276. Chinese TST 2009 and Second Round 2006.

by Virgil Nicula, May 12, 2011, 1:07 AM

PP1. Let $ ABC$ be a triangle with the circumcircle $ w = C(O,R)$ and let $ T$ , $ X$ be the intersections of

$ BC$ , $ w$ respectively with the bisector of the angle $ \angle A$ . Through the vertex $ C$ we draw the perpendicular line to $ BC$

which intersects the external bisector of $ \angle A$ in the point $ P$ . Denote $ K\in OA\cap BP$ . Prove that $ KT\perp BC$ .

Proof 1 (metric). Suppose w.l.o.g. $ b > c$ . Observe that $ \left\|\begin{array}{c} m(\angle KAB) = 90^{\circ} - C \\ \\ m(\angle KAP) = 90^{\circ} - \frac {B - C}{2} \\ \\ m\left(\angle APC\right) = 90^{\circ} - \frac {B - C}{2}\end{array}\right\|$ . Denote $ \left\|\begin{array}{c} D\in BC \\ \ AD\perp BC\end{array}\right\|$ . Therefore,

$ \frac {KB}{KP} = \frac {AB}{AP}\cdot\frac {\sin\widehat {KAB}}{\sin\widehat {KAP}} =$ $ \left|\frac {c\cdot\sin (90^{\circ} - C)}{AP\cdot \cos\frac {B - C}{2}}\right| =$ $ \frac {c\cdot |\cos C|}{DC} = \frac cb$ . In conclusion, $ \frac {KB}{KP} = \frac {TB}{TC}$ $ \implies$ $ KT\parallel PC$ $ \implies$ $ KT\perp BC$ .

Proof 2 (proiective). Denote $ L\in BC\cap AP$ , the diameter $ [XY]$ , the line $ d$ for which $ K\in d$ , $ d\parallel PC$, i.e. $ d\perp BC$ and the points $ U\in AP\cap d$ ,

$ V\in PT\cap LK$ , $ W\in LK\cap PC$ , $ T_1\in d\cap AX$ , $ T_2\in d\cap BC$ . Observe that $ KU=KT_1$ because $ K\in AO$ - the $ A$-median in $ \triangle AXY$

and $ \overline {UKT_1}\parallel \overline {XOY}$ . Thus, the division $ \{B,C;L,T\}$ is harmonically $ \Longleftrightarrow$ the pencil $ P\{B,C;L,T\}$ is harmonically $ \Longleftrightarrow$ the divison $ \{K,W;L,V\}$

is harmonically. Therefore, $ KT_2\parallel WC$ $ \Longrightarrow$ $ KT_2=KU$ $ \Longrightarrow$ $ KT_1=KT_2$ $ \Longrightarrow$ $ T_1\equiv T_2\equiv T$ $ \Longrightarrow$ $ KT\perp BC$ .


PP2 (China S.R. 2006). Solve the system of equations in real numbers $\begin{cases} x-y+z-w=2 \\ x^2-y^2+z^2-w^2=6 \\ x^3-y^3+z^3-w^3=20 \\ x^4-y^4+z^4-w^4=66 \end{cases}$ .

Proof. Can use the substitutions $\left\{\begin{array}{ccc} x+z=a & ; & xz=b\\\\ y+w=c & ; & yw=d\end{array}\right\|$ . Our system becomes $\left\{\begin{array}{cc} a-c=2 & (1)\\\\ a^2-2b-c^2+2d=6 & (2)\\\\ a^3-3ab-c^3+3cd=20 & (3)\\\\ a^4-4a^2b+2b^2-c^4+4c^2d-2d^2=66 & (4)\end{array}\right\|$ .

From $(1)$ , $(2)$ obtain that $\left\{\begin{array}{c} c=a-2\\\\ d=-2a+b+5\end{array}\right\|$ and from $(3)$ , $(4)$ get that $\left\{\begin{array}{c} 5a-2b-14=0\\\\ 5a^2-2ab-10a-b-13=0\end{array}\right\|\iff$ $\left\{\begin{array}{c} 5a-2b=14\\\\ 4a-b=13\end{array}\right\|$ .

Therefore, $a=4$ , $b=3$ and from here obtain that $c=2$ and $d=0$ . In conclusion. all solutions are $\left(\begin{array}{cccc} 1 & 0 & 3 & 2\\\\ 1 & 2 & 3 & 0\\\\ 3 & 0 & 1 & 2\\\\ 3 & 2 & 1 & 0\end{array}\right)$ .
This post has been edited 14 times. Last edited by Virgil Nicula, Nov 22, 2015, 7:24 AM

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excellent! I like your blogs. :wink:
my name is'' tran van lam '', nick name is 'lehungvietbao'. I come from VIETNAM.http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=125553
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by lehungvietbao, Mar 11, 2012, 2:46 PM

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    by the_mathmagician, Jan 17, 2021, 7:43 PM

  • godly blog. opopop

    by OlympusHero, Dec 30, 2020, 6:08 PM

  • long blog

    by MrMustache, Nov 11, 2020, 4:52 PM

  • 372554 views!

    by mrmath0720, Sep 28, 2020, 1:11 AM

  • wow... i am lost.

    369302 views!

    -piphi

    by piphi, Jun 10, 2020, 11:44 PM

  • That was a lot! But, really good solutions and format! Nice blog!!!! :)

    by CSPAL, May 27, 2020, 4:17 PM

  • impressive :D
    awesome. 358,000 visits?????

    by OlympusHero, May 14, 2020, 8:43 PM

72 shouts
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About Owner
  • Posts: 7054
  • Joined: Jun 22, 2005
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  • Blog created: Apr 20, 2010
  • Total entries: 456
  • Total visits: 404395
  • Total comments: 37
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