MHB Solving Problems Without Guess & Check

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The discussion revolves around solving a problem related to the perimeter of a sector without relying on guess and check methods. The perimeter formula is established as P = 2(n + π), leading to the inequality 20 < P < 30. This simplifies to the range for n, specifically 6.86 < n < 11.86, indicating that n can take integer values of 7, 8, 9, 10, or 11. The calculations confirm that these values satisfy the original perimeter constraints. The focus remains on deriving solutions through algebraic manipulation rather than trial and error.
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Can we do this without guess and check? View attachment 6248
 

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I guess that depends, since we aren't given the radius...

-Dan

Edit: I guess the radius was mentioned. (Ahem!) (Angry)
 
The perimeter $P$ of the given sector is:

$$P=2n+n\frac{360}{n}\cdot\frac{\pi}{180}=2(n+\pi)$$

So, we want:

$$20<P<30$$

$$20<2(n+\pi)<30$$

$$10<n+\pi<15$$

Can you continue
 
N can equal 7,8,9,10, or 11?
 
Ilikebugs said:
N can equal 7,8,9,10, or 11?

Well, let's see:

$$10<n+\pi<15$$

$$10-\pi<n<15-\pi$$

Let's use $\pi\approx3.14$:

$$6.86<n<11.86$$

Hence:

$$n\in\{7,8,9,10,11\}\quad\checkmark$$
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...