Sum of Angles

Let $1,2,3,4$ and $5$ be the angles of the irregular pentagram .

By the Exterior Angle Theorem , $\angle FGD=1+3$ because $\angle 1$ and $\angle 3$ are remote angles. It follows that $\angle GFD=2+4$ since it is an exterior angle with remote angles $2$ and $4$ . The sum of the interior angles of $\triangle FGD=1+3+2+4+5=180^\circ$ . But these are the angles of the pentagram. Hence, the sum of the angles of the pentagram is $\boxed{180^\circ}$ .

The figure can be divided into one pentagon and five lateral triangles. The five triangles include the exterior angles of the pentagon twice. The sum of the angles of the five triangles $180°\times5 = 900°$ . The exterior angles of the pentagon $= 360°$ . So, the required angles $= 900° - 360°\times2 = 180°$ .

Solution 1:

We will use that in a triangle the sum of any two angle is equal to the third angle's exterior angle, and that in a triangle the sum of the angles equals to $180°$ .

Then $\beta+\epsilon=\angle BFA$ and $\gamma+\omega=\angle EGA$ .

So $\alpha+\beta+\gamma+\epsilon+\omega=\angle FGA+\angle FAG+\angle AFG=180°$

Solution 2:

We wil use that in a pentagon the sum of the angles is $(5-2)*180°=540°$ , (and that in a triangle the sum of the angles is equal to $180°$ ).

If

• $N°=\alpha+\beta+\gamma+\epsilon+\omega$

• $K°$ equals to the sum of the red angles

then $540°-K°=N°$ .

The red angles are in five triangles, so if the sum of the yellow, green, light blue, dark blue and pink angles is $M°$ , then $5*180°-M°=K°$ .

However we know that $M°$ is $540°$ , because there is a smaller pentagon inside with the yellow, green, light blue, dark blue and pink angles.

So $K°=900°-540°=360°$ , and

$\alpha+\beta+\gamma+\epsilon+\omega=540°-360°=180°$

$\large\dfrac{900 - 540}{2}=\boxed{180}$

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On the assumption that the question has a single solution independent of the regularity of the star - or lack thereof - we might as well just solve it for the easy case where the interior pentagon is regular. In that pentagon, angles are $\frac{540}{5} = 108^{\circ}$ . That would make the base angles of any of the outer isosceles triangles $72^{\circ}$ , so the outer angles would be $180-2\times 72 = 36^{\circ}$ . Five of those sum to $180^{\circ}$ .

Obviously, this doesn't suffice as a demonstration of the general result, but since this question was posed in a way that implies there's only one general solution, it gets the job done.

Trace the star. At each of $n = 5$ points with angle $\alpha_i$ we turn over an angle of $180^\circ - \alpha_i$ . It is also clear that we turn $r = 2$ full revolutions while doing this. Thus $\sum_{i = 1}^n (180^\circ - \alpha_i) = r\cdot 360^\circ,$ $\sum_{i = 1}^n \alpha_i = (n - 2r)\cdot 180^\circ ,$ $\sum_{i = 1}^5 \alpha_i = (5 - 4)\cdot 180^\circ = 180^\circ.$

First, draw a line segment $AC$ (the red one) Notice in triangle $ABC$ , we have $t + u + e + d + b = 180$ . Then notice triangles $AFC$ and $DFE$ . They share an angle $\angle AFC = \angle DFE$ . So the sum of the other two angles in each triangle must be equal. Which means that $t + u = a + c$ . And $a + b + c + d + e = 180$ .

If you took the two bottom points and the top point and stretched the star to be tall and thin, the angle of those three points would approach zero, while the angle of the left and right points would approach 90 degrees, making the answer 180 degrees.

The total of the five angles will remain the same if the figure is fully symmetrical, so we can simply realise that a pentagon's sum of inner angles adds up to 540 degrees (each corner after 3 adds 180 degrees to the total) and then get the angles next to them, which will make two 72 degree angles along with the tips, which means that for a symmetrical figure, all of the tips are (180-72 2) = 32 degrees, so 5 32 = 180 degrees.

Quick headcount solution: you take the triangles formed by two of the opposing marked angles and an angle in the inner pentagram. There are 5 of these, which adds up to 5 * 180. The pentagram has a total of 180 * (5 - 2) = 3 * 180. The marked angles were used two times, while the angles of the pentagram only once, so the sum of the marked angles is (5 * 180 - 3 * 180) / 2 = 2 * 180 / 2 = 180

My quick and lazy answer:

We have 5 of these triangles. The colored angle on each is $180^\circ - (the\ green\ angle + the\ red\ angle)$ .

The sum of all the green angles (going around the whole pentagon) is 360^\circ. Same for the red.

So we have $5\times 180^\circ -360^\circ -360^\circ$ . Or, as I like to call it because I hate arithmetic, $5 \times 180^\circ -4\times 180^\circ=180^\circ$

The internal angles of a regular pentagon are 108 Deg totalling 540 Deg. 540 - 360 = 180 Deg

We can set up following equations:

$m=180^{\circ}-a-b$

$n=180^{\circ}-b-c$

$p=180^{\circ}-c-d$

$q=180^{\circ}-d-e$

$r=180^{\circ}-e-a$

Summing:

$\sum =m+n+p+q+r=5\times 180^{\circ}-2(a+b+c+d+e)$

Notice that $a, b, c, d, e$ are all external angles of internal pentagon and their sum is $360^{\circ}$ (in fact, sum of external angles of any polygon is $360^{\circ}$ ).

Hence, $\sum =m+n+p+q+r=5\times 180^{\circ}-2\times 360^{\circ}=180^{\circ}$ .

Going around the pentagon clockwise - it has a set of external angles = 360. Go around the pentagon anticlockwise and there are another set of external angles = 360. These 2 groups of angles make 2 of the 3 angles in each of the five triangles (call these the "base" angles).

However 5 triangles have 5x180 degrees (or more usefully here, 2.5x360 degrees). So if the 2 "base" angles are taken away, it leaves the coloured angles adding up to 0.5x360 degrees --> 180 degrees.

I found a slick way to solve this really easily...

If we assume that the sum of the five angles is constant (which must be true for one of the given answers to be correct), then do the following:

If the sum of the five angles is constant, then we can manipulate the figure by moving the five marked points any way we want (while preserving the shape) and the sum of the angles should stay the same. Take the bottom two points (dk blue and purple) and move them together so they coincide. This makes the red angle zero, and what we have left is a triangle with a line in the middle (the dk blue angle and the purple angle combine to make a single angle and the red angle is gone). The sum of the angles in a triangle is 180 degrees, so that's the answer.

Notes:

1) If you argue that by making two points coincide, you've changed the type of figure you're dealing with, note that this is a well-posed problem and that you could make the two points arbitrarily close to one another - for simplicity of exposition, I ignored this detail.

2) I admit that this only works if you've been told (or assume) that the sum of the angles is constant. This argument does not prove that the sum is constant - for example, at first glance, it could be true that the sum just approaches 180 as any two points approach each other, but it could be different from 180 in general.