# Thought it was going to be simple! Find Z...

• May 7th 2008, 09:02 PM
stardotstar
Thought it was going to be simple! Find Z...
Hi all,

its a while since I exercised these mental muscles!

I have tried to solve Z in the image below based on the most simplistic trig I can remember but keep finding that I don't have enough information.
(note the pic has been cropped and not to scale anymore - only showing the working part of the SRA3 page I describe in the problem)
The problem is essentially this:

Take a piece of paper 320mm x 450mm (SRA3) and fold one (right angled corner) to the exact centre (160,225) of the sheet and crease it. The crease becomes "Z" in the image below.

I am not concerned about the ratio of the short edge to the long edge in this instance (I understand the relationship within the A series papers) but rather want to know if there is enough information to solve for Z knowing only 160 and 225mm...

• May 8th 2008, 04:13 PM
stardotstar
I find that perhaps c=160/2 which would solve for y...

I also suppose that a line from the centre to the corner would be bisected by Z and therefore provide another known relationship within the kite that has Z as the central axis...

Here is another scan of a scale (but not perfect) diagram of the problem showing as much as I can what is known and what is supposed>?

Note that the image has been slightly cropped at the bottom and left - the total page size is 320mm x 450mm and therefore the centre is known to be at 160x225. The top right corner is assumed to be a right angle and it is folded down to the centre point to create the line Z which is the key length to find.
• May 11th 2008, 02:10 PM
stardotstar
anyone? I am no further than friday with this...

In resorting to measurement it seems that y=238.8 can be found by proving (or assuming) that c= 160/2 implying that the point at which the diagonal Z meets the top page edge is a bisection of that half of the page... Not sure that would necessarily be the case.
• May 11th 2008, 05:16 PM
Reckoner
Hi, stardotstar!

Quote:

Originally Posted by stardotstar
anyone? I am no further than friday with this...

In resorting to measurement it seems that y=238.8 can be found by proving (or assuming) that c= 160/2 implying that the point at which the diagonal Z meets the top page edge is a bisection of that half of the page... Not sure that would necessarily be the case.

I have uploaded a new image with more labels to make it easier for me to describe things.

First, consider the two triangles $\displaystyle \triangle CDH,\ \triangle LKH$. Note that $\displaystyle \overleftrightarrow{CE}\parallel\overleftrightarro w{JL}$, and since $\displaystyle \overleftrightarrow{CL}$ cuts these two parallel lines, you know the alternate interior angles must be equal. Thus, $\displaystyle \angle DCH=\angle KLH$. Furthermore, $\displaystyle \angle CHD=\angle LHK$ since they are vertical angles. Therefore, both triangles have two corresponding angles equal, with equal included sides, so $\displaystyle \triangle CDH\cong\triangle LKH$, and all remaining corresponding sides and angles are equal.

So, you are indeed correct that $\displaystyle \overleftrightarrow{DH}=\overleftrightarrow{HK}$. And, since $\displaystyle \overleftrightarrow{DE}=\overleftrightarrow{KL}=\o verleftrightarrow{CD}$, we now know that $\displaystyle \overleftrightarrow{CD}=\overleftrightarrow{DE}$. None of this is really necessary for finding $\displaystyle z$ though. I just wanted to point them out since you had drawn question marks on the diagram.

Now, since $\displaystyle y = 160 + c$, use the Pythagorean Theorem on $\displaystyle \triangle BAC$:

$\displaystyle 225^2 + c^2 = (160 + c)^2$

$\displaystyle \Rightarrow 50625 + c^2 = c^2 + 320c + 25600$

$\displaystyle \Rightarrow 50625 = 320c + 25600$

$\displaystyle \Rightarrow c = \frac{5005}{64}\approx78.203$

So $\displaystyle c$ is not 80 as you guessed, but it is pretty close.

Now, let's move on to finding $\displaystyle z$:

First, find $\displaystyle x$ by applying the Pythagorean Theorem to $\displaystyle \triangle CFE$:

$\displaystyle 160^2 + \left(\overleftrightarrow{FE}\right)^2 = x^2$

$\displaystyle \Rightarrow160^2 + \left(\overleftrightarrow{FE}\right)^2 = \left(225 - \overleftrightarrow{FE}\right)^2$

$\displaystyle \Rightarrow25600 + \left(\overleftrightarrow{FE}\right)^2 = 50625 - 450\,\overleftrightarrow{FE} + \left(\overleftrightarrow{FE}\right)^2$

$\displaystyle \Rightarrow\overleftrightarrow{FE} = \frac{1001}{18}$

and

$\displaystyle x = 225 - \overleftrightarrow{FE}\Rightarrow x = 225 - \frac{1001}{18} = \frac{3049}{18}\approx169.389$

We know $\displaystyle y = 160 + c = 160 + \frac{5005}{64} = \frac{15245}{64}\approx238.203$.

