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Consider the above pentagon. Suppose that the distance from point $A$ to $BC$ is $a$, the distance from $A$ to $CD$ is $b$, and the distance from $A$ to $DE$ is $c$. In terms of this, how can we find the distance from $A$ to $BE$?

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    "distance from point $A$ to $BC$ is $a$". This is the perpendicular distance or what?2012-11-07
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    @amWhy the distance is the perpendicular distance, yes, in all cases.2012-11-07
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    @EuYu the distance is the perpendicular distance, yes, in all cases2012-11-07
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    @AmWhy not necessarily. Nothing in the question suggests that they are.2012-11-07
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    Actually, your particular image suggests they are equivalent; and since you left out the important specification of what you mean by "distance" of A to ______, I thought it best to have you clarify whether or not the lengths BC, CD, DE were equivalent, etc.2012-11-08
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    @amWhy I also think that they are, but how do we prove it?2012-11-08
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    It can't be proven, given the problem statement. If the objective of the problem were to maximize or minimize the perpendicular distance from A to BE, then we could say something about the optimal lengths of the sides of the pentagon for doing so.2012-11-08

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Hint: express each distance in terms of the radius of the circle and the cosines of the angles subtended at the centre by $AB$, $AC$, $AD$ and $AE$. If I'm not mistaken, you should find that the product of two of the distances is equal to the product of the other two.

EDIT: in fact, if $\beta$ and $\gamma$ are the angles subtended at the centre by $AB$ and $AC$ and the radius is $r$, I find that the distance from $A$ to $BC$ is $2 r |\sin(\beta/2) \sin(\gamma/2)|$. Similarly of course for the other distances.

EDIT (incorporating comment as requested): Given that $d(A,BC)=2r|\sin(\beta/2)\sin(\gamma/2)|$ and similarly $d(A,CD)=2r|\sin(\gamma/2)\sin(\delta/2)|$, $d(A,DE)=2r|\sin(\delta/2)\sin(\epsilon/2)|$ and $d(A,BE)=2r|\sin(\beta/2)\sin(\epsilon/2)|$, we have $$d(A,BC)d(A,DE)=4r^2|\sin(\beta/2)\sin(\gamma/2)\sin(\delta/2)\sin(\epsilon/2)|=d(A,BE)d(A,CD)$$

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    That's what I'm getting, but my derivation a bit cluttered. Considering how nice the relation is, I keep thinking there should be an elegant path to it.2012-11-07
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    @Robert How will that help me to find the distance between A and BE?2012-11-08
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    If you know $xy=zw$ and you know $x,y$ and $z$, how do you find $w$?2012-11-08
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    @Robert but BE is not any of AB, AC, AD, AE.2012-11-08
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    I didn't say it was. I'm talking about the distances from $A$ to $BC$, to $CD$, to $DE$ and to $BE$.2012-11-08
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    @Robert how would we get the answer without $r$ in it? If you could expand on your initial hint toward a solution it would be very much appreciated. I am not able to eliminate r from the answer.2012-11-08
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    Given that $d(A,BC) = 2 r |\sin(\beta/2) \sin(\gamma/2)|$ and similarly $d(A,CD) = 2 r |\sin(\gamma/2) \sin(\delta/2)|$, $d(A,DE) = 2 r |\sin(\delta/2) \sin(\epsilon/2)|$ and $d(A,BE) = 2 r |\sin(\beta/2)\sin(\epsilon/2)|$, we have $$d(A,BC) d(A,DE) = 4 r^2 |\sin(\alpha/2) \sin(\beta/2) \sin(\gamma/2) \sin(\delta/2)|= d(A,BE) d(A,CD)$$2012-11-08
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    Are we supposed to know the lengths of the sides of the pentagon ?2012-11-08
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    @VincentNivoliers: No, we are not.2012-11-08
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    Oh ! right, I should learn how to read before commenting. I finally agree with you on your answer, and then mine is wrong, I'll delete it.2012-11-08
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    @If you can add your comments into a full solution as part of your answer, that will be great- then I can mark it as correct.2012-11-08