Question Number 116983 by Olaf last updated on 08/Oct/20
$$\mathrm{Luigi},\:\mathrm{an}\:\mathrm{Italian}\:\mathrm{cook},\:\mathrm{drops}\:\mathrm{a}\:\mathrm{spaghetti} \\ $$$$\mathrm{which}\:\mathrm{breaks}\:\mathrm{into}\:\mathrm{three}\:\mathrm{pieces}. \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{of}\:\mathrm{making}\:\mathrm{a} \\ $$$$\mathrm{triangle}\:\mathrm{with}\:\mathrm{the}\:\mathrm{three}\:\mathrm{pieces}\:? \\ $$
Commented by mr W last updated on 08/Oct/20
$${i}\:{think}\:{it}'{s}\:\frac{\mathrm{1}}{\mathrm{8}}. \\ $$
Commented by Olaf last updated on 08/Oct/20
$$\mathrm{No}\:\mathrm{sir}… \\ $$
Commented by mr W last updated on 08/Oct/20
Commented by mr W last updated on 08/Oct/20
$${then}\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Commented by Olaf last updated on 08/Oct/20
$$\mathrm{Yes}\:\mathrm{sir}\:!\:\mathrm{Bravo}\:! \\ $$$$ \\ $$$$\mathrm{If}\:{x},\:{y}\:\mathrm{and}\:{z}\:\mathrm{are}\:\mathrm{the}\:\mathrm{3}\:\mathrm{lenghts}\:\mathrm{of} \\ $$$$\mathrm{spaghetti}\:\mathrm{with}\:\mathrm{0}\:\leqslant\:{x},\:{y},\:{z}\:\leqslant\:{l}\:\mathrm{this} \\ $$$$\mathrm{defines}\:\mathrm{an}\:\mathrm{equilateral}\:\mathrm{triangle}\:\mathrm{in}\:\mathrm{space}. \\ $$$$\mathrm{The}\:\mathrm{equation}\:\mathrm{is}\:{x}+{y}+{z}\:=\:{l}. \\ $$$$\mathrm{The}\:\mathrm{area}\:\mathrm{of}\:\mathrm{this}\:\mathrm{triangle}\:\left(\mathrm{all}\:\mathrm{possible}\right. \\ $$$$\left.\mathrm{cases}\right)\:\mathrm{is}\:\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\sqrt{\mathrm{2}{l}}\:=\:\frac{\sqrt{\mathrm{6}}}{\mathrm{4}}{l}. \\ $$$$ \\ $$$$\mathrm{But},\:\mathrm{to}\:\mathrm{make}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{with}\:\mathrm{the}\:\mathrm{3} \\ $$$$\mathrm{pieces}\:\mathrm{we}\:\mathrm{need}\:\mathrm{0}\:\leqslant\:{x},\:{y},\:{z}\:\leqslant\:\frac{{l}}{\mathrm{2}}. \\ $$$$\mathrm{This}\:\mathrm{defines}\:\mathrm{a}\:\mathrm{second}\:\mathrm{equilateral} \\ $$$$\mathrm{triangle}\:\mathrm{in}\:\mathrm{space}\:\left(\mathrm{all}\:\mathrm{probable}\:\mathrm{cases}\right) \\ $$$$\mathrm{with}\:\mathrm{an}\:\mathrm{area}\:\mathrm{a}\:\mathrm{quarter}\:\mathrm{of}\:\mathrm{the}\:\mathrm{first} \\ $$$$\mathrm{one}. \\ $$$$ \\ $$
Answered by mr W last updated on 09/Oct/20
Commented by mr W last updated on 09/Oct/20
$${this}\:{is}\:{how}\:{i}\:{solved}. \\ $$$${say}\:{the}\:{length}\:{of}\:{spagetti}\:{AB}\:{is}\:\mathrm{1}. \\ $$$${it}\:{breaks}\:{randomly}\:{at}\:{point}\:{C}\:\&\:{D} \\ $$$${with}\:{AC}={x}\:{and}\:{AD}={y}. \\ $$$$\mathrm{0}<{x},\:{y}<\mathrm{1} \\ $$$${the}\:{points}\:{in}\:{square}\:\mathrm{1}×\mathrm{1}\:{represent}\: \\ $$$${all}\:{possibilities}.\:{we}\:{only}\:{need}\:{to}\:{treat} \\ $$$${half}\:{of}\:{all}\:{possibilities}:\:{y}\geqslant{x}.\:{that}\:{is} \\ $$$${the}\:{orange}\:{triangle}\:{area}. \\ $$$$ \\ $$$${AC}={x} \\ $$$${CD}={y}−{x} \\ $$$${DB}=\mathrm{1}−{y} \\ $$$${such}\:{that}\:{the}\:{three}\:{pieces}\:{can}\:{form} \\ $$$${a}\:{triangle},\:{we}\:{must}\:{have} \\ $$$${x}+\left({y}−{x}\right)>\mathrm{1}−{y}\:\Rightarrow\:{y}>\frac{\mathrm{1}}{\mathrm{2}}\:\:\:…\left({i}\right) \\ $$$${x}+\left(\mathrm{1}−{y}\right)>{y}−{x}\:\Rightarrow\:{y}−{x}>\frac{\mathrm{1}}{\mathrm{2}}\:\:\:…\left({ii}\right) \\ $$$$\left({y}−{x}\right)+\left(\mathrm{1}−{y}\right)>{x}\:\Rightarrow\:{x}<\frac{\mathrm{1}}{\mathrm{2}}\:\:\:…\left({iii}\right) \\ $$$$\left({i}\right),\left({ii}\right),\left({iii}\right)\:{mean}\:{the}\:{points}\:{in}\:{the} \\ $$$${green}\:{triangle}\:{area}. \\ $$$$ \\ $$$${the}\:{requested}\:{probability}\:{is} \\ $$$${p}=\frac{{area}\:{of}\:{yellow}\:{trangle}}{{area}\:{of}\:{orange}\:{triangle}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Commented by mr W last updated on 09/Oct/20
