High school MCQ (geometry) about maximum of sine of an angle

Question

No calculus is allowed in this question.

Also, it is not allowed to use any compound angle formulas such as

$\sin(A-B) =\sin A \cos B - \cos A \sin B$.

How to find the greatest value of $\sin \angle APB$?

I attempted by consider the difference between angles $\angle OPB$ and $\angle OPA$ but I do now know how to make use of the fact that $P$ makes the angle greatest.

Answer 1

Recall from extended sine-rule, $$\sin \angle APB = \frac{AB}{2R_{\triangle APB}}$$

Since $AB$ is fixed, to maximize $\sin \angle APB$, we have to minimize the circumradius $R_{\triangle APB}$. As the circumcenter $S$ lies on perpendicular bisector of $AB$, $R_{\triangle APB}$ will be least when $SP \perp OC$ as in the following diagram.

Using that $SPOD$ is a rectangle and $A$ is mid of $OB$,

$$\text{max of sin} \angle APB = \frac{1}{3}$$

Answer 2

This is not a geometric solution and may not be the most recommended approach but meets the condition of not using calculus and no compound angle formula.

As $\angle APB \lt 90^0,$ maximizing $\sin \angle APB \ $ will maximize $\angle APB$.

Say $OA = AB = a, OP = k \cdot a \ $ ($k \gt 0)$

Equating the area of $\triangle APB$ and $\triangle OAP$

$\frac{1}{2} AP \cdot BP \sin \angle APB = \frac{1}{2} OP \cdot OA$

$\sin \angle APB = \displaystyle \frac{k \cdot a^2}{\sqrt{a^2+k^2a^2} \cdot {\sqrt{4a^2+k^2a^2}}} = \frac{k}{\sqrt{k^4+5k^2+4}}$

$ = \displaystyle \frac{1}{\sqrt{k^2+\frac{4}{k^2}+5}}$

$\sin \angle APB$ is maximum when $k^2+\frac{4}{k^2}$ is minimum.

Now by A.M-G.M, $k^2+\frac{4}{k^2} \geq 4$

Hence $\sin \angle APB = \displaystyle \frac{1}{3}$.

Answer 3

Another approach:

In figure OC=OB. As canbe seen $\APB$ is maximum when P is at distance $\frac 34 OC$. You can find that:

$AD=\frac{\sqrt 2}4 OB$

$AC=\frac{\sqrt 5}2 OB$

$\sin (\angle ABC)=\frac {AD}{AC}\approx 0.316$

$\sin (\angle APB)>0.316$

So option $\frac 13\approx 0.3333>0.316$ is corect.