190. Very nice ! Coculescu's Contest seniors 2010.

by Virgil Nicula, Dec 12, 2010, 11:00 AM

Proposed problem. Let $ABC$ be a nonisosceles triangle and $E\in(AB)$ , $D\in (AC)$ be two points so that the quadrilateral $BCDE$ is cyclically. Denote

$O\in EC\cap BD$ and the midpoints $N$ , $P$ of the segments $[DE]$ , $[BC]$ respectively. Prove that the line $AO$ is tangent to the circumcircle of $\triangle PON$ .

Proof (official). Denote the midpoint $M$ of the segment $[AO]$ . Then $P\in MN$ - the Gauss' line for the complet quadrilateral $ADOEBC$ . Construct

$\left\|\begin{array}{ccc} N\in (OR) & , & NO=NR\\ P\in (OQ) & , & PO=PQ\end{array}\right\|$ . Observe that $R\in AQ$ . Since $BQCO$ is parallelogram obtain that $CQ=BO$ . Since $\triangle EOD\sim\triangle BOC$

and $\triangle ADE\sim\triangle ABC$ obtain that $\frac{CQ}{EO}=\frac{BO}{EO}=\frac{BC}{ED}=\frac{AC}{AE}$ . Since $\widehat{AEO}\equiv\widehat{ACQ}$ obtain that $\triangle AEO\sim\triangle ACQ$ $\implies$ $\widehat{AOE}\equiv\widehat{AQC}$.

Since $\angle CQP\equiv\angle BOP\equiv\angle EON$ obtain that $\angle AON\equiv\angle OQA\equiv\angle OPN$ $\implies$ the line $AO$ is tangent to the circumcircle of $\triangle PON$ .

Remark. Very nice and interesting problem ! Denote $S\in AO\cap DE$ , $R\in AO\cap BC$ . I"ll prove that the quadrilateral $SNPR$ is cyclically.

Indeed, since division $(A,S,O,R)$ is harmonically and $M$ is midpoint of $[AO]$ obtain that $MO^2=MS\cdot MR$ . Since the line $MO$ is tangent

to circumcircle of $\triangle NOP$ obtain that $MO^2=MN\cdot MP$ . Thus, $MS\cdot MR=MN\cdot MP$ , i.e. quadrilateral $SNPR$ is cyclically. Denote

the Gauss' line $d\equiv\overline{MNP}$ and the intersections $X\in AB\cap d$ , $Y\in AC\cap d$ . Prove easily that $AN$ is tangent to the circumcircle of $\triangle XAY$ .


P1 (ONM Belgia 2004). Let $\triangle ABC$ with $c>b\ ,$ $D\in [AC$ so that $BD=CD$ and $E\in (BC)\ ,$ $F\in AB$ so that $EF\parallel BD$ and $G\in AE\cap BD\ .$ Prove that $\widehat{BCF}\equiv\widehat{BCG}\ .$

Proof 1 (Arab). Let $H\in AD$ such that $GH\parallel BC\ .$ Therefore, $FE\parallel BD\implies \frac{EF}{GB}=\frac{AE}{AG}=\frac{EC}{GH}\implies $ $\triangle CEF\sim \triangle HGB$.

In conclusion, $BCHG$ is an isosceles trapezoid because $BD=CD\ .$. Hence $\angle FCB=\angle FCE=\angle BHG=\angle BCG\ .$ Very nice!

Proof 2 . Let $:\ S\in EF\cap CG\ ,$ $K\in EF\cap AD$ and $M\in EF$ so that $CM\perp CB\ ;\ EB=m\ ,$ i.e. $EC=a-m$ and $DB=DC=d\ .$ Thus, $\frac {EK}{BD}=\frac {CE}{CB}=\frac {a-m}a\iff$

$\boxed{KE=KC=KM=\frac {d(a-m)}a}\ (1)$ and $ME=2\cdot KE=\frac {2d(a-m)}a\ .$ I"ll show that the division $(F,E,S,M)$ is harmonic, i.e. $\boxed{\frac {EF}{ES}=\frac {MF}{MS}}\ (*)\ .$ Therefore,

