A geometry problem by Miraj Shah

First of all, I'll like to make something clear. Neither the solution nor the question is original and hence, I do not take any credit for it! It's just the elegance of the question and more importantly the solution, that compelled me to post this question!

Clarifications

• Re $[ z]$ = Real part of the complex number $z$

• $i=\sqrt{-1}$

• $e^{i\theta} = \cos\theta + i\sin\theta$

Solution:

Now if we observe carefully we can see that the question can be written in the following way as well:

$Re \left \lbrack \large \displaystyle \sum_{r=0}^n \dbinom{n}{r} a^r b^{n-r} e^{i(rB-(n-r)A)} \right \rbrack$

$= Re \left \lbrack \large \displaystyle \sum_{r=0}^n \dbinom{n}{r} (ae^{iB})^r(be^{-iA})^{n-r} \right \rbrack$

$=Re \left \lbrack (ae^{iB}+be^{-iA})^n \right \rbrack$

Now after using the property $e^{i\theta} = \cos\theta +i\sin\theta$ we get;

$=Re \left \lbrack (a\cos(B)+b\cos(A))^n \right \rbrack$

$=c^n$

Therefore the answer is $\boxed{c^n}$