Five congruent circles, with centres C, D, F, G and H, are arranged so that each centre lies on the circumference of at least two other circles, as shown in the attachment below.
a) Let P be the intersection point of line segments AI and BJ. Prove that angle APB is 60 degrees and hence that P lies on the circle with centre C.
My answer: We use Phantom Points. Let $AI$ and the circumcircle of $\triangle ABG$ intersect at $P_1$.
Since $A$, $B$, $G$ and $P_1$ lie on the same circle, this means $\angle AP_1B=\angle AGB=60^\circ$.
Extend $BP_1$ to meet the circumcircle of $\triangle GIJ$ at $J_1$. We have $\angle J_1PI=60^\circ\implies \angle J_1GI=60^\circ$.
Since $\angle JGI=60^\circ$, this implies that $J=J_1$. Therefore, $P_1$ lies on $BJ$, which means that $P_1$ is the intersection of $AI$ and $BJ$.
Therefore, $P=P_1$, which shows that $P$ lies on the $C$-centered circle. This also proves that $\angle APB=60^\circ$.
What's another way of solving this problem?
Since we are only concerning angles, assume circle radius = 2
$\vec{IA} = (1, 3\sqrt3)$
$\vec{JB} = (-4, 2\sqrt3)$
$\cos(∠APB) = {\vec{IA} · \vec{JB} \over |\vec{IA}||\vec{JB}|} = {(1)(-4)+(3\sqrt3)(2\sqrt3) \over \sqrt{1+27}\sqrt{16+12}} = {14\over 28} = {1\over2}$
$∠APB = \cos^{-1} {1\over2} = 60°$
Another way, using slopes
IA slope = $m_1 = 3\sqrt3$
JB slope = $m_2 = {2\sqrt3 \over -4} = {-\sqrt3 \over 2}$
$\tan ∠APB = \large {m_2 - m_1 \over 1+m_2 m_1} = {{-7 \sqrt3 \over 2} \over 1 - {9\over2} } = \sqrt3$
$∠APB = \tan^{-1} \sqrt3 = 60°$