October: 231 & 232 

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Posts: 9

Joined: 11/6/2008
Topic  October: 231 & 232       Flag »  Reply »
View the problems here: http://www.profsurv.com/problemcorner.aspx, then come back to discuss.
  Monday, October 03, 2011 at 10:23:57 AM
John Nolton
Posts: 15

Joined: 10/11/2011
Reply  Re: October: 231 & 232 Flag »  Reply »
I would like to make some comments on problem # 231.
What caught my eye on this problem was the statement by Dr. Bloch; "This is a problem known to be difficult". This might be somewhat true if Dr. Bloch had asked for a strictly GEOMETRICAL SOLUTION as the original problem did. This he did not do. By his last statement in the problem "Without drawing any additional lines, solve for angle X" ( which is angle DEA in his drawing) makes the problem very easy and should not take more than 5 minutes to solve.

What I would like to do for all that look at the problem corner is purpose another similar problem here and see how many readers can solve it using elementary geometry (no trigonometry or any other higher math). I will post the answer here in 2 weeks + or -.

Using Dr. Bloch labeling of the angles in the isosceles triangle (see his problem # 231); angle ACB = 20 degree , angle EAB = 60 degree , angle ABD = 30 degree,
Solve for angle DEA  using only elementary geometry.

Dr. Bloch's solution is long and you do not need three simultaneous equations to solve it.
Another problem with his numerical solution is he list "tan x = 0.364, and x = 20 degrees".
The inverse tangent of 0.364 is 20d 00m 05.4+ seconds not 20 degree's exactly as he showed. 


John Nolton
Tombstone, AZ
  Sunday, October 16, 2011 at 5:28:50 PM
Posts: 1

Joined: 4/24/2009
Reply  Re: October Problem 231 Flag »  Reply »
Problem 231 did not give any numerical lengths specifically because a general solution was required.
I apologize if my intent was not made clearer. 
Thank you for your kind support,
Dr. Benjamin Bloch
  Tuesday, October 18, 2011 at 7:51:34 PM
John Nolton
Posts: 15

Joined: 10/11/2011
Reply  Re: October: 231 & 232 Flag »  Reply »
Dr. Bloch you do not need any length to solve the problem. It is quite clear you did not read my post. Your solution is long and your listed value of tan x= 0.364 is also incorrect.
You have turned a simple GEOMETRIC PROOF into a numerical one.

Hopping for a better solution from you.

John Nolton
Tombstone, AZ
  Wednesday, October 19, 2011 at 11:01:16 AM
John Nolton
Posts: 15

Joined: 10/11/2011
Reply  Re: October: 231 & 232 Flag »  Reply »
More comments and my solution to problem #231 (by John Nolton)

I first came across a similar problem in 1962(one year out of High School and a little over one year in the U.S. Marine Corps) in a book titled "Work This One Out" by L.H. Longley-Cook. They were looking for an elementry geometry solution. I should have payed more attention to my geometry teacher in high school because I did not work this out using geometry. I did use trigonometry and found a very easy solution that even up to today I have not found published anywhere.
Before I give my solution I would like to give some history on this problem.

In "The Mathematical Gazette" Vol. 11, No. 160, Oct. 1922, p. 173 Prof. E.M. Langley purposed the following problem(#644);
ABC is an isosceles triangle. B=C=80 degree. CF at 30 degree to AC cuts AB in F. BE at 20 degree to AB cuts AC in E. Prove angle BEF=30 degree.

In Vol. 11, No. 164, May 1923 of the Gazette, pp 321-323 they give the solution. I should say solutions(6 of them). One of the solutions(the shortest) to this original problem goes like this;
FA/AC=sin30/sin130=1/(2 cos 40)=sin 40/sin 80
=sin BEC/sin ECB=CB/CE=FB/BE;
and angle FAC=FBE;
therefore triangles FAC,FBE are similar, and angle BEF=angle ACF=30 degree.
The Editor noted that "So many solutions of the Problem #644 have been sent in "

This type of problem is known by several different names(do an internet search) one would be Langley's problem another is Adventitious Angles(or Triangles). Nikos Dergiades in 1992 has found 53 Langley's type cases(where all angles are whole degrees). Tom Rike has a interesting paper that was given at the Berkeley Math Circle in 2002 .

My solution in 1962 to this problem goes like this;
Let line A-B be an east-west line and the distance between A&B can be anything, say unit length(1 unit). Assign a Rectangular Coordinate system to the problem and let point A be Zero North and Zero East. Calculate the coordinates of poind B(dah , Zero north and 1.000... east). Solve for the Apex angle's in the triangles ADB and AEB ( another simple mental calculation). Now using either the bearing - bearing (azimuth-azimuth) intersection formulas solve for the coordinates of point D and E ( since I don't remember the formulas I just used the Law of Signs on each triangle). By immediate inspection AD=1*sin 60/sin 40; BE=1*sin 70/sin 30 (please note that you must use the full potential of your calculator, 10 places at least so you will get an accurate answer). Once you have the coordinates on point D and E  do an inverse between the two. The calculated azimuth from point D to point E is 40 degrees (exactly), that's if you did everything correctly. Now subtract from the azimuth D to E (40 degrees)  the azimuth A to E (azimuth A to E is 20 degree) and you get the answer of 20 degree exactly.

I don't see a spell check here so I hope all is OK. If not, live with it.

John Nolton
Tombstone, AZ

PS.  some books that have a Langley problem in them. 1. Work This One Out, by Longley-Cook (1 solution). 2. Challenging Problems in Geometry, by Posamentier and Salkind(6 solutions). 3. Trigonometric Novelties, by Ransom(3 solutions). 4. Mathematical Quickies, by Trigg( 1 solution). 5. The Ticket to Heaven and other Puzzles, by Sole(1 solution).
6. Mathematical Gems II by Honsberger. 7. Mathematical Chestnuts from Around the World by Honsberger. 8. Geometry Revisted by Coxeter and Greitzer(1 solution). 9. 100 Mathematical Curiosities by Ransom. 
  Sunday, October 23, 2011 at 7:09:49 PM

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