Circle Angle Theorems

A and B sit at 330° and 110° on the circle, and P is at 220°. The panel says the angle at P is 70° and the angle at the centre is 140°. Prove the factor of two, rather than reading it — and then find what that proof quietly assumes about where P is.

1. Measure the arc first. Going anticlockwise from A at 330° round to B at 110° sweeps 140°, and that arc is the one P does not sit on. The angle at the centre standing on it is 140° by definition, because a central angle simply is its arc.

   $$\stackrel{⌢}{\mathrm{AB}}=110^{\circ }-330^{\circ }(\text{mod }360)=140^{\circ }$$

2. Now the construction, and it is the whole proof: draw the line from P through the centre O and let it leave the circle at D. Nothing else is added.

3. OA and OP are both radii, so triangle OAP is isosceles and the angles at A and at P are equal. Call that angle α, so ∠OPA = ∠OAP = α.

   $$OA=OP=r\Rightarrow \angle OPA=\angle OAP=\alpha$$

4. The angle ∠AOD is exterior to triangle OAP at O, and an exterior angle equals the sum of the two interior angles it does not touch: ∠AOD = α + α = 2α.

   $$\angle AOD=\alpha +\alpha =2\alpha$$

5. The same argument on the other side gives ∠DOB = 2β, where β = ∠OPB. Add the two, which works while OP produced meets the circle on the arc AB: ∠AOB = 2α + 2β = 2(α + β), and α + β is exactly ∠APB.

   $$\angle AOB=2\alpha +2\beta =2(\alpha +\beta )=2\angle APB$$

6. So the central angle is twice the inscribed one, and here 2 × 70° = 140°. Read back what the proof used: two radii, one isosceles triangle each side, one exterior angle. No length, no coordinate, and no measurement of the arc.

   $$2\times 70^{\circ }=140^{\circ }$$

Both marked points see the chord AB at 60°. Work out what a point off the circle would see instead, then decide where a kicker should stand to get the widest view of the posts.

1. A and B sit at 320° and 80°, so the arc between them that P does not stand on measures 120°. Half of it is 60°, and Q reads the same because it stands on the same arc as P.

   $$\stackrel{⌢}{\mathrm{AB}}=80^{\circ }-320^{\circ }(\mathrm{mod}360)=120^{\circ }⟹\angle APB=\angle AQB=60^{\circ }$$

2. Now the chord itself. The central angle on that arc is twice the inscribed one, and a chord is 2R sin of half the central angle, so AB = 2 sin 60° = √3 on this circle of radius 1.

   $$AB=2R\sin (\angle APB)=2\sin 60^{\circ }=\sqrt{3}=1.7321(R=1)$$

3. Read that line backwards. Hold the chord and let the circle vary: sin of the inscribed angle is AB ÷ 2R, so a smaller circle through A and B gives a larger angle, and the smallest one has AB as a diameter and gives 90°. Which circle you stand on decides everything.

   $$\sin (\angle APB)=\frac{AB}{2R}⟹R\text{smaller}\Rightarrow \angle APB\text{larger},R=\frac{AB}{2}\Rightarrow \angle APB=90^{\circ }$$

4. Off the circle the angle is no longer 60°. From inside, the two lines cut a second arc as well and the angle is half the sum of the two: a little way in from the arc it reads about 71.6°. From outside it is half the difference, about 51.4°. The circle is the boundary between seeing more and seeing less.

   $$\text{inside: }\frac{1}{2}\left(\stackrel{⌢}{\mathrm{AB}}+\stackrel{⌢}{\mathrm{CD}}\right)=71.6^{\circ }>60^{\circ }>51.4^{\circ }=\frac{1}{2}\left(\stackrel{⌢}{\mathrm{AB}}-\stackrel{⌢}{\mathrm{CD}}\right)\text{: outside}$$

5. That settles the kick. Rugby posts stand 5.6 m apart; a try is scored 10 m out from the near post, so from the foot of the run-back line the posts are 10 m and 15.6 m away. Every point that sees them at the same angle lies on one circle through both.

   $$a=10\mathrm{m},b=15.6\mathrm{m},\tan \angle =\frac{(b-a)x}{x^2+ab}$$

6. Each point on the run-back line lies on exactly one such circle, and the angle is largest on the smallest circle that reaches the line at all — the one tangent to it. The touch point is a tangent from the foot, and a tangent length is the geometric mean of the two secant pieces: x² = 10 × 15.6 = 156.

   $$x^2=FA\cdot FB=ab=156⟹x^{\ast }=\sqrt{156}=12.49\mathrm{m}$$

7. So x = √156 = 12.49 m, where the angle is 12.64°. At the 10 m mark it is 12.34°, at 20 m it has fallen to 11.39°, and at 30 m to 9.04°.

   $$\tan {\angle }_{\max }=\frac{b-a}{2\sqrt{ab}}=\frac{5.6}{24.98}=0.2242\Rightarrow 12.64^{\circ };\angle (10)=12.34^{\circ },\angle (20)=11.39^{\circ },\angle (30)=9.04^{\circ }$$