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Topic: Folded card (Read 762 times) |
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NickH
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Folded card
« on: Jun 22nd, 2004, 4:24pm » |
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A piece of card has the shape of a triangle, ABC, with angle ABC a right angle. It is folded once so that:
- B lies on AC; and
- the crease extends from Y on AB to X on BC.
If AB = 9 and BC = 46, find the minimum possible value of the area of triangle YBX.
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towr
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Re: Folded card
« Reply #1 on: Jun 23rd, 2004, 2:22am » |
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I get ::1587/64::
The size of the triangle is ::1/4 |B-B'|*|Y-X|, where YX crosses the middle of the line BB' and is perpendicular to it. After parametrising B' that gives me the above::
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| « Last Edit: Jun 23rd, 2004, 2:51am by towr » |
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Leon
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Re: Folded card
« Reply #2 on: Jun 23rd, 2004, 7:54am » |
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I have zero credibility of course.....but I think I figured this one out, so.....
I get : 20.44 or 20 2/9
looking at the triangle, I think that if you fold point B to AC such that YBX is a right triangle you will minimize the area. The reason I think this is that as you move B down AC towards A, the triangle approaches 1/2 of the original triangle and I can visullay see the area getting larger. As you move B up AC towards C, you can only go so far before Y hits A. That triangle looks larger than the right triangle as well. I think that a righ ttrianlge is the most inefficent triangle in regards to area vs. perimeter as well (just from random calculations, but can't prove it). Also, this is in easy and a right triangle would be the easiest to calculate.
Pulling B over to AC to make a right triangle makes 3 similar triangles. ABC, AYB, BXC.
AB / BC = 9 / 46
AY / YB = x / y
BY / XC = 9 - x / 46 - y
Solving for x and y I get x = 1, y= 46/9. I confirmed this using the equation describing AC which has slope 46/9 if A is at (0,0), which is y = 46/9x.
This means that YB is 46/9 and BX is 8. 1/2 x 46/9 x 8 = 20.44
I'm sure y'all will let me know if I am right or wrong, but I had to give it a shot......::
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towr
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Re: Folded card
« Reply #3 on: Jun 23rd, 2004, 8:55am » |
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I dare you to find a way to fold it so that YBX isn't a right triangle 
As ABC is a right angle, and Y is on AB and X on BC, YBX must be a right angle as well (because it's the same one).
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Leon
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Re: Folded card
« Reply #4 on: Jun 23rd, 2004, 9:49am » |
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on Jun 23rd, 2004, 8:55am, towr wrote:I dare you to find a way to fold it so that YBX isn't a right triangle
As ABC is a right angle, and Y is on AB and X on BC, YBX must be a right angle as well (because it's the same one). |
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Good point. So how is my guess? It's a smaller area than your guess.
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towr
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Re: Folded card
« Reply #5 on: Jun 23rd, 2004, 11:06am » |
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I wouldn't qualify my answer as a guess, the approach and hopefully the implementation of it is perfectly valid, no guessing comes into it.
If you believe your approach is correct, the same should be said for yours. But I'm not convinced it is correct.
With x = 1 and y = 46/9, the distance of B to the fold is (1*46/9)/[sqrt](1 + (46/9)2) ~= 0.98, this should be at least half the distance to the line AC, but that distance is (9*46)/[sqrt](92 + 46 2) ~= 8.83
So B doesn't even come close to being on AC after folding it.
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ThudnBlunder
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Re: Folded card
« Reply #6 on: Jun 25th, 2004, 5:08pm » |
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Quote:I get : 20.44 or 20 2/9
I'm sure y'all will let me know if I am right or wrong, |
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Sorry Leon, they are both wrong. 
Quote:| Pulling B over to AC to make a right triangle makes 3 similar triangles. ABC, AYB, BXC. |
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How can AYB or BXC be triangles, given that "the crease extends from Y on AB to X on BC"?
towr, I can see that |B - B'| and |X - Y| are the diagonals of a cyclic quadrilateral, but how have you gone from a parameterised B' to an optimised B' in one invisible step?
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| « Last Edit: Jun 26th, 2004, 5:43am by ThudnBlunder » |
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Leon
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on Jun 25th, 2004, 5:08pm, THUDandBLUNDER wrote:
How can AYB or BXC be triangles, given that "the crease extends from Y on AB to X on BC"? |
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How else are you going to fold it "once so that B lies on AC and the creases extends from Y on [the original line] AB and X on [the original line] BC"?
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| « Last Edit: Jun 25th, 2004, 10:17pm by Leon » |
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ThudnBlunder
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Re: Folded card
« Reply #8 on: Jun 26th, 2004, 4:13am » |
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on Jun 25th, 2004, 10:17pm, Leon wrote:
How else are you going to fold it "once so that B lies on AC and the creases extends from Y on [the original line] AB and X on [the original line] BC"? |
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The trouble is, you were using the label "B" to represent two different things in the same sentence.
