Submissions and Requests

Solutions are included within the puzzles below. Solvers are listed at the end.

Requests for help

1. An interior designer has an intriguing dilemma. She has to move an amazingly heavy armchair, but the only possible movement is to rotate it through ninety degrees about any of its corners. Can it be moved so it is exactly beside its starting position and facing the same way? [Update 9/18: One reader suggested this problem should be attacked without requiring the angle of rotation always be 90 degrees. Consider this as part (b) of this problem.]
   Solution:
   
   • a) No. To see this, paint the floor in a checkerboard pattern with the chair on a white square and, without loss of generality, assume the chair faces North. Every time the chair is moved, it moves to a square of the opposite color and its orientation toggles between North/South and East/West. So it faces North or South when on white squares and East or West when on black. The squares adjacent to the chair's starting position are black.
   • b) Rotate 30 counterclockwise about B. 60 clockwise about C and finally, 30 counterclockwise about B.

2. An angus is anchored to a 6 meter rope. The rope is attached to the outside corner of a shed measuring 4 meters by 5 meters in a rolling grassy meadow with varying elavations. What area of grass can the cow graze in 1.5 hrs?
   Solution:
   The animal can reach 3/4 of a circle of radius 6m, plus 1/4 of a circle of radius 1m (going around the corner on the 5m side), plus 1/4 of a circle of radius 2m (going around the corner on the 4m side). Total is 27pi + 0.25pi + 1pi = 28.25pi.

3. Prove that: Within a set of triangles having a constant base and constant perimeter, the isosceles triangle has the maximum area.
   Solution:
   Let us consider Heron's formula for area:
   S^2=p(p-a)(p-b)(p-c). p and p-a are constant and
   (p-b)(p-c)=(a-b+c)(a+b-c)=[a^2-(b-c)^2)]/4.
   Hence S^2=k[a^2-(b-c)^2], where k is a constant. a is also constant, so the area is maxim when (b-c)^2=0, that is b=c.
   
   Also, if an ellipse is drawn so that the endpoints of the base of the triangle are the foci and the locus of the third point of the triangle is the ellipse itself, then the solution is clear from the diagram.

4. What is the missing number ?

         30
       26  34
     22   ?  26
   11  18  27  31 

   Solution:
   The missing number is 36. The rule to get a number of the pyramid is: sum the digits of the two adjacent numbers below it and multiply the result by 2. For example, if we take 11 and 18, the result is 2*(1+1+1+8)=2*11=22. To get the missing number we calculate 2*(1+8+2+7)=2*18=36. All the numbers in the pyramid follow this rule.
   This problem or a problem quite similar to this can be found in the book "The mathmagician and the pied puzzler". [Solution from Pedro Romero 10/9/02.]

5. You have 6 pool balls, all identical in size though differently numbered. 5 balls are of the same weight, the 6th is different (i.e. heavier or lighter). You have a set of balancing scales and, using the scales twice only, you have to identify the rogue ball.
   Solution:
   It can't be done. A good analysis of the ball-weighing puzzles can be found at: http://rec-puzzles.org/sol.pl/logic/weighing/balance. This page shows that with two weighings, a rogue ball can only be found among 4.

1. S - - - - S
   Enter a 4 letter word, divisible by 2, to get odd numbers.
   Solution:
   EVEN, creating SEVENS

2. 14 people are eating at their hotel's restaurant. They are seated at two tables which hold 8 and 6 people. If they are seated at these tables for every meal, how many meals are needed to assure that every pair of people sit at the same table at least once? What if the tables instead hold 9 and 5 people?
   Solution:
   One three meal solution for 8/6:
   1: ABCDE FGH : IJK LMN
   2: ABCDE IJK : FGH LMN
   3: ABCDE LMN : FGH IJK
   One four meal solution for 9/5:
   1: 1,2,3,4,5,6,7,8,9
   2: 1,2,3,4,10,11,12,13,14
   3: 5,6,7,8,10,11,12,13,14
   4: 9,10,11,12,13,14,1,2,3

3. a/b + b/c + c/a = 1
   Does a solution exist for a, b, and c belonging to the set of negative and positive integers? Does any solution exist? If so, show a simple example.
   Solution:
   There is no solution with integers. For non-integers, a simple solution is:
   a = 1, b = -1, c = sqrt(2)-1.

4. Simplify the infinite product (1+x)(1+x^2)(1+x^4)(1+x^8)(1+x^16)..., given |x| < 1.
   Solution:
   Multiply by (1-x) to get
   (1-x)(1+x)(1+x^2)(1+x^4)(1+x^8)(1+x^16)..., which is
   (1-x^2)(1+x^2)(1+x^4)(1+x^8)(1+x^16)..., which is
   (1-x^4)(1+x^4)(1+x^8)(1+x^16)..., which is
   (1-x^8)(1+x^8)(1+x^16)...
   And we can see that, continued infinitely, multiplying the infinite product by (1-x) yields 1. Thus, given |x| < 1, the infinite product simplifies to 1/(1-x).

Source: Many sources. Submissions from Denis Borris and Ravi Subramanian.