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kevincan
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Two polar bears are now at points A and B. The two bears are Updated on: 11 Jul 2006, 00:10
Originally posted by kevincan on 09 Jul 2006, 05:03.
Last edited by kevincan on 11 Jul 2006, 00:10, edited 1 time in total.
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Two polar bears are now at points A and B. The two bears are travelling at the same constant speed in different directions, as shown in the drawing attached. Approximately how far apart in miles will the bears be when they are at points A´ and B´, if the distance from A to B is 3 miles ?

(A) 1.6 (B) 1.8 (C) 2.0 (D) 2.2 (E) 2.4



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Two polar bears are now at points A and B. The two bears are 10 Jul 2006, 17:50
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Kevin,

I have attached a file...

I need a hint...please....

C = (3+x) = sqrt(3)*x

but I also feel like this has to do with some ratio...
what am I missing?

thank in advance...
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bear.doc [30.5 KiB]
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Two polar bears are now at points A and B. The two bears are 10 Jul 2006, 18:03
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I got to the same point!

3+x = x*sqrt(3)

=> x = 3/(sqrt(3)-1)

From here I can even find BA' = 6/(sqrt(3)-1)

But I am missing something to get to A'B' distance. Huh?
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Two polar bears are now at points A and B. The two bears are 10 Jul 2006, 18:26
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I got to the same point as both of you...where the distance from B to the intersection of the perpendicular from A' to the horizontal is
3/(sqrt(3)-1)

Couldn't get beyond that...
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Two polar bears are now at points A and B. The two bears are 10 Jul 2006, 23:20
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v1rok
I got to the same point!

3+x = x*sqrt(3)

=> x = 3/(sqrt(3)-1)

From here I can even find BA' = 6/(sqrt(3)-1)

But I am missing something to get to A'B' distance. Huh?
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A'B' - ?

BB' = AA'.

AA' = sqrt(2) * sqrt(3) * x = sqrt(6) * x

BA' = 2*x.

A'B' = BB' - BA' = AA' - BA' = x * (sqrt(6) - 2 ) = 3*(sqrt(6) - 2 ) / (sqrt(3)-1)

But A'B' is not from answer list ....
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Two polar bears are now at points A and B. The two bears are 11 Jul 2006, 00:07
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What if you use these approximations?

sqr root (3)=7/4
sqr. root (5)=9/4
sqr root (6) =2.45=49/20 THESE ARE NICE TO KNOW
sqr root (7)=8/3

3*(sqrt(6) - 2 ) / (sqrt(3)-1) =3(0.45)/(3/4)= 5.4/3=1.8

I've changed the question to include the word "approximately" to make it clear that we are looking for the best answer.

Excellent work! I thought everybody had something against polar bears, and as a Canadian, I was a bit hurt :cry:
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Two polar bears are now at points A and B. The two bears are 11 Jul 2006, 10:38
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Ahhh! The key was that the bears were traveling at THE SAME CONSTANT SPEED, so they should cover the same distance in the same amount of time, so AA'=BB'! I was consumed by the drawing, I completely forgot about the speed condition!
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Two polar bears are now at points A and B. The two bears are 11 Jul 2006, 10:41
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Aha!!! :-)

Good point... Great Question Kevincan!

v1rok
Ahhh! The key was that the bears were traveling at THE SAME CONSTANT SPEED, so they should cover the same distance in the same amount of time, so AA'=BB'! I was consumed by the drawing, I completely forgot about the speed condition!
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Two polar bears are now at points A and B. The two bears are 11 Jul 2006, 16:20
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kevincan
What if you use these approximations?

sqr root (3)=7/4
sqr. root (5)=9/4
sqr root (6) =2.45=49/20 THESE ARE NICE TO KNOW
sqr root (7)=8/3

3*(sqrt(6) - 2 ) / (sqrt(3)-1) =3(0.45)/(3/4)= 5.4/3=1.8

I've changed the question to include the word "approximately" to make it clear that we are looking for the best answer.

Excellent work! I thought everybody had something against polar bears, and as a Canadian, I was a bit hurt :cry:
Show more


I need to read the question more carefully...
nice question...will make a note of the square root values



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