they've gotten..

Returning to #18, where Mark presented the Putnam Prize question

 Shanille O'Keal shoots free throws on a basketball court.  She hits the
 first and misses the second, and thereafter the probability that she hits
 the next shot is equal to the proportion of shots she has hit so far.
 What is the probability she hits exactly 50 of her first 100 shots?

Let P(n, k) be the probability of getting k hits in n throws, in the
space [n, k], 0 < k < n and zero elsewhere.

The transition probabilities to [n, k] are

  (k-1)/(n-1) from [n-1, k-1] and (n-1-k)/(n-1) from [n-1, k]

The P(n, k) recurrence relation is therefore

  P(n, k) = [(k-1)/(n-1)]P(n-1, k-1) + [(n-1-k)/(n-1)]P(n-1, k)

with ICs 

  P(1, 1) = 1
  P(2, 1) = 1

which give 

  P(3, 1) = P(3, 2) = 1/2

Because these probabilities are equal, consider what would happen if
P(n, k) was independent of k. From the recurrence relation, 

   P(n, k) = [(k-1)/(n-1) + (n-1-k)/(n-1)]P(n-1)

           = ((n-2)/(n-1))P(n-1)

and hence all P(n, k) are independent of k. 

Since at each n there are n-1 possible outcomes k with equal probability,
the probability of each must be 1/(n-1).

Hence for n = 100 and k = 50, P(100, 50) = 1/99

QED

Nicely done, Rane.  That's a good proof, as long as you replace "consider
what would happen"  with "by induction". :) 

Here's another from the current Putnam:

Given any five points on a sphere, show that some four of them lie on a
closed hemisphere.

What does "some" mean? Exactly four? Four or more?

Pick any two points.  Draw a great circle though them.  There are two
possible closed hemispheres defined by this circle, each containing
those first two points.  The remaining three points must be in one or
both of the hemispheres.  You can't divide there objects into two sets
without creating at least one set of size two or more, so one of the
hemispheres must contain two of the remaining three points, totalling
four points.

Probably should be written as a proof by contradiction.  But it's too
easy to bother.

What if you exclude the possibiity of a point being in two hemispheres
simultaneously? That is kind of "cheating" as it, in effect, gives you
more points. 

He did.  The first two points determine a circle which can be taken to 
comprise the border of one or the other hemisphere, but not both.  The 
great circle doesn't run between the hemispheres.

OK. It should have read "...one or the other containing those first
two points". 

Not necessarily, though that would have been clearer.  You're taking the 
clause "There are two possible closed hemispheres defined by this circle" 
to refer to the two halves of the sphere.  If instead you take it to 
refer to the two possible hemispheres which contain that circle (not 
simultaneously extant), then the meaning is clear with the wording as is.

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