In a regular pentagram (5-pointed star), the angle in each point is 36 degrees, so the angles in all five points sum to 180 degrees:

regular star

What about an irregular pentagram, such as the following?

irregular star

Now the angles might be all different from each other; the situation is much more complicated. Can you prove that the angles in all five points still sum to 180 degrees?

Restrictions (to make clear that this is neither a maths problem [as opposed to a maths puzzle] nor an exercise in computation or advanced Euclidean geometry):

  • no arithmetic operations allowed (addition, multiplication, ...)
  • you may draw one new line-segment on the star, but no more than that
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    $\begingroup$ Sorry rand, but I think this is just another math problem...("prove", "angles", "sum", "180 degrees") $\endgroup$ – Mark N Jul 9 '15 at 18:35
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    $\begingroup$ @MarkN According to the canonical meta post on this, the sign of a maths puzzle as opposed to problem is to have a clever or elegant solution, often an "aha" moment, an unexpected problem statement, or an unexpected or counterintuitive result. The solution I have in mind definitely has the first of these features, and IMO the last two too. $\endgroup$ – Rand al'Thor Jul 9 '15 at 18:41
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    $\begingroup$ This isn't a math puzzle - it's a logic puzzle. You just usually learn this logic from someone who also teaches math. $\endgroup$ – corsiKa Jul 10 '15 at 14:59

10 Answers 10


$\hskip 1.5in$Isn't it dreamy?

This is an image of an arrow sweeping each of the successive angles in the star. Notice that, after it traces all $5$ angles, its orientation is reversed - meaning it has rotated $180^{\circ}$ and that this must be the sum of the angles. We can do the same thing to the star in your figure, ergo, its angles too sum to $180^{\circ}$.

A Generalization:

We can do the same thing to a figure like this, whose angles sum to $180^{\circ}$: $\hskip 1.5in$enter image description here
We can also do this to a triangle. The important property is this:

There must be no vertices of the star interior to the cone swept out by a ray traversing a given angle.

Satisfying this condition - which basically says that we never have to "ignore" vertices, but instead just rotate the arrow and see what it hits - we find that we can order the vertices in a "clockwise" manner, so that, at each angle, either the head or the tail of the arrow steps to the next vertex in the order (and they alternate which). Obviously, both head and tail will make a full revolution when twice as many angles as vertices have been traced, yielding the desired result.

(One might also express my condition as "assigning the vertices the numbers $1$ through $2n+1$ in clockwise order as seen from a central point, it must be that $1$ connects to $n$ and $n+1$, and all other points are connected analogously")

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    $\begingroup$ (Also, for what it's worth, I really liked this puzzle, even if my answer is not the intended one - I had a good, "Well that's obvious" moment, followed by a few hours of intense head-scratching, trying to figure why it was obvious, followed by "Aha! It was obvious!") $\endgroup$ – Milo Brandt Jul 10 '15 at 2:50
  • $\begingroup$ I take it your comment is a reference to this joke? =) $\endgroup$ – Tyler Seacrest Jul 10 '15 at 3:06
  • $\begingroup$ Accepted because it's even nicer than the answer I was also looking for, and also covers a generalisation. $\endgroup$ – Rand al'Thor Jul 12 '15 at 22:22

Place your pencil on line 1.

enter image description here

Rotate your pencil so it lines up with line 2. You just rotated it counterclockwise by the angle at the top of the pentagram.

Now rotate it counterclockwise again onto line 3. Then again onto line 4, then 5, and finally back to 1. You have just rotated your pencil through all five angles of the pentagram in sequence.

And what happened? The pencil now lies on the same line it started at, pointing in the opposite direction. If you track which direction the pencil points at each step, you can see that in total, you rotated it counterclockwise by one half turn. Whence, $180^\circ$.

  • $\begingroup$ This will be a beautiful proof if you tweak it to exclude the possibility that you have rotated the pencil through some other odd multiple of $180^\circ$. With this heptagram the pencil also ends up pointing the opposite way but has rotated through $540^\circ$ $\endgroup$ – h34 Jul 10 '15 at 1:00
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    $\begingroup$ There is a continuous deformation from the reference pentagram to any deformed pentagram. Thus, the rotation cannot jump from one multiple of 180∘ to another. $\endgroup$ – Christian Mann Jul 10 '15 at 2:50
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    $\begingroup$ Basically, any $\{m:n\}$-gram where $n < \frac m 2$ rotates $360 \times (\frac m 2 - n)$ degrees. $\endgroup$ – Joe Z. Jul 10 '15 at 2:55
  • $\begingroup$ Nice explanation Lopsy ... simple, clean :) I was going to say, take 4 angles and visually start reducing them to 0 .. think about what the star looks like as this happens ... the 5th angle keeps growing to accomodate ... until 4 angles are 0, and the 5th is 180 (ie a straight line) .. :) But I like Lopsy's explanation better .. ;) $\endgroup$ – Ditto Jul 10 '15 at 14:31
  • $\begingroup$ The beauty of this answer is that it doesn't read like a mathematical proof. Anyone can understand it. $\endgroup$ – dennisdeems Jul 11 '15 at 16:36

Here's another proof.

