36. ONM (juniori) - 2010 Iasi, problema 4.

by Virgil Nicula, May 5, 2010, 11:26 PM

Problema este (arhi)cunoscuta. Voi reveni cu o extindere a acesteia.
Quote:
Fie triunghiul isoscel $ABC$ cu $AB=AC$ . Bisectoarea unghiului $\widehat {ABC}$ intalneste latura $[AC]$ in punctul $D$ . Sa se arate ca $BD+DA=BC\ \Longleftrightarrow\ B=40^{\circ}$ .

Metoda 1. $BC$ taie din nou cercul circumscris al $\triangle ABD$ in $E\implies$ $AD=DE=EC$ si $BD+DA=BC$ $\Longleftrightarrow$ $BD+DA=BE+EC$ $\Longleftrightarrow$ $BD=BE$ $\Longleftrightarrow$ $B=40^{\circ}$

Metoda 2. Notam $B=C=2x$ . $m(\angle ADB)=3x$ si $BD+DA=BC$ $\Longleftrightarrow$ $\frac {BD}{AB}+\frac {DA}{AB}=\frac {BC}{AB}$ $\Longleftrightarrow$ $\frac {\sin 4x}{\sin 3x}+\frac {\sin x}{\sin 3x}=\frac {\sin 4x}{\sin 2x}$ $\Longleftrightarrow$

$\sin 4x+\sin x=2\cos 2x\sin 3x$ $\Longleftrightarrow$ $\sin 4x+\sin x=\sin 5x+\sin x$ $\Longleftrightarrow$ $\sin 4x=\sin 5x$ $\Longleftrightarrow$ $9x=180^{\circ}$ $\Longleftrightarrow$ $B=40^{\circ}$ .

Extindere. In $\triangle ABC$ bisectoarea unghiului $\widehat {ABC}$ intalneste $[AC]$ in $D$ . Atunci $BD+DA=BC\ \Longleftrightarrow$

$\cos\frac B2\sin \frac C2=\cos \frac A2\sin \frac {A-C}{2}$ $\Longleftrightarrow \cos \frac A2+\sin\frac C2=\sin\frac {2A-C}{2}+\sin\frac {B-C}{2}$ .

Demonstratie. Notam $B=2x$ si $C=y$ . Procedand analog se obtine relatia $\sin y\sin (2x+y)+\sin x\sin y=\sin (2x+y)\sin (x+y)\Longleftrightarrow$

$\cos 2x-\cos (2x+2y)+\sin x\sin y=\cos x-\cos (3x+2y)\Longleftrightarrow$ $(\cos x-\cos 2x)+[\cos (2x+2y)-\cos (3x+2y)]=$

$\sin x\sin y\Longleftrightarrow$ $\sin \frac {3x}{2}\sin\frac x2+\sin\frac {5x+4y}{2}\sin\frac x2=\sin x\sin y\Longleftrightarrow$ $\sin\frac {3x}{2}+\sin\frac {5x+4y}{2}=2\cos\frac x2\sin y\Longleftrightarrow$

$\sin\frac {3x}{2}+\sin\frac {5x+4y}{2}=\sin \frac {x+2y}{2}+\sin \frac {2y-x}{2} \Longleftrightarrow$ $\left(\sin \frac {3x}{2}-\sin\frac {2y-x}{2}\right)+\left(\sin\frac{5x+4y}{2}-\sin\frac {2y+x}{2}\right)=0\Longleftrightarrow$

$\sin\frac {2x-y}{2}\cos\frac {x+y}{2}+\sin\frac {2x+y}{2}\cos\frac {3(x+y)}{2}=0\Longleftrightarrow$ $\sin\frac {2x-y}{2}+\sin\frac {2x+y}{2}\left(4\cos^2\frac {x+y}{2}-3\right)=0$ $\Longleftrightarrow$

$\sin\frac {2x-y}{2}+\sin\frac {2x+y}{2}\left[2\cos (x+y)-1\right]=0\Longleftrightarrow$ $\sin\frac {2x+y}{2}-\sin\frac {2x-y}{2}=2\cos (x+y)\sin\frac {2x+y}{2}\Longleftrightarrow$

$\boxed {\ \cos x\sin\frac y2=\cos (x+y)\sin\left(x+\frac y2\right)\ }\ \ (*)\Longleftrightarrow$ $\boxed {\ \sin\left(x+\frac y2\right)-\sin\left(x-\frac y2\right)=\sin\left(2x+\frac {3y}{2}\right)-\sin\frac y2\ }\ \mathrm\{O.K.\}$

Observatie Daca $\triangle ABC$ este $A$-isoscel, atunci $y=2x$ si $(*)$ devine $\cos x\sin x=\cos 3x\sin 2x$ $\Longleftrightarrow$

$\cos 3x=\frac 12$ $\Longleftrightarrow$ $x=20^{\circ}$ . Cu $(*)$ se compun frumoase si dificile probleme "slicing".

Alte cazuri particulare.

CP1. Fie triunghiul isoscel $ABC$ cu $CA=CB$ . Bisectoarea unghiului $\widehat {ABC}$ intalneste latura $[AC]$ in punctul $D$ . Sa se arate ca $BD+DA=BC\ \Longleftrightarrow\ C=36^{\circ}$ .

CP2. In triunghiul $ABC$ bisectoarea unghiului $\widehat {ABC}$ intalneste latura $[AC]$ in $D$ si $m(\angle ADB)=60^{\circ}$ . Atunci $BD+DA=BC\ \Longleftrightarrow\ \triangle\ ABC$ este $A$-isoscel.

CP3. In triunghiul $ABC$ bisectoarea unghiului $\widehat {ABC}$ intalneste latura $[AC]$ in $D$ si $B=2C$ . Atunci $BD+DA=BC\ \Longleftrightarrow\ \triangle\ ABC$ este $C$-isoscel.
This post has been edited 26 times. Last edited by Virgil Nicula, Mar 15, 2019, 2:31 PM

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    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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