Analysis of Today’s Putman Problem

November 28th, 2007

I found an interesting problem on the blog Reasonable Deviations.

Question:
Given a plane k, a point P in the plane and a point Q not in the plane, find all points R in k such that the ratio \frac{QP+PR}{QR} is a maximum.

Solution:
First, create a plane that intersects the points Q, P, and R. In addition to these 3-points, I’m going to create a 4th point point called R’ which is perpendicular to the plane.


Using Pythagoras Theorem, we can do the following:
\frac{QP+PR}{QR} = \frac{\sqrt{QR^{\prime 2}+PR^{\prime 2}}+RR^{\prime}+PR^{\prime}}{\sqrt{QR^{\prime 2}+RR^{\prime 2}}}

We can simplify the formula by using the following relations: x = RR^{\prime}, y = QR^{\prime}, z = PR^{\prime}. The important thing to note is that x is the only variable. The other variables are constant.

The formula can be reduced to:

f(x) = \frac{\sqrt{y^{2}+z^{2}}+x+z}{\sqrt{y^{2}+x^{2}}}

The maximum of f(x) will occur when its derivative is 0. Therefore,

\frac{d}{dx}f(x) = \frac{d}{dx}(\frac{\sqrt{y^{2}+z^{2}}+x+z}{\sqrt{y^{2}+x^{2}}}) = 0

\frac{d}{dx}f(x) = \frac{1}{\sqrt{y^{2}+x^{2}}}-\frac{(\sqrt{y^{2}+z^{2}}+x+z)x}{(y^{2}+x^{2})^{\frac{3}{2}}} = 0

\frac{1}{\sqrt{y^{2}+x^{2}}} = \frac{(\sqrt{y^{2}+z^{2}}+x+z)x}{(y^{2}+x^{2})^{\frac{3}{2}}}

y^{2}+x^{2} = (\sqrt{y^{2}+z^{2}}+x+z)x

y^{2} = (\sqrt{y^{2}+z^{2}}+z)

x = \frac{y^{2}}{\sqrt{y^{2}+z^{2}}+z}

Therefore, the point R will a distance x away from R’ defined by the following equation:

RR^{\prime} = \frac{QR^{\prime 2}}{QP+PR^{\prime}}

RR^{\prime} = \frac{QP^2-PR^{\prime 2}}{QP+PR^{\prime}}

RR^{\prime} = \frac{(QP-PR^{\prime})(QP+PR^{\prime})}{QP+PR^{\prime}}

RR^{\prime} = QP-PR^{\prime}

PR=QP

Comment by rod.

Made Wednesday, 28 of November , 2007 at 9:20 am

I’ve just replied to your comment on my blog. Cheers!

Comment by eldila

Made Wednesday, 28 of November , 2007 at 10:17 am

I forgot to include the solution for when R is between P and R’. It should have have a very similar result.

Comment by Matt

Made Wednesday, 28 of November , 2007 at 11:08 am

I like your solution and got something similar using an ugly analytic geometry approach. (Note that your solution can actually be simplified to RR’ = QP – PR’ with one more step.)

Since you already solved it I will give just a sketch: I set my origin to be the orthogonal projection of Q into the plane k, what you are calling R’ above; defined P=(p,0,0), Q=(0,0,q), R=(r,theta,0) using polar coordinates; I guess r=x,q=y,p=z in your notation.) First take partial derivative wrt theta to prove that theta=pi, then you know that PR=p+r and you can take derivatives with r. I end up with a quadratic r^2+2pr-q^2=0 which has only one positive solution r=sqrt(p^2+q^2)-p, or in geometric terms RR’=QP-PR’ which is equivalent to what you have there because (QP-PR’)(QP+PR’)=QP^2-PR’^2=QR’^2 by Pythagoras.

Note that there is one special case where p=0 (P=R’) in which case any choice of theta will work and the maximum is achieved anywhere on a circle of radius PQ around P.

If you don’t mind a small suggestion, the one change I would add to your proof (which is much nicer than mine) is that you should justify how you know that R must lie on the line PR’.

Comment by eldila

Made Wednesday, 28 of November , 2007 at 12:40 pm

Thanks for the input! It is always interesting to see how people solve the same problem.

I wrote this at around midnight so I was sloppy (I wanted to go to bed :D ). And yeah, I could have justified a few things.

Comment by Karim

Made Wednesday, 28 of November , 2007 at 6:31 pm

And simplified again to QP=PR ?

Comment by eldila

Made Wednesday, 28 of November , 2007 at 7:02 pm

That would make it really elegant :D

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