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BMAD

Cross Cornering

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BMAD    61

Let's say I attempted to draw a rectangle with dimensions of L x W.  When I compared the diagonals, I noticed that they were off by 6 cm.  Create an equation so that I can determine how far  and where I should move one of the corners to fix my rectangle.

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gavinksong    11
7 minutes ago, gavinksong said:

Oh, I see. That makes more sense.

In that case, there's one equation that pops into my head.

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Let A be one of the angles of the parallelogram. Using the law of cosines, we can express the difference in the length of the diagonals as follows.

sqrt(L^2 + W^2 + 2*L*W*cos(A)) - sqrt(L^2 + W^2 - 2*L*W*cos(A)) = 6 cm

Seems a bit messy. :wacko:

Here's what it looks like when I solve for the error.

Spoiler

Let x be the angle by which any of the angles in the parallelogram differ from 90 degrees. Then, let H = L^2 + W^2 and E = 2*L*W*sin(x).

sqrt(H + E) - sqrt(H - E) = 6

If we square both sides, we arrive at the following.

(H + E) - sqrt((H + E)(H - E)) + (H - E) = 36

2*H - sqrt(H^2 - E^2) = 36

(2*H - 36)^2 = H^2 - E^2

E^2 = H^2 - (2*H - 36)^2

x = arcsin( sqrt(H^2 - (2*H - 36)^2) / (2*L*W) )

Unless I've made a mistake?

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gavinksong    11

Hmm? I'm a little confused. :/

Do the values L and W tell us something about the defective rectangle? Is only one of the corners off?

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araver    10
On 8/3/2017 at 6:01 PM, BMAD said:

Let's say I attempted to draw a rectangle with dimensions of L x W.  When I compared the diagonals, I noticed that they were off by 6 cm.  Create an equation so that I can determine how far  and where I should move one of the corners to fix my rectangle.

Question:

1. Do we have anything else other than a (marked) ruler (as implied by the 6 cm measurement)?

Second question in a spoiler

Spoiler

After reading the question, I think "moving one of the corners" must mean that already 3 corners form a rectangular triangular, otherwise you wouldn't be able to create a rectangle by moving only one corner.

As such, drawing point A as the point that needs to be moved I understand that one of the diagonals is greater (or smaller) than the other by 6 cm. Please let me know if I understood "off by" correctly.

If so, I understand the problem is to "move" point A to point B (an example in the next figure).

However if one can measure L and W and has managed to draw a 90 degree angle between them, this means perpendiculars can be drawn to create the rectangle safely. How would an equation help?

 CrossCornering.thumb.png.bad82db52b3d19f554bae0e05f0b464b.png

 

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BMAD    61
On 8/6/2017 at 4:09 AM, araver said:

Question:

1. Do we have anything else other than a (marked) ruler (as implied by the 6 cm measurement)?

Second question in a spoiler

  Hide contents

After reading the question, I think "moving one of the corners" must mean that already 3 corners form a rectangular triangular, otherwise you wouldn't be able to create a rectangle by moving only one corner.

As such, drawing point A as the point that needs to be moved I understand that one of the diagonals is greater (or smaller) than the other by 6 cm. Please let me know if I understood "off by" correctly.

If so, I understand the problem is to "move" point A to point B (an example in the next figure).

However if one can measure L and W and has managed to draw a 90 degree angle between them, this means perpendiculars can be drawn to create the rectangle safely. How would an equation help?

 CrossCornering.thumb.png.bad82db52b3d19f554bae0e05f0b464b.png

 

You understand the problem well and yes you only have a ruler. As for the equation, moving the corner of one of the rectangular corners specifically the one opposite of the 90 angle, would adjust the angles adjacent to the 90 degree angle relative to L, W L2, W2. There is an equation that we can create so that we can check to ensure that l=l2 and w=W2 with the assumption that there is a 90 degree angle between L and W. 

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gavinksong    11

So only one of the corners is off. The other three corners form a right triangle with legs of length L and W.

Here's where I'm confused:

Spoiler

The hypotenuse of the right triangle is the square root of L^2 + W^2. Let's call this value H.

All we know is that the other diagonal (the one that extends from the vertex between the two legs of the right triangle to the vertex that is misplaced) is of length H + 6 cm. Then, doesn't that mean that the misplaced vertex can be anywhere on the circle centered on the opposite vertex as shown below?

598c6f616b1c5_ScreenShot2017-08-10at11_35_05PM.png.8883b627a729725ab78d209efc1d29c8.png

Edit: I just remembered the length of the other diagonal can also be H - 6 cm as well. So the misplaced vertex can also lie on a second circle with radius H - 6 cm.

This is where I have been confused. Please help if I misunderstood.

Edited by gavinksong

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BMAD    61

Not anywhere otherwise there is no way someone would confuse the shape for a rectangle as it would lose its fourth side and by symmetry much of that circle is immaterial. 

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gavinksong    11
13 minutes ago, BMAD said:

Not anywhere otherwise there is no way someone would confuse the shape for a rectangle as it would lose its fourth side and by symmetry much of that circle is immaterial. 

True, but do you see where I am confused? If there are multiple possible locations for the fourth vertex, which one do we use when we calculate how far we have to move it?

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BMAD    61

Here is an algorithmic approach to solving this problem that may help with finding the equation. Assume that between L and W is a right angle, use a string to find the length of the hypotenuse. Use that string to arc out the possible endpoints of the other diagonal using the length from the string. Use the same rope to find the length of L or W. Go to the end point of either L or W away from the given ninety degree angle. Arc the line again and where it crosses with the other arc is where you need to put your fourth corner. Done! 

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gavinksong    11
40 minutes ago, BMAD said:

Here is an algorithmic approach to solving this problem that may help with finding the equation. Assume that between L and W is a right angle, use a string to find the length of the hypotenuse. Use that string to arc out the possible endpoints of the other diagonal using the length from the string. Use the same rope to find the length of L or W. Go to the end point of either L or W away from the given ninety degree angle. Arc the line again and where it crosses with the other arc is where you need to put your fourth corner. Done! 

I thought we needed to calculate how far we needed to move the corner that was off? Is it unnecessary to use the fact that one of the diagonals is 6 cm longer than the other?

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BMAD    61
In retrospect, when I thought up the question, I realized that we shouldn't assume that we have a ninety degree angle. In geometric terms, if you've measured so all the sides are the correct length (opposite sides are the same length), then you have a parallelogram, and what I  want to know is how far the angles are from right angles, based on the difference in the diagonals.

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gavinksong    11
1 hour ago, BMAD said:

In retrospect, when I thought up the question, I realized that we shouldn't assume that we have a ninety degree angle. In geometric terms, if you've measured so all the sides are the correct length (opposite sides are the same length), then you have a parallelogram, and what I  want to know is how far the angles are from right angles, based on the difference in the diagonals.

Oh, I see. That makes more sense.

In that case, there's one equation that pops into my head.

Spoiler

Let A be one of the angles of the parallelogram. Using the law of cosines, we can express the difference in the length of the diagonals as follows.

sqrt(L^2 + W^2 + 2*L*W*cos(A)) - sqrt(L^2 + W^2 - 2*L*W*cos(A)) = 6 cm

Seems a bit messy. :wacko:

Edited by gavinksong

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