# CUNY Math Challenge Blog

Solutions, Resources, Further Information & Discussion

## Round 2: Problem 5 Solution

I spent a tiny bit of time on this problem, because the solution did not come immediately to mind.

Ultimately, realizing that being that the angle of incidence equals the angle of reflection helped to come up with a bit of magic that helps visualize the path. Because the angles are equal, we can unwind the path that the puck takes into a straight line by “mirroring” the rink. This will allow us to solve the problem with just a tiny bit of high school trigonometry.

Unwound Path

Essentially what we have done is mapped the bouncing path into a straight one. For the sake of rigorousness we could show that this straight line is a bijective mapping of a bouncing path, but I will assume that it is trivial for our purposes and trust that if you don’t believe me that you trace it out on one of the unused boxes.

This path represents a trajectory that bounces off each of the four walls and arrives in the “goal.” To solve the problem, all we need is the angle $\alpha$. We know that $b=58m$ (because both the puck and goal are 1m away from the walls) and we also know that $h=5+5+10=20m$, therefore $\angle \alpha =\text{tan}^{-1}\left[\frac{20}{58}\right]\approx .3321\text{ rad}$, or since the problem asks for degrees, $19.0256{}^{\circ}$.

What is interesting is that we can modify this method to work with an arbitrary amount of bounces and end up at any point along the rink. Check out some further reading on the topic of Dynamical Billiards.

The Geometry of Billiards (Pre-print version) (PDF, S. Tabachinkov [Penn State])
Running Header: The Diamond System (PDF, R. Miller [UNCC])

March 30, 2009 at 00:04

## Round 2: Problem 4 Solution

with one comment

Potential Meeting Times

This problem tests a bit of geometric probability. Let us represent the hour from 1:00 to 2:00 by a unit square, and let us shade in the areas that the meeting could take place.We arrive at the picture at the right.

Keep in mind that the shaded area represents times that the meeting could not take place, and that it is in fractions of an hour, not minutes, so that we can see for example that would the first person show up exactly at 1:00 $(0,0)$, then the meeting could only take place if the second person shows up within the first quarter of an hour (which is 15 minutes).

Using the diagram, we calculate the area of the two triangles, $A_{\text{Shaded}}=2*\frac{1}{2}*\frac{3}{4}*\frac{3}{4}=\frac{9}{16}$, and subtracting from the total area to get the area of the unshaded stripe we get $A_{\text{Unshaded}}=\frac{7}{16}$.

March 30, 2009 at 00:03

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## Round 2: Problem 3 Solution

It should be relatively obvious what our answer should be here for a few reasons. Firstly, note that we need only to consider a new score of either 0 or 100, because these two scores will affect the new average by the most amount (and anything closer to 50 will affect the new average less than one or both of these scores). Also, it should be clear that the smallest class size will come when the class average was 50, because if the average is greater than 50 a score of 0 will affect the average more, and if the average is less than 50 a score of 100 will affect the average more, and in both cases we are trying to avoid changing the average by more than a point. We therefore should only need to find out at what class size will an average of 50 not be affected by scores of 0 or 100 by more than 1 point, which even without a calculator should be easy to see as being 49 (50 with the sick student).

In case the reasoning above doesn’t satisfy you however, let’s do a tiny bit of work. Let’s define a function $d(oldSize, oldAvg, newScore)=\left|\frac{\text{oldAvg}*\text{oldSize}+\text{newScore}}{\text{oldSize}+1}-\text{oldAvg}\right|$ which will give us the difference in average from the oldAvg in a class size of oldSize after a new score of newScore. Let’s plot the functions $d(x,y,0),d(x,y,100)$ on the 3d plane, along with the function $z=1$ to show us when it is less than 1.

What we get is shown above.

Since d(x,y,0) and d(x,y,100) are monotonous, what we are looking for is the intersection of the three functions that is closest to 0. It may not be totally clear here at the right but that intersection is at (50,49,1) giving us our answer of an average of 50 with a previous class size of 49 (+1) as we reasoned above.

(I apologize but I don’t think I can embed LiveGraphics3D in WordPress but if anyone knows of a way to I will post a rotatable graph.)

