265. Really nice problem !

by Virgil Nicula, Apr 8, 2011, 4:08 PM

Proposed problem. (Shortlist RMO - 2005). Let $ABC$ be a triangle. Consider two points $M\in (BC)$ for

which $\frac{MB}{MC}=\frac{c(b+c)}{b^2}$ and $N\in (AM)$ so that $m\left(\widehat {BNM}\right)=A$ . Prove that $m(\widehat {CNM})=\frac A2$ .

Remarks.

$\blacksquare\ 1^{\circ}.\ b=c\ \Longrightarrow$ Turkey - 1999.

$\blacksquare\ 2^{\circ}.\ b^2=c(b+c)\ \Longrightarrow$ A very nice and difficult problem.

$\blacksquare\ 3^{\circ}$ . Let $ABC$ be a triangle so that exists $p>1$ and $\frac cb =\frac{p-1}{2}$ . Consider two points $M\in (BC)$ for which $\frac{MB}{MC}=\frac{p^2-1}{4}$

and $N\in (AM)$ so that $m\left(\widehat {BAM}\right)=A$ . Prove that $m\left(\widehat {CNM}\right)=\frac A2$ $(\ p=3\Longrightarrow \blacksquare\ 1^{\circ}\ ;\ \ p=\sqrt 5\Longrightarrow \blacksquare\ 2^{\circ}.\ )$

Proof. Denote $:\ D\in (BC)\ ,\ \widehat {BAD}\equiv \widehat {CAD}\ ;\ R\in (AC)\ ,\ MR\parallel AD\ ;\ S\in (AB)\ ,\ MS\parallel AC$ .

$\blacksquare \ 1^{\circ}.\ MS\parallel CA\Longrightarrow \frac{SA}{SB}=\frac{MC}{MB}=\frac{b^2}{c^2+bc}$ $\Longrightarrow \frac{SA}{b^2}=\frac{SB}{c^2+bc}=$ $\frac{c}{b^2+bc+c^2}$ $\Longrightarrow$ $ AS\cdot AB=\frac{b^2c^2}{b^2+bc+c^2};\ \widehat {SBN}\equiv $ $\widehat {ABN}\equiv $

$\widehat {MAC}\equiv$ $ \widehat {SMA}\equiv \widehat {SMN}$ $\Longrightarrow\widehat {SBN}\equiv $ $\widehat {SMN}\Longrightarrow$ $\mathrm{cyclic}\ (BSNM)$ $\Longrightarrow AN\cdot AM=AS\cdot AB$ $\Longrightarrow $ $\overline {\underline {\left|\ AN\cdot AM=\frac{b^2c^2}{b^2+bc+c^2}\ \right| }}\ ;$

$\blacksquare \ 2^{\circ}.\ \frac{MB}{MC}=\frac{c(b+c)}{b^2}\Longrightarrow \frac{MB}{c^2+bc}=\frac{MC}{b^2}=\frac{a}{b^2+bc+c^2}$ $\Longrightarrow $ $MC=\frac{ab^2}{b^2+bc+c^2}\ ;$ $DM=|DC-MC|=\left| \frac{ab}{b+c}-\frac{ab^2}{b^2+bc+c^2}\right|=$

$\frac{abc^2}{(b+c)(b^2+bc+c^2)}\ .$ Therefore, $MR\parallel AD\Longrightarrow \frac{AR}{AC}$ $=\frac{DM}{DC}$ $\Longrightarrow $ $AR=\frac{bc^2}{b^2+bc+c^2}$ $\Longrightarrow$ $ \underline {\overline {\left| \ AR\cdot AC=\frac{b^2c^2}{b^2+bc+c^2}\ \right| }}\ .$

$\blacksquare \ (1^{\circ})\ \wedge \ (2^{\circ})\Longrightarrow AN\cdot AM=AR\cdot AC\Longrightarrow$ $\mathrm{cyclic}\ (MNRC)$ $\Longrightarrow \widehat {CNM}\equiv \widehat {CRM}\equiv \widehat {CAD}\Longrightarrow m(\widehat {CNM})=\frac A2.$

An extension. Let $\triangle ABC$ and the points $\{M,D\}\subset (BC)$ , $N\in (AM)$ so that $\left|\frac {DB}{DC}-\frac {MB}{MC}\right|=\frac {c^2}{b^2}$ and $m\left(\widehat {BNM}\right)=A$ . Prove that $\widehat {CNM}\equiv\widehat {CAD}$ .

