It All Came From Pencils 3

Suppose the angle of the pencil tips are $x$ , it's not hard to show that the interior angles of the polygon ( pencilogon ) are $180^\circ-\frac{x}{2}$ . The exterior angles of the polygon are $180^\circ-(180^\circ-\frac{x}{2})=\frac{x}{2}$ .

Let's consider the situation when the pencils are not yet sharpened.

For all polygons, the sum of all exterior angles is always $360^\circ$ , but in this case, since there are gaps, the sum of the exterior angles must be smaller than $360^\circ$ , so $\frac{x}{2}(n-18)<360^\circ\implies(n-18)<\frac{720^\circ}{x}$ . Here, $x=7^\circ$ , so, $(n-18)<\frac{720}{7}$ $\therefore n<120\frac{6}{7}$ Thus, we know that Hazri originally wants to make a 120- pencilogon , $n=120$ (If $n<120$ , then the gap would be large enough to fit at least 1 pencil, which is not what we wanted).

A 120- pencilogon has exterior angles of $180^\circ-\frac{118\times180^\circ}{120}=3^\circ$ , after he had sharpened all the pencils, the angle of the pencil tips become $(7-m)^\circ$ , $\frac{(7-m)^\circ}{2}=3^\circ$ $\therefore m=1$ Hence, $m+n=121$ .

I think it's safe to assume that given how a pencil sharpener works, each pencil has an isosceles triangle at the end, and then a rectangle for the rest of it. So the interior angle of the attempted pencilogon $=360-7-\frac{180-7}{2}-90=176.5$ . Were this x-pencilogon to be completed, the sum of its interior angles would be: $176.5x=(x-2)180 \color{#D61F06}{\Rightarrow}\color{#333333} x=102.857$ To translate for the word problem, Hazri put down 102 pencils and found that there was not enough room to put down pencil 103. $n-18=102 \color{#D61F06}{\Rightarrow} \color{#3D99F6}{n=120}$ Now one simply works backwards to find what the angle of the pencil should be to make this 120-pencilogon: $\text{Interior angle of the 120-pencilogon} = \frac{(120-2)180}{120}=177$ Let $y$ be the interior angle of the sharpened pencil. $\text{Interior angle of the 120-pencilgon} = 177 = 360-y-\frac{180-y}{2}-90 \color{#D61F06}{\Rightarrow}\color{#333333} y=6=7-m \color{#D61F06}{\Rightarrow} \color{#3D99F6}{m=1}$ $m+n=\color{#3D99F6}{121}\color{#333333}\space\space\space\Box$

The adjoining line of an n-gon takes a 360/n degrees turn. That is the turn centerlines of the pencils take.
Assuming pencil tip as isosceles triangle, its tip will be 2 * 360/n degrees.
So with tip of 7, the pencilogon will need $\dfrac{360}{\frac {7}{ 2}}=102.857$ pencil. Therefore the pencils he has are n=102+18=120. Thus the pencil angle must be 2*{360/120}=6. Implies m=7-6=1. Therefore m+n=121.

In the triangle formed at the apex / tip, the two base angles sum are equal and sum upto 180-7 = 173. Thus each angle is equal to 173/2 = 86.5 . We need to compute each interior angle of the polygon. Int angle = 360 - (90 (from the cross-section of pencil) + 86.5 + 7) = 360 - 183.5 = 176.5.

Now we can estimate number of sides from the formula : Int angle of n-sided polygon = 180 - (360/n1) Therefore, 360 / n1 = 180 - 176.5 = 3.5 n1 = 360/3.5 = 102.86 which is not obviously not possible which indicates that the process of forming the polygon would have stopped here. Thus the original n = 102 (the floor of the above function) + 18 = 120 (which also seems to be the only feasible candidate .near 102 for getting an integral number of sides)

Now , the new interior angle = 180 - 360/120 = 180 - 3 = 177.

If the apex angle is x,

(180-x)/2 + 90 + x = 360-177 = 183

90 - x/2 + 90 + x = 183

180 + x/2 = 183

x/2 = 3

x = 6

7-m = 6 m = 1 n = 120

Therefore, m + n = 121