# CUNY Math Challenge Blog

Solutions, Resources, Further Information & Discussion

## Round 4: Problem 5 Solution

Tricoloring

This is a nice tiling problem in disguise. Let’s ignore the letters and symbols for now and just work with a blank 5×8 board. The 3×1 rectangular pieces are like straight triominoes that we will use to try to tile the board.

Essentially, our problem now is the following: with 13 triomino(e)s, how many different places can we place a monomino on a 5×8 board.

Tricolor the board as shown. Note that we now have 13 red squares, 13 yellow squares, but 14 black squares. Therefore, the monomino must be placed on one of the black squares. However, the monomino cannot be placed on just any black square. A monomino can only be placed on a black square if there is no way to flip our coloring so that the square in our original coloring is no longer black. Flipping the coloring disqualifies every black square besides those 2 on the third row.

We can verify that in fact we may place a monomino on either of these two squares by showing a tiling (try it, it’s trivial). Therefore we conclude that a monomino can only appear on one of these 2 squares. Plugging in these 2 black squares into the original board / cake, we see that on these two squares are the ‘2’ and the ‘9’ of ‘2009.’ These are the only two possible remaining pieces.

April 27, 2009 at 00:05

Posted in Round 4

## 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

Posted in Round 4

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