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

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

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 07:36 AM)Csaba Tizedes Wrote: [ -> ]The crossing height is:

......

This is a better solution

Pauli

As this is a geometry problem that usually arises when the Intercept Theorem is dealt with at school I use this.

Arno

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

(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

Hi,

May be you see attachement for a complete explanation.

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 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 11:46 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.

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

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
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.

8 1/x

12 1/x

+

1/x

==== X

4.8

(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 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 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... :-)