Fun Math Problem 'Cussed Ladders'
09-16-2017, 05:37 AM (This post was last modified: 09-16-2017 07:06 AM by Gamo.)
Post: #1
 Gamo Senior Member Posts: 351 Joined: Dec 2016
Fun Math Problem 'Cussed Ladders'
See this math problem on Popular Scientific magazine look interesting.

Here is the problem:

You were told of two ladders leaning opposite ways between two buildings, one
touching 8 feet above the ground, the other 12. At what height above the ground would they cross?

Which way you will solve for this problem?

Gamo

Attached File(s) Thumbnail(s)

09-16-2017, 06:15 AM (This post was last modified: 09-16-2017 06:19 AM by Paul Dale.)
Post: #2
 Paul Dale Senior Member Posts: 1,404 Joined: Dec 2013
RE: Fun Math Problem 'Cussed Ladders'
At a guess, apply Pythagoras's theorem twice using the edge verticals and the full width across the bottom, call it w. The hypotenuses of these triangles are $$\sqrt {64 + w^2}$$ and $$\sqrt {144 + w^2 }$$.

Look at the similar triangles down from the point x. The sum of their bases is w = a + b. I think this is sufficient information to solve it.

Pauli
09-16-2017, 07:36 AM (This post was last modified: 09-16-2017 07:37 AM by Csaba Tizedes.)
Post: #3
 Csaba Tizedes Member Posts: 287 Joined: May 2014
RE: Fun Math Problem 'Cussed Ladders'
The crossing height is:

$$y=\frac{h_1 · h_2}{h_1+h_2}$$, where $$h_1=8$$ and $$h_2=12$$, therefore $$y=4.8$$

Hint: I wrote the equations of two lines goes through $$(0, h_1) (w, 0)$$ and $$(0, 0) (w, h_2)$$ and I find the intersection. Because I hate geometry - but I like equations...
$$w$$ is the width/distance between the walls

Csaba
09-16-2017, 07:43 AM
Post: #4
 Paul Dale Senior Member Posts: 1,404 Joined: Dec 2013
RE: Fun Math Problem 'Cussed Ladders'
(09-16-2017 07:36 AM)Csaba Tizedes Wrote:  The crossing height is:
......

This is a better solution

Pauli
09-16-2017, 07:52 AM
Post: #5
 Arno K Senior Member Posts: 430 Joined: Mar 2015
RE: Fun Math Problem 'Cussed Ladders'
As this is a geometry problem that usually arises when the Intercept Theorem is dealt with at school I use this.
Arno
09-16-2017, 08:26 AM (This post was last modified: 09-16-2017 12:06 PM by Gamo.)
Post: #6
 Gamo Senior Member Posts: 351 Joined: Dec 2016
RE: Fun Math Problem 'Cussed Ladders'
According to Csaba the answer is wrong and the answer is 4.8 feet. Ha ha ha just kidding Csaba your answer is right. Your equation is very clever way to do it.

According to the article in the magazine said Maybe the simplest way is to lable the distance on the ground from each building to line X as a and b, which allows you to set up two simple ratios based on similar triangles:

12/(a+b)=x/a and 8/(a+b)=x/b

Multiplying both equations through and substituting for the common term yields
what may have been intuitively obvious: b=3/2 a. Now you can rewrite the above equations using 3/2 a in place of b and 5/2 a in place of a+b. Solving both equations for X (the a's cancel) gives the height above the ground at which the ladders intersect: 4.8 feet.

Now you are ready to try the second variation of the cussed ladders. This time, ladder lengths are given as 40 and 30 feet, and they cross at a point 10 feet from the ground. How far apart are the buildings against which the ladders are leaning?
Because in this puzzle you have three dimensions given instead of two, it may at first glance look to be the easier problem. It is not. This one will require that you use your calculator--as well as your head.

Gamo

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09-16-2017, 10:17 AM (This post was last modified: 09-16-2017 01:57 PM by Gilles59.)
Post: #7
 Gilles59 Member Posts: 136 Joined: Jan 2017
RE: Fun Math Problem 'Cussed Ladders'
(09-16-2017 08:26 AM)Gamo Wrote:  Because in this puzzle you have three dimensions given instead of two, it may at first glance look to be the easier problem. It is not. This one will require that you use your calculator--as well as your head.
Gamo

EDIT : OK, there was a contradiction between the picture and the text in the problem 2. Now corrected
09-16-2017, 11:22 AM
Post: #8
 ggauny@live.fr Senior Member Posts: 462 Joined: Nov 2014
RE: Fun Math Problem 'Cussed Ladders'
Hi,

May be you see attachement for a complete explanation.

