Interesting Math Problem

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

Errol Groff

Instructor, Machine Tool Department

H.H. Ellis Technical High School

643 Upper Maple Street Dantieson, CT 06239

New England Model Engineering Society

I was able to do it on my cad program and that took some nudging around. I still can't think of a trig answer for it?

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:

almost any height you want you didnt specify how far awat the bottom is.

Doug

HEH! I did the same thing and did finish, coming up with the amazing answer:

X = ... . . . . . . . . . X !!!

I was too annoyed and bored to try again.

Bob

Well, no. The bottom can only be a fixed distance away. Closer and the top won't touch the building because the cube is holding it away. Further away and it won't touch the cube. Try it with the building and cube drawn on a piece of paper and use a ruler as the ladder.

Bob

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 are: 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 are: real: 92.124028 imaginary: 0.000000 real: 6.252440 imaginary: 0.000000 Base and height of large triangle are:

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

One approach is use of the Pythagoras theorem and ratios of sides of similar triangles. No trig is necessary, just simultaneous equations. X = 116.124

True, CAD will get you the answer but it's just so... inelegant. There's a big difference between getting the answer and solving the problem. Kind of like the difference between buying and making a needed part.

Regards, Marv

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

I did a back of the napkin approximation:

Since 24" is only 1/5 of 120", I figured that I could ignore the difference between the actual radius and the vertical height to get the horizontal length of the large triangle. Since the horizontal distance of the triangle that is 4/5 the size of the real one, the real one would be: 5/4 x 24" = 30"

Solving the python theorem:

120^2 = H^2 + 30^2

H = 116.19

which is not a bad approximation.

Bob

Solve for n.

| |\ | \ | \5+n y| \ | \ |__2__\ | |\ 2| 2| \5-n |_____|__\___ 2 x

Using the Pythagorean theorem:

x^2 = (5-n)^2 - 4 = n^2 - 10n + 21 y^2 = (5+n)^2 - 4 = n^2 + 10n + 21

Using rule of similar triangles:

y/2 = 2/x yx = 4 y^2 * x^2 = 16 (n^2 - 10n + 21) * (n^2 + 10n + 21) = 16 n^4 - 58n^2 + 425 = 0

Solving quadratic equation:

ax^2 + bx + c = 0

x = (-b + Sqrt(b^2 - 4ac))/2a = 49.396

x = (-b - Sqrt(b^2 - 4ac))/2a = 8.604

x = n^2

n = srqt(49.396) or n = sqrt(8.604)

n = 7.0282287953651594252678749478686 (Damn, this can't be right!)

or

n = 2.9332575747792760110661745961244 (Hmmm, maybe)

y^2 = (5+n)^2 - 2^2

y^2 = 7.9332575747792760110661745961244^2 -4

y^2 = 58.936575747792760110661745961237

y = 7.6770160705701769496450099398666

y+2 = 9.6770160705701769496450099398666 ft.

y+2 = 116.1241928468421233957401192784 in.

What a headache.

Wow, I am blown away by the math skills in this group. Marv Klotz's solution is so far beyond my humble math ability that I am embarassed to post the way that I solved the darn thing.

OTOH I got the same answer as the math whizzes so perhaps I am not so dumb after all.

The author of the book gives the answer as 9.6771 feet. I got

9.677002 so I fugure that is probably close enough for government work. Plus there has been so much rain around here lately I think that my ladder sunk into the ground a bit more than his did.

I have posted my answer in the

The Excel file has not shown up yet so I have also posted the files at

My deepest respect to all you math whizzes who obviously paid more attention in high school or college algebra than I did.

Errol Groff

Instructor, Machine Tool Department

H.H. Ellis Technical High School

643 Upper Maple Street Dantieson, CT 06239

New England Model Engineering Society

Errol, thanks for posting this. I used to be a bit of a math whiz but that was years ago and although I did a lot of math-type thinking during my 20 year electronic engineering career, I didn't get to do much actual math. I really find that I enjoy it. I wish more problems like this showed up. Some time ago I posted a shop math problem I encountered, which I found to be fun and challenging. For example, I showed it to my dad's old manager from Boeing, who once got a PhD from MIT in mathematics, and he couldn't solve it. Admittedly, he's been retired a long time now. Lots of guys on this NG solved it within an hour of its being posted. That one is still available at:

if you want to go see it.

Grant Erwin

Errol, you can do that in one line of your spreadsheet by letting Excel's solver do the guessing. ( I tried it to make sure.) Add a cell, let's say N4, that is the difference between the two angles, which would be H4 - I4 in your sheet. Then have Solver adjust cell A4 (wall height) until the number in N4 is 0. Powie, wall height =

116.124028 when Solver accepted a difference between the angles of

-1.11905E-07 (about 0.1 micro degrees) as "close enough". I started with an initial value in A4 of 119.9. It helps to add a constraint that B4 must be

Speaking of acceptable accuracy, when you're talking about a ladder on a wall 3" down from the 120" mark would almost certainly be close enough :-)

GWE

I used Mathcad:

Given (which is a keyword in Mathcad)

sqrt(y^2 + 24^2) + sqrt(x^2 + 24^2) = 120 sqrt( (y+24)^2 + (x+24)^2 ) = 120

Find (y,x)=

92.125, 6.251 which means the ladder touches the wall at 116.125 inches height (and comes out 30.251 inches from the wall).

and also the interchanged solution which you can tell by the drawing is wrong, and two negative solutions which are just theoretical, since Mathcad wouldn't accept x>0, y>0 as constraints for some reason.

with y being the distance from the top edge of the cube to where the ladder touches, and x being the distance from the right edge of the right edge to the ladder.

Doug

Interesting solution. However, using your 2 values of n and substituting into the hypotenuse of the upper small triangle one gets:

5+n = either 12 or 8 (approximately) which is longer than the total ladder. Using the negative roots for n gives the lower triangle a longer hypotenuse that the total ladder.

How is this possible?

Art

| |\ | \ | \5+n y| \ | \ |__2__\ | |\ 2| 2| \5-n |_____|__\___ 2 x

n = 7.0282287953651594252678749478686 (Damn, this can't be right!) or n = 2.9332575747792760110661745961244 (Hmmm, maybe)