Perhaps "... makes 3 similar triangles ABC, AYB', B'XC" would have been better.
But not by much, because in general the angles AYB' and CXB' do not equal 90o.
(We need to find the position of B' on AC which minimises the area of triangle XYB'.)
When they do, we get y/x = (9 - y)/x
where BX = y and BY = x
Hence
y = 4.5
x = 23
and
xy/2 = 51.75, more than twice the actual minimum.
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| « Last Edit: Jun 29th, 2004, 2:44am by ThudnBlunder » |
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towr
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Re: Folded card
« Reply #9 on: Jun 26th, 2004, 9:34am » |
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on Jun 25th, 2004, 5:08pm, THUDandBLUNDER wrote:| towr, I can see that |B - B'| and |X - Y| are the diagonals of a cyclic quadrilateral, but how have you gone from a parameterised B' to an optimised B' in one invisible step? |
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Magic..
And differentiation.
Given B't, you can determine the area of the triangle, A(t). Then find where the derivative is 0. There are several extrema, but it's simple enough to pick out the right one (in my case t had to be between 0 and 1, and only one extremum was in that range).
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| « Last Edit: Jun 26th, 2004, 9:36am by towr » |
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ThudnBlunder
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Re: Folded card
« Reply #10 on: Jun 26th, 2004, 12:39pm » |
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Let AB be along the y-axis
Let BC be along the x-axis
Then the equation of AC is 46y + 9x = 414........[1]
Let YB = q
Let XB = p
So that the crease has equation py + qx = pq....[2]
Let the line BB' have equation y = kx................[3]
(where k = p/q)
From [1] and [3] the coordinates of B' are
B'x = 414 / 46k + 9
B'y = 414k / 46k + 9
YBXB' is a quadrilateral and the two diagonals meet at
(207/46k + 9, 207k/46k + 9)
And this point also lies on the crease ky + x = p
So p = 207(k2 + 1) / (46k + 9)...................[4]
and q = 207(k2 + 1) / (46k + 9)k.................[5]
And we need to minimise pq/2072 = (1/k)[(k2 + 1) / (46k + 9)]2 wrt to k
Magimatica gives (2k - 3)(23k2 + 48k + 3) = 0...........[6]
23k2 + 48k + 3 = 0 has no positive roots.
Hence
k = 3/2
p = 69/8
q = 23/4
and pq/2 = 1587/64 = 24.796875
In general, if AB = r and BC = s
then [6] becomes
sk3 + 3rk2 - 3sk - r = 0
(If this is in Easy, perhaps there's a simpler geometrical method?)
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| « Last Edit: Jun 26th, 2004, 8:50pm by ThudnBlunder » |
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NickH
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Re: Folded card
« Reply #11 on: Jun 26th, 2004, 1:39pm » |
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Quote:| (If this is in Easy, perhaps there's a simpler geometrical method?) |
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If there is a simpler geometrical method, I don't know of it.
Maybe this should be in Medium; I'm not sure. But I thought since it just involves finding the area in terms of k (as you have), or in terms of some trig function of an angle (if you take a different approach), and then differentiating to find the turning point, it's fairly basic. That's not to say it isn't messy, too!
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Leon
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Re: Folded card
« Reply #12 on: Jun 26th, 2004, 2:28pm » |
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on Jun 26th, 2004, 4:13am, THUDandBLUNDER wrote:
The trouble is, you were using the label "B" to represent two different things in the same sentence.
Pehaps "... makes 3 similar triangles ABC, AYB', B'XC" would have been better. |
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Fair enough on the notation. I was following how it was described in the original staetment (ABC triangle folded such that B lies on AC). I will keep it in mind going forward.
on Jun 26th, 2004, 4:13am, THUDandBLUNDER wrote:
But not by much, because in general the angles AYB' and CXB' do not equal 90o.
(We need to find the position of B' on AC which minimises the area of triangle XYB'
When they do, we get y/x = (9 - y)/x
where BX = y and BY = x
Hence
y = 4.5
x = 23
and
xy/2 = 51.75, more than twice the actual minimum.
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I think you are saying here that if AYB' and CXB' were right angles, the area of the triangle would be 51.75. I don't agree (which probalby means I'm wrong ). If that's not wht you are saying....nevermind.
The reason I disagree and how I got there goes back to my original theory, which was that if you fold B to B' such that you get a square with BXB'Y (poorly stated originally of course since this is what I called a "right triangle") you'd have the miniumum area triangle. The crease would bisect it from X to Y. Since this was in easy, I thought it would be a square since it's easy to calculate. This was wrong (as has already been stated) and not so easy for me to calculate at that time (as stated I tried to use similar triangles (since AYB' and CXB' do equal 90o in this case) which I screwed up somehow of course. Instead, I see now that all that has to be done is solve for where B'Y = B'X with B'Y = 46/9AY (using a mirror image of the pic I posted makes it easier for me since, on a grid, A is at (0,0), B at (9,0) and C at (0,46) ) and B'X = 9 - AY.
