2021 WSMO Team Round Problems/Problem 6

Problem

Suppose that regular dodecagon $ABCDEFGHIJKL$ has side length $5.$ The area of the shaded region can be expressed as $a+b\sqrt{c},$ where $c$ is not divisible by the square of any prime. Find $a+b+c$ .

$[asy] size(150); filldraw(polygon(12),grey); filldraw(rotate(75)*(dir(60)--dir(150)--dir(240)--dir(330)--cycle),white); for(int i=30; i<=360; i+=30){ dot(rotate(75)*dir(i)); } label("$C$",dir(45),NE); label("$B$",dir(75),N); label("$A$",dir(105),N); label("$L$",dir(135),NW); label("$K$",dir(165),W); label("$J$",dir(195),W); label("$I$",dir(225),SW); label("$H$",dir(255),S); label("$G$",dir(285),S); label("$F$",dir(315),SE); label("$E$",dir(345),E); label("$D$",dir(375),E); [/asy]$

Proposed by mahaler

Solution

Note that the shaded region consists of four identical trapezoids and a regular dodecagon has interior angles of $\frac{(12-2)\cdot180^\circ}{12} = 150^\circ.$ Consider the trapezoid $\triangle ALCB$ . Let the foot of the perpendicular from $A$ to $LC$ be $X$ . Then $\angle LAX = 60^\circ$ . This means $AX = 5\cos(60^\circ) = \frac{5}{2}\quad\text{ and }\quad LX = 5\sin(60^\circ) = \frac{5\sqrt{3}}{2}$ So, $LC = AX+LX\cdot2 = 5+2\cdot\frac{5\sqrt{3}}{2} = 5 + 5\sqrt{3}.$ The height of the trapezoid is $AX = \frac{5}{2}$ , and the two bases are $5$ and $5+5\sqrt{3}$ . So the area of one trapezoid is $\frac{5}{2}\cdot\frac{5+(5+5\sqrt{3})}{2}=\frac{50+25\sqrt{3}}{4}.$ Multiplying by 4 trapezoids, we get $4\cdot\frac{50+25\sqrt{3}}{4}=50+25\sqrt{3}\Rightarrow50+25+3=\boxed{78}.$ ~pinkpig

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2021 WSMO Team Round Problems/Problem 6

Problem

Solution