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Question Number 199911 by mr W last updated on 11/Nov/23

Commented by mr W last updated on 11/Nov/23

three mirrors build the sides of an equilateral triangle. a laser ray emitted from the midpoint of a side should reach the opposite vertex after multiple reflections as an example in the image above shows. find the angle α (0<α<90°).

$${three}\:{mirrors}\:{build}\:{the}\:{sides}\:{of}\:{an} \\ $$$${equilateral}\:{triangle}.\:{a}\:{laser}\:{ray} \\ $$$${emitted}\:{from}\:{the}\:{midpoint}\:{of}\:{a}\:{side} \\ $$$${should}\:{reach}\:{the}\:{opposite}\:{vertex}\:{after} \\ $$$${multiple}\:{reflections}\:{as}\:{an}\:{example}\: \\ $$$${in}\:{the}\:{image}\:{above}\:{shows}. \\ $$$${find}\:{the}\:{angle}\:\alpha\:\left(\mathrm{0}<\alpha<\mathrm{90}°\right). \\ $$

Commented by Frix last updated on 11/Nov/23

I think for α<(π/2) sin α =((2k)/( (√(4k^2 +3(2n−1)^2 )))) sin α =((2k+1)/( (√(4k(k+1)+12n^2 +1)))) with n, k ∈N^★

$$\mathrm{I}\:\mathrm{think}\:\mathrm{for}\:\alpha<\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{sin}\:\alpha\:=\frac{\mathrm{2}{k}}{\:\sqrt{\mathrm{4}{k}^{\mathrm{2}} +\mathrm{3}\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{2}} }} \\ $$$$\mathrm{sin}\:\alpha\:=\frac{\mathrm{2}{k}+\mathrm{1}}{\:\sqrt{\mathrm{4}{k}\left({k}+\mathrm{1}\right)+\mathrm{12}{n}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\mathrm{with}\:{n},\:{k}\:\in\mathbb{N}^{\bigstar} \\ $$

Commented by mr W last updated on 11/Nov/23

please reason your answer and give a concrete example and verify it, for example the case as shown in the image. thanks!

$${please}\:{reason}\:{your}\:{answer}\:{and}\:{give} \\ $$$${a}\:{concrete}\:{example}\:{and}\:{verify}\:{it},\: \\ $$$${for}\:{example}\:{the}\:{case}\:{as}\:{shown}\:{in}\: \\ $$$${the}\:{image}.\:{thanks}! \\ $$

Commented by Frix last updated on 11/Nov/23

We can draw a tetragonal grid and mark all reflections of the target vertice. All possible paths are straight lines within the grid. The horizontal of the reflections with the start point set to (0, 0) are ((3/2)×(2n−1), (√3)k) or ((3/2)×2n, ((√3)/2)+(√3)k) I might be wrong, writing this while on my way to work and without any possibility to draw it.

$$\mathrm{We}\:\mathrm{can}\:\mathrm{draw}\:\mathrm{a}\:\mathrm{tetragonal}\:\mathrm{grid}\:\mathrm{and}\:\mathrm{mark} \\ $$$$\mathrm{all}\:\mathrm{reflections}\:\mathrm{of}\:\mathrm{the}\:\mathrm{target}\:\mathrm{vertice}.\:\mathrm{All} \\ $$$$\mathrm{possible}\:\mathrm{paths}\:\mathrm{are}\:\mathrm{straight}\:\mathrm{lines}\:\mathrm{within}\:\:\mathrm{the} \\ $$$$\mathrm{grid}.\:\mathrm{The}\:\mathrm{horizontal}\:\mathrm{of}\:\mathrm{the}\:\mathrm{reflections}\:\mathrm{with} \\ $$$$\mathrm{the}\:\mathrm{start}\:\mathrm{point}\:\mathrm{set}\:\mathrm{to}\:\left(\mathrm{0},\:\mathrm{0}\right)\:\mathrm{are} \\ $$$$\left(\frac{\mathrm{3}}{\mathrm{2}}×\left(\mathrm{2}{n}−\mathrm{1}\right),\:\sqrt{\mathrm{3}}{k}\right)\:\mathrm{or}\:\left(\frac{\mathrm{3}}{\mathrm{2}}×\mathrm{2}{n},\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\sqrt{\mathrm{3}}{k}\right) \\ $$$$\mathrm{I}\:\mathrm{might}\:\mathrm{be}\:\mathrm{wrong},\:\mathrm{writing}\:\mathrm{this}\:\mathrm{while}\:\mathrm{on}\:\mathrm{my} \\ $$$$\mathrm{way}\:\mathrm{to}\:\mathrm{work}\:\mathrm{and}\:\mathrm{without}\:\mathrm{any}\:\mathrm{possibility} \\ $$$$\mathrm{to}\:\mathrm{draw}\:\mathrm{it}. \\ $$

