Math Problem ...
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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
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You get a division by zero if you are dealing with vertical lines.
Paul Conrad wrote:
You get a division by zero if you are dealing with vertical lines.
You get division by zero if you deal with horizontal lines. Since tanθ = opposite / adjacent (i.e. vertical over horizontal). But it's only a couple of ifs to sort this out. To be honest I'm suprised that so many people seem to be accepting the incorrect versions here. All you need to know is Pythagoras (a²+o² = h²) to prove that the simpler formulae they've come up with are nonsense.
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 15:56:59 --
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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."
Jun Du wrote:
This is a perfect example that how people make things more complicated than they should be
Maybe it's more complicated but it works. Here's the proof using simple Pythagorean trig and his theorem.
Start with a 3,4,5 triangle with height 3 and width 4 (and hypotenuse 5).
If A is the bottom left-corner, located over an "origin" of (0,0) then it follows that
A = (0,0) (bottom-left corner)
B = (4,3) (top-right corner)
C = (4,0) (bottom-right corner)
θ = Angle BAC (i.e. bottom-left)
Also it can be shown (through Pythagorean Trigonometry) that:
sin(θ)=3/5
cos(θ)=4/5
tan(θ)=3/4So you require 20% of the line AB (in this case it is (0.2*5) = 1).
We can let the co-ordinates of the end point of this line (from A along AB for 20% of AB) be X=(x,y).
It then follows that since we have a hypotenuse equal to 1 and the angle has remained the same that:x = 1*cos(θ) = 4/5
y = 1*sin(θ) = 3/5So from A the new point X which is 20% of the line AB is A+X, i.e: X = (0 + 4/5, 0 + 3/5) = (4/5, 3/5).
If you take you're equations of:
x = a + 0.2 * (c - a)
y = b + 0.2 * (d - b)
You will get the following answers for this case:
x = 0 + 0.2 * (4 - 0) = 0.2 * 2 = 2/5
y = 0 + 0.2 * (3 - 0) = 0.2 * 3 = 9/10
Which is not the same as the ones above.The simplest proof that the previous equations are wrong can be shown quite simply by Pythagoras' Theorem.
If we assume that the triangle formed is of height and width 1 then, your equations show that the height and width of the "new" triangle which provides 20% of AB are both 0.2.
If you use Pythagoras on these you get a hypotenuse of √(0.2²+0.2²) which gives √0.08 which is ~0.2828, not 0.2 as you require.
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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Paul Conrad wrote:
You get a division by zero if you are dealing with vertical lines.
You get division by zero if you deal with horizontal lines. Since tanθ = opposite / adjacent (i.e. vertical over horizontal). But it's only a couple of ifs to sort this out. To be honest I'm suprised that so many people seem to be accepting the incorrect versions here. All you need to know is Pythagoras (a²+o² = h²) to prove that the simpler formulae they've come up with are nonsense.
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 15:56:59 --
Ed.Poore wrote:
You get division by zero if you deal with horizontal lines. 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))
Check your one post because when c-a = 0, the line is vertical :) -- modified at 17:02 Tuesday 13th June, 2006
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Ed.Poore wrote:
You get division by zero if you deal with horizontal lines. 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))
Check your one post because when c-a = 0, the line is vertical :) -- modified at 17:02 Tuesday 13th June, 2006
Oops, maybe your right, did things in a bit of rush because I've got annoyed that people can't see that the "complex" formula is the correct one, you can't get any simpler. Anyway I'm not going to argue just yet, I'm off to have some supper, I'll come back later and see if people have finally accepted it.
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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Though both are correct, I'd go with the simpler way since there is no worries in regards to division by zero. PJC
How can both be correct, I've just posted the proof that they aren't.
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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Oops, maybe your right, did things in a bit of rush because I've got annoyed that people can't see that the "complex" formula is the correct one, you can't get any simpler. Anyway I'm not going to argue just yet, I'm off to have some supper, I'll come back later and see if people have finally accepted it.
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
Ed.Poore wrote:
the "complex" formula is the correct one
The solution by Jun Du is also correct. Tried both your solution and his for several different points and I get the same answers from both. Consider if this problem was being programmed on some weird machine where no square root functions or trig functions were available via hardware, then Jun Du's solution would be best.
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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."
Jun Du wrote:
Signature for today: "Don't discuss math problems with a methematician."
Who's a methematician?
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. - Shog
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Jun Du wrote:
This is a perfect example that how people make things more complicated than they should be
Maybe it's more complicated but it works. Here's the proof using simple Pythagorean trig and his theorem.
Start with a 3,4,5 triangle with height 3 and width 4 (and hypotenuse 5).
If A is the bottom left-corner, located over an "origin" of (0,0) then it follows that
A = (0,0) (bottom-left corner)
B = (4,3) (top-right corner)
C = (4,0) (bottom-right corner)
θ = Angle BAC (i.e. bottom-left)
Also it can be shown (through Pythagorean Trigonometry) that:
sin(θ)=3/5
cos(θ)=4/5
tan(θ)=3/4So you require 20% of the line AB (in this case it is (0.2*5) = 1).
We can let the co-ordinates of the end point of this line (from A along AB for 20% of AB) be X=(x,y).
It then follows that since we have a hypotenuse equal to 1 and the angle has remained the same that:x = 1*cos(θ) = 4/5
y = 1*sin(θ) = 3/5So from A the new point X which is 20% of the line AB is A+X, i.e: X = (0 + 4/5, 0 + 3/5) = (4/5, 3/5).
