Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

enter image description here

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 $BE$?

share|improve this question
1  
"distance from point $A$ to $BC$ is $a$". This is the perpendicular distance or what? –  EuYu Nov 7 '12 at 22:34
    
@amWhy the distance is the perpendicular distance, yes, in all cases. –  George Krasilnikov Nov 7 '12 at 23:16
    
@EuYu the distance is the perpendicular distance, yes, in all cases –  George Krasilnikov Nov 7 '12 at 23:17
    
@AmWhy not necessarily. Nothing in the question suggests that they are. –  George Krasilnikov Nov 7 '12 at 23:46
    
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. –  amWhy Nov 8 '12 at 0:39

1 Answer 1

up vote 3 down vote accepted

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 $$d(A,BC)d(A,DE)=4r^2|\sin(\beta/2)\sin(\gamma/2)\sin(\delta/2)\sin(\epsilon/2)|=d(A,BE)d(A,CD)$$

share|improve this answer
    
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. –  Blue Nov 7 '12 at 23:53
    
@Robert How will that help me to find the distance between A and BE? –  George Krasilnikov Nov 8 '12 at 0:07
    
If you know $xy=zw$ and you know $x,y$ and $z$, how do you find $w$? –  Robert Israel Nov 8 '12 at 0:11
    
@Robert but BE is not any of AB, AC, AD, AE. –  George Krasilnikov Nov 8 '12 at 0:12
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. –  George Krasilnikov Nov 8 '12 at 13:37

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.