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This is from one of those Litton ads in Aviation Week circa 1964--

A rectangular sheet of paper is 24 inches wide left to right, and a bit longer than that top to bottom. Fold the top left corner over to just touch the right edge of the paper, with the fold line running thru the lower left corner. Crease the fold, then unfold the paper; fold the lower right corner to the left edge of the paper in the same way, with the fold line passing thru the upper right corner. Crease that fold too.

The two parallel creases are 4.2 inches apart. How long is the paper, top to bottom?

A rectangular sheet of paper is 24 inches wide left to right, and a bit longer than that top to bottom. Fold the top left corner over to just touch the right edge of the paper, with the fold line running thru the lower left corner. Crease the fold, then unfold the paper; fold the lower right corner to the left edge of the paper in the same way, with the fold line passing thru the upper right corner. Crease that fold too.

The two parallel creases are 4.2 inches apart. How long is the paper, top to bottom?

- petertenthije
**Posts:**3922**Joined:**

The two folds run paralel, so you can use pythagorus

A2 + B2 = C2

A2 + B2 = (4,2*4,2)

A2 + B2 = 17,64

A2 = B2 = 17,64 / 2

A2 = B2 = 8,82

A = B = √8,82

A = B = 2,97

24 + 2,97 = 26,97

Have you really been working at this math question for over 50 years?

A2 + B2 = C2

A2 + B2 = (4,2*4,2)

A2 + B2 = 17,64

A2 = B2 = 17,64 / 2

A2 = B2 = 8,82

A = B = √8,82

A = B = 2,97

24 + 2,97 = 26,97

Have you really been working at this math question for over 50 years?

Attamottamotta!

Quoting petertenthije (Reply 1):A2 + B2 = (4,2*4,2) |

The 4.2 is not a hypotenuse, it is a side. So it should be 4.2^2 + B^2 = C^2. You also cannot assume that the non-hypotenuse sides are equal, so I'm skeptical of that answer.

-Mir

7 billion, one nation, imagination...it's a beautiful day

I get 42.468"

......no I don't. Back to the drawing board.

[Edited 2016-06-05 03:19:20]

......no I don't. Back to the drawing board.

[Edited 2016-06-05 03:19:20]

wheat and dairy can screw up your brain

well the way I see it is by folding the corners in you are making a square with 24 inns sides

The folded creases are 4.2 ins apart

Therefore the top to bottom distance is 24 ins + 4.2 ins + something [A]

Using the 4,2 dimension between the two creases you form a right angle triangle with two angles of 45 degs

therefore the hypotenuse is the [ something dimension { A } ]

Now because you have formed a triangle with 90 degs and two 45 degs angles two sides are equal [4.2 ins]

Therefore = dimension A = the square root of [4.2 squared + 4,2 squared ]

A= square root of 35.28

A = 5.94

Therefore top to bottom paper measures 24 + 5.94 + 29.94 or could we say 30 inches

Please excuse any errors as I have had to do this on bit of paper and I hope i am correct

lttlevc10

[Edited 2016-06-05 02:37:22]

The folded creases are 4.2 ins apart

Therefore the top to bottom distance is 24 ins + 4.2 ins + something [A]

Using the 4,2 dimension between the two creases you form a right angle triangle with two angles of 45 degs

therefore the hypotenuse is the [ something dimension { A } ]

Now because you have formed a triangle with 90 degs and two 45 degs angles two sides are equal [4.2 ins]

Therefore = dimension A = the square root of [4.2 squared + 4,2 squared ]

A= square root of 35.28

A = 5.94

Therefore top to bottom paper measures 24 + 5.94 + 29.94 or could we say 30 inches

Please excuse any errors as I have had to do this on bit of paper and I hope i am correct

lttlevc10

[Edited 2016-06-05 02:37:22]

The answer is 25 inches. Solve a quadratic to find that the cosine of the angle between the crease and the longer edge of the paper is 0.8. Or maybe someone can think of an easier way.

In case it wasn't clear-- the question is, what's the length of the rectangular, unfolded sheet of paper (24 inches wide) that has two creases 4.2 inches apart.

[Edited 2016-06-05 14:08:35]

[Edited 2016-06-05 14:14:55]

[Edited 2016-06-05 14:15:47]

In case it wasn't clear-- the question is, what's the length of the rectangular, unfolded sheet of paper (24 inches wide) that has two creases 4.2 inches apart.

[Edited 2016-06-05 14:08:35]

[Edited 2016-06-05 14:14:55]

[Edited 2016-06-05 14:15:47]

It cannot be 25 inches as it has to be something greater than 24 inches + 4.2 inches

So the answer has got to be greater than 28.2 inches

The angle between the crease and the long side is 45 degs and the cosine of that angle is .707

4.2/.707 = 5.94

24 + 5.94 = 29.94

It is trigonometry really as if you fold the 24 ins side down you make a square all sides being 24 inches and all angles being 90 degs. The crease connects two corners and being a square divides the angle equally [45 Degs]

That makes it's mating angle 135 Degs , now drop a line from the corner of the crease at 90 degs to the crease--- to the other crease-- and that make the angle between the long side and the dropped line 45 degs.

As the formed triangle is a right angle with a 45 degs in one angle then the other angle must be 45 degs. An Isosceles

triangle. The rest is just equations.

