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Difference between revisions of "2023 SSMO Team Round Problems/Problem 15"

(Created page with "==Problem== Consider a piece of paper in the shape of a regular pentagon with sidelength <math>2.</math> We fold it in half. We then fold it such that the vertices of the long...")
 
 
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==Solution==
 
==Solution==
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The main fold line that divides the pentagon in two has length <cmath>\sqrt{5 + 2\sqrt{5}}</cmath> using the formula for the apothem and radius of a regular polygon.
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We can calculate that <cmath>\cos(18^\circ) = \sqrt{\frac{5}{8} + \sqrt{\frac{5}{8}}}</cmath> and that <cmath>\tan(18^\circ) = \sqrt{1 - \frac{2}{\sqrt{5}}}.</cmath>
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Then, the length of the fold line when the outstretching "bump" is unfolded becomes <cmath>\frac{\sqrt{5 + 2\sqrt{5}}}{\cos(18^\circ)} = \frac{1 + \sqrt{5}}{2}.</cmath>
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The longest side length of the bump is then <cmath>\frac{1 + \sqrt{5}}{2} - 1 = \frac{\sqrt{5} - 1}{2}.</cmath>
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Using the Law of Cosines or sine-area formula, we find the area of the bump is <cmath>\frac{1}{16} \sqrt{25 - 10\sqrt{5}}.</cmath>
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The remaining trapezoid has area 
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<cmath>\frac{1 + 1 + \tan(18^\circ) \cdot \frac{\sqrt{5 + 2\sqrt{5}}}{2}}{2} \cdot \frac{1}{2} \cdot \sqrt{5 + 2\sqrt{5}} = \frac{5}{8} \cdot \sqrt{5 + 2\sqrt{5}}.</cmath>
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So the total area is 
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<cmath>\frac{5}{8} \cdot \sqrt{5 + 2\sqrt{5}} + \frac{1}{16} \sqrt{25 - 10\sqrt{5}} = \frac{1}{16} \sqrt{625 + 190\sqrt{5}}.</cmath>
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Thus, the final answer is <cmath>16 + 625 + 190 = \boxed{831}.</cmath>
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~SMO_Team

Latest revision as of 21:38, 9 September 2025

Problem

Consider a piece of paper in the shape of a regular pentagon with sidelength $2.$ We fold it in half. We then fold it such that the vertices of the longest side become the same side. The area of the folded figure can be expressed as $\frac{1}{a}\sqrt{b + c\sqrt{5}}$ where $a, b, c$ are integers and $\gcd(b, c)$ is squarefree. Find $a + b + c.$ (For convenience, note that $\cos(36^\circ) = \frac{1 + \sqrt{5}}{4}$)

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Solution

The main fold line that divides the pentagon in two has length \[\sqrt{5 + 2\sqrt{5}}\] using the formula for the apothem and radius of a regular polygon.

We can calculate that \[\cos(18^\circ) = \sqrt{\frac{5}{8} + \sqrt{\frac{5}{8}}}\] and that \[\tan(18^\circ) = \sqrt{1 - \frac{2}{\sqrt{5}}}.\]

Then, the length of the fold line when the outstretching "bump" is unfolded becomes \[\frac{\sqrt{5 + 2\sqrt{5}}}{\cos(18^\circ)} = \frac{1 + \sqrt{5}}{2}.\]

The longest side length of the bump is then \[\frac{1 + \sqrt{5}}{2} - 1 = \frac{\sqrt{5} - 1}{2}.\]

Using the Law of Cosines or sine-area formula, we find the area of the bump is \[\frac{1}{16} \sqrt{25 - 10\sqrt{5}}.\]

The remaining trapezoid has area \[\frac{1 + 1 + \tan(18^\circ) \cdot \frac{\sqrt{5 + 2\sqrt{5}}}{2}}{2} \cdot \frac{1}{2} \cdot \sqrt{5 + 2\sqrt{5}} = \frac{5}{8} \cdot \sqrt{5 + 2\sqrt{5}}.\]

So the total area is \[\frac{5}{8} \cdot \sqrt{5 + 2\sqrt{5}} + \frac{1}{16} \sqrt{25 - 10\sqrt{5}} = \frac{1}{16} \sqrt{625 + 190\sqrt{5}}.\]

Thus, the final answer is \[16 + 625 + 190 = \boxed{831}.\]

~SMO_Team

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