100 Days Streak problem!

Construction : $AD \perp BC$ at $D$ .

Since the triangle is isosceles , $BD = DC=x$ .Let $DP_i=x_i$ .

$a_i =AP_i^2+P_iB . P_iC \\ \Rightarrow a_i =AP_i^2+(x-x_i)(x+x_i) \\ \Rightarrow a_i =AP_i^2-x_i^2+x^2 \\ \Rightarrow a_i =AD^2+x^2 \dots \text{(Pythagoras Theorem for}1\leq i \leq 100) \\ \Rightarrow a_i = AB^2 = (10)^2 = 100 \\ \Rightarrow\displaystyle\sum_{i=1}^{100} a_i = \Large\boxed{10000}$

Using Stewart's Theorem, we get: $P_{i}C 10^2 + P_{i}B 10^2 = BC(AP_{i} + P_{i}B . P_{i}C)$ $100(P_{i}B +P_{i}C) = BC(AP_{i} + P_{i}B . P_{i}C)$

since $(AP_{i} + P_{i}B . P_{i}C) = a_i$ , then $100(P_{i}B +P_{i}C) = BC . a_{i}$ Consider that $P_{i} B + P_{i}C = BC$ , then $100(P_{i}B +P_{i}C) = BC . a_{i}$ $100BC=BC . a_{i}$

$BC \neq 0$ , then it can be crossed out. Thus, we have $a_i =100$ for any integers of $i$

The value of $\sum_{i=1}^{100} a_i$ is $\sum_{i=1}^{100} 100 = 100*100=10000$

Since there are no restrictions in the question, let us consider all the 100 points at B and infinitesimally close to B. In this case $P_i B$ can be considered as an infinitesimally small value, rounded off to 0. $\sum_{i=1}^{100} a_i\ = \sum_{i=1}^{100}(AP_i^2 +P_iB.P_iC)$ Since the points are at B, $AP_i=AB=10$ $\sum_{i=1}^{100} a_i\ = \sum_{i=1}^{100}AB^2$ $= \sum_{i=1}^{100}100$ $\boxed{100000}$