The Riddler, 2017-08-11 edition

Here’s a link to the problems. I usually try to solve Riddlers with code if I can, but this week’s problems were relatively straightforward, so no code was necessary.

Riddler Express

The most efficient way to do this is to represent each staffer as an unsigned 7-bit binary number, with each bit representing knowledge/no-knowledge of one of the seven fake stories that you and your chief of staff have come up with:

    \[ \begin{tabular}{|c|c c c c c c c|}   \hline              & S0     & S1     & S2     & S3     & S4     & S5     & S6     \\ \hline   Staffer 0  & 0      & 0      & 0      & 0      & 0      & 0      & 0      \\   Staffer 1  & 0      & 0      & 0      & 0      & 0      & 0      & 1      \\   Staffer 2  & 0      & 0      & 0      & 0      & 0      & 1      & 0      \\   Staffer 3  & 0      & 0      & 0      & 0      & 0      & 1      & 1      \\   \vdots     & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\   Staffer 99 & 1      & 1      & 0      & 0      & 0      & 1      & 1      \\   \hline \end{tabular} \]

Armed with this, we can pinpoint the miscreant staffer by tracking each story’s status. For example, if no stories were leaked, we know that staffer 0 is the culprit since she had no knowledge of any stories. Similarly if only S0 (story 0) was leaked, the leaker must be staffer 64 (64 in decimal = 1000000 in binary), since he was the only one that was told only S0. Thus, by intelligently distributing distinct combinations of 7 stories to all 100 staffers, we can determine who the pesky leaker is.

In any case, don’t forget to watch your chief-of-staff as well 😉

Riddler Classic

Imagine gluing together five tetrahedrons as shown on the FiveThirtyEight webpage (displayed below for convenience):

However, instead of looking at the resulting object from top-down, imagine holding it flush in your palm and raising it to eye level such that your new perspective is horizontal. If you then flatly slice the object along the edge closest to you (parallel to your palm), you can easily see that the slice is made of five isosceles triangles – one for each tetrahedron. The figure below should help visualize this:

The blue vertical plane shows were our horizontal cut was made. This plane will allow us to determine the central angle created by the five-tetrahedron setup. To compute this angle, we first need to determine the length of the slice as projected onto a face opposite the edge closest to us:

Armed with all side lengths of each slice (\frac{\sqrt{3}s}{2}, \frac{\sqrt{3}s}{2}, and s), we can now find the central angle:

Basic trigonometry tells us that the central angle is 2*sin^{-1}(\frac{1}{\sqrt{3}})\approx70.5288\dots degrees. There are five of these altogether, bringing the total central angle to approximately 352.6439 degrees. Thus, the empty space has an angle of 7.3561 degrees (0.1284 radians).