# CUNY Math Challenge Blog

Solutions, Resources, Further Information & Discussion

## Round 4: Problem 2 Solution

Here’s a geometric construction proof:

• Extend $AF$ and $BC$ so that they intersect at an exterior point $\text{X}$.
• Extend $FE$ and $CD$ in a similar fashion so they intersect at another exterior point, $\text{Y}$, forming quadrilateral $XFYC$.
• $\angle XAB \cong \angle XBA \cong \angle YED \cong \angle YDE = 60^{\circ}$ since each external angle is supplementary with the 60 degree internal angle.
• $\angle X \cong \angle Y = 60^{\circ}$, so $\triangle XAB, \triangle YDE$ are equilateral triangles.
• $AB = XB, DE = YE$ in equilateral triangles.
• $XFYC$ is a parallelogram (both pairs of opposite angles are congruent), so $FY = XC$.
• $FE + EY = AB + BC$, so $FE + ED = AB + BC$ by equilateral triangles as shown above.

April 27, 2009 at 00:02

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## 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 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 1: Problem 5 Solution

Cross Section

Pictured at the left is the cross-section of the three spheres. To solve the problem, let us note that the spheres of radii 3 and 5 are similar, and they both have the same center. Therefore, any area traced by the radius-1 ball on the outer sphere will project itself onto the inner ball. We must then discern at what ratio the projection will be in. To determine the ratio, consider what type of inking is occurring on the outer sphere. The black ball colors in an area of the outer sphere’s surface. Therefore the amount covered, in this case an area of 1, should be proportionate with the surface areas of the two spheres. For a sphere of radius r the surface area A = 4 π r2. For our purposes then we need only to compare r2 for each of the spheres, arriving at our answer of 9/25.

March 16, 2009 at 00:04

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## Round 1: Problem 1 Solution

with one comment

There has developed a set of terminology to describe these objects. These tiling objects have been named polyomino(e)s (as generalized forms of dominoes), or in this specific case, pentomino(e)s. This specific arrangement is usually referred to as the F-pentomino, for its resemblance to the letter F (all of the pentominos are named after the letters they look like, though I’d agree that this one is a bit of a stretch).

With that in mind, there are a number of useful sites with information on pentominos and plane tilings. I won’t list all of them because they are relatively easy to find, but dmoz.org links to a whole directory’s worth of sites dedicated to polyominos. It can be found here. Of particular interest is Gerard’s Polyomino Solution Page, with more than 60 polyomino puzzles and solutions.

For our purposes, however, this problem should not pose too much difficulty beyond simply trying out various rotations of the tiles. There’s a simple solution with fewer pieces; an example can be found at the NGfL here (warning: PDF) along with solutions for all of the other pentominos. I’ll link to a nice larger solution as well because, well, I like the colors.

One point of confusion for this problem is that tilings of the plane (or tessellations as they may be called) do not need to be regular (rectangular). In fact there are only three polygons that form regular tessellations. See the page at MathWorld for further info. Rather, the critical point here is that the shape formed can be attached to itself at any end (including top and bottom), meaning that we could continuously attach replications of the shape to itself to tile an infinitely large area.

Please post any comments, corrections, interesting points, or alternate solutions that you may have found in the comments!