IMO Shortlist 1985 problem 21
Dodao/la:
arhiva2. travnja 2012. The tangents at
![B](/media/m/c/e/e/ceebc05be717fa6aab8e71b02fe3e4e3.png)
and
![C](/media/m/5/a/b/5ab88f3f735b691e133767fe7ea0483c.png)
to the circumcircle of the acute-angled triangle
![ABC](/media/m/a/c/7/ac75dca5ddb22ad70f492e2e0a153f95.png)
meet at
![X](/media/m/9/2/8/92802f174fc4967315c2d8002c426164.png)
. Let
![M](/media/m/f/7/f/f7f312cf6ba459a332de8db3b8f906c4.png)
be the midpoint of
![BC](/media/m/5/0/0/5005d4d5eac1b420fbabb76c83fc63ad.png)
. Prove that
(a)
![\angle BAM = \angle CAX](/media/m/4/d/c/4dc0b78e664d2810b4d2896fb97ce8d9.png)
, and
(b)
%V0
The tangents at $B$ and $C$ to the circumcircle of the acute-angled triangle $ABC$ meet at $X$. Let $M$ be the midpoint of $BC$. Prove that
(a) $\angle BAM = \angle CAX$, and
(b) $\frac{AM}{AX} = \cos\angle BAC.$
Izvor: Međunarodna matematička olimpijada, shortlist 1985