BdPhO 2016 Category C Set 4 Problem 10

Question

In the above picture, the floor is a mirror. So, $\angle AOP = \angle POB$ and $AO + BO = 12cm$, find $y$.
I started with $OB = 12-AO$ but that is leading me to too much complexity that I can't handle. How should I start? Full solution of any hint will be helpful.

Note: This is a problem from BdPhO 2016. But as it is truly math related so I posted it here.

Answer 1

Reflect segment $OB$ along $DE$ to obtain $B'$. So $BE=EB'=3$.

Also $BO=B'O\implies AO+OB=AO+OB'=AB'=12$.

$AB'$ is a straight line since $\angle AOD=\angle BOE =\angle B'OE$.

$C$ and $C'$ are points on line $AD$ such that $AC\perp CB$, $AC'\perp C'B'$.

So $CD=DC'=3$, $AC=y-3$.

Notice $CB=C'B'$. Consider $\triangle ACB$ and $\triangle AC'B'$, we get

$$6^2-(y-3)^2=12^2-(y+3)^2$$

Then solve for $y$.

Answer 2

The $\angle (AO, floor)=\angle (BO, floor)=x$.

Then $y=AO \sin x$ and $3=BO \sin x$ so

$$AO+BO=12 \rightarrow (y+3)\cdot\sin x=12 \quad (1)$$

Also , in the triangle $ABC$:

$$[(AO+BO)\cos x]^2=6^2-(y-3)^2$$

$$(12\cos x)^2=6^2-(y-3)^2 \quad (2)$$

Can you finish using $(1)$ and $(2)$?

Answer 3

Hint:

Take $\angle AOB=2\theta$ and OB=x. Then we have, $$x\cos\theta=3$$ From cosine rule in $\triangle AOB$ $$6^2=(12-x)^2+x^2-2(12-x)(x)\cos2\theta$$ Solve for $x$ and $cos\theta$ using above two eqs. To get $y$: $$y=(12-x)\cos\theta$$