Walking Under Ladders -- II

Let the two parts of the width be $c$ and $w-c$ from left to right. By similar triangles we get:

$w-c=\dfrac{w}{A} \\ c=\dfrac{w}{B}$

Adding them we get $1=\dfrac{1}{A}+\dfrac{1}{B}$ .

Now, by the Pythagorean Theorem we know that:

$A^2+w^2=4 \implies A=\sqrt{4-w^2} \\ B^2+w^2=9 \implies B=\sqrt{9-w^2}$ .

Substituting them into the first equation we get:

$1=\dfrac{1}{\sqrt{4-w^2}}+\dfrac{1}{\sqrt{9-w^2}}$

After simplifying we get:

$w^8-22w^6+163w^4-454w^2+385=0$

That is a bi-quartic equation in $w^2$ . Using Newton Raphson method we get $w \approx 1.2311$ .

Nice problem, you should also try this problem .

Let width w of alley = x+y where the horizontal h meets ground

Now B^2 +w^2 = 9 and A^2 +w^2 = 4

B^2 - A^2 = 5 ----------------------------(1)

By similar triangles,

A/w= h/y so y= w/A since h=1--------(2)

B/w = h/x so x = w/B since h = 1-------(3)

x+y = w

So w/A+w/B = w -----------------------------from (2), (3)

1/A+1/B= 1

A+B= AB

B= A/(A-1)---------------------------------(4)

Putting (4) into (1)

(A/A-1)^2 - A^2 = 5

A^4-2A^3+5A^2-10A +5 = 0

Solving quartic, A = 1.576

width w = 1.231

Draw a line parallel to the base through the point where the ladders intersect. Divide the base into x and W - x, where x is to the left of the vertical line to the intersection and W - x is to the right . By similar triangles, x/(W - x) = 1/(B - 1), so B = W/x. Also, by similar triangles, x/(W - x) = (A - 1)/1, so A =W/(W - x). Now, A^2 + W^2 = 4 and B^2 + W^2 = 9, so B^2 - A^2 = 5. Substituting for A and B, W^2/x^2 - W^2/(W - x)^2 = 5, or W^4 - 2W^3 x =5x^4 - 10W x^3 + 5x^2 W^2. Transposing and letting x = rW, 5 r^4 W^4 - 10 r^3 W^4 +5 r^2 W^4 +2 r*W^4 - W^4 = 0. Dividing by W^4, 5r^4 -10r^3 +5r^2 + 2r - 1 =0. = 1/r = Solving by Newton-Raphson with an initial guess of r_0 = 0.4, r= .36553405. Then B = W/x = 1/r = 2.735723252. Then W^2 = 9 - B^2 = 1.515818288, and W = 1.231185724.

$For ~the~two~ladder~problem, \dfrac 1 A +\dfrac 1 B= \dfrac 1 h= 1.\\ \implies ~B=\dfrac A {A-1}............(1)\\ Again~ solving~ right~ \angle ed~\Delta s, W^2=a^2 - B^2 =b^2 - A^2.\\ Substituting ~ (1) ~in~ this~we~ get ~a~quartic~ in ~A,\\ A^4-2A^3+(1-A)^2(a^2-b^2)=0.\\ Solving~A=2.7357.\\ \therefore~ W=\sqrt{b^2 - A^2}=.231222 .$