Doing it now I find that a square would be 7.527 by 7.527 giving an area of the triangle XB'Y of 28.328 (this is were you say I am wrong and it is 51.75).
I'm sure you find my thought process highly interesting 
I have not delved into your solution yet so I will try to see what sense I can make of it to determine the smallest triangle.
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| « Last Edit: Jun 26th, 2004, 2:28pm by Leon » |
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ThudnBlunder
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Re: Folded card
« Reply #13 on: Jun 26th, 2004, 3:26pm » |
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Quote:| this is were you say I am wrong... |
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And this is where I say you are right!
What is the maximum area?
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| « Last Edit: Jun 26th, 2004, 3:43pm by ThudnBlunder » |
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Leon
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Re: Folded card
« Reply #14 on: Jun 26th, 2004, 4:32pm » |
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on Jun 26th, 2004, 3:26pm, THUDandBLUNDER wrote:
And this is where I say you are right!
What is the maximum area?
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B' on AC such that X = C. I'll have to work on the math though.
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ThudnBlunder
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Re: Folded card
« Reply #15 on: Jun 26th, 2004, 9:36pm » |
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Quote:| B' on AC such that X = C. I'll have to work on the math though. |
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Maybe this time our answers will agree. 
:Putting q = 9 in [5] gives (k2 + 1) / (46k + 9) = k/23
So 23k2 + 9k - 23 = 0
k = 0.8233079691...
Hence p = 9k = 7.409771722...
And pq/2 = 33.34397275...
Putting p = 46 in [4] gives (k2 + 1) / (46k + 9) = 2/9
So 9k2 - 92k - 9 = 0
k = 10.31912962...
Hence q = p/k = 4.457740303...
And the maximum = pq/2 = 102.528027...
This is slightly less than half the total area, as one might expect considering 46 >> 9
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| « Last Edit: Jun 26th, 2004, 11:56pm by ThudnBlunder » |
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Leon
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Re: Folded card
« Reply #16 on: Jun 27th, 2004, 7:34pm » |
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on Jun 26th, 2004, 4:32pm, Leon wrote:
B' on AC such that X = C. I'll have to work on the math though. |
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I like to keep things as simple as possible. Mostly because the most math I had was one year of college Calc, which I remember almost nothing of, so I have no idea what "parametrising" is (and I haven't looked at T&B's methodology yet). As such, to calculate the area of the triangle I used simple methods:
All numbers except final are rounded.
In Triangle ABC: AB = 9, BC = 46 and AC = 46.872. Area = 207.
Fold B to B' on AC such that Y (which lies on BC) = C. X lies somewhere between the ends of AB.
Line CB' = 46 (since C = Y, all of BC was folded onto AC). Line B'A = 0.872.
A line, starting at point N on XA, is perpendicular to AB and runs to B'.
The area of XB'C is the (area of ABC - area of AXB')/2.
Triangle ANB' is similar to triangle ABC. The length of line AN is 0.167 and line B'N is 0.856.
Line AB = AN + NX + BX. 9 = 0.167 + NX + BX thererfore BX = 8.833 - NX.
B'X2 = B'N2 + NX2
B'X = BX and B'N = 0.872, so
(8.833 - NX)2 = 0.8722 + NX2
..........NX = 4.375
Area of AXB' = area of ANB' + area of XNB' = 1/2 (0.872 x 4.375) + 1/2 (0.167 x 0.872) = 1.980.
(Area ABC (207) - 1.98 ) / 2 = 102.528027.....
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ThudnBlunder
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Re: Folded card
« Reply #17 on: Jun 27th, 2004, 8:04pm » |
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Well done, Leon! Are answers agree and I am pretty certain that they are correct.
'Parameterisation' is when we express two or more variables in terms of a third.
For example, let x = p2 and y = 2p
Then p = y/2
So x = (y/2)2
and y2 = 4x
This means that every point on the curve y2 = 4x can be represented by the point (p2, 2p).
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| « Last Edit: Jun 27th, 2004, 8:50pm by ThudnBlunder » |
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Leon
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Re: Folded card
« Reply #18 on: Jun 27th, 2004, 8:34pm » |
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on Jun 27th, 2004, 8:04pm, THUDandBLUNDER wrote:Well done, Leon! Are answers agree and I am pretty certain that they are correct.
'Parameterisation' is when we express two variables in terms of a third.
For example, let x = p2 and y = 2p
Then p = y/2
So x = (y/2)2
and y2 = 4x
This means that every point on the curve y2 = 4x can be represented by the point (p2, 2p).
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So I do know what it is. I didn't know what I knew.
Thanks T&B.
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ThudnBlunder
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Re: Folded card
« Reply #19 on: Jun 28th, 2004, 8:53am » |
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What is the average area?
I make it 57.91842141...
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