An $n$ -gon is a polygon with $n$ sides.

Consider a 1000-gon that is convex and has all its 1000 vertices on a circle. Let the internal angles of this polygon (in degrees) be $a_1, a_2, \ldots, a_{999}, a_{1000}$ in that order. Then what is the sum of all the odd-numbered values: $a_1+a_3+a_5+\cdots+a_{997}+a_{999}?$

If you believe that the sum varies with the shape of the polygon, enter 0. If you believe it is constant, just enter the sum.

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Hint:
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Here is a diagram for
$n=6.$
Now, do it for
$n=1000.$

The answer is 89820.

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Solution 1Divide the polygon into cyclic quadrilaterals.

There will be two on the ends, similar to $ABCD$ and $AFGH$

then a total of $\frac12(1000-6)=497$ quadrilaterals of the type $ADEF$ , each with one vertex at $A$ .

Within each quadrilateral, the sum of the angles at $A$ and at the point opposite point $A$ will be $180^\circ$ .

Together all the angles at $A$ will add up to the angle $BAH$ ,

the angles opposite will add up to the rest of “every other” angle along the perimeter.

Sum in degrees $=(2+497)\times 180^\circ=\boxed{89820}$

...

Solution 2Treat all angles which are to be added equally from the start.

Let $a$ be one of the angles to be added.

Corresponding central angle is $b=2a$

$c=360^\circ-b=360^\circ-2a$

$a=\frac12(360^\circ-c)=180-\frac c2$

Sum of all $a$ ’s, and there are $\frac{1000}2=500$ of them, is $500\times180^\circ$ minus $\frac12$ sum of $c$ ’s.

The sum of all angles $c=360^\circ$

Sum of all $a$ ’s $=500\times180^\circ-\frac12 \times360^\circ=\boxed{89820}$