NZMO Round One

Solution:

Let the original rectangle be $ABCD$. Let the fold line be $XY$ with point $X$ lying on side $AB$ and point $Y$ lying on side $CD$. Consider the fold such that point $D$ lands on point $B$. Let $P$ be the location of corner $A$ after the fold.

The two-sheets-thick area is $\triangle BXY$ which should be $80\%$. By symmetry, triangles $\triangle PXB$ and $\triangle CYB$ are congruent with equal hypotenuses

$$BX = BY.$$

Triangles $\triangle PXB$ and $\triangle CYB$ have equal area and their total area is $20\%$ of rectangle $ABCD$. Therefore $\triangle BYC$ has area $10\%$. Hence the ratio of the areas $|BYC|:|BXY|$ is $10\% : 80\%$. Both triangles have height $BC$ equal to the height of rectangle $ABCD$. Therefore their bases are also in the ratio $1:8$, i.e. $XB = 8YC$.

$$\therefore BY = XB = 8YC.$$

Now we apply Pythagoras' to $\triangle BCY$ to get

$$BC^{2} = BY^{2} - CY^{2} = (8CY)^{2} - CY^{2} = 63CY^{2}.$$

Therefore $BC = \sqrt{63}\, CY$.

Combining this with $CD = CY + YD = CY + YB = 9CY$ we get

$$BC:CD = \sqrt{7}:3.$$