"Errol Groff" wrote in message
...
Ladder_Problem.PDF is an interesting math problem I found in the book
Machine Shop Trade Secrets by James A Harvey.

Simple similar triangles problem. Well, not perfectly simple, but not
impossible.


















Let the top point be A, lower left-hand corner C, far right corner E, and
the middle left point B, middle bottom point D and middle hypotenuse, F.
A
|\
B-F
| |\
C-D-E

Line segments BC, CD, DF and BF are given as 24.000 inches (I had no idea
ladders could be positioned so exactly). Segment AE is given as 120". Let
AC = X. Given the straight lines, parallel lines and right angles, find X.

There are three basic right triangles defined, ACE, ABF and FDE. We know
the base of ABF is 24", the left side of FDE is 24" and the hypotenuse of
ACE is 120".

BF = 24 DF = 24 X = 24 + AB
AB = X - 24 DE = CE - 24 CE = 24 + DE
AF = 120 - EF EF = 120 - AF AE = 120

BC = BF = CD = DF = 24

They are similar triangles because of the equal angles, thus:
AB/X = AF/AE = CD/CE (top over whole)
DF/X = EF/AE = DE/CE (bottom-right over whole) and
AB/DF = AF/EF = CD/DE (top over bottom-right)

Now substitute like an M.F....

(X-24)/X = (120-EF)/120 = 24/CE
24/X = (120-AF)/120 = (CE-24)/CE
(X-24)/24 = (120-EF)/EF = 24/DE

Split the parenthesis:

1 - 24/X = 1 - EF/120 = 24/CE
24/X = 1 - AF/120 = 1 - 24/CE
X/24 - 1 = 120/EF - 1 = 24/DE

(Bottom left relation)
X/24 = 120/EF = X = 120/24*EF = X = 120/24*(120 - AF)

(Center by center right relation)
-1\\ + AF/120 = -1\\ + 24/CE = AF = 24*120/CE

X = 120\\\/24*(120\\\ - 24*120\\\/CE) = 1/(24 - 24*24/CE)
= 1/[24 - 24*24/(24+DE)] = [1/1] / [24*(24+DE) - 24*24]/(24+DE)
= (24 + DE) / 24*24\\\\\ + 24*DE - 24*24\\\\\
= 24/24*DE + DE/24*DE = X = 1/DE + 1/24

Um, and so on... it seems I'm too tired to finish this tonight.

Tim

--
"California is the breakfast state: fruits, nuts and flakes."
Website: http://webpages.charter.net/dawill/tmoranwms