The lower corner of a leaf in a book is folded over so as to just reach the inner edge of the page. The fraction of which folded over if the area of the folded part is minimum is

1. Setting up the picture

Take the page width as 1 unit for simplicity.
Bottom-left corner = ((0,0)), bottom-right corner = ((1,0)).
Suppose after folding, that corner lands at the point ((0,,y)) on the left edge.

2. The crease (fold line)

The segment being folded is from ((1,0)) to ((0,y)).
Mid-point $M\Bigl(\tfrac{1}{2},\,\tfrac{y}{2}\Bigr)$
Slope of this segment is $-y$, so the crease is perpendicular to it and has slope $\tfrac{1}{y}$.

Equation of crease:

3. Where does the crease meet the edges?

• Bottom edge ((Y=0)) $0-\frac{y}{2}=\frac{1}{y}\Bigl(X-\frac{1}{2}\Bigr) \;\Rightarrow\; X=\frac{1}{2}-\frac{y^{2}}{2}=r$
• Right edge ((X=1)) \;\Rightarrow\; Y=\frac{y}{2}+\frac{1}{2y}=s$$

The folded region is the triangle with vertices $P(1,0),\;R(r,0),\;S(1,s).$

4. Area of the folded triangle

Base on the bottom = $1-r$, height on the right = $s$.

5. Minimising the area

Differentiate $A$ (or the simpler $g(y)=\dfrac{(1+y^{2})^{2}}{y}$):

6. Fraction of width folded

The part of the bottom edge inside the fold = $1-r=\dfrac{1+y^{2}}{2}$. Plug $y^{2}=\dfrac{1}{3}$:

Thus (\dfrac{2}{3}) of the page-width is folded when the triangle’s area is the least.

What is happening?

Imagine you have a rectangular page. You take the bottom-right corner and fold it so that this very tip just touches the left (inner) edge of the page. The little flap you fold is a triangle.

What do we want?

We can choose how far up the left edge the tip touches. Some choices make a big triangle, some give a small triangle. We want the choice that makes the smallest triangle possible.

What do we have to find?

After we make the best (smallest) fold, what fraction of the bottom edge actually got folded?
Answer: two-thirds ( (\dfrac{2}{3}) ) of the width is inside the fold.

Step-by-step solution

1. Let the page width be 1.
2. Let the corner land at ((0,y)).
3. Crease equation and intercepts:
   • Mid-point of the move: $M\bigl(\tfrac12,\,\tfrac y2\bigr)$.
   • Slope of (PP'): $-y$. Slope of crease: $\tfrac1y$.
   • Bottom intercept (R(r,0)): $r=\tfrac12-\tfrac{y^{2}}2$.
   • Right intercept (S(1,s)): $s=\tfrac y2+\tfrac1{2y}$.
4. Folded-triangle area: $A(y)=\frac12(1-r)\,s=\frac{(1+y^{2})^{2}}{8y}.$
5. Minimise: $\frac{dA}{dy}=0\;\Rightarrow\;3y^{2}=1\;\Rightarrow\;y^{2}=\frac13.$
6. Width folded: $1-r=\frac{1+y^{2}}2=\frac{1+\tfrac13}{2}=\frac23.$

[\boxed{\text{Fraction folded}=\dfrac{2}{3}}]