, 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,$$\text{10 (1 + n 100 ) (1 + 2n 100 ) = 10 (1 + 3.14n 100 ) (1 + n 100 ) (1 + 2n 100 ) = (1 + 3.14n 100 ) (100 + n)(100 + 2n) = 100(100 + 3.14n) (multiplying by}$100
$\cdot$ 100$\text{) 10000 + 300n + 2n 2 = 10000 + 314n 2n 2 - 14n = 0 2n(n-7) = 0}$$Since$n > 0$, then it must be the case that$n = 7$. Solution 1: Let$BC = x$and$CD = y$. Join$A$to$C. [[IMAGE0]] Using the cosine law in

$\triangle$
ABC$, we obtain$$\text{AC 2 = AB 2 + BC 2 - 2(AB)(BC) ( ABC) = 16 + x 2 - 8x (45 ) = 16 + x 2 - 8x ! ( 1 2 ) = 16 + x 2 - 4 2 x}$ Using the cosine law in

$\triangle$ ADC$, we obtain$$\text{AC 2 = AD 2 + DC 2 - 2(AD)(DC) ( ADC) = 16 + y 2 - 8y (135 ) = 16 + y 2 - 8y ! (- 1 2 ) = 16 + y 2 + 4 2 y}$$Equating expressions for$AC^2$, we obtain$$\text{16 + x 2 - 4 2 x = 16 + y 2 + 4 2 y x 2 - y 2 - 4 2 x - 4 2 y = 0 (x+y)(x-y) - 4 2 (x+y) = 0 (x+y)(x-y - 4 2 ) = 0}$$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$AB$. [[IMAGE1]] 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, $BC - CD = BC - CP = BP = 4\sqrt{2}$.