Math Problem ...
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You should change your name to OneFifthMan.
Urm... I don't get it :((
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Urm... I don't get it :((
0.5 = 1/2. 0.2 = 1/5. Get it???? :laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh: :laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh: :laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh: :laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh: :laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh: :^)
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0.5 = 1/2. 0.2 = 1/5. Get it???? :laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh: :laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh: :laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh: :laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh: :laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh::laugh: :^)
But 20% = 0.2. I hope. :^) -- modified at 13:03 Tuesday 13th June, 2006 ;)
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Hey Guys, I have a math issue, I know how you like brain teasers so I thought I would post it here, say I have two points ab & cd, how do I find the point 20% of distance between these two points on the same slope ? Regards Ray "Je Suis Mort De Rire" Blogging @ Keratoconus Watch
I'm no math whiz, but can't you just:
.2(a+c)=e .2(b+d)=f ef=20% between ab & cdThat was stupid. I thought about it, and I think this would work:a+(.2(c-a))=e b+(.2(d-b))=f ef=20% between ab & cdIf that isn't right, then I'm too dumn to figure it out without more coffee. ---------- Some problems are so complex that you have to be highly intelligent and well informed just to be undecided about them. - Laurence J. Peters -- modified at 13:11 Tuesday 13th June, 2006 -
Hey Guys, I have a math issue, I know how you like brain teasers so I thought I would post it here, say I have two points ab & cd, how do I find the point 20% of distance between these two points on the same slope ? Regards Ray "Je Suis Mort De Rire" Blogging @ Keratoconus Watch
1. Translate the point to the origin (remember the translation) 2. Get the position vector of the second point 3. Calculate the length of the vector 3. Normalize the vector 4. Multiply each component with 0.2 x the length from #3 5. Use the inverse translation -- modified at 13:06 Tuesday 13th June, 2006 Oh, I was too late
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Hey Guys, I have a math issue, I know how you like brain teasers so I thought I would post it here, say I have two points ab & cd, how do I find the point 20% of distance between these two points on the same slope ? Regards Ray "Je Suis Mort De Rire" Blogging @ Keratoconus Watch
Ray Kinsella wrote:
the point 20% of distance between these two points on the same slope
If you mean 20% of the distance from the point (a, b), then (e, f)=(a+0.2*(c-a), b+0.2*(d-b)) Otherwise (e, f)=(a+0.8*(c-a), b+0.8*(d-b)) - It's easier to make than to correct a mistake.
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Make a vector between the points [c-a,d-b], normalise the vector (divide it by it's magnitude) and then multiply that by 0.2. Then add this onto [a,b] and you should have that. Now watch me get that all wrong and look a fool :)
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If I'm reading the question right (ie 20% of the way from AB to CD or vice versa) I don't think the vector should be normalized, so it's just (CD-AB)*0.2+AB
Oh spanner! I think you're right. P.S: Please don't tell my boss I got a question like this wrong! :-O
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1. Translate the point to the origin (remember the translation) 2. Get the position vector of the second point 3. Calculate the length of the vector 3. Normalize the vector 4. Multiply each component with 0.2 x the length from #3 5. Use the inverse translation -- modified at 13:06 Tuesday 13th June, 2006 Oh, I was too late
don't get it let a,b=2,3 c,d=6,6 2. vector = 4,3 3. magnitude |v| = 5 4. = <4, 3> huh ? Regards Ray "Je Suis Mort De Rire" Blogging @ Keratoconus Watch
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Ray Kinsella wrote:
the point 20% of distance between these two points on the same slope
If you mean 20% of the distance from the point (a, b), then (e, f)=(a+0.2*(c-a), b+0.2*(d-b)) Otherwise (e, f)=(a+0.8*(c-a), b+0.8*(d-b)) - It's easier to make than to correct a mistake.
yep ... figured that out a minute ago .... brain hurts ... too long since school ... thanks to all. Regards Ray "Je Suis Mort De Rire" Blogging @ Keratoconus Watch
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don't get it let a,b=2,3 c,d=6,6 2. vector = 4,3 3. magnitude |v| = 5 4. = <4, 3> huh ? Regards Ray "Je Suis Mort De Rire" Blogging @ Keratoconus Watch
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But 20% = 0.2. I hope. :^) -- modified at 13:03 Tuesday 13th June, 2006 ;)
HalfWayMan wrote:
20% = 0.2
20% = 0.2 = 2/10 = 1/5 ;P
Formula 1 - Short for "F1 Racing" - named after the standard "help" key in Windows, it's a sport where participants desperately search through software help files trying to find actual documentation. It's tedious and somewhat cruel, most matches ending in a draw as no participant is able to find anything helpful. - Shog9
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gotcha now ... thanks ... Regards Ray "Je Suis Mort De Rire" Blogging @ Keratoconus Watch
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yep ... figured that out a minute ago .... brain hurts ... too long since school ... thanks to all. Regards Ray "Je Suis Mort De Rire" Blogging @ Keratoconus Watch
That won't work since you're actually getting the distance that's Sqrt(2(0.2*0.2)) along the line. Think of it as a right-angled triangle and you're using getting 0.2 along the x and 0.2 along the y so you're actually getting closer to 0.3 along the line. Give me 5 mintues and I'll write out the solution (I hope I'm right :rolleyes:)
Formula 1 - Short for "F1 Racing" - named after the standard "help" key in Windows, it's a sport where participants desperately search through software help files trying to find actual documentation. It's tedious and somewhat cruel, most matches ending in a draw as no participant is able to find anything helpful. - Shog9
Last modified: 13 June 2006 13:25:47 --
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Hey Guys, I have a math issue, I know how you like brain teasers so I thought I would post it here, say I have two points ab & cd, how do I find the point 20% of distance between these two points on the same slope ? Regards Ray "Je Suis Mort De Rire" Blogging @ Keratoconus Watch
Ok, a reference image is provided @ http://ed.poore.googlepages.com/question.bmp[^] Basically: Let A=(a,b) Let B=(c,d) Let P=(x,y) (the point where you want to find) |AB| = √((c - a)²+(d - b)²)
θ = arctan((d - b) / (c - a))
x = a + (0.2*|AB|*cos(θ)) y = b + (0.2*|AB|*sin(θ))
Some basic trig and pythagoras, you can do this by vectors, but it essentially boils down to the same thing if I remember correctly (better remember correctly since I might need it for my exam on Friday).
