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2 solutions
Nice. My (non-rigorous) argument was that the setup yielded a unique α , and then recognize the famous olympiad problem setup.
Great Nice! t r i g b a s h . A more straightforward solution : There is a cyclic quadrilateral outside the figure, which will greatly reduce this trigonometry stuff to a mere equation which would yield the value of α .
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I can paint a "similar" quadrilateral with α = 1 0 . 5 º , and you can do it too. And I can paint "another similar" quadrilateral with α = 1 1 º . Try it, and you get it....
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? ? ? ? ? ? ? What do you mean? The value of α is constant. R i g h t ? Then how can you p a i n t or draw an α which equals other than 10 degrees?
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@Vishwash Kumar Γξω – No, the value of α is not constant . You just only have to paint. Try or think it in your mind
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@Guillermo Templado – Can you define p a i n t ? Is it similar to d r a w or c o n s t r u c t i o n ?
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@Vishwash Kumar Γξω – Rohit, to clarify, can you be explicit about what you did?
Guillermo, you're not getting your point across, and it's not clear to me exactly what you are trying to say. Simply saying "You just only have to paint. Try or think it in your mind" does not tell us anything.
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@Calvin Lin – @Calvin Lin What do you think is α variable ? Guillermo Templado says this.
Yes! Its 11:55 here, I will write a solution tomorrow that is based on some constructions.
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@Vishwash Kumar Γξω – I suspect what he is saying is that "Even if we put it in a cyclic quad, that is not enough information to hunt down the specific value. We could have numerous constructions that yield possible configurations".
That is why I asked you to be "explicit about what you did". You might have done it via a certain way that gets around the concern about "not enough information", or you might have done it in such a way that introduces assumptions about the configuration.
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@Calvin Lin – Exact, Calvin."Even if we put it in a cyclic quad, that is not enough information to hunt down the specific value. We could have numerous constructions that yield possible configurations". There is not enough information. Start at point B . Make an angle 2 α ,approximately 2 ⋅ 1 0 . 5 . Later 3 α . When you are at C make an angle 2 α and 6 α ... And you can continue untill you can create a "similar" quadrilateral. Think it, it's very easy...
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@Guillermo Templado – If you look at my solution, the angle α must satisfy the equation ( sin 4 α sin 8 α − cos α sin 3 α sin 6 α ) 2 = 4 cos 2 α cos 2 2 α sin 2 3 α sin 2 6 α I then made the assumption that 8 α < 9 0 ∘ , which forced how I should take the square root of this equation, and ended up with an equation with a single solution. If we relax that restriction (as we would have to for your proposed solution of 1 1 . 5 ∘ ), we would have to consider the two equations sin 4 α sin 8 α − cos α sin 3 α sin 6 α = ± 2 cos α cos 2 α sin 3 α sin 6 α Since ∠ C A D = 4 α we must have 1 1 α < 1 8 0 ∘ . The only solution to either of the above equations for this range of α is α = 1 0 ∘ .
You say the construction is "easy". It is easy to start to construct a diagram like this, drawing angles of 2 α , 3 α at B and 2 α , 6 α at C . But to be exact about the final angle α at D - that is another matter...
@Guillermo Templado – From my current understanding of your statement, I disagree with "Think it, it's very easy". Echoing Mark, there is a lot of constraint on the angles, and it's not as simple as "draw this initial angle of 11, draw this line there", because we have to connect it back and obtain the final angle.
FWIW, I would not be surprised if there was a cyclic quad construction to this problem. As I stated I recongized this as the famous "hardest easy geometry problem", which does have a variety of synthetic solutions.
@Guillermo Templado – I have my solution based on cyclic quadrilaterals. Which finally yields a simple trigonometric equation and a single value of α .
I was looking for a Euclidean solution to this problem and thought of drawing a line at an angle 'α' to DB such that it meets BC in P & AC in Q. Now, it's easy to see that quad. APCD is concyclic. All that remains to be done is to show that tr. AQP or tr. DQC is equilateral or that AP//DC. Can anyone help me prove this?
A h ! I would love to. Do you need my help?
@ Rohit Camfar - Yes, please.
Not elegant, but effective...
Angle-chasing gives us that ∠ C A D = 4 α . If C D = X , then the Sine Rule gives A D = sin 4 α X sin 6 α B D = sin 3 α X sin 8 α B A = sin 4 α sin 2 α X sin 6 α sin α = 2 cos α sin 4 α X sin 6 α Applying the Cosine Rule, 4 cos 2 α sin 2 4 α sin 2 6 α 4 cos 2 α sin 2 3 α sin 2 6 α ( sin 4 α sin 8 α − cos α sin 3 α sin 6 α ) 2 = = = = sin 2 4 α sin 2 6 α + sin 2 3 α sin 2 8 α − 2 cos α sin 3 α sin 4 α sin 6 α sin 8 α sin 2 3 α sin 2 6 α + sin 2 4 α sin 2 8 α − 2 cos α sin 3 α sin 4 α sin 6 α sin 8 α ( sin 4 α sin 8 α − cos α sin 3 α sin 6 α ) 2 + sin 2 α sin 2 3 α sin 2 6 α 4 cos 2 α ( 1 − 4 sin 2 α cos 2 α ) sin 2 3 α sin 2 6 α = 4 cos 2 α cos 2 2 α sin 2 3 α sin 2 6 α Since the diagram makes it clear that 8 α is acute, we see that sin 4 α sin 8 α − cos α sin 3 α sin 6 α is positive, and hence this equation becomes sin 4 α sin 8 α − cos α sin 3 α sin 6 α 2 cos α sin 4 α sin 8 α sin 2 α sin 4 α sin 8 α = = = 2 cos α cos 2 α sin 3 α sin 6 α ( 2 cos 2 α + cos 2 α ) sin 3 α sin 6 α = ( 3 − 4 sin 2 α ) sin 3 α sin 6 α sin 2 3 α sin 6 α Plotting y = sin 2 x sin 4 x sin 8 x − sin 2 3 x sin 6 x for 0 < x < 8 9 0 ∘ shows that there is one solution to this equation where 8 α is acute. Now sin 2 0 ∘ sin 4 0 ∘ sin 8 0 ∘ = = = cos 1 0 ∘ sin 2 0 ∘ sin 4 0 ∘ = 2 1 cos 1 0 ∘ ( cos 2 0 ∘ − cos 6 0 ∘ ) = 4 1 cos 1 0 ∘ ( 2 cos 2 0 ∘ − 1 ) 4 1 cos 1 0 ∘ ( 4 cos 2 1 0 ∘ − 3 ) = 4 1 ( 4 cos 3 1 0 ∘ − 3 cos 1 0 ∘ ) = 4 1 cos 3 0 ∘ = 8 1 3 sin 2 3 0 ∘ sin 6 0 ∘ so we see that this single solution is α = 1 0 ∘ .