Quadratic equation paradox

$x^2+x+1=0$

Let's rewrite our equation in two equivalent forms:

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

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

We get:

$x\cdot(-x^2)=-1$

$x^3=1$

$x=1$

Triangle paradox

Consider an arbitrary triangle $ABC$. Let $m$ be the perpendicular bisector of $AC$ (it divides $AC$ into two equal segments) and let $l$ be the angle bisector (it divides angle $B$ into two equal angles); let $O$ be the 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 hypotenuse $l_1$ and the leg $h_1$ is equal to the right-angled triangle with the hypotenuse $l_2$ and the leg $h_2$ (because $h_1=h_2$ and $l_1=l_2$). This means that $\beta_1=\beta_2$. So we come to the result that $A=C$ (because $\alpha_1=\alpha_2$ and $\beta_1=\beta_2$). Similarly, we can prove that $A=B$. Finally, we have proved that any triangle $ABC$ is equilateral: $A=B=C=60^{\circ}$

Derivative paradox

Consider any smooth vector function $\overrightarrow{r(t)}$ such that neither the vector nor its derivative is ever equal to zero. We denote the vector's length by $r=|\overrightarrow{r}|$.

$r^2=\overrightarrow{r}\cdot\overrightarrow{r}$

$2r\cdot\dot{r}=2\overrightarrow{r}\cdot\dot{\overrightarrow{r}}$

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

We see that the cosine of the angle between $\overrightarrow{r}$ and its derivative $\dot{\overrightarrow{r}}$ is equal to 1.

So at any time $t$, the vector $\overrightarrow{r}$ is parallel to its derivative $\dot{\overrightarrow{r}}$. 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 an ellipse is $S=\pi ab$. Let's try to derive the formula for this area:

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

$\displaystyle=\int\limits_0^{2\pi}\frac12(a^2\cos^2\varphi+b^2\sin^2\varphi)d\varphi=\frac{\pi}{2}(a^2+b^2)$

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

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