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RE: A little geometry problem - 11/8/2008 9:45:06 PM

corsetgirl

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quote:

ORIGINAL: rexrgisformidoni

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

Looking at that problem gives me a headache. All I could think of is a former, old, retired military math teacher in my high school. Not a good mental picture of this man because he was scary! Maybe that is why I still have a fear of these problems today.

< Message edited by corsetgirl -- 11/8/2008 9:52:32 PM >

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RE: A little geometry problem - 11/8/2008 9:49:24 PM

michaelOfGeorgia

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quote:

ORIGINAL: blacksword404

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.

one full day, if all sides are equal

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RE: A little geometry problem - 11/8/2008 10:44:16 PM

aDayIntheLife

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Clarifying that "A" is a path 45 degrees away from "B" and that "A" turns back toward "B" with a second 45 degree angle, this not being clear in the original problem, but insinuated by the fact that "A" does in fact intersect "B" on the second day, so "A" must turn right (toward) and not left (away) from "B"

Then you have:

Both "A" and "B" start at the origin of the graph and B journies on the x-axis, while "A" moves up toward (1+i) on this graph. Once "A" begins its second day it goes 45 degrees again, headed toward the x-axis.

The distance is:

To get the value of “c” or the distance of one day of travel for “B” you must use algebra and take the square root of each side of the equation:

√ (a2 + b2 ) = √ c2

√ (a2 + b2 ) = c

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RE: A little geometry problem - 11/9/2008 5:56:58 AM

FullCircle

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quote:

ORIGINAL: came4U
If they could walk non-stop, considering the earth is approx. 25040 mi in circumfrence it would at a pace of 3ish miles per hour (8346 hrs total)...take 347 days. They would have intersect at the 225 degrees point except the one guy who has an extra day of walking because he turned around and had lost a day. Depending on which guy, one can come up upon the other and beat him to that intersection.

They'd need a boat at some point.

quote:

*edit to add: if the assassin was the guy walking straight, he can beat the other guy to the intersect (considering he kept going), if the assassin was was the one who went at the 45 deg angle, he travelled two days to return back to the origional starting point, he only has to return to that point on the 299th day to get his mark.

A = Assassin
B = Bin Laden
From the above it can be seen that A and B will never meet.

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RE: A little geometry problem - 11/9/2008 6:02:28 AM

FullCircle

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quote:

ORIGINAL: michaelOfGeorgia
one full day, if all sides are equal

An isosceles triangle has two equal sides
An equilateral triangle has three equal sides and each angle would have to be 60 degrees not 45 degrees.

Apart from this I see nothing wrong with your solution.

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RE: A little geometry problem - 11/9/2008 8:30:18 AM

came4U

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They'd need a boat at some point.

I did say:

quote:

If they could walk non-stop

did I not???

quote:

A = Assassin
B = Bin Laden
From the above it can be seen that A and B will never meet.

huh?

Do the math. They can meet, is the point of my post, if one does a lot of walking, the other lives their life to only wait at a much later date. Depending on which one is the assassin and which is not, one can outsmart the other either way.

besides:

quote:

An isosceles triangle has two equal sides
An equilateral triangle has three equal sides and each angle would have to be 60 degrees not 45 degrees.

*any angle from the same starting point would eventually meet any given round surface (such as the earth). Even if both pass the equator....given the bulge and influx of area at the equator, due time is made for both (since the earth is more oval in that section). Time and space do not adjust for the area counted if walking straight lines at any degree crossing the equator.

Since this particular riddle refers to the Sadar Market, it indeed does cross the equator at 45 degrees. From India heading right at 45, or a any given n,s,w,e, position, it must cross the equator. The calculated time is the same for both.

The answer would be the same wheather it be at a 7deg. posting or a 65deg. etc. Ignore the angle degree and work on the actual circumfrence value. Ignore also describing angles as if it were some halo to a math math paper. You can jot down diagrams all you want, you just don't get it.

so the guys's question:

quote:

I need to know how long A walked to intersect with B.

depends on the above calculations which I have already answered in assumption of a pace of 3mi per hour.

< Message edited by came4U -- 11/9/2008 9:09:45 AM >

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RE: A little geometry problem - 11/10/2008 12:05:57 AM

came4U

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Dammit, I almost woke up screaming in my sleep ...I forgot one major thing...first of all, *the total walk is 14,458 hours, not 8346 as I orig. posted. No clue how I got that number. Ignore it anyways. It isn't relevant. I also said 299th day, wtf? That was mighty mental of me, no clue what I meant by that either. It should be 'if B returns after 345 days to the origional spot, he can wait for A or...

The two can also criss-cross at the half way point on the opposite side of the earth duh me. Remember, they are walking non stop, 3 miles an hour making it 72 miles/day. This time, A is ahead of B, and can wait for B at mile 12520, this would be the 173rd day. B, being 144 miles behind, would arrive on the 175th day. I previously only accounted for and calculated for them meeting at the same point they started. Sorry about that. In either scenerio someone gets away or someone gets killed.

< Message edited by came4U -- 11/10/2008 1:11:57 AM >

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RE: A little geometry problem - 12/6/2008 2:36:22 PM

MrAlex43

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16hrs58min and 23 secs

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RE: A little geometry problem - 12/6/2008 5:00:49 PM

windchymes

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RE: A little geometry problem - 12/6/2008 5:03:42 PM

Raechard

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Are both person A and person B walking at the same speed and how long are person B's legs as a ratio of Person A's?

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