, 2026

Xander's rope is $10 $\text{}$
m}$ long.

Since Yasmin's rope is $n$% longer
than Xander's rope, then the length of Yasmin's rope is $\$10\left(1 + \dfrac{n}{100}\right)\text{}$
m}$.

Since Zhe's rope is $(2n)$% longer
than Yasmin's rope, then the length of Zhe's rope is $\$10\left(1 + \dfrac{n}{100}\right)\left(1 + \dfrac{2n}{100}\right)\text{}$ m}$.

Since Zhe's rope is $(3.14n)$% longer
than Xander's rope, then the length of Zhe's rope can also be written as
$\$10\left(1 + \dfrac{3.14n}{100}\right)\text{}$
m}$.

Therefore, $$\begin{align*} 10\left(1 + \dfrac{n}{100}\right)\left(1 + \dfrac{2n}{100}\right) & = 10\left(1 + \dfrac{3.14n}{100}\right)\\ \left(1 + \dfrac{n}{100}\right)\left(1 + \dfrac{2n}{100}\right) & = \left(1 + \dfrac{3.14n}{100}\right)\\ (100 + n)(100 + 2n)& = 100(100 + 3.14n) & \text{(multiplying by $100 \cdot 100$)}\\ 10000 + 300n + 2n^2 & = 10000 + 314n \\ 2n^2 - 14n & = 0\\ 2n(n-7) & = 0\end{align*}$$ Since $n > 0$, then it must be the case that
$n = 7$.
Solution 1:

Let $BC = x$ and $CD = y$. Join $A$ to $C$.

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Using the cosine law in $\$\triangle$
ABC$, we obtain$$$$\begin{align*} AC^2 & = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC) \\ & = 16 + x^2 - 8x\cos(45\degree) \\ & = 16 + x^2 - 8x\!\left(\frac{1}{\sqrt{2}}\right) \\ & = 16 + x^2 - 4\sqrt{2}x\end{align*}$$ Using the cosine law
in $\triangle ADC$, we obtain $$\begin{align*} AC^2 & = AD^2 + DC^2 - 2(AD)(DC)\cos(\angle ADC) \\ & = 16 + y^2 - 8y\cos(135\degree) \\ & = 16 + y^2 - 8y\!\left(-\frac{1}{\sqrt{2}}\right) \\ & = 16 + y^2 + 4\sqrt{2}y\end{align*}$$$$Equating expressions for \$AC^2\$, we obtain$$$$\begin{align*} 16 + x^2 - 4\sqrt{2}x & = 16 + y^2 + 4\sqrt{2}y \\ x^2 - y^2 - 4\sqrt{2}x - 4\sqrt{2}y & = 0\\ (x+y)(x-y) - 4\sqrt{2}(x+y) & = 0\\ (x+y)(x-y - 4\sqrt{2}) & = 0\end{align*}$$$$$$ Since $x > 0$ and $y > 0$, then $x + y > 0$. Thus, $x - y - 4\sqrt{2} = 0$ and so $BC - CD = x - y = 4\sqrt{2}$.

Solution 2:

Let point $P$ be on $BC$ so that $AP$ is perpendicular to $AB$.

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To see why $P$ is on $BC$ (and not some extension of $BC$) first observe that isosceles $\triangle BAD$ has $\$\angle$ ADB = $\angle$ ABD < $\angle$ ABC =
$45\degree$$, so$$\$\angle$ BAD =
$180\degree - \angle$ ADB - $\angle$ ABD > $180\degree - 45\degree - 45\degree = 90\degree$$$$$ Therefore, $\angle BAD$ is obtuse.

Now suppose $P$ were on some
extension of $BC$. Since $\angle BAD$ is obtuse and $\angle BAP=90\degree$, $AP$ must intersect $CD$ at some point $M$, and so $AM<AP$. However, $AP=AB=4$ since $\triangle BAP$ is is a right-isosceles
triangle, which means in $\$\triangle$
AMD$, we have that$AM$ is
not the longest side while it is opposite obtuse $\angle ADM$. This is impossible, so we
conclude that $P$ must be on $BC$.

It was mentioned above that $\$\triangle$
BAP$is right-angled and isosceles, with$AP = AB = 4$which means that$BP = $\sqrt{2}AB = 4\sqrt{2}\$.$

Since $\angle BPA = 45\degree$ and
$BPC$ is a straight angle, then $\$\angle$ CPA = $180\degree - \angle$ BPA =
$135\degree$$. Therefore,$$\angle$ APC =
$\angle$ ADC$.

Since $AP=AD=4$, then $\triangle APD$ is isosceles, and so $\angle APD = \angle ADP$. Then $$\angle CPD = \angle APC - \angle APD = \angle ADC - \angle ADP = \angle CDP$$ Since $\angle CPD = \angle CDP$, then $\triangle CPD$ is isosceles, and so $CD = CP$. Thus, $BC - CD = BC - CP = BP =
$4\sqrt{2}\$.$