Triangle $ ABC$ is inscript in a circumference $ \Gamma$. A chord $ MN=1$ of $ \Gamma$ intersects the sides $ AB$ and $ AC$ at $ X$ and $ Y$ respectively, with $ M$, $ X$, $ Y$, $ N$ in that order in $ MN$. Let $ UV$ be the diameter of $ \Gamma$ perpendicular to $ MN$ with $ U$ and $ A$ in the same semiplane respect to $ MN$. Lines $ AV$, $ BU$ and $ CU$ cut $ MN$ in the ratios $ \frac{3}{2}$, $ \frac{4}{5}$ and $ \frac{7}{6}$ respectively (starting counting from $ M$). Find $ XY$
Problem
Source: Argentina IMO TST Problem 2
Tags: ratio, geometry, algebra proposed, algebra
04.09.2009 02:57
With analitycal geometry the value is 1/11
06.11.2011 07:31
castigioni wrote: With analitycal geometry the value is 1/11 you did get the answer,but it can be solved by just some ratio substitutions,not analytical.
06.11.2011 07:33
shall I give some hint here? the key step(at least it seems to me) is to notice if chord $AB,MN$ intersect at $X$,then $\frac{MX}{XN}=\frac{MA*MB}{AN*BN}$.
06.12.2011 22:57
1/11 is the correct answer. Sorry I plugged in a wrong number and came up with a different numerical result. Please delete my post above. Thanks.
13.12.2011 18:40
http://voducdien.wordpress.com/2011/12/02/problem-2-of-argentine-mathematical-olympiad-2008/