Question Number 163148 by ajfour last updated on 04/Jan/22
Commented by mr W last updated on 05/Jan/22
Commented by mr W last updated on 05/Jan/22
$${only}\:{some}\:{isosceles}\:{triangles}\:{are} \\ $$$${possible},\:{right}? \\ $$
Commented by mr W last updated on 05/Jan/22
$${a}\:{mirror}\:{labyrinth}\:{sir}? \\ $$
Commented by ajfour last updated on 05/Jan/22
$${sides}\:{of}\:{inner}\:{walls}\:{of}\:{triangular} \\ $$$${fence}\:{are}\:{mirrored},\:{A}\:{ray}\:{of} \\ $$$${light}\:{from}\:{top}\:{vertex}\://\:{to} \\ $$$${plane}\:{of}\:\bigtriangleup\:{has}\:{to}\:{get}\:{back}\:{symm}- \\ $$$${etrically}\:{after}\:{shown}\:{reflections}; \\ $$$${find}\:\theta. \\ $$
Answered by mr W last updated on 05/Jan/22
Commented by mr W last updated on 05/Jan/22
$${nice}\:{question}! \\ $$
Commented by mr W last updated on 05/Jan/22
$$\alpha=\mathrm{90}°−\theta \\ $$$$\beta=\mathrm{90}°−\frac{\phi}{\mathrm{2}} \\ $$$$\alpha+\mathrm{2}\beta=\mathrm{180}° \\ $$$$\mathrm{90}°−\theta+\mathrm{2}\left(\mathrm{90}°−\frac{\phi}{\mathrm{2}}\right)=\mathrm{180}° \\ $$$$\theta+\phi=\mathrm{90}° \\ $$$$\Rightarrow\theta=\mathrm{90}°−\phi \\ $$$$\theta>\mathrm{0}\:\Rightarrow\phi<\mathrm{90}° \\ $$$$\theta<\frac{\phi}{\mathrm{2}}\:\Rightarrow\phi>\mathrm{60}° \\ $$$${only}\:{isosceles}\:{triangles}\:{with}\: \\ $$$$\mathrm{60}°<\phi<\mathrm{90}°\:{are}\:{possible}. \\ $$
Commented by mr W last updated on 05/Jan/22
Commented by ajfour last updated on 05/Jan/22
$${Thank}\:{u}\:{sir}. \\ $$
Commented by Tawa11 last updated on 06/Jan/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$