## Poll

9 votes (45%) | |||

7 votes (35%) | |||

4 votes (20%) | |||

2 votes (10%) | |||

6 votes (30%) | |||

2 votes (10%) | |||

3 votes (15%) | |||

2 votes (10%) | |||

9 votes (45%) | |||

6 votes (30%) |

**20 members have voted**

November 25th, 2020 at 9:39:09 AM
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Follow up question:

For arbitrary rectangle with Height=H, Width=W......(oriented with H>=W), What range can the area ratio (Blue/Total) of the folded result have? And for what H,W is the Blue area 50%?

For arbitrary rectangle with Height=H, Width=W......(oriented with H>=W), What range can the area ratio (Blue/Total) of the folded result have? And for what H,W is the Blue area 50%?

November 25th, 2020 at 10:22:48 AM
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Wiz,

Your second set disproves your earlier statement about the permissible number of evens.

Dog Hand

Your second set disproves your earlier statement about the permissible number of evens.

Dog Hand

November 25th, 2020 at 4:45:14 PM
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Quote:chevyFollow up question:

For arbitrary rectangle with Height=H, Width=W......(oriented with H>=W), What range can the area ratio (Blue/Total) of the folded result have? And for what H,W is the Blue area 50%?

I get a general equation for the ratio of H/W = sqrt((ratio+1)/(3ratio -1)) .

For a ratio of 0.5, H/W = sqrt(3)

If H=W, ratio = 1 (same as folding a square in half).

If H >> W, the ratio approaches 1/3.

I heart Crystal Math.

November 25th, 2020 at 7:14:55 PM
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Quote:CrystalMath

I get a general equation for the ratio of H/W = sqrt((ratio+1)/(3ratio -1)) .

For a ratio of 0.5, H/W = sqrt(3)

If H=W, ratio = 1 (same as folding a square in half).

If H >> W, the ratio approaches 1/3.

I agree!

November 26th, 2020 at 7:40:24 AM
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The ratio of the area of that pentagon where the paper overlaps to the entire pentagon is 1/3

Sorry, didn't notice that the problem was a day old.

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

November 26th, 2020 at 11:01:07 AM
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Happy (US) Thanksgiving!

Turkey = ?

Cornucopia = ?

Native American = ?

Mayflower = ?

Pilgrim = ?

Turkey = ?

Cornucopia = ?

Native American = ?

Mayflower = ?

Pilgrim = ?

Have you tried 22 tonight? I said 22.

November 26th, 2020 at 1:17:31 PM
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Quote:Gialmere

Turkey = ?

Cornucopia = ?

Native American = ?

Mayflower = ?

Pilgrim = ?

Just the numbers this time...

Turkey = 6

Cornucopia = 8

Native American = 7

Mayflower = 11

Pilgrim = 29

November 26th, 2020 at 4:21:01 PM
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Quote:chevy

***I "think" the red/blue diagram is not to scale. It makes blue appear as right angle. I "think" the line from A,D to the center of opposite side is the perpendicular.***

Call the center of the rectangle Y. Call the ends of the fold (X,Z). So side of blue triangle opposite A,D is XYZ with Y at midpoint and (A,D) to Y is perpendicular to XZ.

Area of Blue = .5 * XZ * AY = XY * AY

Area of rectangle = 2*Area of Blue + 2* Area of Red = 4

Area of Pentagon = Area of rectangle - Area of Blue = 4-Area of Blue

Ratio = Area of Blue / (4-Area of Blue)

From Trig using the rectangle diagram with XYZ added accordingly....

AD = sqrt(4^2+1^2)=sqrt(17)

AY = AD/2 = sqrt(17)/2

Call theta angle given by CAD, then

tan(theta) = DC/AC = 1/4

Theta is also angle for XAY, so

tan(theta) = XY/AY

XY = AY * tan(theta) = sqrt(17)/2 * (1/4) = sqrt(17)/8

Thus Area of Blue = XY * AY = [sqrt(17)/8] * sqrt(17)/2 = 17/16

And

ratio = Area of Blue / (4-Area of Blue)

= (17/16) / (4 - 17/16)

=17/47

I agree!

Welcome to the WoV forum, by the way.

That was a tough problem and only one member answered it correctly (you), so please consider yourself invited to the prestigious "Beer Club." This means I owe you a beer should we ever meet.

It's not whether you win or lose; it's whether or not you had a good bet.

November 26th, 2020 at 8:02:17 PM
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Quote:ThatDonGuy

Just the numbers this time...

Turkey = 6

Cornucopia = 8

Native American = 7

Mayflower = 11

Pilgrim = 29

Correct!

-----------------------

Have you tried 22 tonight? I said 22.

November 26th, 2020 at 10:31:01 PM
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Quote:Wizard

I agree!

Welcome to the WoV forum, by the way.

That was a tough problem and only one member answered it correctly (you), so please consider yourself invited to the prestigious "Beer Club." This means I owe you a beer should we ever meet.

Thanks for the "Beer Club" honor.

In fairness, CrystalMath answered my followup about the general HxW rectangle with a formula I agree with, so I assume he too had the original problem solved, just never posted.

Hopefully that doesn't preclude my membership in said club.