Formula 1 - Short for "F1 Racing" - named after the standard "help" key in Windows, it's a sport where participants desperately search through software help files trying to find actual documentation. It's tedious and somewhat cruel, most matches ending in a draw as no participant is able to find anything helpful. - Shog9
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Ok, a reference image is provided @ http://ed.poore.googlepages.com/question.bmp[^] Basically: Let A=(a,b) Let B=(c,d) Let P=(x,y) (the point where you want to find) |AB| = √((c - a)²+(d - b)²)
θ = arctan((d - b) / (c - a))
x = a + (0.2*|AB|*cos(θ)) y = b + (0.2*|AB|*sin(θ))
Some basic trig and pythagoras, you can do this by vectors, but it essentially boils down to the same thing if I remember correctly (better remember correctly since I might need it for my exam on Friday).
Formula 1 - Short for "F1 Racing" - named after the standard "help" key in Windows, it's a sport where participants desperately search through software help files trying to find actual documentation. It's tedious and somewhat cruel, most matches ending in a draw as no participant is able to find anything helpful. - Shog9
This is a perfect example that how people make things more complicated than they should be:omg: This is how it works:
x = a + (0.2*|AB|*cos(θ))) = a + 0.2*(c-a);
y = b + (0.2*|AB|*sin(θ))) = b + 0.2*(d-b);BTW, I like your "θ". How come It shows "θ" in my editing panel, but not in the post? ...maybe because of my Linux (I'm using RedHat 8 at work). Signature for today: "Don't discuss math problems with a methematician."
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Ok, a reference image is provided @ http://ed.poore.googlepages.com/question.bmp[^] Basically: Let A=(a,b) Let B=(c,d) Let P=(x,y) (the point where you want to find) |AB| = √((c - a)²+(d - b)²)
θ = arctan((d - b) / (c - a))
x = a + (0.2*|AB|*cos(θ)) y = b + (0.2*|AB|*sin(θ))
Some basic trig and pythagoras, you can do this by vectors, but it essentially boils down to the same thing if I remember correctly (better remember correctly since I might need it for my exam on Friday).
Formula 1 - Short for "F1 Racing" - named after the standard "help" key in Windows, it's a sport where participants desperately search through software help files trying to find actual documentation. It's tedious and somewhat cruel, most matches ending in a draw as no participant is able to find anything helpful. - Shog9
just look at that happy happy math! whee! x2 cos(θ4) :) Cleek | Image Toolkits | Thumbnail maker
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But 20% = 0.2. I hope. :^) -- modified at 13:03 Tuesday 13th June, 2006 ;)
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Ok, a reference image is provided @ http://ed.poore.googlepages.com/question.bmp[^] Basically: Let A=(a,b) Let B=(c,d) Let P=(x,y) (the point where you want to find) |AB| = √((c - a)²+(d - b)²)
θ = arctan((d - b) / (c - a))
x = a + (0.2*|AB|*cos(θ)) y = b + (0.2*|AB|*sin(θ))
Some basic trig and pythagoras, you can do this by vectors, but it essentially boils down to the same thing if I remember correctly (better remember correctly since I might need it for my exam on Friday).
Formula 1 - Short for "F1 Racing" - named after the standard "help" key in Windows, it's a sport where participants desperately search through software help files trying to find actual documentation. It's tedious and somewhat cruel, most matches ending in a draw as no participant is able to find anything helpful. - Shog9
You get a division by zero if you are dealing with vertical lines.
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This is a perfect example that how people make things more complicated than they should be:omg: This is how it works:
x = a + (0.2*|AB|*cos(θ))) = a + 0.2*(c-a);
y = b + (0.2*|AB|*sin(θ))) = b + 0.2*(d-b);BTW, I like your "θ". How come It shows "θ" in my editing panel, but not in the post? ...maybe because of my Linux (I'm using RedHat 8 at work). Signature for today: "Don't discuss math problems with a methematician."
Though both are correct, I'd go with the simpler way since there is no worries in regards to division by zero. PJC
