Law of Reflection

Geometry Level 3

A light ray emitted from a vertex of this isosceles triangle reflects 6 times before reaching another vertex.

What is the total length of the path of this ray?


The answer is 5.411.

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5 solutions

David Vreken
Jun 30, 2018

Since the ray reflects 6 6 times, it follows a straight line through a series of 7 7 reflected congruent triangles, as shown below. Let A A be the base angle that the ray starts from, B B be the base angle that the ray ends on, O O be the vertex angle, C C be the the angle the ray ends on in the last reflected triangle, and A O B = x \angle AOB = x .

Since there are 7 7 congruent angles at O O , A O C = 7 x \angle AOC = 7x . Since O A C \triangle OAC is an isosceles triangle with a vertex angle of 7 x 7x , O A C = 180 ° 7 x 2 \angle OAC = \frac{180° - 7x}{2} , and since we are given that B A C = 60 ° \angle BAC = 60° , B A O = 180 ° 7 x 2 + 60 ° \angle BAO = \frac{180° - 7x}{2} + 60° .

However, we also know that since A O B \triangle AOB is an isosceles triangle with a vertex angle of x x , B A O = 180 ° x 2 \angle BAO = \frac{180° - x}{2} .

Therefore, we have B A O = 180 ° 7 x 2 + 60 ° = 180 ° x 2 \angle BAO = \frac{180° - 7x}{2} + 60° = \frac{180° - x}{2} , and this solves to x = 20 ° x = 20° , which means B A O = 80 ° \angle BAO = 80° and O A C = 20 ° \angle OAC = 20° .

Using the law of sines on A O B \triangle AOB gives A O sin 80 ° = 1 sin 20 ° \frac{AO}{\sin 80°} = \frac{1}{\sin 20°} , which means A O = sin 80 ° sin 20 ° AO = \frac{\sin 80°}{\sin 20°} , and using the law of sines on A O C \triangle AOC gives A C sin 140 ° = sin 80 ° sin 20 ° sin 20 ° \frac{AC}{\sin 140°} = \frac{\frac{\sin 80°}{\sin 20°}}{\sin 20°} , which means A C = sin 140 ° sin 80 ° sin 20 ° sin 20 ° AC = \frac{\sin 140° \sin 80°}{\sin 20° \sin 20°} .

Using the sine double angle formula, A C = 2 sin 70 ° cos 70 ° sin 80 ° 2 sin 10 ° cos 10 ° sin 20 ° AC = \frac{2 \sin 70° \cos 70° \sin 80°}{2 \sin 10° \cos 10° \sin 20°} , and using the product-to-sum formulas and the identity sin x = sin ( 180 ° x ) \sin x = \sin (180° - x) , A C = 2 sin 70 ° ( sin 30 ° + sin 10 ° ) 2 sin 10 ° ( sin 30 ° + sin 10 ° ) AC = \frac{2 \sin 70° (\sin 30° + \sin 10°)}{2 \sin 10° (\sin 30° + \sin 10°)} , which simplifies to A C = sin 70 ° sin 10 ° 5.411 AC = \frac{\sin 70°}{\sin 10°} \approx \boxed{5.411} .

i got sin70 / sin10 the other way.

albert yiyi - 2 years, 11 months ago

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Very nice! I like your way better.

David Vreken - 2 years, 11 months ago

Could somebody please tell me the name of software he draw the picture?

Leon Cheng - 2 years, 11 months ago

That's so awesome at so many levels. Thanks for sharing this explanation!

Yassine Alouini - 2 years, 11 months ago

how does it follow straight line ac? would you suggest basic knowledge I need to know to solve it?

ridoy k - 2 years, 9 months ago

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In the law of reflection, the angle of incidence equals the angle of reflection, so the image of an object's reflection appears to be in a straight line. One trick for solving questions with reflections is to keep the path of light straight but have it go through the reflected space instead (in this case, the space is the triangles).

