# What are the coordinates of the points a and b and minimize the length of the hypotenuse of a right triangle that is formed in the first quadrant by the x-axis, the y-axis, and a line through the point (1,2) where point a is at (0,y) and point b is at (x,0)?

Feb 28, 2015

$\left({x}_{0} , {y}_{0}\right) = \left(1 , 2\right)$

Suppose that the base point along the Y-axis is some scaled factor, $s$, of ${y}_{0}$ beyond ${y}_{0} \setminus$

That is the point on the Y-axis (0,y_a) is at
$\left(0 , {y}_{0} + s \cdot \left({y}_{0}\right)\right)$
or $\left(0 , 2 + 2 s\right)$ in our particular case

The point on the X-axis (x_b, 0) must satisfy the equivalent slope ratios:

${x}_{b} / \left(2 + 2 s\right) = \frac{1}{2 s}$

$\rightarrow$ ${x}_{b} = 1 + \frac{1}{s}$

The square of the ladder length based on the scaling factor can be expressed as
${L}^{2} \left(s\right) = {\left({x}_{b}\right)}^{2} + {\left({y}_{a}\right)}^{2}$
$= {\left(1 + \frac{1}{s}\right)}^{2} + {\left(2 + 2 s\right)}^{2}$

which simplifies to
${L}^{2} \left(s\right) = {s}^{- 2} + 2 {s}^{- 1} + 5 + 8 s + 4 {s}^{2}$

We want to minimize $L \left(s\right)$
Since $L \left(s\right)$ is required to go through $\left(2 , 1\right)$, $L \left(s\right)$ must be greater than $1$
and we can simplify our effort with the same results by minimizing ${L}^{2} \left(s\right)$

$\frac{d \left({L}^{2} \left(s\right)\right)}{\mathrm{ds}}$
 = ((-2)s^(-3) + (-2)s^(-2) + (8 + 8s)
$= \frac{- 2}{s} ^ 3 + \frac{- 2}{s} ^ 2 + 8 s + 8$

Setting the derivative to zero to find the minimum length:
$\frac{d \left({L}^{2} \left(s\right)\right)}{\mathrm{ds}} = 0$ gives
$\frac{- 2}{s} ^ 3 + \frac{- 2}{s} ^ 2 + 8 s + 8 = 0$

multiplying by $\left({s}^{3}\right)$ [valid since s can not be $0$]
$- 2 - 2 s + 8 {s}^{4} + 8 {s}^{3} = 0$

factoring out $\left(s + 1\right)$ [again valid since $s$ can not be $- 1$]
$\left(- 2\right) \left(s + 1\right) + \left(s + 1\right) \left(8 {s}^{3}\right) = 0$
$\left(- 2\right) + 8 {s}^{3} = 0$
$8 {s}^{3} = 2$
${s}^{3} = \frac{2}{{2}^{3}} = \frac{1}{{2}^{2}}$
$s = \frac{1}{4} ^ \left(\frac{1}{3}\right)$

Recalling that
${y}_{a} = 2 + 2 s$ and ${x}_{b} = 1 + \frac{1}{s}$

we have
${y}_{a} = 2 + \frac{2}{4} ^ \left(\frac{1}{3}\right)$
$= 3.26$ (approx.)
and
${x}_{b} = 1 + {4}^{\frac{1}{3}}$
$= 2.59$ (approx.)

No sense stopping now.
From our earlier formula
${L}^{2} \left(s\right) = {s}^{- 2} + 2 {s}^{- 1} + 5 + 8 s + 4 {s}^{2}$

${L}^{2} \left(s = \frac{1}{4} ^ \left(\frac{1}{3}\right)\right) = {\left(\frac{1}{4} ^ \left(\frac{1}{3}\right)\right)}^{- 2} + 2 {\left(\frac{1}{4} ^ \left(\frac{1}{3}\right)\right)}^{- 1} + 5 + 8 \left(\frac{1}{4} ^ \left(\frac{1}{3}\right)\right) + 4 {\left(\frac{1}{4} ^ \left(\frac{1}{3}\right)\right)}^{2}$

${L}^{2} \left(s = \frac{1}{4} ^ \left(\frac{1}{3}\right)\right) = 17.32$ (approx.)

$L \left(s = \frac{1}{4} ^ \left(\frac{1}{3}\right)\right) = 4.16$ (approx.)

Congratulations to anyone who made it this far!
I hope there is a simpler method, but I couldn't find it.
I also hope that this is somewhere approaching correct and I didn't mess up.