The Sum of Angles

Geometry Level 3

Given a 100-side polygon $A_{1}A_{2}...A_{100}$ which is inscribed in a circle. Besides that, $A_{100}A_{99}\geq A_{98}A_{97} \geq A_{96}A_{95} \geq ... \geq A_{4}A_{3} \geq A_{2}A_{1}$ and $A_{99}A_{98}\leq A_{97}A_{96} \leq A_{95}A_{94} \leq A_{3}A_{2} \leq A_{1}A_{100}$ . Find (in degree) $\large \angle A_{2}+\angle A_{4} +\angle A_{6}+\space...\space+\angle A_{98}+ \angle A_{100}$

The answer is 8820.

1 solution

Chan Tin Ping
Nov 13, 2017

Statement: For any $2n$ side polygon ( let it as $A_{1}A_{2}…A_{2n}$ ) which inscribed in a circle, $\sum_{t=1}^{n} \angle A_{2t}= \frac{1}{2}(2n-2)(180°)=180°(n-1)$ which is half of the sum of all interior angle.

Proof (use induction) : For n = 2, which is a inscribed quadrilateral, it's obvious that the statement is true. Let n=k is true for the statement above. Consider the case n=k+1, which is a inscribed 2(k+1)-side polygon. Let it as $A_{1}A_{2}…A_{2k}A_{2k+1}A_{2k+2}$ . Draw the line $A_{2k}A_{1}$ , we got a inscribed 2k-side polygon $A_{1}…A_{2k}$ and a inscribed quadrilateral $A_{2k}A_{2k+1}A_{2k+2}A_{1}$ .

For the 2k-side polygon, we can get $\sum_{t=1}^{k-1}\angle A_{2t} + \angle A_{2k-1}A_{2k} A_{1} = 180°(k-1).$ For the inscribed quadrilateral, $\angle A_{1}A_{2k}A_{2k+1} + \angle A_{2k+1}A_{2k+2}A_{1}=180°$ . Plus these two summation of angles together, we can conclude that the above statement is correct. (QED)

Hence, the answer is $\frac{1}{2}(100-2)(180°)=8820°$ . The weird inequality is just misleading others~~.

The Sum of Angles

The answer is 8820.

1 solution

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