If you take you're equations of:
x = a + 0.2 * (c - a)
y = b + 0.2 * (d - b)
You will get the following answers for this case:
x = 0 + 0.2 * (4 - 0) = 0.2 * 2 = 2/5
y = 0 + 0.2 * (3 - 0) = 0.2 * 3 = 9/10
Which is not the same as the ones above.The simplest proof that the previous equations are wrong can be shown quite simply by Pythagoras' Theorem.
If we assume that the triangle formed is of height and width 1 then, your equations show that the height and width of the "new" triangle which provides 20% of AB are both 0.2.
If you use Pythagoras on these you get a hypotenuse of √(0.2²+0.2²) which gives √0.08 which is ~0.2828, not 0.2 as you require.
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
Ed.Poore wrote:
x = 0 + 0.2 * (4 - 0) = ... = 2/5
.2 * 4 = .8
Ed.Poore wrote:
y = 0 + 0.2 * (3 - 0) = 0.2 * 3 = 9/10
.2 * 3 = .6 Cleek | Image Toolkits | Thumbnail maker
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Ed.Poore wrote:
x = 0 + 0.2 * (4 - 0) = ... = 2/5
.2 * 4 = .8
Ed.Poore wrote:
y = 0 + 0.2 * (3 - 0) = 0.2 * 3 = 9/10
.2 * 3 = .6 Cleek | Image Toolkits | Thumbnail maker
Chris Losinger wrote:
.2 * 4 = .8
Chris Losinger wrote:
.2 * 3 = .6
Last time I checked, these are the correct answers.
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Chris Losinger wrote:
.2 * 4 = .8
Chris Losinger wrote:
.2 * 3 = .6
Last time I checked, these are the correct answers.
yup. he was right up until the last little bit of arithmetic. Cleek | Image Toolkits | Thumbnail maker
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Ed.Poore wrote:
the "complex" formula is the correct one
The solution by Jun Du is also correct. Tried both your solution and his for several different points and I get the same answers from both. Consider if this problem was being programmed on some weird machine where no square root functions or trig functions were available via hardware, then Jun Du's solution would be best.
Paul Conrad wrote:
The solution by Jun Du is also correct.
:wtf: How? Can you provide some proof!
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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Paul Conrad wrote:
The solution by Jun Du is also correct.
:wtf: How? Can you provide some proof!
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
Can't you just accept the fact someone else came up with an easier solution than yours (and yours is correct, too)? You can't even get horizontal and vertical lines correct. Division by zero occurs when you have a vertical line. I knew this BEFORE you were even born. Mr. Losinger even pointed out math errors in your one post.
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Can't you just accept the fact someone else came up with an easier solution than yours (and yours is correct, too)? You can't even get horizontal and vertical lines correct. Division by zero occurs when you have a vertical line. I knew this BEFORE you were even born. Mr. Losinger even pointed out math errors in your one post.
If I've understood the post correctly he wants to get 20% of the line between the two points. Agree? If so then how can you simply take 20% of the width of the "triangle" formed and "20%" of the height to get a hypotenuse that is 20% of the original triangle? If you use Pythagoras it'll show you that if you do that you'll actually get a hypotenuse that's ~28% of the original rather than 20%.
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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Can't you just accept the fact someone else came up with an easier solution than yours (and yours is correct, too)? You can't even get horizontal and vertical lines correct. Division by zero occurs when you have a vertical line. I knew this BEFORE you were even born. Mr. Losinger even pointed out math errors in your one post.
Oooooooooopppppppppppssssssssssssssss :doh::doh::doh::doh::doh: :-O:-O:-O:-O:-O Now I know I should go to bed, sorry I've completely ballsed this up havn't I, just got a pencil and paper out to work it out, forgot to substitute for cos and sin :doh: Combination of too much revision and a late night.
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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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
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If I've understood the post correctly he wants to get 20% of the line between the two points. Agree? If so then how can you simply take 20% of the width of the "triangle" formed and "20%" of the height to get a hypotenuse that is 20% of the original triangle? If you use Pythagoras it'll show you that if you do that you'll actually get a hypotenuse that's ~28% of the original rather than 20%.
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
Though it does not qualify as an actual mathematical proof, I've tested your approach and Jun Du's approach in Excel with 10-15 different values and I get the same results. Remember that: θ = arctan( y / x ) x = cos( θ ) y = sin( θ ) from Inverse Tangent[^] You wrote: θ = arctan((d - b) / (c - a)) x = a + (0.2*|AB|*cos(θ) y = b + (0.2*|AB|*sin(θ) and based on the link from MathWorld, (c-a) = cos( θ ) and (d-b) = sin( θ ) PJC
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yup. he was right up until the last little bit of arithmetic. Cleek | Image Toolkits | Thumbnail maker
Chris Losinger wrote:
he was right up until the last little bit of arithmetic.
Yep, I've now confirmed that I've been a complete and utter nincompoop, see http://www.codeproject.com/lounge.asp?msg=1530321#xx1530321xx[^] for my explanation and excuses. I now realise I was barking up the wrong tree :doh:
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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Though it does not qualify as an actual mathematical proof, I've tested your approach and Jun Du's approach in Excel with 10-15 different values and I get the same results. Remember that: θ = arctan( y / x ) x = cos( θ ) y = sin( θ ) from Inverse Tangent[^] You wrote: θ = arctan((d - b) / (c - a)) x = a + (0.2*|AB|*cos(θ) y = b + (0.2*|AB|*sin(θ) and based on the link from MathWorld, (c-a) = cos( θ ) and (d-b) = sin( θ ) PJC
I just wrote it out on paper, basically if you substitute adjacent / hypotenuse in place of cos then the root will cancel out and you'll get what Jun Du gave. I'm just too exhausted from 12 hours revision plus a 2 hour shooting walk with my father and the dog. I stand corrected and apologise if I was dogmatic :-O.
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
