Buildings and ladders, how far apart?

[Fusion_power]

There are two buildings along the side of a street. These buildings are separated by an unknown distance. Each building has a ladder that goes from the base of one building to the wall of the other. One ladder is 60 feet long and the other ladder is 45 feet long. Where the two ladders cross there is a third ladder that drops straight to the ground. It is 15 feet long.

How far apart are the buildings?

Fusion

[Wacky]

Are the ladders straight? Do the buildings have straight vertical walls?
Without such restrictions, there is no unique solution.

[Fusion_power]

The ladders are presumed to be straight, the walls are presumed to be vertical, the buildings are presumed to meet the ground at true 90 degree angles.

the unknowns are the distance separating the two buildings and the height each ladder reaches on the wall.

[c00ler]

The distance is approximately 39.0493163163478 feet.
It's non-analytical brute-force answer.

Later I'll calculate the real value of the distance, and I'll post "programmer's" and "mathematician's" solutions.

[Jwb52z]

Am I wrong in thinking that you could do this problem with geometry?

[Fusion_power]

Cooler is correct, it is a bit over 39 feet. I would be interested in seeing his math.

A working knowledge of both trigonometry and geometry helps to solve it.

[c00ler]

I'm posting solutions

!WARNING! if you are far from math skip mathemetician's solution

L-distance between buildings
a-length of the ladder #1 (45 ft)
b-length of the ladder #2 (60 ft)
h-length of the ladder #3 (15 ft)
if a=b then L=sqrt(a^2-4*h^2) else a<>b

L is divided into 2 parts by the ladder #3
Parts are L*(h/sqrt(a^2-L^2)) and L*(h/sqrt(b^2-L^2)) ft long.(It's simple geometry)
So L=L*h/sqrt(a^2-L^2)+L*h/(b^2-L^2)
or 1/h=1/sqrt(a^2-L^2)+1/(b^2-L^2)=ff(L)

Programmer writes a short program solving ff(L)=1/h and gets L~=39.0493163163478 ft

Mathematicial solves it like that:
Substitute y=1/h, A=a^2, B=b^2, x=L^2
y=1/sqrt(A-x)+1/sqrt(B-x)
Square both parts
y^2=1/(A-x)+2/(sqrt(A-x)*sqrt(B-x))+1/(B-x)
Substitute Y=y^2
Y-1/(A-x)-1/(B-x)=2/(sqrt(A-x)*sqrt(B-x))
Square and substitute m=1/(A-x), n=1/(B-x)
Y^2+m^2+n^2-2yn-2ym+2mn=4mn
Y^2+(m-n)^2=2y(m+n)
Square
Y^4+2*Y^2*(m-n)^2+(m-n)^4=4*Y^2*(m+n)^2
Y^4+2*Y^2*(m-n)^2+(m-n)^4=4*Y^2*((m-n)^2+4mn)

Let's look at m-n
m-n=1/(A-x)-1/(B-x)=(B-A)/((A-x)*(B-x))=(B-A)*mn
Substitute k=4/(B-A)
4mn=k*(m-n)

Y^4+2*Y^2*(m-n)^2+(m-n)^4=4*Y^2*(m-n)^2+4*Y^2*k*(m-n)
Substitute t=m-n (t>0)
Y^4-2*Y^2*t^2+t^4=4*Y^2*k*t
(Y^2-t^2)^2=4*Y^2*k*t
Square root
abs(Y^2-t^2)=2*Y*sqrt(kt)

Let's prove Y>t
Y=m+n+2/sqrt(mn)
t=m-n
n>0 so Y>t

Y^2-t^2=2*Y*sqrt(kt)
Substitute z=sqrt(t)
Y^2-z^4=2*Y*sqrt(k)*z
z^4+2*Y*sqrt(k)*z-Y^2=0
Substitute w=2*Y*sqrt(k) (g>0)
z^4+w*z-Y^2=0
F(z)=0
There are only 2 real solutions and only one of them is positive (not very hard to prove)
So F(z)=(z^2+pz+q)*(z^2+rz+n)
but F(z) doesn't contain z^3 so r=-p
F(z)=(z^2+pz+q)*(z^2-pz+n)=z^4+(q+n-p^2)*z^2+p*(n-q)*z+qn
So
q+n=p^2 (1)
p*(n-q)=w (2)
qn=-Y^2 (3)

square (2)
p^2*(n-q)^2=w^2
P^2*((n+q)^2+4*nq)=w^2
With the help of (1),(3)
w^2=p^2*(p^4+4*Y^2)
p^6+4*Y^2*p^2-w^2=0
The lasy substitute p^2=g
g^3+4*Y^2*g-w^2=0
Unsubstitute w^2=4*(Y^2)*k
g^3+4*Y^2*g-4*(Y^2)*k
It has only one real solution(easy to prove)
It is equal to cube_root(u)+cube_root(v) where u,v are the solutions of
u+v=4*(Y^2)*k
uv=-(4*Y^2*(1/3))^3
(easy to check and to prove)

u=2*Y^2*(k+sqrt(k^2+16*(Y^2)/27))
v=2*Y^2*(k-sqrt(k^2+16*(Y^2)/27))
So we know u,v => we know g => we know p
Now we'll find n,q
n=(p^2+w/p)/2
q=(p^2-w/p)/2

(z^2+pz+q)*(z^2-pz+n)=0 has only 2 real solutions (because F(z)=0 has 2 real solutions)
and n<q
So z^2+pz+q=0 has 2 real solutions and z^2-pz+n=0 has no real solutions
now we solve z^2+pz+q=0
and take the greatest solution (one is positive, the other is negative)
z=(-p+sqrt(p*p-4*q))/2;
So we've found z
t=z^2
And t is found
To find x solve 1/(A-x)-1/(B-x)=t
x=A-2/(sqrt(t^2+kt)+t)
And L=sqrt(x)~=39.0493163163478 ft
That's the end of mathematician's solution.

That's mathematician's pseudo programmed solution:
a0:=45; or input(a0)
b0:=60; or input(b0)
if b0<a0 then change b0,a0
y0:=1/15; or input(y0)
a:=a0*a0;
b:=b0*b0;
y:=y0*y0;
k:=4/(b-a);
w:=2*y*sqrt(k);
u:=2*y*y*(k+sqrt(k*k+16*y*y/27));
v:=2*y*y*(k-sqrt(k*k+16*y*y/27));
p:=sqrt(exp(ln(u)*(1/3))-exp(ln(-v)*(1/3)));
q:=(p*p-w/p)/2;
z:=(-p+sqrt(p*p-4*q))/2;
t:=z*z;
x:=a-2/(sqrt(t*t+k*t)+t);
l:=sqrt(x); and output(l)

P.S. I'm not sure if programmer's program is slower than mathematician's one. Can someone calculate about 1000 digit's of answer using both programs and compare speeds of programs?