1. ## Crazy puzzle

After spending several hours on this one, I've yet to come up with anything that really led anywhere. I wonder what I'm missing?
I don't have the original problem statement anymore, only the diagram I drew of it, so I'll try and describe the problem as best I can from that.

I've attached a simple diagram as well with the names I gave to all the values.

Point L is due north of Point A
Point B is due east of Point A
Point C is due west of Point A
The distance from Point A to Point B to Point L is twice the distance from Point A to Point L
The distance from Point A to Point C to Point L is three times the distance from Point A to Point L
The distance from Point B to Point C is 39 km more than the distance from Point A to Point L
What is the distance between points A and L?

Using the names on the attached diagram, these are the equations I immediately saw:
$b+h=2x$ (given in problem)
$c+k=3x$ (given in problem)
$b+c=x+39$ (given in problem)
$x^2=h^2-b^2=k^2-c^2$ (pythagorean theorem)

But this still leaves us with four equations and five unknowns. My usual routine at this point for complicated systems of equations is to write down as many equations as possible and see which ones might be useful.
So I decided to get the angles involved as well. Let the capital letters denote the angle at their position (so capital B refers to the angle between b and h, capital C refers to the angle between c and k, and capital L refers to the angle between h and k).
$B+C+L=180$
(sum of angles around the large triangle is 180)

$arcsin(b/h)+arcsin(c/k)=L$
(sum of the two smaller angles at L equals the larger angle at L)

$C+arcsin(c/k)=B+arcsin(b/h)=90$
(sum of the angles around each of the small triangles is 180)

$h/sin(C)=k/sin(B)=(b+c)/sin(L)$
(law of sines for the large triangle)

$(b+c)^2=h^2+k^2-2hk*cos(L)$
(law of cosines for the large triangle - We could easily substitute x+39 for b+c from the first set of equations; not sure which form is more useful)

Well, it looks like I've now got 9 equations, and 9 unknowns after introducing the angles into the problem.
But I still can't figure out what substitutions to perform in order to eliminate just one variable. Am I missing something, or is it just so complex that I'm just getting lost? HOW DO YOU DO THIS!?

Will

2. Originally Posted by Fox2013
$b+h=2x$ (given in problem)
$c+k=3x$ (given in problem)
$b+c=x+39$ (given in problem)
$x^2=h^2-b^2=k^2-c^2$ (pythagorean theorem)
In fact, you have five equations (count the = signs!).

Step 1. Enter the equations in WolframAlpha:
b+h=2*x, c+k=3*x, b+c=x+39, x^2=h^2-b^2, x^2=k^2-c^2
Step 2. Eliminate the solutions with x = 0.

Step 3. Profit!

3. Originally Posted by Fox2013
After spending several hours on this one, I've yet to come up with anything that really led anywhere. I wonder what I'm missing?
I don't have the original problem statement anymore, only the diagram I drew of it, so I'll try and describe the problem as best I can from that.

I've attached a simple diagram as well with the names I gave to all the values.

Point L is due north of Point A
Point B is due east of Point A
Point C is due west of Point A
The distance from Point A to Point B to Point L is twice the distance from Point A to Point L
The distance from Point A to Point C to Point L is three times the distance from Point A to Point L
The distance from Point B to Point C is 39 km more than the distance from Point A to Point L
What is the distance between points A and L?

Using the names on the attached diagram, these are the equations I immediately saw:
$b+h=2x$ (given in problem)
$c+k=3x$ (given in problem)
$b+c=x+39$ (given in problem)
$x^2=h^2-b^2=k^2-c^2$ (pythagorean theorem)

But this still leaves us with four equations and five unknowns. My usual routine at this point for complicated systems of equations is to write down as many equations as possible and see which ones might be useful.
So I decided to get the angles involved as well. Let the capital letters denote the angle at their position (so capital B refers to the angle between b and h, capital C refers to the angle between c and k, and capital L refers to the angle between h and k).
$B+C+L=180$
(sum of angles around the large triangle is 180)

$arcsin(b/h)+arcsin(c/k)=L$
(sum of the two smaller angles at L equals the larger angle at L)

$C+arcsin(c/k)=B+arcsin(b/h)=90$
(sum of the angles around each of the small triangles is 180)

$h/sin(C)=k/sin(B)=(b+c)/sin(L)$
(law of sines for the large triangle)

$(b+c)^2=h^2+k^2-2hk*cos(L)$
(law of cosines for the large triangle - We could easily substitute x+39 for b+c from the first set of equations; not sure which form is more useful)

Well, it looks like I've now got 9 equations, and 9 unknowns after introducing the angles into the problem.
But I still can't figure out what substitutions to perform in order to eliminate just one variable. Am I missing something, or is it just so complex that I'm just getting lost? HOW DO YOU DO THIS!?

Will

$b+h=2x\Rightarrow\ b=2x-h$

$c+k=3x\Rightarrow\ c=3x-k$

$b+c=x+39$

$x^2=h^2-b^2$

$x^2=k^2-c^2$

you can write x in terms of k and x in terms of h by using the equations for b and c
and substituting them into the Pythagoras theorem equations.

$x^2=k^2-(3x-k)^2\Rightarrow\ x^2=k^2-\left[9x^2-6xk+k^2\right]$

$\Rightarrow\ x^2=6xk-9x^2\Rightarrow\ 10x^2=6xk\Rightarrow\ 10x=6k$

$\Rightarrow\ x=\frac{6k}{10}$

Similarly $x^2=h^2-b^2=h^2-(2x-h)^2$

leads to $x=\frac{4h}{5}$

Hence we can write h in terms of k and vice versa

Then $b+h=2x$ and $c+k=3x$ become

$b+h=2x$ and $c+\frac{4h}{3}=3x$

Hence adding these gives $b+c+\frac{7h}{3}=5x$

Then we can solve for x using $b+c=x+39$

$b+c=x+39$

$b+c=5x-\frac{7}{3}h=5x-\frac{7}{3}\ \frac{5}{4}x$

Solving for x gives x=36.