IMO Shortlist 1999 problem G4
Dodao/la:
arhiva2. travnja 2012. For a triangle
![T = ABC](/media/m/8/6/e/86eaae9e42134554d572bd89c472b993.png)
we take the point
![X](/media/m/9/2/8/92802f174fc4967315c2d8002c426164.png)
on the side
![(AB)](/media/m/f/d/b/fdb25e507dcaa6d071fde79eef99da95.png)
such that
![AX/AB=4/5](/media/m/d/c/7/dc727e51137e03af5da237144eadcdb2.png)
, the point
![Y](/media/m/3/b/c/3bc24c5af9ce86a9a691643555fc3fd6.png)
on the segment
![(CX)](/media/m/d/a/7/da74315b0a0501ed6da41c4d90b20267.png)
such that
![CY = 2YX](/media/m/c/3/1/c313d3080af5b4df61f5ec1849ba4cd5.png)
and, if possible, the point
![Z](/media/m/7/9/4/794ff2bd637e30ea27e50e57eecd0b76.png)
on the ray (
![CA](/media/m/a/a/e/aaec86bc003cfdb64d54116a4cabd387.png)
such that
![\widehat{CXZ} = 180 - \widehat{ABC}](/media/m/6/4/f/64ffbbbec1305320c3e419615a6a2c7c.png)
. We denote by
![\Sigma](/media/m/c/a/b/cab3b5d26e6dabb66878a77b1791db45.png)
the set of all triangles
![T](/media/m/0/1/6/016d42c58f7f5f06bdf8af6b85141914.png)
for which
![\widehat{XYZ} = 45](/media/m/a/7/1/a71dc19639fc04ba532734221303ac18.png)
. Prove that all triangles from
![\Sigma](/media/m/c/a/b/cab3b5d26e6dabb66878a77b1791db45.png)
are similar and find the measure of their smallest angle.
%V0
For a triangle $T = ABC$ we take the point $X$ on the side $(AB)$ such that $AX/AB=4/5$, the point $Y$ on the segment $(CX)$ such that $CY = 2YX$ and, if possible, the point $Z$ on the ray ($CA$ such that $\widehat{CXZ} = 180 - \widehat{ABC}$. We denote by $\Sigma$ the set of all triangles $T$ for which
$\widehat{XYZ} = 45$. Prove that all triangles from $\Sigma$ are similar and find the measure of their smallest angle.
Izvor: Međunarodna matematička olimpijada, shortlist 1999