AoPS Academy
such that for any 
:
such that 
for all 
be a triangle with circumcircle 
and let 
be the 
-excircle. Let 
and 
be the intersection points of and . Let 
and 
be the projections of onto the tangent lines to at and respectively. The tangent line at to the circumcircle of the triangle 
intersects the tangent line at to the circumcircle of the triangle 
at a point 
. Prove that 
.
, prove out of
,
,
1/3 and one 
than 1/3,
![\[[P_1PP_2]=a\]](http://latex.artofproblemsolving.com/d/a/1/da1ff8901331b92d934e934e6448d22c3590380a.png)
,![\[[P_2PP_3]=b\]](http://latex.artofproblemsolving.com/5/1/c/51cc40443ff770bca34a40fd517098c6495fe7d7.png)
,![\[[P_1PP_3]=c\]](http://latex.artofproblemsolving.com/5/f/c/5fcf1612fd1d3efab8fa69aff655215619ada7a7.png)
.![\[P_1/PQ_1=[P_1PP_3]/[PQ_1P_3]=[P_1PP_2]/[PP_2Q_1]=(a+c)/b\]](http://latex.artofproblemsolving.com/3/e/5/3e53db1d8ffb37e707c4219fd49ed2d5bab62641.png)
![\[P_2P/PQ_2=(a+b)/c\]](http://latex.artofproblemsolving.com/b/5/4/b54d25994d83e6bd53b73bda3b4a4c00c79edfcf.png)
and ![\[P_3P/PQ_3=(b+c)/a\]](http://latex.artofproblemsolving.com/5/a/7/5a76e36e0beeadff3940e70157ef58b30c7a1993.png)
![\[a+c>2b ..(1)\]](http://latex.artofproblemsolving.com/0/7/1/071eea2d523079ab158883f4b26ee7a60a71d4ce.png)
![\[a+b>2c ..(2)\]](http://latex.artofproblemsolving.com/6/e/9/6e9937edab7441f4ef1b2e81cb46076ebfb707e3.png)
![\[b+c>2a ..(3)\]](http://latex.artofproblemsolving.com/e/a/d/eada085deec551d2850a0b85b6d3137000136063.png)
which contradicts with (3).Contradiction.
, and we are finished.
, 
, 
, 
, 
, 
, and 
to 
, 
, 
, 
, 
, 
, and , respectively, where is the overall center of mass. Then, we have that 
, 
, and 
along cevians 
, 
, and 
, in that order. Now, suppose FTSOC that 
, i.e.,
But summing gives 
, contradiction, so at least one of the 
are 
. Analogously, at least one of the are 
, hence proven. 
![\[
\begin{cases}
x - y z = 1,\\[2pt]
y - z x = 2,\\[2pt]
z - x y = 4.
\end{cases}
\]](http://latex.artofproblemsolving.com/b/2/d/b2d94e49491a71f86109103f985fe3f238cf4d95.png)
and a point 
within the triangle. Lines 
intersect the opposite sides in points 
respectively. Prove that, of the numbers ![\[ \dfrac{P_1P}{PQ_1}, \dfrac{P_2P}{PQ_2}, \dfrac{P_3P}{PQ_3} \]](http://latex.artofproblemsolving.com/9/3/e/93e17213c1d74012c3c596e1a5ab19c1361f0581.png)
and at least one is 
![$\frac{P_{1}P}{PQ_{1}}=\frac{[PP_{1}P_{2}]}{[PP_{2}Q_{1}]}= \frac{[PP_{1}P_{3}]}{[PP_{2}Q_{1}]}=\frac{[P_{1}P_{2}P_{3}]-[PP_{2}P_{3}]}{[PP_{2}P_{3}]}$](http://latex.artofproblemsolving.com/1/3/9/13990a8093f6259d25456b6ca2187e49827c975a.png)
We get similar results for 
and 
.
.![$\frac{[P_{1}P_{2}P_{3}]-[PP_{2}P_{3}]}{[PP_{2}P_{3}]}+\frac{[P_{1}P_{2}P_{3}]-[PP_{1}P_{3}]}{[PP_{1}P_{3}]}+\frac{[P_{1}P_{2}P_{3}]-[PP_{1}P_{2}]}{[PP_{1}P_{2}]}$](http://latex.artofproblemsolving.com/c/9/7/c97c70641fca92821796e4990fe12fb04300db61.png)
![$\frac{[P_{1}P_{2}P_{3}] }{[PP_{2}P_{3}]}+\frac{[P_{1}P_{2}P_{3}] }{[PP_{1}P_{3}]}+\frac{[P_{1}P_{2}P_{3}] }{[PP_{1}P_{2}]}-3 = [P_{1}P_{2}P_{3}][\frac{1 }{[PP_{2}P_{3}]}+\frac{1 }{[PP_{1}P_{3}]}+\frac{1 }{[PP_{1}P_{2}]}]-3$](http://latex.artofproblemsolving.com/7/6/d/76dfdb4ff5d438de1f33059217b451e043e01c7b.png)
.
.![$[PP_{2}P_{3}], [PP_{3}P_{1}] , [PP_{1}P_{2}]$](http://latex.artofproblemsolving.com/d/e/6/de682704074e71857417bf44915bb602badcf811.png)
Without loss of generality let it be ![$[PP_{2}P_{3}] \geq \frac{[P_{1}P_{2}P_{3}]}{3}$](http://latex.artofproblemsolving.com/4/2/a/42af968353c317eaef83f49d002241f442d14465.png)
.![$3\geq \frac{[P_{1}P_{2}P_{3}]}{[PP_{2}P_{3}]}$](http://latex.artofproblemsolving.com/e/b/5/eb57cd10d1b95cbcbdae6f496df0a42afc12ab33.png)
w
and we are done.