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.

b) The line segment EF and circle with centre C intersect at F; let Q be their second point of intersection. Prove that Q also lies on the circle with centre P which passes through C.

enter image description here

I'm really stuck on problem b); I tried using chords AB and PQ and the cyclic quadrilateral ABPQ to show that PCQ is an equilateral triangle and, thus, to prove that Q lies on the circle with centre P. However, I realised that my proof led to a dead end... How would I solve this question?

Any help would be extremely appreciated!


Assume all circles have radius of 2, $C=(0,0)$

$A=(1,\sqrt3), I=(0,-2\sqrt3)$

AP slope, $ m_1 = 3\sqrt3$

EF slope = ${\sqrt3 \over 5}, E=(4,0) → y={\sqrt3 \over 5}(x-4)$

Q is intersection of EF, and circle C → $ x^2 + y^2 = 4$

Solving for Q, we get $Q=({13\over7}, {-3\sqrt3 \over 7})$

AQ Slope, $m_2 ={\sqrt3 + 3\sqrt3/7 \over 1-13/7} = {-5 \over \sqrt3 } $

$∠PAQ = \tan^{-1} {m_2 - m_1 \over 1+m_2 m_1} = \tan^{-1} {-14 / \sqrt3 \over 1-15} = \tan^{-1} {1 \over \sqrt3 } = 30°$

$$∠PCQ = 2 ∠PAQ = 60° → PQ = PC$$

Realized a simpler proof, without solving a quadratic.
Let $M_{FQ}$ be mid-point of FQ

Line CM perpendicular to EF → $y = {-5 \over \sqrt3}x$
Solving for $M_{FQ}: \large y= {\sqrt3 \over 5}(x-4)= {-5 \over \sqrt3}x → x={3\over7}$
$$Q = 2M_{FQ} - F = 2\times({3\over7}, {-5\sqrt3 \over 7}) - (-1, -\sqrt3) = ({13\over7}, {-3\sqrt3 \over 7})$$

For line AI, solve for $M_{AP} :\large y = 3\sqrt3(x-1) + \sqrt3 = {-x \over 3\sqrt3} → x={9\over14} $
$$P = 2M_{AP} - A = 2\times({9\over14}, {-\sqrt3 \over 14}) - (1, \sqrt3) = ({2\over7}, {-8\sqrt3 \over 7})$$

$$PQ = \sqrt{({11\over7})^2 + ({5\sqrt3 \over 7})^2} = 2 = PC$$


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