Difference between revisions of "2021 WSMO Accuracy Round Problems/Problem 5"
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==Solution 1== | ==Solution 1== | ||
| − | + | <asy> | |
size(150); | size(150); | ||
draw(polygon(8)); for(int i=45; i<=360; i+=45){ dot(rotate(22.5)*dir(i)); } | draw(polygon(8)); for(int i=45; i<=360; i+=45){ dot(rotate(22.5)*dir(i)); } | ||
| − | label(" | + | label("$D$",dir(337.5),SE); |
| − | label(" | + | label("$C$",dir(22.5),NE); |
| − | label(" | + | label("$B$",dir(67.5),N); |
| − | label(" | + | label("$E$",dir(292.5),S); |
| − | label(" | + | label("$F$",dir(247.5),S); |
| − | label(" | + | label("$G$",dir(202.5),W); |
| − | label(" | + | label("$H$",dir(157.5),NW); |
| − | label(" | + | label("$A$",dir(112.5),N); |
| − | label(" | + | label("$O$",(0,0),S); |
dot((0,0),red); | dot((0,0),red); | ||
draw(dir(112.5)--dir(67.5)--(0,0)--cycle,red); | draw(dir(112.5)--dir(67.5)--(0,0)--cycle,red); | ||
dot(dir(112.5),red); | dot(dir(112.5),red); | ||
dot(dir(67.5),red); | dot(dir(67.5),red); | ||
| − | label(" | + | label("$5$",(0,1.05)); |
draw(anglemark(dir(67.5),(0,0),dir(112.5)),red); | draw(anglemark(dir(67.5),(0,0),dir(112.5)),red); | ||
| − | label(" | + | label("$45^{\circ}$",(0,0.25),N); |
| − | + | </asy> | |
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> | 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> | ||
~pinkpig | ~pinkpig | ||
Latest revision as of 10:55, 11 July 2022
Problem
Suppose regular octagon 
has side length 
If the distance from the center of the octagon to one of the sides can be expressed as 
where 
and 
is not divisible by the square of any prime, find 
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]](http://latex.artofproblemsolving.com/6/0/f/60fcfeefd166a492098ce2e8f61d6ef068b0a2ae.png)
Let the center of the octagon be 
We will focus on triangle 
Let 
From the Law of Cosines on triangle 
we find that ![\[x^2+x^2-2x^2\cdot\cos{(45^{\circ})}=5^2=25\implies\]](http://latex.artofproblemsolving.com/3/6/5/3652539210eabaa5d65ad44a6af1b0dbc3caad5a.png)
![\[(2-\sqrt{2})x^2=25\implies x^2=\frac{25}{2-\sqrt{2}}=\frac{25(2+\sqrt{2})}{2}.\]](http://latex.artofproblemsolving.com/b/c/8/bc85dc3ce9f7a0d65c23972883f4f9eb504533c0.png)
Now, let the distance from the center of the octagon to one of its sides be 
This means that ![\[[AOB]=\frac{5h}{2}.\]](http://latex.artofproblemsolving.com/f/e/d/fedb7e4a3eb88633fa8bf02723258e76fea7583f.png)
In addition, from the sine area formula, ![\[[AOB]=\frac{1}{2}\sin{\angle{AOB}}\cdot AO\cdot BO=\frac{x^2\sqrt{2}}{4}=\frac{25(\sqrt{2}+1)}{4}.\]](http://latex.artofproblemsolving.com/9/4/f/94fcd874995d2f4fcb82df28ffaee91bc02ef29f.png)
Therefore, we have ![\[\frac{5h}{2}=\frac{25(\sqrt{2}+1)}{4}\implies h=\frac{5+5\sqrt{2}}{2}\Longrightarrow5+5+2+2=\boxed{14}.\]](http://latex.artofproblemsolving.com/b/b/0/bb06f735e22b1bb4bbf0b8a201b1aad78b1e4a5a.png)
~pinkpig
Solution 2
Note that the area of a polygon with 
sides, 
side length, and 
apothem (distance from the center to one of the sides) can be expressed as 
Applying this formula, we get ![\[(8\cdot 5\cdot l)/2=40l/2=20l.\]](http://latex.artofproblemsolving.com/1/6/d/16dd48f55dda99efdfdc67ec0dd05e6adbff9b9a.png)
Now, we need something to equate to this. Remember that the area of a regular octagon with side length is 
This means that the area of octagon is 
Therefore, the answer is ![\[l=\frac{50+50\sqrt{2}}{20}=\frac{5+5\sqrt{2}}{2}\implies \boxed{14}.\]](http://latex.artofproblemsolving.com/a/d/9/ad91f9bbbbe17b7b9b14f2ca78c7613d26059cd2.png)
~captainnobody