Again, use Pythagorean's Theorem to find $\displaystyle z$:

$\displaystyle x^2 + y^2 = z^2$

$\displaystyle \Rightarrow z^2 = \left(\frac{3049}{18}\right)^2 + \left(\frac{15245}{64}\right)^2 = \frac{28344726649}{331776}$

$\displaystyle \Rightarrow z = \sqrt{\frac{28344726649}{331776}} = \frac{3049\,\sqrt{3049}}{576}\approx292.290$

Now, I could have very easily made a mistake somewhere, so I would appreciate it if someone could point out any errors. However, I hope I at least gave you some ideas on what to do here (and I don't mind providing a complete answer since it seems like this is for a personal project and not schoolwork; is that right?).
• May 13th 2008, 06:08 PM
stardotstar
Thank you for the solution; I will follow your working and see if I arrive at the same results.

This is indeed my own project. This is actually a theory I have about testing if a piece of paper for a printing press is square and correct dimensions using the simplest method.

A single fold of the corner to a dot printed on the centre of a (theoretically) perfect sra3 sheet (160,225) should yield a measurement (give or take) of z that is equivalent to the number solved. Any deviation from this would be caused by incorrect setup of the press, corner out of square or wrong sheet dimensions. In fact I am not even sure it is a good way to do this but got stumped trying to solve what I thought was a simple problem :-)

Thanks for the help.

Will
• May 28th 2008, 06:47 PM
stardotstar
I have continued to work on this a little bit at at time - having been given another solution by a colleague.

I am following your approach Reckoner and concur with everything about the similarity of CDH and HKL; that much I could follow.

I immediately ran into trouble understanding how you start with the assumption that

y=160+c

I must be missing a concept or working that applies to these triangles but at best I find that

y^2=225^2+c^2

I fail to see why y = 160+c

and since cd=de=jk=kl=80 and you find that c<>80 I can't seem to draw any conclusion about bc...
• May 28th 2008, 07:15 PM
Reckoner
Quote:

Originally Posted by stardotstar
I have continued to work on this a little bit at at time - having been given another solution by a colleague.

I am following your approach Reckoner and concur with everything about the similarity of CDH and HKL; that much I could follow.

I immediately ran into trouble understanding how you start with the assumption that

y=160+c

I must be missing a concept or working that applies to these triangles but at best I find that

y^2=225^2+c^2

$\displaystyle y = \overline{AL} = \overline{BE} = 160 + c$

This is because $\displaystyle \triangle ACH\cong\triangle ALH$ (two sides and an included angle), and $\displaystyle \overleftrightarrow{BE}\parallel\overleftrightarro w{AL}$

You may also have trouble seeing that $\displaystyle x = 225 - \overline{FE}$, and I apologize for not clarifying.

$\displaystyle \triangle FCH\cong\triangle FLH$ (2 sides and an included angle), which implies that $\displaystyle x = \overline{LF} = \overline{LE}-\overline{FE} = \overline{AB} - \overline{FE} = 225 - \overline{FE}$ (since $\displaystyle \overleftrightarrow{LE}\parallel\overleftrightarro w{AB}$).

Is that clearer? I'm sorry I left out such a big step; in my working I just thought it was obvious that $\displaystyle y$ would be equal to $\displaystyle \overline{AL}$, since that was the edge of the paper that was folded over.
• May 28th 2008, 07:38 PM
stardotstar
Reckoner, thank you for that - as soon as you mentioned the "obvious" bit it dawned on me just how obvious that is! :lol:

Funny how I missed that relationship but began the whole right triangle process with the assumption that because the corner of the page is a 90' then the angle formed at the centre would have to be and so on...

That fact means that the kite CFLA is symmetrical (perhaps all kites are along at least one axis?) and composed of two (necessarily by virtue of the fold I made to construct the problem) similar right triangles ALF and ACF.

Back to my own working now on that.

Thank you for the assistance - it amazes me how much I miss when its just staring me in the face.(Headbang)