Proof. Denote $\frac {MB}{MC}=m$ , $\frac {DB}{DC}=p$ , where $|m-p|=\frac {c^2}{b^2}$ , the points $R\in (AC)\ ,\ MR\parallel AD$ and $S\in (AB)\ ,\ MS\parallel AC$ .

$\blacksquare \ 1^{\circ}.\ MS\parallel CA\Longrightarrow \frac{SA}{SB}=\frac{MC}{MB}=\frac 1m$ $\Longrightarrow \frac{SA}{1}=\frac{SB}{m}=$ $\frac{c}{1+m}$ $\Longrightarrow$ $ AS\cdot AB=\frac{c^2}{1+m};\ \widehat {SBN}\equiv $ $\widehat {ABN}\equiv $

$\widehat {MAC}\equiv$ $ \widehat {SMA}\equiv \widehat {SMN}$ $\Longrightarrow\widehat {SBN}\equiv $ $\widehat {SMN}\Longrightarrow$ $\mathrm{cyclic}\ (BSNM)$ $\Longrightarrow AN\cdot AM=AS\cdot AB$ $\Longrightarrow $ $\overline {\underline {\left|\ AN\cdot AM=\frac{c^2}{1+m}\ \right| }}\ ;$

$\blacksquare \ 2^{\circ}.\ \frac{MB}{MC}=m\Longrightarrow \frac{MB}{m}=\frac{MC}{1}=\frac{a}{1+m}$ $\Longrightarrow $ $MC=\frac{a}{1+m}\ ;$ $DM=|DC-MC|=\left| \frac{a}{1+p}-\frac{a}{1+m}\right|=$

$\frac{a|m-p|}{(1+p)(1+m)}\ .$ Therefore, $MR\parallel AD\Longrightarrow \frac{AR}{AC}$ $=\frac{DM}{DC}$ $\Longrightarrow $ $AR=\frac{b|m-p|}{1+m}$ $\Longrightarrow$ $ \underline {\overline {\left| \ AR\cdot AC=\frac{b^2|m-p|}{1+m}\ \right| }}\ .$

$\blacksquare \ (1^{\circ})\ \wedge\ (2^{\circ})\ \wedge\ |m-p|=\frac {c^2}{b^2}$ $\Longrightarrow AN\cdot AM=AR\cdot AC\Longrightarrow$ $\mathrm{cyclic}\ (MNRC)$ $\Longrightarrow \widehat {CNM}\equiv \widehat {CRM}\equiv \widehat {CAD}\Longrightarrow m(\widehat {CNM})=\frac A2.$
This post has been edited 17 times. Last edited by Virgil Nicula, Nov 22, 2015, 8:42 AM

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  • orzzzzzzzzz

    by mathMagicOPS, Jan 9, 2025, 3:40 AM

  • this css is sus

    by ihatemath123, Aug 14, 2024, 1:53 AM

  • 391345 views moment

    by ryanbear, May 9, 2023, 6:10 AM

  • We need virgil nicula to return to aops, this blog is top 10 all time.

    by OlympusHero, Sep 14, 2022, 4:44 AM

  • :omighty: blog

    by tigerzhang, Aug 1, 2021, 12:02 AM

  • Amazing blog.

    by OlympusHero, May 13, 2021, 10:23 PM

  • the visits tho

    by GoogleNebula, Apr 14, 2021, 5:25 AM

  • Bro this blog is ripped

    by samrocksnature, Apr 14, 2021, 5:16 AM

  • Holy- Darn this is good. shame it's inactive now

    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
Blog Stats
  • Blog created: Apr 20, 2010
  • Total entries: 456
  • Total visits: 404396
  • Total comments: 37
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