MHB Discover the Best Paper Folding Techniques: AG = 30cm and More Tips!"

[Ilikebugs]

View attachment 6183 So I know that AG is 30cm, what else do I do?

 

[I like Serena]

So I know that AG is 30cm, what else do I do?

Hey Ilikebugs! ;)

Let's draw some helping lines and attach some variables:
\begin{tikzpicture}[shorten >=1pt, shorten <=1pt, font=\large, scale=0.2]

\def\w{40};
\def\b{30};
\def\x{75/2};
\def\y{35/4};

\draw[ultra thick, blue, pattern=fivepointed stars, pattern color=blue]
(0,\b) coordinate (A) -- (0,0) coordinate (B) -- node[below] {$y$} (\y,0) coordinate (E) -- cycle;
\draw[ultra thick, dashed]
(E) -- (\w,0) coordinate (C) -- (\w,\b) coordinate (D) -- node[below] {$y$} (\w-\y,\b) coordinate (F);
\draw[ultra thick, blue]
(E) -- node[above left] {$x/2$} (\w / 2, \b /2) node[right,gray] {M} -- node[above left] {$x/2$} (F) -- ({\b^2 / (\w - \y)}, {\b + \b * \y/(\w - \y)}) coordinate (G) -- (A);
\draw[gray] (A) -- node[above right] {$25$} (\w / 2, \b /2) -- (C);
\draw[gray] (A) -- (F) -- (\w-\y,0) coordinate (H) -- node[above, black] {$40-2y$} (E);
\draw[gray] (\w-\y,0) rectangle (\w - \y - 2, 2);

\node

at (A) {A};
\node

at (B) {B};
\node

at (C) {C};
\node

at (D) {D};
\node[below] at (E) {E};
\node[above right] at (F) {F};
\node

at (G) {G};
\node[below, gray] at (H) {H};

\draw[triangle 45-triangle 45] ([yshift=-4cm] B) -- node[below] {$40$} ([yshift=-4cm] C);
\draw[triangle 45-triangle 45] ([xshift=-4cm] A) -- node

{$30$} ([xshift=-4cm] B);

\end{tikzpicture}

We're looking for $x$... can we find a couple of equations that include $x$? (Wondering)​

 

[Ilikebugs]

uhh I don't know

 

[I like Serena]

Well... we have for instance the right triangle $\triangle EFH$ with sides $40-2y$, $30$, and $x$.
How about applying the Pythagorean theorem? (Wondering)

 

[Ilikebugs]

x= 2500-160y+4y^2 ?

 

[I like Serena]

Good! (Nod)
Do add a square on the left hand side: $x^2= 2500-160y+4y^2$.

Now how about applying Pythagoras to $\triangle ABE$?
And to $\triangle AEM$?

 

[Ilikebugs]

(AE)^2= 900+y^2 or 625 + x^2/4?

 

[I like Serena]

Good! (Nod)

So we have:
$$x^2= 2500-160y+4y^2 \\
900+y^2 = 625 + x^2/4$$

Can we solve $x$ from these 2 equations?

 

[Ilikebugs]

Well I can't

 

[I like Serena]

Let's see... (Thinking)
$$x^2= 2500-160y+4y^2 \tag 1$$
$$900+y^2 = 625 + x^2/4 \tag 2$$
Eliminate $x$ by substituting (1) in (2) to find:
$$900+y^2 = 625 + (2500-160y+4y^2)/4 \\
900+y^2 =625 + 625-40y+y^2 \\
40y=625 + 625 - 900 =350$$
$$y=\frac{35}{4}\tag 3$$
Find $x$ now by substituting (3) in (2):
$$900+\Big(\frac{35}{4}\Big)^2 = 625 + x^2/4 \\
900+\Big(\frac{35}{4}\Big)^2 - 625 = x^2/4 \\
x^2 = 4\left(900+\Big(\frac{35}{4}\Big)^2 - 625\right)
=3600 - 2500 +\frac{35^2}{4} = 1100+\frac{1225}{4} = \frac{4400+1225}{4}=\frac{5625}{4} \\
x=\frac{75}{2}
$$
So the length of the crease is $\frac{75}{2}$. (Cool)

 

[Greg]

Alternatively, using Pythagoras, $30^2+y^2=(40-y)^2\implies y=\frac{35}{4}$.
$(40-2y)^2+30^2=\left(40-\frac{35}{2}\right)^2+900=x^2\implies x=\frac{75}{2}$.