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2023 SSMO Team Round Problems/Problem 15

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 $\cos(36^\circ) = \frac{1 + \sqrt{5}}{4}$)

[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ real xmin = -2.464957433726653, xmax = 3.328270917519617, ymin = -1.0176780156925203, ymax = 3.975434729820146; /* image dimensions */ /* draw figures */ filldraw((0,0)--(2,0)--(2.618033988749895,1.9021130325903064)--(1,3.077683537175253)--(-0.6180339887498947,1.9021130325903073)--cycle, opacity(0.4)+lightblue,lightblue); filldraw((0.3090169943749479,2.853169548885461)--(0.690983005625053,2.853169548885461)--(1.1545084971874735,1.4265847744427302)--(-0.30901699437494734,0.9510565162951536)--(-0.6180339887498947,1.9021130325903073)--(0.190983005625053,2.48989828488278)--cycle, opacity(0.4)+palegreen,green); draw((xmin, 0.32491969623290623*xmin + 1.0514622242382672)--(xmax, 0.32491969623290623*xmax + 1.0514622242382672), dashed); /* line */ draw((xmin, -3.0776835371752562*xmin + 4.979796569765563)--(xmax, -3.0776835371752562*xmax + 4.979796569765563), dashed); /* line */ /* dots and labels */ clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]

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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