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You are in a square (eg room), and you know the distances from a point on the floor to three of the corners is (going clockwise) 2ft, 6ft, and 9ft.
How far is it to the 4th corner?
How far is it to the 4th corner?
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Re: Distances from corners of a square
Sun, October 19, 2008 - 12:32 PMIf you draw two lines perpendicular to the sides through "yourself," it will create right triangles with the hypotenuses 2, 6, 9 and c. We want to know c.
Let a and b be the other sides of c's triangle, going Pythagorean on them:
1) c^2 = a^2 + b^2
2) 9^2 = a^2 + X^2 (because they share side X)
3) 2^2 = Y^2 + b^2 (because they share side Y)
4) 6^2 = X^2 + Y^2 (doesn't share any sides with c's triangle but does with the other two).
a^2 = 81 - X^2, from 2
b^2 = 4 - Y^2, from 3
c^2 = 81 - X^2 + 4 - Y^2, using 1
= 81 + 4 - (X^2 + Y^2)
= 85 - 36, using 4
= 49
c = 7 -
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Re: Distances from corners of a square
Sun, October 19, 2008 - 12:52 PMbecause they share the sides a and b, not X and Y. X and Y are the other sides.
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Re: Distances from corners of a square
Mon, October 20, 2008 - 1:27 AMWell done.
So what is the general equation, given the distances of p, q, and r to 3 of the corners, to find s to the 4th corner? -
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Re: Distances from corners of a square
Wed, October 22, 2008 - 5:42 AMAn alternative to finding the distance to the 4th corner is to find the size of the square room.
Karl -
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Re: Distances from corners of a square
Fri, October 24, 2008 - 6:05 PMWell done Karl, that WAS my next question.
Given p, q, r, and (can be calculated) s what is the size of the square?
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Re: Distances from corners of a square
Wed, October 22, 2008 - 7:23 AMUsing the logic above: SQRT(p^2 + r^2 - q^2)
Do we have to stay in the room for this to work? (my guess, no)
What does an imaginary solution mean? (my guess, that you can't create the appropriate square with p, r and q) -
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Re: Distances from corners of a square
Fri, October 24, 2008 - 6:07 PMI'm not sure about staying inside the room, I'd have to do some algebra to confirm that, but it does seem reasonable.
I did however discover that:
p²+r² = q²+s²
is also true for a rectangle too. -
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Re: Distances from corners of a square
Fri, October 24, 2008 - 9:59 PMyeah. I didn't see that. But the derivation doesn't use the "squareness," so it makes sense. -
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Re: Distances from corners of a square
Fri, October 24, 2008 - 10:13 PMA quick drawing seems to point to getting the same sort of triangles when the point is outside. So I think it works there as well. -
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Re: Distances from corners of a square
Sat, October 25, 2008 - 6:02 AMYes, It works from outside- it just means the "altitudes" may have negative values.
As for rectangles- you don't even need the right angles, so it works for parallellograms, too.
No luck working out the size of the square, so far, but it may be possible there isn't a unique solution. Just knowing the distance to the corners may not be enough to determine the length of a side. -
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Re: Distances from corners of a square
Sat, October 25, 2008 - 1:41 PMIf its a square, my guess is the area is fixed. I'm not sure about rectangles and parallelograms yet. -
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Re: Distances from corners of a square
Sun, October 26, 2008 - 2:37 PMThe area of all parallograms (squares, rectangles, rhombi, etc) and trapozoids are given by:
A = L H
Where L is the length of the longest side and H is the shortest distance between the the parallel sides (one of which was used to get L). Given a point is one of these quadralaterials (those listed above) and the ordered set of distances from the point to the corners then the area of the quadralaterial is unique. -
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Re: Distances from corners of a square
Sun, October 26, 2008 - 3:35 PMNot sure I buy that. If p = q = r = s, then your suggestion would imply the area of any parallelogram you can inscribe in a circle of a given length has the same area (the point is in the center). I don't think that is true.
If 1 = p = q = r = s, then a square of sides SQRT(2) would match. But if you set the height to SQRT(2)/x, x > 1 centered in the circle, the width would have to be x to match that size - which contradicts the initial assumption of the distances being 1. -
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Re: Distances from corners of a square
Sun, October 26, 2008 - 3:41 PM2/x, x > 1 - not SQRT(2)/x. Still works.
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Re: Distances from corners of a square
Mon, October 27, 2008 - 11:20 AM"Not sure I buy that. If p = q = r = s, then your suggestion would imply the area of any parallelogram you can inscribe in a circle of a given length has the same area (the point is in the center). I don't think that is true. "
'Tis not what I said.
First off, the diagonals of a rhombus are not equal, so that parallelogram is excluded from your example.
