Reflect on this...
A square A B C D of side length k contains unit circles at each of corners B and D such that each circle is tangent to the square at precisely two points. A ray of light emanating from point A reflects off each circle and then returns to A , creating a path in the shape of an equilateral triangle.
There is a unique value of k for which this scenario can occur. Find ⌊ 1 0 0 0 0 ⋅ k ⌋ .
Note: "Reflecting" means that the angle of incidence equals the angle of reflection.
The answer is 66639.
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2 solutions
using trig we get k = (1 + sin 15°)/tan 15° + cos 15°+1 Is there some way I can add a figure? Describing this in just words will be pretty long winded :-(
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I understand how you feel; I've put off posting my solution because I don't know how to add a picture and, as the saying goes, a picture is worth a thousand words. :) I'm glad, though, that you were able to solve the problem; I was beginning to wonder if anyone would.
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To add a picture, upload it onto the internet and then use the markdown code of
I've added a screenshot of the video, so you can refer to the syntax.
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@Calvin Lin – Thank you Calvin! That really looks neat!
Thank you for that interesting problem Brian! I enjoyed solving it. Is it ok if I post the solution on my YouTube channel (https://www.youtube.com/user/UjjwalRane) and post the link to it here? By the way this reminded me of a similar problem with a 'find the shortest path' twist. I will post that on Brilliant. Thanks again :-)
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@Ujjwal Rane – Yes, of course. I look forward to seeing your solution. :)
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@Brian Charlesworth – It is up there now Please see (use full screen): https://www.youtube.com/watch?v=IUDJ9M3QZc8
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@Ujjwal Rane – That's great! The graphics and your explanation work so well together. Thanks for posting the video. :) I've made an edit to my 'solution' directing people to your video. I hope that it's o.k. that I did this; since they may not read all the comments they might not otherwise be aware of it.
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@Brian Charlesworth – The pleasure is all mine Brian! And kind of you to point to the video solution. Thank you!
Here's a graphic by which k can be computed through trigonometry
Reflect on this
From this we can see that
k = 1 + 2 C o s ( 1 5 ) + C o t ( 1 5 ) = 6 . 6 6 3 9 0 2 4 6 . . .
This is the same as Ujjwal's solution.
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Yup, that's it in a nutshell. :) falzoon's original problem was more general, whereby the triangle was isosceles with the angle at A being θ . So instead of 1 5 degrees we would be dealing with 2 9 0 − θ .
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Hello... M glad I solved this on first shot. Very nice problem!! But I solved it for equilateral triangle. And I am lazy enough to not solve it for general isosceles triangle. Can you please mention formula for general angle?
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@Pranjal Jain – Well, with θ as defined above, we would have
k = 1 + 2 cos ( 4 5 − 2 θ ) + cot ( 4 5 − 2 θ ) .
We could rewrite this in a number of ways, I suppose. My preferred way is
k = 1 + sec ( θ ) + tan ( θ ) + 2 + 2 sin ( θ ) .
Note that if the radii of the two circles is only specified as r then we would just scale k by the factor r .
The exact value for k is ( 2 1 ) ( 6 + 2 + 2 3 + 6 ) , which comes out to 6 . 6 6 3 9 0 2 5 . . . . Thus ⌊ 1 0 0 0 0 ∗ k ⌋ = 6 6 6 3 9 . Hopefully I'll have time this weekend to type out my solution method.
Edit: Please check out the excellent video solution posted by @Ujjwal Rane. My thanks to Ujjwal for posting this.
Screenshot at 5:06