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Difference between revisions of "2021 WSMO Accuracy Round Problems/Problem 5"

(Created page with "==Problem== Suppose regular octagon <math>ABCDEFGH</math> has side length <math>5.</math> If the distance from the center of the octagon to one of the sides can be expressed a...")
 
 
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==Problem==
 
==Problem==
 
Suppose regular octagon <math>ABCDEFGH</math> has side length <math>5.</math> If the distance from the center of the octagon to one of the sides can be expressed as <math>\frac{a+b\sqrt{c}}{d}</math> where <math>\gcd{(a,b,d)}=1</math> and <math>c</math> is not divisible by the square of any prime, find <math>a+b+c+d.</math>
 
Suppose regular octagon <math>ABCDEFGH</math> has side length <math>5.</math> If the distance from the center of the octagon to one of the sides can be expressed as <math>\frac{a+b\sqrt{c}}{d}</math> where <math>\gcd{(a,b,d)}=1</math> and <math>c</math> is not divisible by the square of any prime, find <math>a+b+c+d.</math>
 +
 
==Solution 1==
 
==Solution 1==
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<asy>
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size(150);
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draw(polygon(8)); for(int i=45; i<=360; i+=45){ dot(rotate(22.5)*dir(i)); }
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label("$D$",dir(337.5),SE);
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label("$C$",dir(22.5),NE);
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label("$B$",dir(67.5),N);
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label("$E$",dir(292.5),S);
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label("$F$",dir(247.5),S);
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label("$G$",dir(202.5),W);
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label("$H$",dir(157.5),NW);
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label("$A$",dir(112.5),N);
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label("$O$",(0,0),S);
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dot((0,0),red);
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draw(dir(112.5)--dir(67.5)--(0,0)--cycle,red);
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dot(dir(112.5),red);
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dot(dir(67.5),red);
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label("$5$",(0,1.05));
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draw(anglemark(dir(67.5),(0,0),dir(112.5)),red);
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label("$45^{\circ}$",(0,0.25),N);
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</asy>
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Let the center of the octagon be <math>O.</math> We will focus on triangle <math>AOB.</math> Let <math>AO=OB=x.</math> From the Law of Cosines on triangle <math>AOB,</math> we find that <cmath>x^2+x^2-2x^2\cdot\cos{(45^{\circ})}=5^2=25\implies</cmath><cmath>(2-\sqrt{2})x^2=25\implies x^2=\frac{25}{2-\sqrt{2}}=\frac{25(2+\sqrt{2})}{2}.</cmath> Now, let the distance from the center of the octagon to one of its sides be <math>h.</math> This means that <cmath>[AOB]=\frac{5h}{2}.</cmath> In addition, from the sine area formula, <cmath>[AOB]=\frac{1}{2}\sin{\angle{AOB}}\cdot AO\cdot BO=\frac{x^2\sqrt{2}}{4}=\frac{25(\sqrt{2}+1)}{4}.</cmath> Therefore, we have <cmath>\frac{5h}{2}=\frac{25(\sqrt{2}+1)}{4}\implies h=\frac{5+5\sqrt{2}}{2}\Longrightarrow5+5+2+2=\boxed{14}.</cmath>
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~pinkpig
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==Solution 2==
 
Note that the area of a polygon with <math>n</math> sides, <math>s</math> side length, and <math>l</math> apothem (distance from the center to one of the sides) can be expressed as <math>(nsl)/2.</math> Applying this formula, we get <cmath>(8\cdot 5\cdot l)/2=40l/2=20l.</cmath> Now, we need something to equate to this. Remember that the area of a regular octagon with side length <math>s</math> is <math>2s^2(1+\sqrt{2}).</math> This means that the area of octagon <math>ABCDEFGH</math> is <math>50+50\sqrt{2}.</math> Therefore, the answer is <cmath>l=\frac{50+50\sqrt{2}}{20}=\frac{5+5\sqrt{2}}{2}\implies \boxed{14}.</cmath>
 
Note that the area of a polygon with <math>n</math> sides, <math>s</math> side length, and <math>l</math> apothem (distance from the center to one of the sides) can be expressed as <math>(nsl)/2.</math> Applying this formula, we get <cmath>(8\cdot 5\cdot l)/2=40l/2=20l.</cmath> Now, we need something to equate to this. Remember that the area of a regular octagon with side length <math>s</math> is <math>2s^2(1+\sqrt{2}).</math> This means that the area of octagon <math>ABCDEFGH</math> is <math>50+50\sqrt{2}.</math> Therefore, the answer is <cmath>l=\frac{50+50\sqrt{2}}{20}=\frac{5+5\sqrt{2}}{2}\implies \boxed{14}.</cmath>
 +
~captainnobody

Latest revision as of 09:55, 11 July 2022

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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