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Find $x$, such there exsit four point $A,B,C,D$ on the plane,such $$|AB|=\sin{x},|BC|=\cos{x},|CD|=\tan{x},|DA|=\cot{x},|AC|=\sec{x},|BD|=\csc{x}$$

I think it's nice problem. I'm looking for an arbitrary quadrilateral six relations，and this six trigonmetry have some relations.such $$\sin^2{x}+\cos^2{x}=1.\tan{x}\cot{x}=1,$$$$\dfrac{1}{\sin{x}}=\csc{x},\dfrac{1}{\cos{x}}=\sec{x},1+\cot^2{x}=\csc^2{x},1+\tan^2{x}=\sec^2{x}$$

2 Answers 2

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There are no such $x\in\mathbb R$.

We may suppose that $$0\lt x\lt \frac{\pi}{2},\quad A(0,0),\quad B(\sin x,0),\quad C(c_1,c_2),\quad D(d_1,d_2)$$ with $c_2\ge 0$.

Since $C(c_1,c_2)$ where $c_2\ge 0$ is both on the circle $X^2+Y^2=|AC|^2$ and on the circle $(X-|AB|)^2+Y^2=|BC|^2$, we get $$c_1=\frac{\sec^2x+\sin^2x-\cos^2x}{2\sin x}=\frac{\frac{(1-t)(t+2)}{t+1}}{2\sin x},\quad c_2=\frac{\sqrt{X_1}}{2\sin x}$$ where $$t=\cos(2x)$$ and $$X_1=(2\sin x\sec x)^2-(\sec^2x+\sin^2x-\cos^2x)^2=\frac{t(1-t)(t^2+3t+4)}{(t+1)^2}$$

Also, since $D(d_1,d_2)$ is both on the circle $X^2+Y^2=|AD|^2$ and on the circle $(X-|AB|)^2+Y^2=|BD|^2$, we get $$d_1=\frac{\cot^2x+\sin^2x-\csc^2x}{2\sin x}=\frac{-\frac{t+1}{2}}{2\sin x},\quad d_2=\pm\frac{\sqrt{X_2}}{2\sin x}$$ where $$X_2=(2\sin x\cot x)^2-(\cot^2x+\sin^2x-\csc^2x)^2=\frac{(t+1)(7-t)}{4}$$

Now multiplying the both sides of $$\tan^2x=(c_1-d_1)^2+(c_2-d_2)^2$$ by $4\sin^2x$ gives $$\begin{align}&4\sin^2x\tan^2x=\left(\frac{(1-t)(t+2)}{t+1}+\frac{t+1}{2}\right)^2+\left(\sqrt{X_1}\mp \sqrt{X_2}\right)^2\\\\&\implies 4\cdot \frac{1-t}{2}\cdot \frac{1-t}{t+1}-\left(\frac{(1-t)(t+2)}{t+1}+\frac{t+1}{2}\right)^2-X_1-X_2=\mp 2\sqrt{X_1X_2}\\\\&\implies (t-2)(t+3)=\mp 2\sqrt{X_1X_2}\end{align}$$

Squaring the both sides gives

$$(t-2)^2(t+3)^2=4\cdot\frac{t(1-t)(t^2+3t+4)}{(t+1)^2}\cdot \frac{(t+1)(7-t)}{4}$$

Multiplying the both sides by $t+1$ gives

$$2t^4+t^3-3t^2-t+9=0$$ which can be written as $$2\left(t^2+\frac t4-1\right)^2+\frac{7}{8}t^2+7=0$$

This has no real solutions since the LHS is always positive.

Hence, there are no $x\in\mathbb R$ satisfying the conditions given in the question.

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    if the problem change this $\{|AB|,|BC|,|CD|,|DA|,|AC|,|BD|\}=\{\sin{x},\cos{x},\tan{x},\cot{x},\sec{x},\csc{x}\}$,there exsit $x?$ – 2017-02-20
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    @inequality: I think that we can apply the same method as my answer. – 2017-02-20
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    dear @mathlove,It is said my teacher,that if this case there exsit $x$ such it – 2017-02-20
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    @inequality: Such points may exist, but don't know. (Applying the same method should work but it must be very tedious.) – 2017-02-20
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This is an attempt at another solution which agrees with mathlove's answer, for $0 \leq x \leq \pi/2$.

If $x<\pi/6$, then $\sec x + \tan x < \cot x$, violating triangle inequality for $A,C,D$.

If $x>\pi/4$, then $\sin x + \cos x < \sec x$, also violating triangle inequality for $A,B,C$.

No time now to see if this works by using other triangles.