# About the area of the region where the paper is twofold when you double a piece of paper in the shape of a triangle

Suppose that you have a piece of paper in the shape of a triangle $ABC$ whose area is $S_0$ and that the area of the region where the paper is twofold when you double the paper in two along a line is represented as $S$. Then, supposing that $BC=a\ge CA=b\ge AB=c$, what is the max of the ratio $\frac{S}{S_0}$?

Here is my expectation.

My expectation :

If a point $(\frac ba, \frac ca)$ is in the region Nomber 1, then the max of $\frac{S}{S_0}$ would be $\frac{b}{a+b}.$

If a point $(\frac ba, \frac ca)$ is in the region Nomber 2, then the max of $\frac{S}{S_0}$ would be $\frac{c}{b+c}.$

If a point $(\frac ba, \frac ca)$ is in the region Nomber 3, then the max of $\frac{S}{S_0}$ would be $\frac{a^2}{2a^2+b^2-c^2}.$

The following is the figure of each Number where $s=\frac ba, t=\frac ca$

Then, here is my question.

Question: Is my expectation true? If it is true, could you tell me how to prove that? If it is not true, please tell me the answer you get with a proof.

I really need a rigorous proof; for me it's so hard that I can't do anything.

Remark : This question has been asked previously on math.SE without receiving any answers.

Motivation : I'm interested in this question because I've heard the following.

"A bisector of an apex angle doesn't always make $\frac{S}{S_0}$ the max."$\ \ \ \cdots(\star)$

Letting $S_A, S_B, S_C$ be each area of the region where the paper is twofold when you double the triangle $ABC$ along a bisector of each apex angle $CAB, ABC, BCA$ respectively, then we can easily get the following : $$\frac{S_A}{S_0}=\frac{c}{b+c}, \frac{S_B}{S_0}=\frac{c}{c+a}, \frac{S_C}{S_0}=\frac{b}{a+b}.$$

After struggling to find a special line by using computer, I got the following:

Let $P$ be a point which internally divides the line segment $CB$ into $a^2+b^2-c^2 : a^2$. Also, let $S_L$ be the area of the region where the paper is twofold when you double the triangle $ABC$ along a line $L$ which satisfies the follwoing two:

1. $L$ passes through a point $P$.

2. $L$ is perpendicular to the line $BC$.

Then, we get with a tedious calculation $$\frac{S_L}{S_0}=\frac{a^2}{2a^2+b^2-c^2}.$$

By using this, we can see that $(\star)$ is true.

Let's consider the case when $a=10, b=6, c=5.$ Then, we can get $$\frac{S_L}{S_0}=\frac{100}{211}=\frac{26400}{211\cdot 3\cdot 8\cdot 11}.$$ On the other hand, we can see $$\frac{S_A}{S_0}=\frac{5}{11}=\frac{25320}{211\cdot 3\cdot 8\cdot 11}, \frac{S_B}{S_0}=\frac{1}{3}=\frac{18568}{211\cdot 3\cdot 8\cdot 11}, \frac{S_C}{S_0}=\frac{3}{8}=\frac{20889}{211\cdot 3\cdot 8\cdot 11}.$$

Hence, now $(\star)$ is proven to be true.

However, the above question still remains unsolved without a proof. After the above argument, I reached the following expectation.

My expectation: The max of $\frac{S}{S_0}$ would be one of the following three:$$\frac{b}{b+c}, \frac{b}{a+b}, \frac{a^2}{2a^2+b^2-c^2}.$$

This seems to be true because of the following figure where $\frac ba=s, \frac ca=t$ and where $O(0,0), D(\frac12,\frac12), E(1,1), F(1,0), G(\frac{1}{\sqrt2}, 1-\frac{1}{\sqrt2}), H(\frac{1}{\sqrt2}, \frac{1}{\sqrt2}), I(\alpha, {\alpha}^2).$

Note that $0.81\lt \alpha\lt 0.82$ is the positive real solution of $x^4+x^3-1=0$ and that

$$\frac{b}{b+c}=\frac{t}{s+t}, \frac{b}{a+b}=\frac{s}{1+s}, \frac{a^2}{2a^2+b^2-c^2}=\frac{1}{2-s^2-t^2}.$$

You can see the following three curves meet at a point $I$: $$t=\frac{\sqrt{s^2+4}-s}{2}, t=s^2, t=\sqrt{s^2+1-\frac 1s}.$$

As a result, my expectation can be described as I worte at the top with more details.

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A very special case is discussed at web.maths.unsw.edu.au/~mikeh/webpapers/paper163.pdf – Gerry Myerson Sep 26 '13 at 1:03
@GerryMyerson: Thank you very much. – mathlove Sep 26 '13 at 12:24