Problem using triangular inequality

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CarlosP

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Let l be a line, O be a point of l, and P, Q be two points outside l located in the same semiplane bounded by l. Determine from among all pairs of points A, B belonging to l such that O belongs to AB and | OB | = 2 | OA |, the one for which the expression | BQ | + 2 | AP | take the maximum value.
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Can't solve that problem, can someone give me some clue how to solve this ? Thanks :)
 
Points O, P, Q are fixed. Points A and B can move with a constraint | OB | = 2 | OA | . We want to maximize | BQ | + 2 | AP | .
If A and B keep moving away from O the value of | BQ | + 2 | AP | increases. Not sure there is a maximum...
 
lev888 said:
Points O, P, Q are fixed. Points A and B can move with a constraint | OB | = 2 | OA | . We want to maximize | BQ | + 2 | AP | .
If A and B keep moving away from O the value of | BQ | + 2 | AP | increases. Not sure there is a maximum...
Click to expand...
sorry but in the question says : " the one for which the expression | BQ | + 2 | AP | take the maximum value " , so i assume there are a maximum value.
 
CarlosP said:
sorry but in the question says : " the one for which the expression | BQ | + 2 | AP | take the maximum value " , so i assume there are a maximum value.
Click to expand...
I see that. That's why I wrote "We want to maximize | BQ | + 2 | AP | "
But either I misunderstand something or you didn't copy the problem correctly.
 
lev888 said:
I see that. That's why I wrote "We want to maximize | BQ | + 2 | AP | "
But either I misunderstand something or you didn't copy the problem correctly.
Click to expand...
You're right , i didn't copy the problem correctly. we want the minimum value ! when i translate to english i thought that said "maximum" instead of "minimum"
 
I think you should find a good candidate for the minimum and then apply the triangular inequality to prove that any other solution would increase the value of | BQ | + 2 | AP | .
 
lev888 said:
I think you should find a good candidate for the minimum and then apply the triangular inequality to prove that any other solution would increase the value of | BQ | + 2 | AP | .
Click to expand...
Hmm can you help me?

We want minimize | BQ | + 2 | AP | , but to minimize |BQ| then B should be the most close to point O. But to minimize |AP| we right the triangular inequality : |AP| < |B'P| + |B'A| , but |B'A| will be |OA| since we put B the closest possible to O ( B' is reflected in point O ). So, if , |AP| < |PO| + |OA| , then we are saying that A should be the closest possible to O. Is that right ?
 
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"to minimize |BQ| then B should be the most close to point O "

Are you sure? Look at the diagram. As B moves towards O, where is BQ shortest?

Don't quite follow the second part - |AP| < |B'P| + |B'A|. Why are you considering BP?
 
But my problem is : that image is only a example to help people understand the problem and try visualize. So the point P and Q could be in different locations in the same semiplane bounded by l. So how i prove where is that points A,B that satisfy this problem?
 
lev888 said:
"to minimize |BQ| then B should be the most close to point O "

Are you sure? Look at the diagram. As B moves towards O, where is BQ shortest?

Don't quite follow the second part - |AP| < |B'P| + |B'A|. Why are you considering BP?
Click to expand...
Yes you were right in that point. B should be close to point O, just need be under Q. But i don't know how i translate this to "mathematical language" and prove where are that points.

The second part i just trying to make a triangle to use the triangular inequality. Thats why i reflect B across line l.
 
CarlosP said:
But my problem is : that image is only a example to help people understand the problem and try visualize. So the point P and Q could be in different locations in the same semiplane bounded by l. So how i prove where is that points A,B that satisfy this problem?
Click to expand...
For different P and Q we will get different solutions for A and B. So, I am assuming, we need to deal with this specific diagram.
 
CarlosP said:
Yes you were right in that point. B should be close to point O, just need be under Q. But i don't know how i translate this to "mathematical language" and prove where are that points.

The second part i just trying to make a triangle to use the triangular inequality. Thats why i reflect B across line l.
Click to expand...
Yes, BQ is shortest when BQ is normal to the line l. But is this the minimum for |BQ| + 2 |AP| ?
Visualize, what happens if we move A and B closer to O. |BQ | increases, |AP| decreases. But what about |BQ| + 2 |AP|?
 
lev888 said:
Yes, BQ is shortest when BQ is normal to the line l. But is this the minimum for |BQ| + 2 |AP| ?
Visualize, what happens if we move A and B closer to O. |BQ | increases, |AP| decreases. But what about |BQ| + 2 |AP|?
Click to expand...
Then we need find the closest possible point A such that |AP| is the minimum distance, because when |BQ| increases , 2|AP| decreases more then |BQ|. Right ?
 
Yes, the factor of 2 indicates that we'll "benefit" more from decreasing |AP|. I don't know whether it's true for any A and B. Assuming it is true, what's the solution for min |AP|?
 
lev888 said:
Yes, the factor of 2 indicates that we'll "benefit" more from decreasing |AP|. I don't know whether it's true for any A and B. Assuming it is true, what's the solution for min |AP|?
Click to expand...
See if my thoughts were right please:
Since reflections preserve distances, minimize |AP| is the same as minimizing |A'P'| (A' is the reflection of A in the line l, and P' is the simetry of P in the line l.
Considering the triangle P'A'Q, |P'A| + |A'Q| < |P'Q|. Then A' belongs to P'Q. (So we find point A 'and consequently point A and point B to solve the question).
 
Sorry, please clarify where exactly A' and P' are. Or better post an image.
 
You wrote: "Then we need find the closest possible point A such that |AP| is the minimum distance"
So, where should we put A to minimize |AP|?
This would give us a potential solution.
 
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