In the case of a trapezoid, I did make a mistake. The area A = H S + (L- S) H / 2 where H is the height between the parallel sides, S is the length of the short parallel side and L is the length of the long parallel side. -
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Re: Distances from corners of a square
Mon, October 27, 2008 - 7:23 PMYup. I should have said rectangles.
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Re: Distances from corners of a square
Mon, October 27, 2008 - 6:23 AMThe problem is interesting when applied to an equilateral triangle.
I've read that formula for determining the length of the side of the triangle given the 3 distances to the vertices is exactly the same as determining the distance to one vertex given the distances to the other two vertices and the length of the side of the triangle.
Karl
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Re: Distances from corners of a square
Tue, October 28, 2008 - 4:43 PMFrancis> Just knowing the distance to the corners may not be enough to determine the length of a side.
It is possible, and took me a while to figure it out...
Given the distances to the corners as p,q,r,s (going clockwise), and the (unknown) side of the square is w.
Then draw perpendicular lines from the point inside the square to the sides of the square, and add labels x and y as smaller lengths on the sides of the square from the perpendicular line to the corner where line p is attached, so that there is a triangle labeled with p as the hypotenuse and sides x and y.
There are also triangles: q,(w-x),y and r,(w-x),(w-y) and s,x,(w-y)
Just so you know how I've labeled things, see image: mathematicsontribe.tribe.net/pho...b011
Obviously we know from simple Pythagoras:
1) p² = x² + y²
2) q² = (w-x)² + y²
3) r² = (w-x)² + (w-y)²
4) s² = x² + (w-y)²
Subtract above 1) from 2) we get:
q² - p² = (w-x)² + y² - x² - y²
= (w² - 2wx + x²) + y² - x² - y²
= w² - 2wx
Solve for x:
x = (p² - q² + w²) / 2w
Subtract above 1) from 4) we get:
s² - p² = x² + (w-y)² - x² - y²
= x² + (w² - 2wy + y²) - x² - y²
= w² - 2wy
Solve for y:
y = (p² - s² + w²) / 2w
Substitute x and y into 1) above we get:
p² = ((p² - q² + w²) / 2w)² + ((p² - s² + w²) / 2w)²
4p²w² = (p² - q² + w²)² + (p² - s² + w²)²
= p²(p² - q² + w²) - q²(p² - q² + w²) + w²(p² - q² + w²) + p²(p² - s² + w²) - s²(p² - s² + w²) + w²(p² - s² + w²)
= p²p² - p²q² + p²w² - q²p² + q²q² - q²w² + w²p² - w²q² + w²w² + p²p² - p²s² + p²w² - s²p² + s²s² - s²w² + w²p² - w²s² + w²w²
= 2w²w² + 4p²w² - 2q²w² - 2s²w² + p²p² - 2p²q² + q²q² + p²p² - 2p²s² + s²s²
Simplified to:
2w²w² - 2(q² + s²)w² + (p² - q²)² + (p² - s²)² = 0
To solve for w (the side of the square), use:
(-b ± sqrt(b² - 4ac)) / 2a
where:
a= 2
b= -2(q² + s²)
c= (p² - q²)² + (p² - s²)²
Notice we didn't even need to know the 4th length (in this is labeling "r") anyway!
PS Sorry I used w²w² instead of w^4 - I cannot see a power of 4 in my font. -
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Re: Distances from corners of a square
Tue, October 28, 2008 - 5:44 PM"Simplified to:
2w²w² - 2(q² + s²)w² + (p² - q²)² + (p² - s²)² = 0
To solve for w (the side of the square), use:
(-b ± sqrt(b² - 4ac)) / 2a
where:
a= 2
b= -2(q² + s²)
c= (p² - q²)² + (p² - s²)²
Notice we didn't even need to know the 4th length (in this is labeling "r") anyway!
PS Sorry I used w²w² instead of w^4 - I cannot see a power of 4 in my font. "
Alas, you've made an error for
w^2 = (-b ± sqrt(b² - 4ac)) / 2a
w = Sqrt[(-b ± sqrt(b² - 4ac)) / 2a]
Yes, there four possible solutions two of which are complex, one is negative and only one is positive. -
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Re: Distances from corners of a square
Tue, October 28, 2008 - 5:48 PMA correction, instead of "there four possible solutions two of which are complex, one is negative and only one is positive. "
It should read:
"in general, there four possible solutions two of which are complex, one is negative or zero, and only one is positive. "
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Re: Distances from corners of a square
Tue, October 28, 2008 - 6:27 PMYes you are right, I should have said the quadratic formula can be used to solve for w².
It was difficult enough to transcribe my scribble into something that flowed and was readable on the screen!
But yes... Thank you for pointing that out.
While we can find integer solution for all values in the equation: p² + r² = q² + s²
eg p=2, q=6, r=9, s=7
Can anyone find a solution where the side of the square (ie w) is an integer too?
(ie as well as p, q, r, and s being integers)
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