Proposed Solutions to Posed Problems

Spring issues of MathNEXUS included several nonstandard problems-of-the-week, sent to me by regular readers of MathNEXUS. Some readers even responded, as shown below...

First, re-read Problem 1 (Question 2) before reading the response below...

C.S. (Mount Vernon, WA) wrote: The next solutions to tan(n)>n are n=122925461 and n=534483448. I can find more if you'd like. I made a smart-search algorithm using Mathematica which revealed A LOT of interesting features of the real numbers mod pi/2.

Even though the solution above is over 534 million I only had to search (or test) 1000 values. Takes seconds in Mathematica. My guess is there are infinite solutions.

In a later e-mail, C.S. added: They're taking a little longer to find... 3,083,975,227. What I can't guarantee is that I am finding all of them up to this last one. My algorithm simply finds the next optimal one (but I'm fairly confident it is). Much to think about.

Second, re-read Problem 2 (Question 1) before reading the response below...

M.J. (Bellingham, WA) Tell C.S., my first thought on the answer to his x^x^x... is that it converges on (.5,1]. But, now I am thinking maybe it converges from (.5 to1.3 or 1.4). Clearly x=2 diverges and I think so does 1.5, but haven't checked. I learned something while reading this week that really surprised me and I think it relates to this question is a weird way. I will be interested in his answer and how he determined it.

Third, re-read Problem before reading the response below...

Solution proposed by D.K. (Fife, WA): Suppose right triangle ABC has right angle B. Use the perimeter of 324 cm, Pythagoreans theorem and trial and error to find AB = a = 81, BC = b = 108 and we know the AC = c = 135. Suppose x, y, and z are points of tangency to circle with center i on AC, AB, and BC respectively. As we know points of tangency are perpendicular to the radius, they can be used as heights of triangles.

Since all radii are congruent, xi = yi = zi = r (radius). Divide triangle ABC into three smaller triangles: AiB, BiC and AiC. Then area of triangle ABC = sum of areas of triangles AiB, BiC, and AiC. Or,