Quadratic equation paradox


Let's rewrite our equation in two equivalent forms:

$1)\quad x\cdot(x+1)=-1$

$2)\quad x+1=-x^2$

We get:




Triangle paradox

Consider an arbitrary triangle $ABC$. Let $m$ be a perpendicular bisector of $AC$ (divides $AC$ on two equal intervals) and $l$ - angle bisector (divides $B$ on two equal angles), $O$ - intersection of $m$ and $l$.
Now we can see that $\alpha_1=\alpha_2$ and $l_1=l_2$ (because $m$ is a perpendicular bisector). $h_1=h_2$ (because $l$ is an angle bisector). We see that the right-angled triangle with the hipotenuse $l_1$ and the cathetus $h_1$ is equal to the right-angled triangle with the hipotenuse $l_2$ and the cathetus $h_2$ (because $h_1=h_2$ and $l_1=l_2$). This means that $\beta_1=\beta_2$. So we come tho the result that $A=C$ (because $\alpha_1=\alpha_2$ and $\beta_1=\beta_2$). Analogically we can prove that $A=B$.
Finally we proved that any triangle $ABC$ is equilateral: $A=B=C=60^{\circ}$

Derivation paradox

Consider any smooth vector function $\overrightarrow{r(t)}$ s.t. the vector and it's derivation are never equal to zero. We denote vector's length as $r=|\overrightarrow{r}|$



$2|\overrightarrow{r}|\cdot \dot{|\overrightarrow{r}|}=2|\overrightarrow{r}|\cdot{|\dot{\overrightarrow{r}}|}\cdot \cos(\angle\overrightarrow{r}\dot{\overrightarrow{r}})$

We see that cosinus of the angle between $\overrightarrow{r}$ and it's derivation $\dot{\overrightarrow{r}}$ is equal to 1.

So at any time $t$ vector $\overrightarrow{r}$ is parallel with it's derivation $\dot{\overrightarrow{r}}$. Nonsense! Where is the mistake?

Ellipse paradox

Consider any ellipse

$x=a\cos\varphi,\quad y=b\sin\varphi$

We know that the area of ellipse is $S=\pi ab$. Let's try to derive the formula for this area:

$\displaystyle\pi ab=S=\int\limits_0^{2\pi}\frac12r^2d\varphi=\int\limits_0^{2\pi}\frac12(x^2+y^2)d\varphi$


So we see that $\displaystyle ab=\frac{a^2+b^2}{2}$

It means that $a=b$, i.e any ellipse is a circle.