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Fixing It

Stage: 5 Challenge Level: Challenge Level:1
$\mathbf{A}$ and $\mathbf{B}$ are fixed points on a circle, while $\mathbf{RS}$ is a variable diameter. $\mathbf{P}$ is the intersection of $\mathbf{AR}$ and $\mathbf{BS}$. We make the conjecture that the locus of $\mathbf{P}$ isa a second circle through the points $\mathbf{A}$ and $\mathbf{B}$.

Edward Wallace of Graveney School, Tooting sent a very good proof of this conjecture.

First consider the two cases illustrated in the given diagrams. It is impossible to have $\mathbf{R}$ and $\mathbf{S}$ both on the minor arc of the circle. The other two cases, where $\mathbf{R}$ and $\mathbf{S}$ exchange positions, and $\mathbf{P}$ is outside the original circle, are proved similarly.

Case 1

Both $\mathbf{R}$ and $\mathbf{S}$ are on the major arc $\mathbf{AB}$ so that $\mathbf{P}$ is inside the original circle.

Case 2

$\mathbf{S}$ is on the minor arc $\mathbf{AB}$ and $\mathbf{R}$ on the major arc so that $\mathbf{P}$ is outside the original circle.

In Case 1, $\angle\mathbf{ARB}$ is unchanging (invariant) as $\mathbf{R}$ moves on the circumference because angles subtended on the circumference of a circle by a fixed arc of the circle are equal. Hence ($\angle \mathbf{ARB} = m^o$ (constant). Also $\angle \mathbf{RBS}=90^o$ because the angle subtended by a diameter is a right angle.

It follows that $\angle \mathbf{RPB}=(90 - m)^o$ because the angles of the triangle $\mathbf{BRP}$ add up to $180^o$.

Hence $\angle \mathbf{APB}$ is constant and equal to $(90 + m)^o$ because angles on a line add up to $180^o$. This is a necessary and sufficient condition for P to lie on the circumference of a circle of which $\mathbf{AB}$ is a chord. Let us call this arc $\mathbf{AP_{1}B}$ to emphasise the fact that, in this case, $\mathbf{P_1}$ is inside the original circle.

In Case 2, exactly the same argument applies, giving $\angle \mathbf{ARB} = m^o$ (the same value as in Case 1) and $\angle\mathbf{RBS}=90^o$, but in this case $\angle\mathbf{APB}$ is the same angle as $\mathbf{RPB}$ so $\angle\mathbf{APB}=(90 - m)^o$ (again constant).

As before, this is a necessary and sufficient condition for $\mathbf{P}$ to lie on the circumference of a circle of which $\mathbf{AB}$ is a chord. We call this arc $\mathbf{AP_{2}B}$ to emphasise the fact that now $\mathbf{P_2}$ is outside the original circle.

Note that:
$\angle\mathbf{AP_{1}B} + \angle\mathbf{AP_{2}B} = (90 + m)^o + (90 - m)^o = 180^o$ and hence $\mathbf{AP_{1}BP_{2}}$ is a cyclic quadrilateral which confirms that the locus of $\mathbf{P}$ is a single circle (and not two arcs of different circles).

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