35. Nice problem from Arqady (Eduard_Tur).

by Virgil Nicula, May 4, 2010, 2:35 AM

Quote:
Let $ABC$ be a triangle with the incircle $C(I,r)$ . Denote $D\in AI\cap BC$ , $E\in BI\cap CA$ and $F\in CI\cap AB$ . Prove that $\widehat{IED}\equiv\widehat {IFD}\ \implies\ b=c \ \vee\ A=120^{\circ}$ .

Method 1. Suppose w.l.o.g. that $b\ne c$ . Denote the incircle $C(I,r)$ . Observe that : $\odot\ \left\|\ \begin{array}{c} r=IE\cdot\sin \widehat{AEB}=IE\cdot\sin\left(C+\frac B2\right)\\\\ r=IF\cdot\sin \widehat {AFC}=IF\cdot\sin\left(B+\frac C2\right)\end{array}\ \right\|$ $\Longrightarrow$ $\boxed {\ \frac {IE}{IF}=\frac {\sin \left(B+\frac C2\right)}{\sin\left(C+\frac B2\right)}\ }\ (1)$ .

Denote $x=m(\angle IED)=m(\angle IFD)$ . Apply the Sinus' theorem in the triangles : $\odot\ \left\|\ \begin{array}{ccc} \triangle\ IED & \implies & \frac {ID}{\sin x}=\frac {IE}{\cos \left(\frac C2+x\right)}\\\\ \triangle\ IFD & \implies & \frac {ID}{\sin x}=\frac {IF}{\cos\left(\frac B2+x\right)}\end{array}\ \right\|$ $\implies$ $\boxed {\ \frac {IE}{IF}=\frac {\cos \left(\frac C2+x\right)}{\cos\left(\frac B2+x\right)}\ }\ (2)$ .

From the relations $(1)\wedge (2)$ obtain $\frac {\sin \left(B+\frac C2\right)}{\sin\left(C+\frac B2\right)}=\frac {\cos \left(\frac C2+x\right)}{\cos\left(\frac B2+x\right)}$ $\Longleftrightarrow$ $\sin \left(B+\frac C2\right)\cdot\cos\left(\frac B2+x\right)=\sin\left(C+\frac B2\right)\cdot\cos \left(\frac C2+x\right)$ $\Longleftrightarrow$

$\sin\left(B+\frac {B+C}{2}+x\right)+\sin\left(\frac {B+C}{2}-x\right)=$ $\sin\left(C+\frac {B+C}{2}+x\right)+\sin\left(\frac {B+C}{2}-x\right)$ $\Longleftrightarrow$ $\sin\left(B+\frac {B+C}{2}+x\right)=$ $\sin\left(C+\frac {B+C}{2}+x\right)$ $\Longleftrightarrow$

$B=C\ \ \vee\ \ 2(B+C)+2x=180^{\circ}$ $\Longleftrightarrow$ $x=A-90^{\circ}$ because $B\ne C$ . Apply in $\triangle BFC$ an well-known property $\frac {DB}{DC}=\frac {FB}{FC}\cdot\frac {\sin \left(\angle DFB\right)}{\sin\left(\angle DFC\right)}$ . Thus,

$\frac cb=\frac {\sin\frac C2}{\sin B}\cdot\frac {\sin\left(A+\frac C2-x\right)}{\sin x}$ $\Longleftrightarrow$ $\frac {\sin C}{\sin B}=\frac {\sin\frac C2}{\sin B}\cdot\frac {\sin\left(90^{\circ}+\frac C2\right)}{\sin\left(A-90^{\circ}\right)}$ $\Longleftrightarrow$ $-2\cos\frac C2\cos A=-cos \frac C2$ $\Longleftrightarrow$ $\cos A=-\frac 12$ $\Longleftrightarrow$ $A=120^{\circ}$ .

Method 2. Denote the point $L\in (AC)$ for which $AL=AF$ . This point $L\not\equiv E$ because $b\ne c$ . Observe that $\widehat{ILD}\equiv\widehat {IED}$ , i.e. the quadrilateral $ILED$

is cyclic. Thus, $AI\cdot AD=AL\cdot AE=AF\cdot AE$ , i.e. $\boxed {\ AE\cdot AF=AI\cdot AD\ }$ . This relation is equivalently to $\frac {(b+c)\cdot AD^2}{2s}=$ $\frac {b^2c^2}{(a+b)(a+c)}$ $\Longleftrightarrow$

$\frac {b+c}{2s}\cdot\frac {4bcs(s-a)}{(b+c)^2}=\frac {b^2c^2}{(a+b)(a+c)}$ $\Longleftrightarrow$ $(b+c-a)(a+b)(a+c)=bc(b+c)$ $\Longleftrightarrow$ $a^2=b^2+c^2+bc$ $\Longleftrightarrow$ $\cos A=-\frac 12$ $\Longleftrightarrow$ $A=120^{\circ}$ .

Remark. Denote for an interior point $M$ with the barycentrical coordinates $(x,y,z)$ w.r.t. $\triangle ABC$ the intersections $D\in AM\cap BC$ , $E\in BM\cap CA$

and $F\in CM\cap AB$ . Then $AE\cdot AF=AM\cdot AD$ $\Longleftrightarrow$ $\boxed {bcyz(y+z)+a^2yz(x+y)(x+z)=(x+y)(y+z)(z+x)\left(b^2z+c^2y\right)}$ .
Quote:
Similar problem. Let $ABC$ be a triangle with the $A$-exincenter $I_a$ and $b\ne c$ .

Denote $D\in AI_a\cap BC$ , $E\in BI_a\cap CA$ and $F\in CI_a\cap AB$ .

Prove that $\boxed {\ \widehat {DEB}\equiv\widehat {DFC}\ \Longleftrightarrow\ AE\cdot AF=AD\cdot AI_a\ \Longleftrightarrow\ A=120^{\circ}\ }$ .
This post has been edited 19 times. Last edited by Virgil Nicula, Nov 27, 2015, 8:41 AM

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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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