A little geometry problem (Full Version)

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blacksword404 -> A little geometry problem (10/30/2008 9:41:20 PM)

I have a problem i need some help on.


If you have two people who are standing in the same spot. They both walk off. One straight ahead and the other to the right 45 degrees. The one who went straight is A and the other is B. B walks for a full day and then turns back towards his original direction 45 degress and walks for another full day. After he finishes his path intersects with the path of A. It creates an iscoceles triangle. I need to know how long A walked to intersect with B.



CalifChick -> RE: A little geometry problem (10/30/2008 10:13:13 PM)

I'm not sure if I'm visualing this correctly.  If I am, this is what I came up with:

I'm getting a right-triangle with the hypotenuse equaling one day (the first day that B walked - that is the only known length).  Since you have two interior angles equaling 45 degrees, the third (unknown) interior angle has to be 90 degrees, since the interior angles of all triangles have to equal 180.

In a right triangle that is also an isosceles triangle, the hypotenuse equals the length of the leg times the square root of 2.

So the hypotenuse, which is "one day" (1) is equal to the unknown length (U) times the square root of 2.

1 = U x (square root of 2)
1 = U x (1.189)
1/1.189 = U
.84 = U

The unknown length is .84 of a day, or 20.16 hours.

Or I could be completely wrong.  I'm still not sure I'm visualizing correctly.


Cali



Kirata -> RE: A little geometry problem (10/30/2008 10:24:10 PM)

quote:

ORIGINAL: CalifChick

I'm not sure if I'm visualing this correctly.

Unh, I kinda don't think so. If B bears right 45 degrees from A's path, and then (whenever) bears left 45 degrees, his path becomes parallel to A and they ain't never gonna meet.

K.
 
 



rexrgisformidoni -> RE: A little geometry problem (10/30/2008 10:24:15 PM)

I suck at math, but I bet they had a nice walk.



CalifChick -> RE: A little geometry problem (10/30/2008 10:30:10 PM)

quote:

ORIGINAL: Kirata

quote:

ORIGINAL: CalifChick

I'm not sure if I'm visualing this correctly.

Unh, I kinda don't think so. If B bears right 45 degrees from A's path, and then (whenever) bears left 45 degrees (back toward A), his path becomes parallel to A and they ain't never gonna meet.


It would have to be 135 degrees to become parallel; a straight line less the degree already used (180-45).



Cali



blacksword404 -> RE: A little geometry problem (10/30/2008 10:39:26 PM)

quote:

ORIGINAL: CalifChick

I'm not sure if I'm visualing this correctly.  If I am, this is what I came up with:

I'm getting a right-triangle with the hypotenuse equaling one day (the first day that B walked - that is the only known length).  Since you have two interior angles equaling 45 degrees, the third (unknown) interior angle has to be 90 degrees, since the interior angles of all triangles have to equal 180.

In a right triangle that is also an isosceles triangle, the hypotenuse equals the length of the leg times the square root of 2.

So the hypotenuse, which is "one day" (1) is equal to the unknown length (U) times the square root of 2.

1 = U x (square root of 2)
1 = U x (1.189)
1/1.189 = U
.84 = U

The unknown length is .84 of a day, or 20.16 hours.

Or I could be completely wrong.  I'm still not sure I'm visualizing correctly.


Cali



I think you are right on the math. And you are visualizing it right. But i think Kirata is right about the angle. I'll just have to change the return angle of B to 90 degrees. Thanks for the help Cali and Evil one. 



Kirata -> RE: A little geometry problem (10/30/2008 11:00:54 PM)

Right 45 and then left 45 puts him parallel. I figured it was a joke.

then turns back towards his original direction 45 degress 

I'm not sure what "his original direction" might be interpreted to mean in this context. He can't turn "back" 45 degrees. I do see what you mean, though, I think. He bears off from a full 180 degree reversal by 45 degrees toward A? But if that's the case, and he walks for a second full day, he would cross A's path and continue on past it. The point at which he would intersect A's path in that scenario, however, creates a right triangle with two 45 degree angles. A would have to walk 0.71 days (17.04 hours) to reach that intersect.

K.





Hippiekinkster -> RE: A little geometry problem (10/30/2008 11:16:26 PM)

A never intersects Bs path. And if you want to make a triangle, B would have to turn greater than 45 degrees back to the path of A. If B turned 45degrees to the right, then turned 45 degrees to the left, after the left turn, his path would parallel As.