David Vreken - 2 years, 9 months ago
Albert Yiyi
Jun 4, 2018

Hint:

The exact answer is sin 7 0 sin 1 0 . \text{The exact answer is } \frac{ \sin 70^{\circ} }{ \sin 10^{\circ} } .

Wow,that's a really beautiful problem

Roberto Gomide - 3 years ago

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thank you!

albert yiyi - 3 years ago

I can see what are the brilliant

Head White - 2 years, 11 months ago

Could somebody please tell me the name of software he draw the picture?

Leon Cheng - 2 years, 11 months ago

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1st picture, i use geogebra

2nd picture, i use good o' mspaint.

albert yiyi - 2 years, 11 months ago

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Thanks a lot: )

Leon Cheng - 2 years, 11 months ago

How did you figure out the angle is 20?

Ahmed Almubarak - 2 years, 11 months ago

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albert yiyi - 2 years, 11 months ago

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great job, thanks

Ahmed Almubarak - 2 years, 11 months ago

Dats brilliant man👍

Shizue Izawa - 2 years, 11 months ago
Vinod Kumar
Jul 3, 2018

Another way of angle calculation as 20°

If the acute angle of the triangle is theta, then, after each reflection from the slanted side of the triangle, the ray which began with 60° angle with base of the triangle, loses this angle by (theta).

Therefore, if after 3 reflections from the slanted sides of the triangle, the ray has become parallel to the base of the triangle, i.e. 0°, we have the relation,

3*(theta)=60°,

giving theta=20°.

Length calculations are given by the following sum 'S=5.411',

S= 1+2{cos (60°)+cos(40°)+cos(20°)}

can you please explain this "If the acute angle of the triangle is theta, then, after each reflection from the slanted side of the triangle, the ray which began with 60° angle with base of the triangle, loses this angle by (theta)."

Ansen Antony - 2 years, 11 months ago

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Because of the law of reflection, the angle between incident ray and normal equals the angle between the reflected ray and normal... And the normal is off from the triangle's base by half its vertex angle.

C . - 2 years, 9 months ago

Can you please put a reference on how did you get the final sum?

Roberto Aponte - 2 years, 11 months ago
Stewart Gordon
Jul 7, 2018

In an octagon, the angles add up to 108 0 1080^\circ . Since we know that B A H = A H G = 6 0 \angle BAH = \angle AHG = 60^\circ , the other angles in the octagon A B C D E F G H ABCDEFGH must equal 16 0 160^\circ . Thus the base angle of the original triangle is half of this, 8 0 80^\circ , and the vertical angle is 2 0 20^\circ . This further tells us that in triangle A O H AOH , the vertical and base angles are 14 0 140^\circ and 2 0 20^\circ respectively.

Using the sine rule on triangle A B O ABO , since A B = 1 AB = 1 , we deduce that A O = sin 8 0 sin 2 0 AO = \frac{\sin 80^\circ}{\sin 20^\circ} . We can use the sine rule again on A O H AOH , giving us that A H = A O sin 14 0 sin 2 0 = sin 8 0 sin 14 0 sin 2 2 0 5.411 . AH = \frac{AO \sin 140^\circ}{\sin 20^\circ} \\ = \frac{\sin 80^\circ \sin 140^\circ}{\sin^2 20^\circ} \\ \approx \boxed{5.411}.

Lance Kuanwu
Jul 6, 2018

I will do a step by step breakdown: First notice the angle of light relative to horizontal decrease by same amount after every reflection. 6 reflection = decrease 6 times. Notice that the light from its initial position to its final position, the angle changed by 120 degrees (vector diagram or just visualize) Thus each reflection the angle changes by 120/6= 20 degrees.

Next, know that when the mirror angle changes by x degrees, the reflection changes by 2x degrees. This can be proven or just google. Thus we know the bottom most line must have moved up 20/2=10 degrees relative to the horizontal. Then look at the bottom picture and you can figure out everything using trigonometry =)

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