Thank you Andrei Lazanu, age 14, from School No. 205, Bucharest, Romania for this solution.

1. The ratio of the lengths of A4 paper is v2 to 1. So, let the side of an A4 paper have lengths x and xv2.

Now, I write the lengths of the sides:

AB = x

AD = (x \sqrt{2})

AE = (x \sqrt{2 / 2})

The ratio of sides EF to AE is:

EF / AE = x / (x \sqrt{2} / 2) = 2 / \sqrt{2} = \sqrt{2}

Continuing is the same manner, I observe that the smaller and the bigger A-paper all have the same proportions of length.

2. The angles of a regular pentagon must all be 108°.
a) First I looked at the possibility of working with A4 paper.

In the problem, I have to put A over O. So, triangles ARF and ORF are congruent, AM = MO and AO is perpendicular on FR.

As before, I note the smaller side of the paper (AB), with length x. Consequently, AD is

x \sqrt{2}

.
As P is on the middle of AD, AP =

x \sqrt{2} / 2

. OP is x/2.

Triangle AOP is a right-angled triangle. As AP is parallel to EO, angles PAO and EOA are equal and, from the ratio of the side lengths of the paper, we get angle PAO =

\tan^{-} 1

(1 \sqrt{2})

= 35.264 degrees.

As triangle MEO is also a right-angled triangle, angle EOM + angle MEO =

90^{\circ}

, so angle MEO is 54.736 degrees (to 3 decimal places).

But the figures (rectangle and then the pentagon) are symmetrical in respect to EO, so angle RES is two times angle MEO.

Angle RES = 2

*

MEO =

109 . 472^{\circ}

The result is not

108^{\circ}

. The error is not so important, it is of the order of 1.4%

\frac{109 . 5^{\circ} - 108^{\circ}}{108^{\circ}} = 1.4 %

b) Now I have to calculate the new dimensions of the paper, so that I obtain a regular pentagon.

To make a regular pentagon in this way the ratio of the side lengths of the paper would have to be equal to tan 54⁰ , that is in the ratio 1.376 to 1 (to 4 significant figures) rather than 1.414 to 1 as in A4 paper.

In the figure angle RES has a measure of 108°, XY is parallel to PQ and XK and YL are angle bisectors of angles RXY and SYX respectively.

Angles EXY and EYX are both equal to 36°. Consequently angles RXY and XYS both have a measure of 144°. Now, I calculate angle KXY:

KXY = RXY/2 = 72°

Now, I calculate angles EXK and EYL (the angles of the pentagon):

EXK = EYL = 72° + 36° = 108°

I observe that the other two remaining angles are also congruent (by symmetry), both having a measure of 108°.

So, I proved that in this case I obtain a regular pentagon.