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angles between diagonals

One needs to get it right.

What I have tried: can do it (see my comment below), would like a simpler solution. Will post a complete solution after a while.

Any constructive feedback would be appreciated!

$\bf{Added:}$ My attempt, using trigonometry.

The projections of the top and bottom vertices on the horizontal diagonal divide it in ratios $\frac{\tan 10^{\circ}}{\tan 20^{\circ}}$, and $\frac{\tan 30^{\circ}}{\tan 50^{\circ}}$. So let's show that these ratios are equal, equivalently

$$\tan 10^{\circ} = \tan 20^{\circ} \tan 30^{\circ} \tan 40^{\circ}$$

The equation $\tan x = \tan 2x \tan 3x \tan 4x$ appears in some high school math journal in the 60's.

Write $\tan x = \frac{e^{i x} - e^{-i x}}{i(e^{i x} + e^{ix})} = \frac{z^2-1}{i(z^2+1)}$, so, with $f(z) = \frac{z^2-1}{z^2+1}$ we need to solve

$$f(z)+ f(z^2) f(z^3) f(z^4)=0$$

The LHS factors as

$$\frac{2 (z-1)(z+1) (z^{12}-z^6+1)}{(z^2+1) (z^4-z^2+1) (z^8+1)}$$ and now notice that $z=e^{\frac{2\pi i}{36}}$ is a root of the cyclotomic polynomial $\Phi_{36}(z) = z^{12}-z^6+1$.

Conclusion: the ratios are equal, the top and bottom vertices project onto the same point of the diagonal, and therefore, the diagonals are perpendicular.

A geometric proof would be nice to have.

$\bf{Added:}$

@timon92: brought this post to my attention. @Andrew Hwang had an intepretation almost identical to this question; should undelete his answer.

Disclosure: discovered this cute question by accident/luck, while playing with Geogebra some 15 years ago. Now the question itself has to do with adventitious quadrilaterals, specifically those with orthogonal diagonals, a classic subject. Yes, there are some other ones.

orangeskid
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    What's the question? – Misha Lavrov Jan 07 '24 at 15:43
  • @Misha Lavrov: the problem is, one needs to get a right angle – orangeskid Jan 07 '24 at 16:03
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    @peterwhy: right angle $\iff \frac{\tan 10^{\circ}}{{\tan 20^{\circ}}}=\frac{\tan 30^{\circ}}{{\tan 50^{\circ}}}$, and Bob's your uncle – orangeskid Jan 07 '24 at 16:14
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    What is one allowed to do? Either it is or it is not a right angle and there's nothing you can do to make it or prevent it from being so. As a question, .... it isn't a question. – fleablood Jan 07 '24 at 16:41
  • @fleablood: by any means – orangeskid Jan 07 '24 at 17:53
  • If you want to find the values of the “outside” angles so that the 4 “inner” angles are right, then use the property of right triangles having complementary angles – Тyma Gaidash Jan 07 '24 at 18:27
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    see here: https://math.stackexchange.com/questions/1624651/what-is-the-geometry-behind-frac-tan-10-circ-tan-20-circ-frac-tan-30 – timon92 Jan 08 '24 at 00:21
  • @timon92: Interesting! Andy Hwang had a very similar interpretation, should have not deleted his answer. The equality itself is older than most of us here. So is your proof purely geometric then? If I get to reopen maybe you can re-adapt it for this. – orangeskid Jan 08 '24 at 00:32
  • Are you just looking for a proof of $\frac{\tan 10^{\circ}}{{\tan 20^{\circ}}}=\frac{\tan 30^{\circ}}{{\tan 50^{\circ}}}$? If not, can you please clarify? This is rather unclear. – Brian Tung Jan 08 '24 at 01:37
  • Note that Andrew Huang's answer in the linked question is not a proof; it is just a pictorial depiction of the equality (which is all that question asked for). – Brian Tung Jan 08 '24 at 01:39
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    @Brian Tung: the question is about geometry. Got a proof using trig, would like an alternate proof, preferably with geometry. – orangeskid Jan 08 '24 at 02:15
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    @orangeskid: Thanks for the context! I'll give it some thought; I like the problem. I might recommend that you put some of this clarification in the title and/or further up in the post. The title is not informative, and the first few paragraphs do not make it clear (to me, at least) what you desire. – Brian Tung Jan 08 '24 at 05:57

1 Answers1

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enter image description here

Consider two equilateral triangles ( in red) with a common vertex, one rotation of the other by $40^{\circ}$. There are formed two isosceles triangles with base angles $40^{\circ}$ (equal sides in blue). Now draw the perpendicular bisector of one of the equilateral triangle ( in yellow). We get another isosceles triangle with base angle $40^{\circ}$. Now conclude that we have an isosceles triangle with base angle $70^{\circ}$. The purple quadrilateral has the required shape.

orangeskid
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