3
$\begingroup$

I know this is easy to lots of you but I am just struggling to find the solution. How can I prove the sum of the highlighted angles in the following 3 times 3 square grid is equal to $\pi$?

Picture

$\endgroup$
3
  • 4
    $\begingroup$ Answers here and here. $\endgroup$ – dtldarek Jan 1 '17 at 21:24
  • $\begingroup$ Cool, feel free to mark as duplicated. Thanks! $\endgroup$ – kchpchan Jan 1 '17 at 22:13
  • $\begingroup$ This is a question from Brilliant. Brilliant problems cannot be discussed or solved anywhere else. $\endgroup$ – Soha Farhin Pine Jan 2 '17 at 20:13
2
$\begingroup$

Here's a fun little proof without words. Sorry for the horrible art skills.

$\endgroup$
3
$\begingroup$

Hint: Apply the arctan addition formula to angles $\arctan(1), \arctan(2), \arctan(3)$.

$\endgroup$
0
$\begingroup$

The first angle is $\pi /4$ because his tangent is equal to $1$ so we must show that $a+b=3\pi /4$. Looking to the picture we have $\tan a= 2$ and $\tan b= 3$ so:

$$\tan(a+b)=\frac{\tan a + \tan b}{1-\tan a \tan b}=\frac{2+3}{1-2\cdot3}=-1$$

and once $a+b \le \pi$ then $a+b=3\pi /4$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.