Attached File(s)
deuxechellesdansuncouloir.pdf (Size: 106.15 KB / Downloads: 46)

Gérard.
09-16-2017, 11:46 AM (This post was last modified: 09-16-2017 11:50 AM by Gamo.)
Post: #9
 Gamo Senior Member Posts: 351 Joined: Dec 2016
RE: Fun Math Problem 'Cussed Ladders'
Sorry for the second picture that was an approximate drawing to virtualize in pictures.

Thank You ggauny@live.fr

Your pdf explain very well. Look like you use the Ratio Comparison technic.

For the second problem I do not have the answer because the article didn't provide any steps and answer.

Anyone got some other ways or idea are very welcome to share with other here.

Thank You

Gamo
09-16-2017, 11:49 AM (This post was last modified: 09-16-2017 12:11 PM by Gerson W. Barbosa.)
Post: #10
 Gerson W. Barbosa Senior Member Posts: 1,073 Joined: Dec 2013
RE: Fun Math Problem 'Cussed Ladders'
(09-16-2017 08:26 AM)Gamo Wrote:  Now you are ready to try the second variation of the cussed ladders. This time, ladder lengths are given as 40 and 30 feet, and they cross at a point 10 feet from the ground. How far apart are the buildings against which the ladders are leaning?
Because in this puzzle you have three dimensions given instead of two, it may at first glance look to be the easier problem. It is not. This one will require that you use your calculator--as well as your head.

Done that before, except that then the height was 15 ft instead of 10 ft. After this change, I find the new distance is now 26.0328775442 ft (previously 15.9876490088 ft).

Old thread on this subject .
09-16-2017, 11:54 AM (This post was last modified: 09-16-2017 12:09 PM by Gilles59.)
Post: #11
 Gilles59 Member Posts: 136 Joined: Jan 2017
RE: Fun Math Problem 'Cussed Ladders'
(09-16-2017 11:46 AM)Gamo Wrote:  Sorry for the second picture that was an approximate drawing to virtualize in pictures.

For the second problem I do not have the answer because the article didn't provide any steps and answer.

Anyone got some other ways or idea very welcome to share with other here.

Thank You

Gamo

Hi Gamo, I think that what seems wrong in the second picture is that the lengths 40 and 30 feet have to refer to the length of the ladders and not like in the first problem to the touching point of the ladder on the wall.
09-16-2017, 12:10 PM
Post: #12
 Gamo Senior Member Posts: 351 Joined: Dec 2016
RE: Fun Math Problem 'Cussed Ladders'
Hello Gilles59

Thanks to let me know about the error on the picture. The length is suppose to be of the length of the ladders itself not the height of each side of the building where the ladder touch.

I make the updated of the picture.

Thanks

Gamo
09-16-2017, 05:18 PM (This post was last modified: 09-16-2017 05:51 PM by Gerson W. Barbosa.)
Post: #13
 Gerson W. Barbosa Senior Member Posts: 1,073 Joined: Dec 2013
RE: Fun Math Problem 'Cussed Ladders'
For the second problem, just submit solve [(c+d)/(√(d*(d+2*c)))=b/(√((a*(b-c-d)/b-c)*((a*(b-c-d)/b-c)+2*c))+√(d*(d+2*c)))] for d to W|A. You'll get four very long expressions for d. One of them might give you a suitable real answer (make a = 40, b = 30 and c = 10). Use that result to compute e = (a*(b-c-d)/b-c). Your final answer will be √(d*(d+2*c)) + √(e*(e+2*c)). I think I won't check this out :-)

http://m.wolframalpha.com/input/?i=solve...9%5D+for+d
09-16-2017, 06:50 PM
Post: #14
 pier4r Senior Member Posts: 1,737 Joined: Nov 2014
RE: Fun Math Problem 'Cussed Ladders'
(09-16-2017 05:18 PM)Gerson W. Barbosa Wrote:  http://m.wolframalpha.com/input/?i=solve...9%5D+for+d

I guess you broke it.

