Thank you Andrei Lazanu, age 14, from School No. 205, Bucharest,
Romania for this solution. 
<p>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.</p>
<p><img src="mcprob1/image001.gif" width="150" height="186" /></p>







 
Now, I write the lengths of the sides:
<br />
<br />

<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow><mi>AB</mi><mo>=</mo><mi>x</mi></mrow></math>
<br />

<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow><mi>AD</mi><mo>=</mo><mo stretchy="false">(</mo><mi>x</mi><msqrt><mrow><mn>2</mn></mrow></msqrt><mo stretchy="false">)</mo></mrow></math>
<br />

<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow><mi>AE</mi><mo>=</mo><mo stretchy="false">(</mo><mi>x</mi><msqrt><mrow><mn>2</mn><mo>/</mo><mn>2</mn></mrow></msqrt><mo stretchy="false">)</mo></mrow></math>
<br />
<br />
The ratio of sides EF to AE is:
<br />
 
<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow><mi>EF</mi><mo>/</mo><mi>AE</mi><mo>=</mo><mi>x</mi><mo>/</mo><mo stretchy="false">(</mo><mi>x</mi><msqrt><mrow><mn>2</mn></mrow></msqrt><mo>/</mo><mn>2</mn><mo stretchy="false">)</mo><mo>=</mo><mn>2</mn><mo>/</mo><msqrt><mrow><mn>2</mn></mrow></msqrt><mo>=</mo><msqrt><mrow><mn>2</mn></mrow></msqrt></mrow></math>

<!----><!---->
<p>Continuing is the same manner, I observe that the smaller and
the bigger A-paper all have the same proportions of length.</p>
<p><b style='mso-bidi-font-weight:normal'>2.</b> The angles of a
regular pentagon must all be 108&#x00B0;.<br />
<i style='mso-bidi-font-style:normal'>a)</i> First I looked at the
possibility of working with A4 paper.</p>
<p><img src="mcprob1/image004.gif" width="261" height="365" /></p>
<p>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.</p>







 
As before, I note the smaller side of the paper (AB), with length x. Consequently, AD is 
<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow><mi>x</mi><msqrt><mrow><mn>2</mn></mrow></msqrt></mrow></math>.
<br />
As P is on the middle of AD, AP =  
<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow><mi>x</mi><msqrt><mrow><mn>2</mn></mrow></msqrt><mo>/</mo><mn>2</mn></mrow></math>. OP is x/2.
<br />
<br />
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 = 
<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow>
<msup><mrow><mi>tan</mi></mrow><mrow><mo>-</mo></mrow>
</msup>
<mn>1</mn></mrow></math> 
<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow><mo stretchy="false">(</mo><mn>1</mn><msqrt><mrow><mn>2</mn></mrow></msqrt><mo stretchy="false">)</mo></mrow></math> = 35.264 degrees.
<br />
<br />
As triangle MEO is also a right-angled triangle, angle EOM + angle MEO = 90°, so angle MEO is 54.736 degrees (to 3 decimal places).
<br />
<br />
But the figures (rectangle and then the pentagon) are symmetrical
in respect to EO, so angle RES is two times angle MEO.
<br />
<br />
Angle RES = 2 
<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow><mo>*</mo></mrow></math> MEO = 109.472°
<br />
<br />
The result is not 108°. The error is not so important, it is of the order of 1.4%
<br />
<br />
<br />
<table width="100%"><tr><td align="center">
    <math xmlns="http://www.w3.org/1998/Math/MathML">
    <mstyle displaystyle="true"><mrow>
<mfrac><mrow><mn>109</mn><mo>.</mo><mn>5</mn>°<mo>-</mo><mn>108</mn>°</mrow>
<mrow><mn>108</mn>°</mrow>
</mfrac>
<mo>=</mo><mn>1</mn><mo>.</mo><mn>4</mn><mi>%</mi></mrow>
    </mstyle></math>
</td></tr></table>
<br />


<!----><!---->
<p>b) Now I have to calculate the new dimensions of the paper, so
that I obtain a regular pentagon.</p>
<p>To make a regular pentagon in this way the ratio of the side
lengths of the paper would have to be equal to tan 54<sup>0</sup>,
that is in the ratio 1.376 to 1 (to 4 significant figures) rather
than 1.414 to 1 as in A$ paper.</p>
<p><img src="mcprob1/image007.gif" width="488" height="240" /></p>
<p>In the figure angle RES has a measure of 108&#x00B0;, XY is
parallel to PQ and XK and YL are angle bisectors of angles RXY and
SYX respectively.</p>
<p>Angles EXY and EYX are both equal to 36&#x00B0;. Consequently
angles RXY and XYS both have a measure of 144&#x00B0;. Now, I
calculate angle KXY:</p>
<p>KXY = RXY/2 = 72&#x00B0;</p>
<p>Now, I calculate angles EXK and EYL (the angles of the
pentagon):</p>
<p>EXK = EYL = 72&#x00B0; + 36&#x00B0; = 108&#x00B0;</p>
<p>I observe that the other two remaining angles are also congruent
(by symmetry), both having a measure of 108&#x00B0;.</p>
<p>So, I proved that in this case I obtain a regular pentagon.</p>
<br />
<br />