March 30, 2009 at 00:02

Posted in Round 2

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## Round 2: Problem 2 Solution

Let us quickly examine a similar, more well known problem, and then modify the strategy to fit our needs.

Numbers such that $f(n)g(n)=n$, where $f(n)$ is the sum of the digits of $n$ and $g(n)$ is the product of the digits of $n$, are known as sum-product numbers. There are a total of three positive sum-product numbers, and they are 1, 135, and 144.

There is a simple proof attributed to D. Wilson that can be found on the MathWorld page proving that there are no other sum-product numbers which we can modify to solve our problem.

We attempt to find numbers $n$ such that $6f(n)g(n)=n^2$. Let us say that $n$ has $d$ digits.

Being that $n$ has $d$ digits, $10^{d-1}\leq n$, and so it follows that $100^{d-1}\leq n^2$. In addition, $f(n)\leq 9d$ and $g(n)\leq 9^d$, and so $6f(n)g(n)\leq 6d9^{d+1}$. Since we want $n^2=6f(n)g(n)$, we have $100^{d-1}\leq n^2=6f(n)g(n)\leq 6d9^{d+1}$. With a bit of algebra (or a calculator to plot) we can see that $100^{d-1}\leq 6d9^{d+1}$ only when $d<5$. We therefore have now proven that no number with 5 or more digits could be a number in our desired set. Since we now have a small enough sample size to brute-force, we can now throw the first 9999 numbers into a loop and see which satisfy our condition. My program below is in Python which (other than the function definitions at the top) should be readable or understandable enough even for those who don’t know the language; if not, what we are doing at this point is trivial enough to write in whatever language you’d like.

def f(n):
return sum([int(x) for x in str(n)])

def g(n):
count=1
for x in str(n):
count*=int(x)
return count

for x in range(1,10000):
if 6*f(x)*g(x)==x**2:
print(x)



This results in our answer which is just one number: 48.

We could optimize the program by considering an even smaller subset (I think we can probably discard anything besides multiples of 6 in fact), but the program as is for me finishes in .215 seconds.

March 30, 2009 at 00:01

## Round 2: Problem 1 Solution

Orange Pile

Geometry, especially in 3D, is definitely anything but fun for me personally, but with a tiny trick we can solve this problem quite easily. Let us call the center of the bottom three oranges A, B and C, and let us call the center of the top orange D. By connecting A, B, C and D we can form polygon ABCD, which is a tetrahedron (a pyramid with three sides above the base).
It is given in the problem that the radii of all four oranges is 2, and so it should be relatively clear that our tetrahedron has side length of 4 (on all 3 sides of the base, as well as the “diagonal” edges). It should also be clear that as soon as we find the height of our tetrahedron, that we can add 2 to get from the bottom of the pile to the base of the tetrahedron because the base of the tetrahedron is seated on the center of the three base oranges, which have radius 2. Also, we can then add 2 more to get from the vertex of the tetrahedron to the top of the topmost orange because the top vertex of the tetrahedron reaches to the center of the top orange.
What’s left then is to find the height of the tetrahedron and then we are done. In general, the spacial height (i.e. from the center of the base to the vertex) of a tetrahedron is given by the formula $h=\frac{s\sqrt{6}}{3}$, so plugging in 4 for s we get $h_{\text{ABCD}}=\frac{4\sqrt{6}}{3}$, leaving us with a final answer of $h_{\text{pile}}=4+\frac{4\sqrt{6}}{3}$.
Comments, corrections, and further discussion are always welcome and encouraged in the comments!

March 30, 2009 at 00:00

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## Round 2 is almost done. Get your answers in!

Don’t forget, round 2 is over Sunday night at 11:59 p.m. Submit your responses on the official site. The answers that we’ve got will be up right then at midnight, so check back to see if you’ve got what we’ve got once again.

It seems someone didn’t heed my warning, so let’s try this again: submit earlier than the deadline in case something goes wrong and you need to email support to check if your submission was received. The math contest is not the best example of functional code that I’ve ever seen if I may say so myself. Blame it on the economy.

March 26, 2009 at 11:20