Marvin W. Klotz wrote:

On Sun, 29 May 2005 01:05:13 GMT, Errol Groff wrote:


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

The solution I used to find the answer will be posted as
Ladder_Problem_Answer.PDF

http://metalworking.com/DropBox/

Errol Groff

Instructor, Machine Tool Department

H.H. Ellis Technical High School
643 Upper Maple Street
Dantieson, CT 06239

New England Model Engineering Society
www.neme-s.org



This isn't a trig problem. Rather it's an algebra problem that, surprisingly,
requires the solution to a quartic equation. However, by some clever
selection of variables, the quartic can be separated into two quadratics.

b = side of box
l = length of ladder
y = height of smallest triangle
x = base of medium sized triangle

Then the base of the large triangle is x+b and its height is y+b.

From similar triangles, we have:

y/b = b/x = x*y = b^2 = y = b^2/x (1)

Applying Pythagoras to the large triangle:

(x+b)^2 + (y+b)^2 = l^2

or:

x^2 + 2*b*x + b^2 + y^2 + 2*b*y + b^2 = l^2

Substituting y = b^2/x from (1) yields:

x^2 + 2*b*x + b^2 + b^2/x^2 + 2*b*b^2/x + b^2 = l^2 (2)


Now define:

a = x + b^2/x (3)

so:

a^2 = x^2 + 2*b^2 + b^4/x^2

Then (2) becomes:

a^2 + 2*b*a - l^2 = 0

Solve this quadratic for a.

Now substitute a into (3) and solve the resulting quadratic to find x.
a = x + b^2/x (3)
x^2 - a*x + b^2 = 0

Grinding through the numbers (I used a program I wrote):

Length of ladder [25] ? 120
Side of box [6] ? 24

Solutions to:
+1.0000 * x^2 +48.0000 * x^1 -14400.0000 = 0
a
real: 98.376468 imaginary: 0.000000
real: -146.376468 imaginary: 0.000000

a selected = 98.3765

Solutions to:
+1.0000 * x^2 -98.3765 * x^1 +576.0000 = 0
a
real: 92.124028 imaginary: 0.000000
real: 6.252440 imaginary: 0.000000

Base and height of large triangle a
116.1240, 30.2524 or 30.2524, 116.1240

So the value labeled X in the PDF is 116.124

Regards, Marv

Home Shop Freeware - Tools for People Who Build Things
http://www.geocities.com/mklotz.geo


I just wrote equations and combined until I got:

x^2 + 576/(1-24/x)^2 = 14400 and I solved that numerically to get x = 116.124"

but I didn't see any elegant way to get this algebraically.

GWE