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 $BE$?
Cyclic Pentagon
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1"distance from point $A$ to $BC$ is $a$". This is the perpendicular distance or what? – 2012-11-07
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0@amWhy the distance is the perpendicular distance, yes, in all cases. – 2012-11-07
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0@EuYu the distance is the perpendicular distance, yes, in all cases – 2012-11-07
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0@AmWhy not necessarily. Nothing in the question suggests that they are. – 2012-11-07
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0Actually, 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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0@amWhy I also think that they are, but how do we prove it? – 2012-11-08
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0It 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
1 Answers
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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0That'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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0@Robert How will that help me to find the distance between A and BE? – 2012-11-08
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0If you know $xy=zw$ and you know $x,y$ and $z$, how do you find $w$? – 2012-11-08
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0@Robert but BE is not any of AB, AC, AD, AE. – 2012-11-08
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0I 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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0@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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0Given 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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0Are we supposed to know the lengths of the sides of the pentagon ? – 2012-11-08
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0@VincentNivoliers: No, we are not. – 2012-11-08
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0Oh ! 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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1@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