Day 3: Inverted Thinking

Geometric inversion is not necessary, but a property given by it is useful.

Notice that, with respect to the circle, $P$ is the inverse of $R$ and $A$ is the inverse of itself. Hence we can use the similar triangle theorem in inversion to show that $OAP$ is similar to $ORA$ .

Therefore $\angle OAP = \angle ORA$ . But since $AOB$ is isosceles, $\angle OAP = 15^{\circ}$ and therefore $\angle ARP = \angle ORA = \boxed{15^{\circ}}.$

Power of P = AP $\times$ PB = OQ $^{2}$ - OP $^{2}$

OQ $^{2}$ - OP $^{2}$ = PQ $^{2}$ [ Since $\triangle$ OPQ is a right triangle

Again, we observe $\triangle$ OPQ $\sim$ $\triangle$ ROQ So, PQ $^{2}$ = OP $\times$ PR Therefore, OP $\times$ PR = AP $\times$ PB

This implies that the points O , A , R , B are concyclic. So $\angle$ ARO = $\angle$ ARP = $\angle$ ABO

Since $\angle$ ABO =150°, $\angle$ ABO = $\angle$ BAO = 15°

So the answer is 15

Brace yourselves.

Let the angles $QOP=\alpha, APR=\beta, AOR=\gamma, ARO=\delta$ . We want to find $\delta$ and know that $\angle OAB=15$ (I'll omit degree signs).

Let's set the scale so that $OQ=1$ . Then, from basic trigonometry, $OP=\cos\alpha, OR=\frac{1}{\cos\alpha}$ . From the law of sines for $\triangle AOP$ , we get $\sin\beta=\frac{\sin15}{\cos\alpha}$ .

$\gamma$ can be expressed easily from $\triangle AOP$ as $\beta-15$ .

From the law of cosines for $\triangle AOR$ , we obtain $AR^2=1+\frac{1}{\cos^2\alpha}-2\frac{\cos\gamma}{\cos\alpha}=1+\frac{1}{\cos^2\alpha}-\frac{2}{\cos\alpha}(\cos\beta\cos15+\sin\beta\sin15)$ using the identity for $\cos{x+y}$ ; we can further simplify it to $AR^2=1+\frac{1}{\cos^2\alpha}-\frac{2}{\cos\alpha}\left(\cos\beta\cos15+\frac{\sin^2{15}}{\cos\alpha}\right)=1+\frac{1-2\sin^2{15}}{\cos^2\alpha}-\frac{2\cos\beta\cos15}{\cos\alpha}=1+\frac{\cos^2{15}-\sin^2{15}}{\cos^2\alpha}-\frac{2\cos\beta\cos15}{\cos\alpha}\,.$

Next step: law of sines again, for $\triangle AOR$ . Using the expression for $AR^2$ , we get $1+\frac{\cos30}{\cos^2\alpha}-\frac{2\cos\beta\cos15}{\cos\alpha}=\frac{\sin^2\gamma}{\sin^2\delta}=\left(\frac{\sin\beta\cos15-\cos\beta\sin15}{\sin\delta}\right)^2=\left(\frac{\cos15}{\cos\alpha}-\cos\beta\right)^2\frac{\sin^2{15}}{\sin^2\delta}\,.$

Expanding the right side, we get $\left(\frac{\cos^2{15}}{\cos^2\alpha}+\cos^2\beta-\frac{2\cos15\cos\beta}{\cos\alpha}\right)^2\frac{\sin^2{15}}{\sin^2\delta}=\left(\frac{\cos^2{15}}{\cos^2\alpha}+1-\sin^2\beta-\frac{2\cos15\cos\beta}{\cos\alpha}\right)^2\frac{\sin^2{15}}{\sin^2\delta}=\left(\frac{\cos^2{15}-\sin^2{15}}{\cos^2\alpha}+1-\frac{2\cos15\cos\beta}{\cos\alpha}\right)^2\frac{\sin^2{15}}{\sin^2\delta}\,;$

notice that we got the same expression as on the left side, just multiplied by $\frac{\sin^2{15}}{\sin^2\delta}$ . Since the left side is non-zero, we can divide by it and get $\sin{15}=\sin{\delta}$ ; it's clear that $0 \le \delta \le 90$ , so $\delta=15$ . The only significant angle was that $15$ degree one, so this is easily generalizable.

Why do it nicely if you can use horrible analytics! :D