A geometry problem by SKYE RZYM

If $O$ is the centre of the circle: We can apply this to the entire circle: Then clearly $2\angle A + 2\angle B + 2\angle C + 2\angle D + 2\angle E + 2\angle F + 2\angle G + 2\angle H + 2\angle I = 360^\circ$ .

So $\angle A + \angle B + \angle C + \angle D + \angle E + \angle F + \angle G + \angle H + \angle I = 180^\circ$ .

An inscribed angle has a measure = 1/2 of its arc. Since all of the angles have a total arc of 360 then the sum of the angles is 180.

Ordinary 9-sided polygon would have sum of internal angles $(n-2)\times180^\circ$ . Self-intersecting polygon has that minus the number of times the line goes around the center times $360^\circ$ . In this case

$(9-2)\times180^\circ-3\times360^\circ=7\times180^\circ-6\times180^\circ=180^\circ$

A classic solution worth mentioning: Take a pencil and place it on line HD. Rotate about H to HC, rotate about C to CG, and so on. The pencil traverses the star, and upon returning to HD, it has rotated 180 degrees. Since every time it rotated, it added the angle of rotation to the net rotation, the sum of the angles must be 180 degrees.

As you trace out the pattern without lifting your pencil, at every point the slope of your line increases by the included angle of the star points. Thus the slope of the penultimate line relative to the first line drawn is just the sum of all of the preceding included angles. When you arrive at the penultimate point, join it to the start position thus leaving two angles undefined, one at the start point and one at the end point. The sum of the included angles of a triangle including these two undefined angles and the last sloping line must equal 180 degrees. But this is just the sum of all of the included angles of the points. This proof does not depend on the number of points nor does it require the points to lie on a circle. QED.

At point A (for instance), we turn over an angle of $180^\circ - \angle A$ . The total angle over which we turn when tracing the path is $180\ n - \sum \angle_i,$ where $n = 9$ is the number of angles and $\angle_i$ runs through each of them.

Since we end up where we started, this total angle must be a multiple of $360^\circ$ . If $m$ is the number of full revolutions, then $360\ m = 180\ n - \sum \angle_i$ or $\sum \angle_i = 180\ n - 360\ m.$

Tracing the path carefully shows that $m = 4$ . But for this multiple choice question we even don't need this. After all, $180\ n - 360\ m = 180(n - 2m)$ is a multiple of 180 degrees, and there is only one option given for which this is true.

The answer is $180(9 - 2\cdot 4) = 180\cdot 1 = 180^\circ.$

There are 9 points. Only 2 of the solutions were divisible by 9 with a whole number, 90 and 180. Eyeballed it and saw the surface area of the shaded part was more than a quarter of the circle which leaves 180

It's a 9 sided polygon that crosses itself 4 times. So $(9-2\cdot 4)(180)=(9-8)\cdot 180=180$

This is a modified formula of the sum of the interior angles of and exterior polygon being $(n-2)\cdot 180 \iff 180n-360$

Since it crosses itself 3 times, and the fourth closes itself, then we multiply 360 with 4. Generally $180n-360c \iff 180(n-2c)$ with n sides and c crosses/closures.

$\overline{BF}$ can be seen as a rotation of $\overline{AF}$ $F^\circ CW$ . Similarly, $\overline{BG}$ can be seen as a rotation of $\overline{BF}$ $B^\circ CW$ . Continuing this pattern eventually gets you back to $\overline{AF}$ after 9 rotations. Therefore since $\overline{AF}$ is a rotation of itself at $180^\circ CW$ , and this rotation was the sum of the rotations of $F^\circ CW$ , $B^\circ CW$ , ... $E^\circ CW$ , and $A^\circ CW$ .

$\therefore \angle A + \angle B + \ldots + \angle H + \angle I = \boxed{180^\circ}$

Total circle is 2×pi. Therefore total angles makes half of it; pi.

A method from my school days for seeing that the angles in a triangle add up to $180^\circ$ is to lay a pencil along one of the sides and rotate it about each of the vertices in turn. In the course of the operation, the pencil rotates through $180^\circ$ in total. I essentially applied the same process to this star.

It is tempting to call the solution 180° but it isn't. I went to the trouble to actually draw it up a number of times and came sometimes close to 180° but with large deviations. I suspect there is a condition missing to make it 180°