Wikis are great, Contribute :)
09-17-2017, 12:22 AM
Post: #15
 Gerson W. Barbosa Senior Member Posts: 1,073 Joined: Dec 2013
RE: Fun Math Problem 'Cussed Ladders'
(09-16-2017 06:50 PM)pier4r Wrote:
(09-16-2017 05:18 PM)Gerson W. Barbosa Wrote:  http://m.wolframalpha.com/input/?i=solve...9%5D+for+d

I guess you broke it.

That's a convoluted solution, not to be taken seriously. RMollov's solution in the 2013 thread gives the distance in only one step, but its closed form version is also lengthy:

http://m.wolframalpha.com/input/?i=solve...+%5D+for+x
09-17-2017, 10:37 AM
Post: #16
 ggauny@live.fr Senior Member Posts: 462 Joined: Nov 2014
RE: Fun Math Problem 'Cussed Ladders'
hi,

Sir Gamo

I am not sure understand what you tell me. May be it is when we know the hight intersection.

May be you see the attachement new.

Attached File(s) Thumbnail(s)

Gérard.
09-17-2017, 06:50 PM
Post: #17
 Vtile Senior Member Posts: 376 Joined: Oct 2015
RE: Fun Math Problem 'Cussed Ladders'
8 1/x
12 1/x
+
1/x
==== X
4.8
09-19-2017, 10:04 AM
Post: #18
 AlexFekken Member Posts: 151 Joined: May 2016
RE: Fun Math Problem 'Cussed Ladders'
(09-17-2017 12:22 AM)Gerson W. Barbosa Wrote:  That's a convoluted solution, not to be taken seriously. RMollov's solution in the 2013 thread gives the distance in only one step, but its closed form version is also lengthy:

Why wouldn't you take that seriously? You get a quartic in x^2 and solving that symbolically requires using the solution of a cubic resolvent auxiliary equation (hence the cube roots). Of course a numerical solution and using explicit values for all 3 parameters avoids all that. I see no inconsistency.
09-19-2017, 03:54 PM (This post was last modified: 09-19-2017 04:00 PM by Gerson W. Barbosa.)
Post: #19
 Gerson W. Barbosa Senior Member Posts: 1,073 Joined: Dec 2013
RE: Fun Math Problem 'Cussed Ladders'
(09-19-2017 10:04 AM)AlexFekken Wrote:
(09-17-2017 12:22 AM)Gerson W. Barbosa Wrote:  That's a convoluted solution, not to be taken seriously. RMollov's solution in the 2013 thread gives the distance in only one step, but its closed form version is also lengthy:

Why wouldn't you take that seriously? You get a quartic in x^2 and solving that symbolically requires using the solution of a cubic resolvent auxiliary equation (hence the cube roots). Of course a numerical solution and using explicit values for all 3 parameters avoids all that. I see no inconsistency.

The problem is mine is not a straightforward solution. If a quartic equation has to be solved, then it's better to choose RMollov's, which gives the answer we are looking for without further calculation. I don't believe my final expression, after fully simplified, would be any shorter than the one obtained through the other quartic equation.
For practical purposes it's better to use a numerical solver than those beautiful (and lengthy) exact solutions.

That's what I did then:

http://farm3.staticflickr.com/2860/11828...6eef_o.jpg

A somewhat winding road towards the desired destination.
09-19-2017, 10:19 PM
Post: #20
 AlexFekken Member Posts: 151 Joined: May 2016
RE: Fun Math Problem 'Cussed Ladders'
(09-19-2017 03:54 PM)Gerson W. Barbosa Wrote:  For practical purposes it's better to use a numerical solver than those beautiful (and lengthy) exact solutions.

If this was about actual ladders then I would agree.

But the aim of a puzzle is to stretch mind and in that scenario I prefer, and get a lot more satisfaction from, an exact solution (or even just finding out if an exact solution is possible) and preferably finding it without the help of a machine.
Otherwise you could just get some (usually obvious) lower and upper bounds for the solution and use a generic numerical solver after writing out the equation to solve (in this case basically 1/h1 + 1/h2 = 1/h, which we already had from the first puzzle). Now where's the fun in that?

And in this particular case there was also the explicit challenge: "This one will require that you use your calculator--as well as your head". Not true... :-)
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