Is there some kind of difficult problem you were trying to solve?



blacksword404 -> RE: A little geometry problem (10/30/2008 11:26:02 PM)

quote:

ORIGINAL: Kirata

Right 45 and then left 45 puts him parallel. I figured it was a joke.

then turns back towards his original direction 45 degress 

I'm not sure what "his original direction" might be interpreted to mean in this context. He can't turn "back" 45 degrees. I do see what you mean, though. He bears off from a full 180 degree reversal by 45 degrees toward A. But if that's the case, and he walks for a second full day, he would cross A's path and continue on past it.
 
The point at which he would intersect A's path, however, creates a right triangle with two 45 degree angles. A would have to walk 0.71 days (17.04 hours) to reach that intersect.

K.





It's for a story i am working on. A man coming back from the Sadar. He thinks he may be being followed so he attempts to lose the tail by riding off 45 degrees. Camping for a while then turning back toward his original path. And walking the same distance which will allow him to come in behind his assailant.



Kirata -> RE: A little geometry problem (10/30/2008 11:29:20 PM)

quote:

ORIGINAL: blacksword404

It's for a story i am working on. A man coming back from the Sadar. He thinks he may be being followed so he attempts to lose the tail by riding off 45 degrees. Camping for a while then turning back toward his original path. And walking the same distance which will allow him to come in behind his assailant.

I don't care why you're doing this to us! [:D]

K.



BossyShoeBitch -> RE: A little geometry problem (10/30/2008 11:31:41 PM)

quote:

ORIGINAL: CalifChick

I'm not sure if I'm visualing this correctly.  If I am, this is what I came up with:

I'm getting a right-triangle with the hypotenuse equaling one day (the first day that B walked - that is the only known length).  Since you have two interior angles equaling 45 degrees, the third (unknown) interior angle has to be 90 degrees, since the interior angles of all triangles have to equal 180.

In a right triangle that is also an isosceles triangle, the hypotenuse equals the length of the leg times the square root of 2.

So the hypotenuse, which is "one day" (1) is equal to the unknown length (U) times the square root of 2.

1 = U x (square root of 2)
1 = U x (1.189)
1/1.189 = U
.84 = U

The unknown length is .84 of a day, or 20.16 hours.

Or I could be completely wrong.  I'm still not sure I'm visualizing correctly.


Cali



I love you...[sm=hearts.gif]



GreedyTop -> RE: A little geometry problem (10/30/2008 11:33:09 PM)

*eyes glaze over*  Sword.. darlin.. you know I like you.. but this is just CRUEL!!



lemmebeYourMine -> RE: A little geometry problem (10/30/2008 11:45:57 PM)

NO you don't, it's a ******* story and who the **** is going to stop and do the math. Pull a number out of your hat. Sheesh. 34 hours? 75? Whatever amount of time you deem fit works for all but 10% of the rest of the world, and they are too busy doing math problems to read your story.

Just my two cents worth.

Cuz I know I wouldn't care.

lemmebe.



blacksword404 -> RE: A little geometry problem (10/30/2008 11:54:58 PM)

quote:

ORIGINAL: GreedyTop

*eyes glaze over*  Sword.. darlin.. you know I like you.. but this is just CRUEL!!


Lol. What did i do? I just needed to make sure of the math before i wove it into the story. Don't worry no math in the story.



GreedyTop -> RE: A little geometry problem (10/30/2008 11:57:32 PM)

math *TWITCH TWITCH TWITCH*



blacksword404 -> RE: A little geometry problem (10/31/2008 12:05:40 AM)

quote:

ORIGINAL: GreedyTop

math *TWITCH TWITCH TWITCH*


Oh no. She cracked. *dangle shiny keys in front of her eyes*



MadAxeman -> RE: A little geometry problem (10/31/2008 12:07:01 AM)

So it's a thrilling story of our hero being followed by an assailant on a piece of graph paper?
Why can't he hide behind a feckn tree like a regular person?



GreedyTop -> RE: A little geometry problem (10/31/2008 12:07:24 AM)

SPAAAAAAAAARRRRRRRRRRRRRRRRRRRRRRRKLIEEEEEEEEEEEEEEEEEEEESSSSSSSSSS!!!!!!!!!!!!!!!!!!!



persephonee -> RE: A little geometry problem (10/31/2008 12:10:14 AM)

hey dar GT....whatcha doin....



GreedyTop -> RE: A little geometry problem (10/31/2008 12:12:16 AM)

huh? oooh...Pers.... even better than sparklies!!   *tacklegrope*



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