UNFORTUNATELY,   UNLIKE  BONES,  BEHAVIOR  DOES  NOT  BECOME  FOSSILIZED.  —  RICHARD  LEAKEY

Correct answers came from  Delmar Burkitt,  Andrzej Derdzinski,  Jeff Gerken,  Bob Kaplan,  Alma Litten,  Julia Minturn,  Carla Nuenke,  and  Sally Yocom.
 
 
The second puzzle proved a bit deceptive.  Each merger, no matter how done, decreases the number of companies by one.  Hence 105 mergers are needed.  However, the first merger begins on the first Friday and the last (#105) occurs on Friday 104 weeks later.  So 104 is the answer.  Only  Andrzej  and  Bob  had the correct answer.
 
 
The correct answer to the third puzzle was 1500 Democrats, 1300 Republicans, 1100 Independents with 6100 not voting.  Correct were:  Delmar,  Andrzej,  Jeff,  and  Carla.
 
 
Finally,  Andrzej,  Jeff,  and  Steve  submitted answers to the overlapping circles puzzle submitted by Doug McDonald. The challenge was to find when the area of overlap of two circles is equal to half the area of one of the circles. Here is an analysis of the solutions:

From Andrzej J. Derdzinski:

I took a look at the overlapping circles problem.  The answer is both simple and disappointing.  Simple because it can be obtained by setting up an equation in a most straightforward fashion, and disappointing because the answer (the distance d between the centers or, more precisely, the ratio d/r of that distance to the common radius r) does not have an explicit expression in terms of elementary functions.

One way of stating the answer is:  Take the smallest positive angle a (measured in radians) such that a = cos a.  This number must exist, and lie between 0 and /2, since the value of y=x - cos x is negative (y=-1) at x=0 and positive (y=/2) at x=/2, and so, by reasons of continuity, we must have y=0 at some intermediate x=a.

The ratio d/r at which the overlapped and nonoverlapped parts have equal areas then equals d/r=cos b, with b=a/2 + /4.

To see that this is the case, one may denote P,Q,R the center of Circle "B", one of the intersection points of the two circles and, respectively, the midpoint between their centers, and denote b the angle QPR expressed in radians.

Then one-quarter of the overlap area equals the difference between the area of the circular wedge QPR and the area of the triangle QPR, i.e., br²/2-r²(cos b sin b)/2, and we want this number to be one-eighth of r², which gives the equation b-cos b sin b = /4.  Rewritten in terms of a=2b + /2, this equation reads a=cos a.

The fact that we do not have an elementary formula for this angle a (or the ratio d/r) makes it no less "explicit" than other numbers that are already entrenched in common usage (such as , e, etc.).  We could imagine giving it a name, using standard procedures to find its decimal approximations, etc., just like it once happened with (when someone asked the similar question:  what exactly is the area of a circle of radius one?).

Jeff Gerken sent this solution:

For the circle problem, the area of overlap will be equal to the non-overlapped area when the arc (theta) subtended by overlap satisfies the equation (with theta in degrees):

     (*)/180 - sin()=/2.

The solution to this equation is approximately 132.348 degrees.

Steve Herrick's solution:

The area of the overlap is the sum of the areas of the arcs described by DAE and DBE minus the area of the rhombus ADBE.

The area of arc DAE is r²*DAE/2 in radians.
The area of arc DBE is equal to DAE so their sum is r²*DAE/ or r²DAE.
The area of the rhombus is side BE*height AF.
To determine length of AF, we use the sin of angle AEF which equals AF/AE.
Angle AEB is equal to -DAE thus we get r²sin(-DAE).

So we want to know the size of angle DAE when r²DAE-r²sin(-DAE)=*r²/2.
Dividing by r², we get
DAE-sin(-DAE)=/2.
This solves out at app. 2.3 radians or 132 degrees.

What I was looking for was the length of AB.
We know that the cos(DAC)=AC/AD and AB=2AC and DAC=DAE/2
So AB=2r*cos(DAE/2)
So AB=.808*r

S I N Y E C E K I        Q J        P F C Q K Q E Z J        J F V N Q S F V N        K P E A Q V K        A Q C H.       

— O P Q F H P Q W Y        J W Y F C C Q V K       

Here is a crossword-type puzzle where you have to put a digit in each of the eight unshaded squares.

To help you I should give you clues for each of the three "across" numbers and each of the two 3-digit "down" numbers.  In fact the clue for each of these five numbers is they are "divisible by 18".

Now there are a number of possible solutions, but if you were told the position of one specific even number there would be a unique answer.  What is that answer?

There is an arsonist, a burglar, a con-man, and a double-agent whose happy hobby is stamp-collecting.  Arnold, in fact, has twice as many as Bernard, who has twice as many as Clarence, who has twice as many as Dennis.  The arsonist has more mint issues than anyone with more stamps than the burglar, and also has more penny blacks than anyone with more stamps than the con-man.  The con-man has 1085 more stamps than the double-agent.

The police have been inquiring which is which.  What are the matches?