IMO Shortlist 2004 problem G5
Dodao/la:
arhiva2. travnja 2012. Let
![A_1A_2A_3...A_n](/media/m/7/8/4/784a2de79932508630c92be3ea666e59.png)
be a regular n-gon. Let
![B_1](/media/m/5/d/9/5d9518a7c0ead344571aac61b51bb25c.png)
and
![B_n](/media/m/9/4/9/949aba6eae76461763dec24a3a6fa974.png)
be the midpoints of its sides
![A_1A_2](/media/m/2/1/9/219e9d359659bac5a9acdb7632b15563.png)
and
![A_{n-1}A_n](/media/m/1/3/a/13a1da720f1cf2e5d6aa8251887ad8e8.png)
. Also, for every
![i\in\left\{2;\;3;\;4;\;...;\;n-1\right\}](/media/m/b/1/c/b1c82c713220f895d11faaba408f6319.png)
, let
![S](/media/m/c/6/3/c63593c3ec0773fa38c2659e08119a75.png)
be the point of intersection of the lines
![A_1A_{i+1}](/media/m/3/9/1/3915b469e03689fcd8a897aa4abf5370.png)
and
![A_nA_i](/media/m/e/4/5/e45227ee45fa4b4f32b952cdfb81a737.png)
, and let
![B_i](/media/m/1/4/5/14587d3e0ae49b15b1042914a7f802f4.png)
be the point of intersection of the angle bisector bisector of the angle
![\measuredangle A_iSA_{i+1}](/media/m/9/3/7/93764b63192d11c76082a43b988c3cb4.png)
with the segment
![A_iA_{i+1}](/media/m/8/1/7/8177b14dd5ed7502c7574dad25dee836.png)
.
Prove that:
![\sum_{i=1}^{n-1} \measuredangle A_1B_iA_n=180^{\circ}](/media/m/a/5/5/a55bd036234d8324c060809c3988e979.png)
.
%V0
Let $A_1A_2A_3...A_n$ be a regular n-gon. Let $B_1$ and $B_n$ be the midpoints of its sides $A_1A_2$ and $A_{n-1}A_n$. Also, for every $i\in\left\{2;\;3;\;4;\;...;\;n-1\right\}$, let $S$ be the point of intersection of the lines $A_1A_{i+1}$ and $A_nA_i$, and let $B_i$ be the point of intersection of the angle bisector bisector of the angle $\measuredangle A_iSA_{i+1}$ with the segment $A_iA_{i+1}$.
Prove that: $\sum_{i=1}^{n-1} \measuredangle A_1B_iA_n=180^{\circ}$.
Izvor: Međunarodna matematička olimpijada, shortlist 2004