Label the points as shown, and draw the line-segment CD. Use A, B, etc. to denote the angles we are asked to find the sum of.

enter image description here


$\angle ADC + \angle DCA + A = 180^\circ$ (angles in a triangle)

So it is sufficient to prove that

$\angle ADC + \angle DCA = B + C + D + E$


$\angle ADC = D + \angle BDC$ and $\angle DCA = C + \angle ECD$

So it is sufficient to prove that

$\angle BDC + \angle ECD = B + E$

which is obviously true because

the LHS is the complement of $\angle DFC$ and the RHS is the complement of $\angle EFB$, where $\angle DFC$ and $\angle EFB$ are equal because vertically opposite.

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    $\begingroup$ This is the answer I was looking for. $\endgroup$ – Rand al'Thor Jul 9 '15 at 18:57
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    $\begingroup$ So, pretty much, you can distill this solution to 2 rules: angles in triangles = 180 and opposite angles of 2 intersecting lines are equal. $\endgroup$ – JonTheMon Jul 9 '15 at 19:12
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    $\begingroup$ @randal'thor This solution also involves addition, so would not comply with your restrictions, or you should alter your restrictions. $\endgroup$ – fibonatic Jul 10 '15 at 5:39
  • $\begingroup$ Yeah i was going to say this is like not -the- but one of the most math-ish answers on here. The absence of arithmetic does not mean it isn't math... $\endgroup$ – Spacemonkey Jul 10 '15 at 21:13

The sum of the interior angles of a pentagon is always 540°.

The angle of each exterior point is always the sum of the two adjacent interior angles - 180°. We can say this since, given internal angles A and B, the angles of the triangle are 180 - A, 180 - B, X. By definition of angles of a triangle, X is equal to $180 - (180 - A) - (180 - B) = A + B - 180$.

Each interior angle of the pentagon is used twice, and there are 5 points, so $(2 \times 540) - (5 \times 180) = 180°$

  • $\begingroup$ I believe this is grade 9 geometry peaking its head out... $\endgroup$ – Mark N Jul 9 '15 at 18:34
  • $\begingroup$ This is more complicated than the proof I was thinking of. I may edit the question to restrict the possible proofs a bit more, but I'll still give you a +1. Could you justify your second sentence? Also I don't understand what the third sentence is saying. $\endgroup$ – Rand al'Thor Jul 9 '15 at 18:36
  • $\begingroup$ If we let A and B be two adjacent interior angles of the pentagon, then the angle of the point in the triangle is 180 - (180-A + 180-B) = A + B - 180 $\endgroup$ – dzastergamer Jul 9 '15 at 18:37
  • $\begingroup$ +1 Nice proof, but it would be cool if you could use a pic or 2, or even a gif! $\endgroup$ – JLee Jul 9 '15 at 18:46
  • $\begingroup$ I think it's possible to generalise this proof to show that the angles at the points of any n-gram sum to $180^\circ$ provided the shape connects each point to two adjacent points on the n-gon. (Note that the unicursal hexagram doesn't meet the connection criterion; nor does the hexagram formed from two triangles; and only one of the two unicursal heptagrams does.) $\endgroup$ – h34 Jul 9 '15 at 19:29

Here's another neat proof, this time by induction. We can make the pentagram by starting with the regular one and successively moving four of the points. So it is sufficient to prove that

moving a point in a pentagram doesn't change the sum of the angles at the points


move point A to A' and call both the angle at A and the angle at A' the top angle

We get this:

Diagram 1

It is sufficient to prove that

the change in the top angle and the changes in the angles at C and D sum to zero.

On this new diagram

Diagram 2

we show

the change in the top angle as $x-w$ and the changes at D and C as $-y$ and $z$,

and we need to prove that

$x-w-y+z=0$, or in other words, that $x+z=w+y$,

which is obvious, as before, because

LHS and RHS are the complements of vertically opposite angles at G.


Another approach:

Starting with the regular star, We know that $A + B + C + D + E = 180^{\circ}$. Now let's draw a line segment as shown in the diagram.


