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 10:51, 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
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
~captainnobody