$\frac {EF}{ES}=\frac {EF}{BG}\cdot\frac {BG}{ES}=\frac {AK}{AD}\cdot \frac {CB}{CE}=$ $\frac {AC+CK}{AC+CD}\cdot \frac {CB}{CE}=\frac {b+\frac {d(a-m)}a}{b+d}\cdot\frac a{a-m}\implies$ $\boxed{\frac {EF}{ES}=\frac {a(b+d)-md}{(a-m)(b+d)}}\ (2)\ .$ Apply the Menelaus' theorem to the transversal $\overline{AEG}$

and $\triangle BCD\ :\ \frac {AC}{AD}\cdot \frac{GD}{GB}\cdot\frac {EB}{EC}=1\iff$ $\boxed{\frac {GB}{GD}=\frac b{b+d}\cdot \frac m{a-m}}\ (3)\ \implies$ $\frac {EF}{EK}=\frac {GB}{GD}\ \stackrel{1\wedge 3}{\implies}\ EF$ $=\frac {d(a-m)}a\cdot \frac {bm}{(b+d)(a-m)}\implies$ $\boxed{EF=\frac {mbd}{a(b+d)}}\ (4)\ .$

Observe that $\frac {SE}{SK}=\frac {GB}{GD}\ \stackrel{3}{=}\ \frac {bm}{(b+d)(a-m)}\implies$ $\frac {SE}{bm}=\frac {SK}{(b+d)(a-m)}=$ $\frac {EK}{bm+(b+d)(a-m)}=$ $\frac {d(a-m)}{a(ab+ad-md)}\implies$ $\boxed{SE=\frac {mbd(a-m)}{a(ab+ad-md)}}\ (5)\ .$ Hence

$\frac {MF}{MS}=\frac {ME+EF}{ME-ES}=$ $\frac { \frac {2d(a-m)}{a} +\frac {mbd}{a(b+d)} } { \frac {2d(a-m)}{a}-\frac {mbd(a-m)}{a(ab+ad-md)} }=$ $\frac {2(a-m) +\frac {mb}{b+d} } {2(a-m)-\frac {mb(a-m)}{ab+ad-md} }=$ $\frac {2(a-m) +\frac {mb}{b+d} } {(a-m)\cdot\left[2-\frac {mb}{ab+ad-md}\right] }=$ $\frac {2(ab+ad-md)-mb}{b+d}\cdot\frac {ab+ad-md}{(a-m)[2(ab+ad-md)-mb]}=$

$\frac {ab+ad-md}{(b+d)(a-m)}\implies$ $\boxed{\frac {MF}{MS}=\frac {a(b+d)-md}{(b+d)(a-m)}}\ (6)\ .$ In conclusion, the relations $(2)$ and $(6)$ prove that $\frac {EF}{ES}=\frac {MF}{MS}\ ,$ i.e. the division $(F,E,S,M)$ is harmonic. From an well

known property obtain that the rays $[CE\perp [CM\implies$ the ray $[CE$ is the bisector of the angle $\widehat{FCS}\ ,$ i.e. $\widehat{BCF}\equiv\widehat{BCG}\ .$

Remark. Another case: "Let $\triangle ABC$ with $c<b\ ,$ $D\in (AC)$ so that $BD=CD$ and $E\in (BC)\ ,$ $F\in AB$ so that $EF\parallel BD$ and $G\in AE\cap BD\ .$ Prove that $\widehat{BCF}\equiv\widehat{BCG}$ ".
This post has been edited 56 times. Last edited by Virgil Nicula, Aug 17, 2016, 2:19 PM

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21 bis. Some problems with complex numbers.
21. Probleme de geometrie (Yetty & I).
20. O ineg.cu suma de AG/AI .
19. Own proposed problems in CRUX Mathematicorum - Canada.
18. O problema a lui Toshio Seimiya.
17. O inegalitate "tare" intr-un triunghi.
16. Length of the (interior) angle bisector (10 proofs).
15. Inequalities stronger than Panaitopol's.
14. Opinii si comentarii relativ la inv. rom.
13. Inegalitatea de la Sorin Borodi.
12. Inegalitatea I. Maftei & S. Radulescu (2).
11. Inegalitatea I. Maftei & S. Radulescu (1).
10. Walker's inequality and its extension.
9. Inegalitatea HO+HA+HB+HC >= 3R .
8. The first problem Shortlist ONM 2010 Romania.
7. Some interesting "slicing" problems.
6. Interesting problems from JBMO.
5. Theoretical preliminary for inequalities.
4. Remarkable geometrical inequalities (2).
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    by OlympusHero, Dec 30, 2020, 6:08 PM

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

  • 372554 views!

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

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

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