Commented by Frix last updated on 11/Nov/23

I think the image shows the first case with k=1∧n=2 sin α =(2/( (√(31)))) ⇒ tan α =((2(√3))/9) ⇒ α≈21.05°

$$\mathrm{I}\:\mathrm{think}\:\mathrm{the}\:\mathrm{image}\:\mathrm{shows}\:\mathrm{the}\:\mathrm{first}\:\mathrm{case}\:\mathrm{with} \\ $$$${k}=\mathrm{1}\wedge{n}=\mathrm{2} \\ $$$$\mathrm{sin}\:\alpha\:=\frac{\mathrm{2}}{\:\sqrt{\mathrm{31}}}\:\Rightarrow\:\mathrm{tan}\:\alpha\:=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{9}}\:\Rightarrow\:\alpha\approx\mathrm{21}.\mathrm{05}° \\ $$

Commented by mr W last updated on 11/Nov/23

correct!

$${correct}! \\ $$

Answered by ajfour last updated on 11/Nov/23

Answered by ajfour last updated on 11/Nov/23

α=tan^(−1) ((2/( (√3))))

$$\alpha=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\right) \\ $$

Commented by mr W last updated on 12/Nov/23

yes, thanks sir!

$${yes},\:{thanks}\:{sir}! \\ $$

Answered by mr W last updated on 11/Nov/23

Commented by mr W last updated on 11/Nov/23

red mirror is the first mirror which reflects the light ray and the brown one is the last mirror. between them all three mirrors maybe hit by the light ray one or more times. we can find that in the red mirror the destination point (red dot) must lie on the blue line and the possible positions are blue marked. we can find out that the angle α is α=tan^(−1) ((2(√3))/n) where n is an odd number which is equal to or larger than 3. n means also how many times the light ray is reflected before it reaches the destination.

$${red}\:{mirror}\:{is}\:{the}\:{first}\:{mirror}\:{which} \\ $$$${reflects}\:{the}\:{light}\:{ray}\:{and}\:{the}\:{brown} \\ $$$${one}\:{is}\:{the}\:{last}\:{mirror}.\:{between}\:{them} \\ $$$${all}\:{three}\:{mirrors}\:{maybe}\:{hit}\:{by}\:{the}\:{light} \\ $$$${ray}\:{one}\:{or}\:{more}\:{times}. \\ $$$${we}\:{can}\:{find}\:{that}\:{in}\:{the}\:{red}\:{mirror} \\ $$$${the}\:{destination}\:{point}\:\left({red}\:{dot}\right)\:{must} \\ $$$${lie}\:{on}\:{the}\:{blue}\:{line}\:{and}\:{the}\:{possible} \\ $$$${positions}\:{are}\:{blue}\:{marked}. \\ $$$${we}\:{can}\:{find}\:{out}\:{that}\:{the}\:{angle}\:\alpha\:{is} \\ $$$$\alpha=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{2}\sqrt{\mathrm{3}}}{{n}}\:{where}\:{n}\:{is}\:{an}\:{odd}\:{number} \\ $$$${which}\:{is}\:{equal}\:{to}\:{or}\:{larger}\:{than}\:\mathrm{3}. \\ $$$${n}\:{means}\:{also}\:{how}\:{many}\:{times}\:{the} \\ $$$${light}\:{ray}\:{is}\:{reflected}\:{before}\:{it}\:{reaches} \\ $$$${the}\:{destination}. \\ $$

Commented by mr W last updated on 11/Nov/23

Commented by mr W last updated on 11/Nov/23

this example shows when the light is reflected 3 times.

$${this}\:{example}\:{shows}\:{when}\:{the}\:{light} \\ $$$${is}\:{reflected}\:\mathrm{3}\:{times}. \\ $$

Commented by mr W last updated on 11/Nov/23

Commented by mr W last updated on 11/Nov/23

this example shows the case given in the question where the light ray is reflected 9 times.

$${this}\:{example}\:{shows}\:{the}\:{case}\:{given} \\ $$$${in}\:{the}\:{question}\:{where}\:{the}\:{light}\:{ray} \\ $$$${is}\:{reflected}\:\mathrm{9}\:{times}. \\ $$

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