Proposed Solutions to Posed Problems
Spring issues of MathNEXUS included several nonstandard problemsoftheweek, sent to me by regular readers of MathNEXUS. Some readers even responded, as shown below...
First, reread 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 smartsearch 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 email, 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, reread 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, reread 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,
0.5 ab = 0.5 a (yi) + 0.5 b (zi) + 0.5 c (xi)
0.5 ab = 0.5 ar + 0.5 br + 0.5 cr
0.5 ab = 0.5r(a + b + c)
ab = r(a +b +c)
[ab]/[a+b+c] = r
[(81)(108)]/[81+108+135] = r
8748/324 = 27 = r
Thanks for these responses....If others want to respond, please do so....
Hint: Why not send in your own solution?
Solution Commentary:
