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Difference between revisions of "2021 WSMO Team Round Problems/Problem 6"

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

Latest revision as of 13:16, 9 September 2025

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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