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I'm wondering whether it's possible to solve for $x^{\circ}$ in terms of $a^{\circ}$ and $b^{\circ}$ given that $ABCD$ is a parallelogram. In particular, I'm wondering if it's possible to solve it using only "elementary geometry". I'm not sure what "elementary geometry" would usually imply, but I'm trying to solve this problem without trigonometry.

Is it possible? Or if it's not, is there a way to show that it isn't solvable with only "elementary geometry" techniques?

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I am pretty certain it cannot be solved only by simple angle chasing, as $\angle BCD = y$ keeps popping us as a second unknown, and all the triangles I can find in the picture just keeps telling me that $a+b+x+y = 180^\circ$ –  Arthur Dec 16 '11 at 4:57
Well, it is possible to prove that $a$ and $b$ define $ABCD$ up to scaling using just the Euclidean axioms, so it should be possible to compute $x$ using the same axioms, I think :) –  Alexei Averchenko Dec 16 '11 at 5:07
Obviously the information given is enough to find $x$. If you can prove it's actually necessary to use the sine or cosine function, that would be a surprising result. I'll state two suspicions. Then I'll crash for the night an look at this tomorrow. (1) Wildberger's rational trigonometry will handle this neatly and maybe it's a nice propaganda example for that program; (2) recall the (see Parallelogram law). –  Michael Hardy Dec 16 '11 at 7:51
What makes it a parallelogram is that the diagonals bisect each other. So, if P is the center, we somehow have to use the fact that |AP| = |PD|. Maybe draw the perpendiculars from C to AD and from P to CD and look at the resulting four right triangles. –  marty cohen Dec 18 '11 at 6:18
It really depends by what you mean. It was explained in the answers why $x$ cannot be a simple expression in $a$ and $b$, but at the same time, "finding" x is the same problem as constructing the median in the triangle $ACD$, given the angles of the triangle... This can be done easily geometrically, even with the ruler and compass. –  N. S. Dec 20 '11 at 0:28
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4 Answers

up vote 3 down vote accepted

Let me rephrase your question first: Can the angle $x$ in the diagram be expressed in terms of angles $a$ and $b$ using only addition, subtraction, multiplication, division, exponentiation, $\pi$, and auxiliary rational numbers, and no trig or inverse trig functions? (For example, is $x$ an expression like $(a+2b)^{3/2}$?)

Here is strong evidence that the answer is no. If $a=\frac{\pi}{2}$ and $b=\frac{\pi}{3}$ (the two "simplest" angles to play with) then allowing ourselves access to trigonometry, we can show that $$x=2\arctan\left(\sqrt{13}-2\sqrt{3}\right)=\arctan\left(\frac{\sqrt{3}}{6}\right)$$

So you would have to believe that this expression is equal to a combination of $\frac{\pi}{2}$ and $\frac{\pi}{3}$ using only addition, subtraction, multiplication, division, exponentiation, $\pi$, and auxiliary rational numbers. I highly doubt this. (The transcendence of $\pi$ and $\arctan$ might be used to even prove that it is not possible - but that's just my conjecture.)

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I believe $x = \arctan{\frac{1}{2\sqrt{3}}}$ in that example. I'm not sure if that can be expressed as a combination of $\frac{\pi}{2}$ and $\frac{\pi}{3}$ though. –  Herman Chau Dec 18 '11 at 10:11
@Herman Those two values are equal, but your version is definitely in a simpler form. I'll edit the question to include it - thanks! –  alex.jordan Dec 18 '11 at 19:28
The ratio of your angle and $\pi$ is transcendental, by Gelfond-Schneider and some details. –  Will Jagy Dec 18 '11 at 22:03
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In the case $a = \pi/2$, $x = \arccos \left( 2\,{\frac {\sin \left( b \right) }{\sqrt {4-3\, \cos^2 \left( b \right) }}} \right)$. This is not an algebraic function of $b$, because its derivative is $\frac{dx}{db} = \frac{2}{3 \cos^2 b - 4}$ for $-\pi/2 < b < \pi/2$, and $\cos(b)$ is not an algebraic function.

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The example by alex.jordan does finish the matter, and similar ones may be constructed. We have an angle $$ \theta = \arctan \left( \frac{1}{\sqrt{12}} \right) $$ and we wish to know whether $ x = \frac{\theta}{\pi} $ is the root of an equation with rational coefficients.

Well, $$ e^{i \theta} = \sqrt{\frac{12}{13}} + i \sqrt{\frac{1}{13}} $$ Next, $\cos 2 \theta = 2 \cos^2 \theta - 1 = \frac{11}{13}.$ So, by Corollary 3.12 on page 41 of NIVEN we know that $2 \theta$ is not a rational multiple of $\pi.$ So, neither is $\theta,$ and $$ x = \frac{\theta}{\pi} $$ is irrational.

Now, the logarithm is multivalued in the complex plane. We may choose $$ \log(-1) = \pi i. $$ With real $x,$ we have chosen $$ (-1)^x = \exp(x \log(-1)) = \exp(x\pi i) = \cos \pi x + i \sin \pi x. $$ With our $ x = \frac{\theta}{\pi}, $ we have $$ (-1)^x = e^{i \pi x} = e^{i \theta} = \sqrt{\frac{12}{13}} + i \sqrt{\frac{1}{13}} $$ The right hand side is algebraic.

The Gelfond-Schneider Theorem, Niven page 134, says that if $\alpha,\beta$ are nonzero algebraic numbers, with $\alpha \neq 1$ and $\beta$ not a real rational number, then any value of $\alpha^\beta$ is transcendental.

Taking $\alpha = -1$ and $\beta = x,$ which is real but irrational. We are ASSUMING that $x$ is algebraic over $\mathbb Q.$ The assumption, together with Gelfond-Schneider, says that $ (-1)^x$ is transcendental. However, we already know that $ (-1)^x = \sqrt{\frac{12}{13}} + i \sqrt{\frac{1}{13}} $ is algebraic. This contradicts the assumption. So $x = \theta / \pi$ is transcendental, with $ \theta = \arctan \left( \frac{1}{\sqrt{12}} \right) $

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On taking into account the triangles ACP and ABP, where P is the point where the diagonals intercept, we have, from sines' law: $\sin{x}/\sin{a} = \sin{\theta}/\sin{b}$, $\theta$ being the angle between the smaller diagonal and the line AB.

Then $\sin{\theta} = \sin{x} \, \dfrac{\sin{b}}{\sin{a}}$.

Now, note that, in triangle ACP, the sum of the inner angles is $\,x+a+b+\theta = 180^{\circ}$, hence $\sin{\theta} = \sin{(x+a+b)}$.

Therefore: $\sin{x} \, \dfrac{\sin{b}}{\sin{a}} = \sin{(x+a+b)} = \sin{x} \, \cos{(a+b)} + \sin{(a+b)} \, \cos{x}$.

This implies that $\cot{x} = \dfrac{\sin{b}}{\sin{a} \, \sin{(a+b)}} - \cot{(a+b)}$.

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I've edited your answer to reflect community guidelines and improve readability. –  Lord_Farin Oct 9 '13 at 19:26
Thanks to Lord_Farin for editing. –  Dr. Fabio M. S. Lima Oct 9 '13 at 21:28
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