Question-163148

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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 △ has to get back symm- etrically after shown reflections; find θ.

$${sides}\:{of}\:{inner}\:{walls}\:{of}\:{triangular} \\ $$$${fence}\:{are}\:{mirrored},\:{A}\:{ray}\:{of} \\ $$$${light}\:{from}\:{top}\:{vertex}\://\:{to} \\ $$$${plane}\:{of}\:\bigtriangleup\:{has}\:{to}\:{get}\:{back}\:{symm}- \\ $$$${etrically}\:{after}\:{shown}\:{reflections}; \\ $$$${find}\:\theta. \\ $$

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

α=90°−θ β=90°−(φ/2) α+2β=180° 90°−θ+2(90°−(φ/2))=180° θ+φ=90° ⇒θ=90°−φ θ>0 ⇒φ<90° θ<(φ/2) ⇒φ>60° only isosceles triangles with 60°<φ<90° are possible.

$$\alpha=\mathrm{90}°−\theta \\ $$$$\beta=\mathrm{90}°−\frac{\phi}{\mathrm{2}} \\ $$$$\alpha+\mathrm{2}\beta=\mathrm{180}° \\ $$$$\mathrm{90}°−\theta+\mathrm{2}\left(\mathrm{90}°−\frac{\phi}{\mathrm{2}}\right)=\mathrm{180}° \\ $$$$\theta+\phi=\mathrm{90}° \\ $$$$\Rightarrow\theta=\mathrm{90}°−\phi \\ $$$$\theta>\mathrm{0}\:\Rightarrow\phi<\mathrm{90}° \\ $$$$\theta<\frac{\phi}{\mathrm{2}}\:\Rightarrow\phi>\mathrm{60}° \\ $$$${only}\:{isosceles}\:{triangles}\:{with}\: \\ $$$$\mathrm{60}°<\phi<\mathrm{90}°\:{are}\:{possible}. \\ $$

Commented by mr W last updated on 05/Jan/22