Note that $B, D, E$ remain unchanged! From our observations, we see that $Y = C - X$ and $Z = A + X$.

Thus the sum of the points of our new star $ZBYDE = Z + B + Y + D + E = (A+ X) + B + (C-X) + D + E = 180^{\circ}$.

So we can continue to draw segments and create new stars (and rescale them) until we reach the shape desired.

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    $\begingroup$ Nice, but can you maybe add something to make it more intuitive that you can make a general irregular pentagram by a sequence of moves of one point along one of the lines through that point and rescalings. $\endgroup$ – h34 Jul 9 '15 at 20:37
  • $\begingroup$ I could try if only geometry didn't hurt my brain so much D: $\endgroup$ – dzastergamer Jul 10 '15 at 1:42

It's inevitable that some arithmetic must be done--the intended conclusion is a quantitative one, after all--so the challenge should not be to hide the arithmetic, nor to call it by some other name, but to make it obvious and dead simple. The following argument reduces the arithmetic to observing that five is one more than four (and that a whole is twice a half, a fact used in passing).

The star winds twice around its center, and therefore anyone traversing it will have to turn two full circles (four half circles). All turning occurs only at the vertices, where the maximum amount is a complete about-face of one-half of a circle. For five vertices that would be five half-circles, or one more half-circle than is actually turned: 180 degrees. The deficiency between this maximum and the amount of turning that actually is accomplished is precisely the sum of the interior angles, QED.

This approach is the one taken in modern (i.e., post-18th century) mathematics. It generalizes to arbitrary figures of arbitrary dimensions drawn within other figures that themselves can be curved. It is known as the Gauss-Bonnet Theorem.


There is a circle based theorem that states "The measure of an inscribed angle is half the measure of the arc it intercepts." This means that for angle x , the arc it intercepts will be 2x. Inscribed angles

Now, if you inscribe the star into a circle, you get this:

enter image description here

Labeling the previous drawing, you get this;

enter image description here

With this theorem, we know that angle ∠1=c/2, ∠2=d/2, ∠3=e/2, ∠4=a/2, and ∠5=b/2. If we distribute that, we get ∠1+∠2+∠3+∠4+∠5=(a+b+c+d+e)/2 . Furthermore, Because the measures of all arcs in a circle add up to 360, we know that a+b+c+d+e=360 . Finally, using the substitution property, we get ∠1+∠2+∠3+∠4+∠5=360/2 , or ∠1+∠2+∠3+∠4+∠5 = 180 . Thus the sum of all angles is 180.

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    $\begingroup$ There's one flaw in your argument: not every pentagram can be inscribed in a circle. $\endgroup$ – lirtosiast Jul 12 '15 at 15:11
  • $\begingroup$ @ThomasKwa Can you give me an example? $\endgroup$ – user1812 Jul 12 '15 at 15:14
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    $\begingroup$ @user1812 just move any one point on your example into or out of the circle. It only takes three points to define a circle, and a pentagram has five. $\endgroup$ – Kevin Jul 12 '15 at 16:26

This proof in a sense is nothing but counting the degree of angles.

Recall that a pentagon, whether regular or irregular, has its internal angles sum to 540. Also, the angles of an intersection of 2 straight lines sum to 360, where also the opposite angles are congruent.

Consider the 5 points of the central pentagon, the points where the intersection of 2 lines occurs. Around these 5 points are 360 x 5 = 1800 degrees total, and 5 x 4 = 20 angles to count.

Of the 20 angles, 5 are from the pentagon, 5 more are congruent to those. So this accounts for 540 + 540 = 1080 degrees. The remains of the 1800 - 1080 = 720 degrees come from inside the 5 triangles.

5 triangles contain 5 x 180 = 900 degrees worth of interior angles. 720 of those degrees are at the corners of the pentagon/triangle/intersection.

This leaves at the tips of the star 900 - 720 = 180 degrees.

Edit: The arithmetic here is simply shorthand for angle addition and subtraction, same as is done in other answers.


The central Pentagon as A,B,C,D,E contains 540 DEGREES

Sum the 5 PAIRS supplementary angles ie. 2(180-A)+2(180-B)+2(180-C)+2(180-D)+2(180-E)= 1800 2(540) = 720 This 720 degrees represents the 'base' angles of the 5 triangles Which totals 5*180=900 900-720 = 180 (the 5 angles being sought.

The five Triangles at the points sum to 5*180 = 900

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    $\begingroup$ The question specifically asks to prove it without using arithmetic operations. $\endgroup$ – f'' Feb 18 '16 at 1:11

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