http://www.mathlinks.ro/Forum/viewtopic.php?t=15773 . darij

shobber wrote: Problem six is an extremely hard one, probably the hardest ever appeared in the China TST history. Only 3 of the 6 team members solved this problem. Congrats for making a big step for being on the Chinese team.

orl wrote: shobber wrote: Problem six is an extremely hard one, probably the hardest ever appeared in the China TST history. Only 3 of the 6 team members solved this problem. Congrats for making a big step for being on the Chinese team. I got this information from a book. It provided a elementary trig proof, which is also hard to come up with. The book says we can turn this problem into proving: \[\tan{\angle{ADN}}=\frac{\sin{C}-\sin{B}}{1-\cos{B}-\cos{C}}\] Most guys managed to get here, but failed to go deeper.

A very hard and nice problem! I have nearly spent 2 hours on it. WLOG, assume $b\ge a\ge c$. We will start with 2 lemmas: Lemma 1 $\angle ACN=\angle IOD$. Proof: Let D' be the reflection of D in BC, D'' the antipode of D. By similar triangle, $BD^2=DM\cdot DD''\Rightarrow DI^2=2DM\cdot DO=DD'\cdot DO$. Thus, $\angle IOD=\angle D'ID=\angle PID+\angle IPD=180^\circ-\angle IDP=\angle ACN$. Lemma 2 Let X and Y be the points on rays BA and CA respectively such that BX=CY=a. Then $XY\perp OI$ Proof: problem posted before by me Now we apply a homothety to $\triangle AXY$ centered at A with the ratio $\frac{AC}{AY}$. Suppose the image is ACZ. Then consider the spiral smilarity of $\triangle ACZ$ centered at C through an angle $\angle ZCB$ (anticlockwisely) and ratio $\frac{BC}{AC}$. Suppose the image is $\triangle CBN'$. Clearly $N'\in(O)$. We have $\frac{BN'}{CN'}=\frac{a-c}{b-a}\Rightarrow bBN'+cCN'=a(BN'+CN')$. But the Ptolemy Theorem yields $bBN'+cCN'=aAN'$. Therefore BN'+CN'=AN'. Note that $90^\circ=\angle(XY,IO)=\angle(ZC,IO)=\angle(BC,IO)+\angle BCZ=90^\circ-\angle IOD+\angle ACN'$. Thus, $\angle IOD=\angle ACN'$. By Lemma 1, N=N', so we also have AN=BN+CN. For the case $b\ge c\ge a,a\ge b\ge c$, etc. The proof is essentially the same so I don't want to repeat, which is extremely tedious and boring.

mecrazywong wrote: Lemma 2 Let X and Y be the points on rays BA and CA respectively such that BX=CY=a. Then $XY\perp OI$ Proof: problem posted before by me Do you have a link or idea for proof this lemma ?

I have a really nice solution . I made it yesterday as all know we must prove $\sin AP=\sin BP + \sin CP$ we know that $\frac{XM}{XI}=\frac{1}{2}=\frac{TM}{TI}(=\sin \frac{A}{2}).\frac{\sin AP -(\frac{B-C}{2})}{\sin AP}\Leftrightarrow \sin AP = 2.\sin \frac{A}{2}.\sin AP -(\frac{B-C}{2})=\sin PB + \sin PC$

In early 90's, this problem seemed very hard, but it's not that hard nowaday. I saw a very smart and simple solution, maybe the official solution, on a Chinese journal by using the area means.

Dear Mathlinkers, I have also this reference http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=323883 Sincerely Jean-Louis

Let $a=x^2$, $b=y^2$, $c=z^2$, $d=-yz$, $i=-xy-yz-zx$, $m=\frac{y^2+z^2}2$, $p=y^2+z^2+xy+yz+zx$. Then, \begin{align*} n&=\frac{d-p}{d\overline p-1}\\ &=\frac{-(y+z)(x+y+z)}{-yz\left(\frac1{y^2}+\frac1{z^2}+\frac1{xy}+\frac1{yz}+\frac1{zx}\right)-1}\\ &=\frac{-xyz(y+z)(x+y+z)}{(y+z)(xy+yz+zx)}\\ &=\frac{xyz(x+y+z)}{xy+yz+zx}. \end{align*} Therefore, $$n-a=\frac{xyz(x+y+z)-x^2(xy+yz+zx)}{xy+yz+zx}=\frac{x(yz-x^2)(y+z)}{xy+yz+zx},$$so $AN=\frac{|yz-x^2||y+z|}{|xy+yz+zx|}.$ Therefore, $|yz-x^2|^2|y+z|^2=-\frac{(yz-x^2)^2(y+z)^2}{x^2y^2z^2}$, so $ANxyz|xy+yz+zx|=\pm i(yz-x^2)(y+z)$. Therefore, there exists a sign $\varepsilon_A$ such that $ixyz|xy+yz+zx|\varepsilon_AAN=y^2z+yz^2-x^2y-x^2z$, so taking the cyclic sum implies $\varepsilon_AAN+\varepsilon_BBN+\varepsilon_CCN$, so one of the three segments is the sum of the other two.