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ABC is an equilateral triangle with side 6, and its circumcircle has radius R…

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$ABC$ is an equilateral triangle with side $6$ , and its circumcircle has radius $R$ . A point $P$ moves along the arc $BC$ of the circumcircle that does not contain $A$ , and $P$ is never at $B$ or $C$ . The figure shows one position of $P$ with the three segments $PA$ , $PB$ and $PC$ . As $P$ moves, the three lengths change, but some relations between them stay true all the time. This problem finds them and uses them.

An equilateral triangle ABC drawn inside its circumcircle, with A at the top, B at the bottom left and C at the bottom right. A point P lies on the lower arc between B and C, nearer to B. The segments PA, PB and PC are drawn in blue.

Prove that $PA = PB + PC$ for every position of $P$ on the arc.
Prove that $PB^2 + PB \cdot PC + PC^2 = 36$ . Deduce that $PB \cdot PC = PA^2 - 36$ .
Suppose $PB = 2$ . Find $PC$ and $PA$ .
Find the greatest possible value of $PB + PC$ , and say where $P$ is when it happens. Then find the greatest possible area of triangle $PBC$ .
Prove that $PA^2 + PB^2 + PC^2$ has the same value for every position of $P$ on the arc, and find that value.

Give each answer that is not a whole number to two decimal places. You may use the fact that the angles in the same segment of a circle are equal, and that opposite angles of a cyclic quadrilateral add up to $180^\circ$ . Each part may use the results of the earlier parts.

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Prove (a) by cutting $PA$ into two pieces equal to $PB$ and $PC$ . The cosine rule then gives (b), and (c) to (e) follow from (a) and (b).

(a) The angles $APB$ and $ACB$ stand on the same arc $AB$ , so $\angle APB = \angle ACB = 60^\circ$ . In triangle $ABP$ the angle at $B$ is $\angle ABC + \angle CBP > 60^\circ$ , the angle at $P$ is $60^\circ$ , so the angle at $A$ is less than $60^\circ$ . The longer side faces the larger angle, so $PB < AB < PA$ . Mark $Q$ on segment $PA$ with $PQ = PB$ . This $Q$ lies inside the triangle: $PA$ crosses $BC$ at a point $X$ , and in triangle $PBX$ the angle at $B$ is $\angle PBC = \angle PAC < 60^\circ$ while the angle at $X$ is $120^\circ - \angle PBC > 60^\circ$ , so $PX < PB = PQ$ . Triangle $BPQ$ is isosceles with a $60^\circ$ angle at $P$ , so it is equilateral: $BQ = BP$ and $\angle QBP = 60^\circ$ . Now $\angle ABQ = 60^\circ - \angle QBC = \angle CBP$ , and $BA = BC$ , $BQ = BP$ . So triangles $ABQ$ and $CBP$ are congruent (two sides and the angle between them), and $AQ = CP$ . Hence $PA = PQ + QA = PB + PC$ . (Ptolemy's theorem for $ABPC$ gives the same: $PA \cdot BC = PB \cdot AC + PC \cdot AB$ , and all three sides are $6$ .)

(b) $ABPC$ is cyclic, so $\angle BPC = 180^\circ - \angle BAC = 120^\circ$ . The cosine rule in triangle $PBC$ gives $36 = PB^2 + PC^2 - 2\,PB \cdot PC\cos 120^\circ = PB^2 + PC^2 + PB \cdot PC$ . Then $PA^2 = (PB + PC)^2 = PB^2 + PC^2 + 2\,PB \cdot PC = 36 + PB \cdot PC$ , so $PB \cdot PC = PA^2 - 36$ .

(c) With $PB = 2$ , (b) gives $4 + 2\,PC + PC^2 = 36$ , so $PC^2 + 2\,PC - 32 = 0$ and $PC = -1 + \sqrt{33} \approx 4.74$ (the other root is negative). By (a), $PA = 2 + PC = 1 + \sqrt{33} \approx 6.74$ .

(d) $PB + PC = PA$ , and $PA$ is a chord, so it is at most the diameter $2R$ . For an equilateral triangle of side $6$ , $R = \dfrac{6}{\sqrt{3}} = 2\sqrt{3}$ , so $PB + PC \le 4\sqrt{3} \approx 6.93$ . Equality holds when $PA$ is a diameter, that is, when $P$ is the midpoint of the arc $BC$ , with $PB = PC = 2\sqrt{3}$ . The area of triangle $PBC$ is $\dfrac12 PB \cdot PC \sin 120^\circ = \dfrac{\sqrt{3}}{4}\left(PA^2 - 36\right)$ , which is greatest when $PA$ is greatest: $\dfrac{\sqrt{3}}{4}(48 - 36) = 3\sqrt{3} \approx 5.20$ .

(e) By (b), $PB^2 + PC^2 = 36 - PB \cdot PC$ and $PA^2 = 36 + PB \cdot PC$ . Adding, $PA^2 + PB^2 + PC^2 = 72$ , the same for every $P$ on the arc.

Check. Placing the triangle in coordinates and moving $P$ along the arc in small steps confirms (a), (b) and (e) at every step, the values in (c) and the two greatest values in (d).

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