Difference between revisions of "2021 WSMO Accuracy Round Problems/Problem 5"
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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 | ||
Revision as of 09:51, 6 January 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
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
