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2021 WSMO Accuracy Round Problems/Problem 5

Revision as of 09:55, 11 July 2022 by Pinkpig (talk | contribs)

Problem

Suppose regular octagon $ABCDEFGH$ has side length $5.$ If the distance from the center of the octagon to one of the sides can be expressed as $\frac{a+b\sqrt{c}}{d}$ where $\gcd{(a,b,d)}=1$ and $c$ is not divisible by the square of any prime, find $a+b+c+d.$

Solution 1

[asy] size(150); draw(polygon(8)); for(int i=45; i<=360; i+=45){ dot(rotate(22.5)*dir(i)); } label("$D$",dir(337.5),SE); label("$C$",dir(22.5),NE); label("$B$",dir(67.5),N); label("$E$",dir(292.5),S); label("$F$",dir(247.5),S); label("$G$",dir(202.5),W); label("$H$",dir(157.5),NW); label("$A$",dir(112.5),N); label("$O$",(0,0),S); dot((0,0),red); draw(dir(112.5)--dir(67.5)--(0,0)--cycle,red); dot(dir(112.5),red); dot(dir(67.5),red); label("$5$",(0,1.05)); draw(anglemark(dir(67.5),(0,0),dir(112.5)),red); label("$45^{\circ}$",(0,0.25),N); [/asy] Let the center of the octagon be $O.$ We will focus on triangle $AOB.$ Let $AO=OB=x.$ From the Law of Cosines on triangle $AOB,$ we find that \[x^2+x^2-2x^2\cdot\cos{(45^{\circ})}=5^2=25\implies\]\[(2-\sqrt{2})x^2=25\implies x^2=\frac{25}{2-\sqrt{2}}=\frac{25(2+\sqrt{2})}{2}.\] Now, let the distance from the center of the octagon to one of its sides be $h.$ This means that \[[AOB]=\frac{5h}{2}.\] In addition, from the sine area formula, \[[AOB]=\frac{1}{2}\sin{\angle{AOB}}\cdot AO\cdot BO=\frac{x^2\sqrt{2}}{4}=\frac{25(\sqrt{2}+1)}{4}.\] Therefore, we have \[\frac{5h}{2}=\frac{25(\sqrt{2}+1)}{4}\implies h=\frac{5+5\sqrt{2}}{2}\Longrightarrow5+5+2+2=\boxed{14}.\] ~pinkpig

Solution 2

Note that the area of a polygon with $n$ sides, $s$ side length, and $l$ apothem (distance from the center to one of the sides) can be expressed as $(nsl)/2.$ Applying this formula, we get \[(8\cdot 5\cdot l)/2=40l/2=20l.\] Now, we need something to equate to this. Remember that the area of a regular octagon with side length $s$ is $2s^2(1+\sqrt{2}).$ This means that the area of octagon $ABCDEFGH$ is $50+50\sqrt{2}.$ Therefore, the answer is \[l=\frac{50+50\sqrt{2}}{20}=\frac{5+5\sqrt{2}}{2}\implies \boxed{14}.\] ~captainnobody

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