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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.

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    $OP$ is perpendicular to the floor? – 2017-01-05
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    @ArnaldoNascimento As the floor is a mirror .. I forgot to say. Yes, $OP$ is perpendicular to floor. – 2017-01-05
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    Is AB = 6 given? – 2017-01-05
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    @jnyan See the picture -_- – 2017-01-05

3 Answers 3

2

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 $y$.

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    You should take $y = AD$, but you have taken $y = AC'$. So, this is nearly wrong solution. Then $y$ will be $9$. Also $AC = y-3$ and so on. -_- But this is the easiest of all solutions :) (y) Now please change your picture. It is misleading. – 2017-01-05
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    OK I will correct it. – 2017-01-05
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enter image description here

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 $(2)$?

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    Good job, though I believe there might be a typo in RHS of $(2)$. shouldn't it be $6^2-(y-3)^2$? – 2017-01-05
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    @SiongThyeGoh I was also thinking this – 2017-01-05
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    @SiongThyeGoh, thank you! Fixed! – 2017-01-05
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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 $y$: $$y=(12-x)\cos\theta$$