It would be much easier if i could draw it, but I do not know how to get it onto the site

littlevc10

So the answer has got to be greater than 28.2 inches

The angle between the crease and the long side is 45 degs and the cosine of that angle is .707

4.2/.707 = 5.94

24 + 5.94 = 29.94

It is trigonometry really as if you fold the 24 ins side down you make a square all sides being 24 inches and all angles being 90 degs. The crease connects two corners and being a square divides the angle equally [45 Degs]

That makes it's mating angle 135 Degs , now drop a line from the corner of the crease at 90 degs to the crease--- to the other crease-- and that make the angle between the long side and the dropped line 45 degs.

As the formed triangle is a right angle with a 45 degs in one angle then the other angle must be 45 degs. An Isosceles

triangle. The rest is just equations.

It would be much easier if i could draw it, but I do not know how to get it onto the site

littlevc10

Quoting vc10 (Reply 8):The angle between the crease and the long side is 45 degs |

If you do that, with the crease passing thru the lower left corner of the rectangular paper, the upper left corner of the paper will be beyond the right edge of the paper. (Since the paper is longer, top to bottom, than it is wide.)

Quoting timz (Reply 7):Solve a quadratic to find that the cosine of the angle between the crease and the longer edge of the paper is 0.8. |

With what? You need two sides and the enclosed angle to use the law of cosines, and you only know one side.

-Mir

7 billion, one nation, imagination...it's a beautiful day

- einsteinboricua
**Posts:**8477**Joined:**

My brain right now

"You haven't seen a tree until you've seen its shadow from the sky."

Law of cosines isn't involved.

The equation to solve is

24 times (cos of the angle) = [12 divided by (cosine of the angle)] plus 4.2

Solution: cosine is 0.8 exactly-- a 3-4-5 right triangle.

If anyone's not clear on what the question is, I'll rephrase it. Start with a rectangle ABCD, with A in the top left corner, B top right, C lower right, D lower left. Sides AB and CD are 24 inches, the other two are longer than that.

Draw a circular arc, centered at D, radius equal to AD, that intersects side BC at E. Draw in the line that bisects angle ADE.

Draw another circular arc, same radius, centered at B, that intersects side AD at F. Draw in the line that bisects angle CBF.

The two bisectors are parallel, 4.2 inches apart. How long is the rectangle, top to bottom?

[Edited 2016-06-06 12:06:22]

The equation to solve is

24 times (cos of the angle) = [12 divided by (cosine of the angle)] plus 4.2

Solution: cosine is 0.8 exactly-- a 3-4-5 right triangle.

If anyone's not clear on what the question is, I'll rephrase it. Start with a rectangle ABCD, with A in the top left corner, B top right, C lower right, D lower left. Sides AB and CD are 24 inches, the other two are longer than that.

Draw a circular arc, centered at D, radius equal to AD, that intersects side BC at E. Draw in the line that bisects angle ADE.

Draw another circular arc, same radius, centered at B, that intersects side AD at F. Draw in the line that bisects angle CBF.

The two bisectors are parallel, 4.2 inches apart. How long is the rectangle, top to bottom?

[Edited 2016-06-06 12:06:22]

- dragon6172
**Posts:**1128**Joined:**

Quoting timz (Reply 13): |

Agree with what you've written regarding the folds. Not quite sure on the equation you are using to find the cos of the angle. I'm not saying it is wrong, just wondering if you can provide some more detail on it so I can wrap my head around what it is you are doing. It has been a while since my last trig class.

Phrogs Phorever

- VapourTrails
**Posts:**3939**Joined:**

Don't understand 99% of it, perhaps moreso ~ 30 years ago. Takes me back. I feel nostalgic. Dare I say it, maybe more of these threads.. I am enjoying the narrative that goes with it!

Quoting timz (Thread starter):How long is the paper, top to bottom? |

Is this African or European paper?

"In this present crisis, government is not the solution to our problem - government IS the problem." - Ronald Reagan

Comments made here are my own and are not intended to represent the official position of Alaska Air Group

Comments made here are my own and are not intended to represent the official position of Alaska Air Group

Quoting dragon6172 (Reply 14):Not quite sure on the equation you are using to find the cos of the angle. |

Yeah, I confess, I'm stumped.

Once you have the rectangle with the two creases, add a line from the upper left corner of the rectangle to the right crease, perpendicular to the creases. The angle between that line and the top of the rectangle is the same as the angle between each crease and its side of the rectangle, obviously, and that's the angle whose cosine we can calculate.

How many inches along the added line from the rectangle's upper left corner to the right crease? 24 times the cosine-- that becomes one side of the equation.

How many inches along the added line from the upper left corner to the left crease? 12 divided by the cosine-- so that, plus 4.2, becomes the other side of the equation.

Where did 12 divided by the cosine come from? Look the diagram over and you'll get it-- it's just similar triangles. (You do need to add one more line, from the midpoint of the top edge of the rectangle to the intersection of the left crease and the line perpendicular to it.)

How many inches along the added line from the rectangle's upper left corner to the right crease? 24 times the cosine-- that becomes one side of the equation.

How many inches along the added line from the upper left corner to the left crease? 12 divided by the cosine-- so that, plus 4.2, becomes the other side of the equation.

Where did 12 divided by the cosine come from? Look the diagram over and you'll get it-- it's just similar triangles. (You do need to add one more line, from the midpoint of the top edge of the rectangle to the intersection of the left crease and the line perpendicular to it.)

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