140. A synthetical property <==> linear/angled relation.

by Virgil Nicula, Oct 4, 2010, 5:35 AM

PP1. Let $\triangle ABC$ with the incenter $I$ and the circumcenter $O$ . Denote $M\in AB\cap IO$ , $N\in AC\cap IO$ . Prove that $BMNC$ is cyclically $\iff$ $h_a=R+r$ .

PP2. In $\triangle ABC$ with $c>b$ the incircle touches $AB$ , $BC$ at $E$ , $D$ respectively, the exterior $A$-angle bisector meet $BC$ in $L$ and the

$B$-exincircle touches $BC$ in $F$ . Prove that $LC=2\cdot AE\iff $ $AD=AF\iff C=90^{\circ}+\frac B2\iff c^2=b^2+ac$ .

PP3. Let $\triangle ABC$ , the midpoint $M$ of $[BC]$ and the foot $N$ of the $A$-angle bisector. Prove that $A=90^{\circ}+m\left(\widehat{MAN}\right)\iff 2\cos A=1-\frac bc$ .


Proof of PP1. $BMNC$ is cyclically $\iff$ $MN\parallel AA$ $\iff$ $MN\perp OA$ $\iff$ $IO\perp OA$ $\iff$ $IA^2=IO^2+AO^2$ . From the well-known relations

$s\cdot AI^2=bc(s-a)$ , $OI^2=R^2=2Rr$ obtain $MN\perp OA$ $\iff$ $bc(s-a)=sR^2+s\left(R^2-2Rr\right)$ $\iff$ $abc(s-a)=2Rsa(R-r)$ $\iff$

$4RS(s-a)=2Rsa(R-r)$ $\iff$ $2sr(s-a)=as(R-r)$ $\iff$ $2r(s-a)=a(R-r)$ $\iff$ $\boxed{\frac {b+c}{a}=\frac Rr}$ $\iff$ $r(a+b+c)=a(R+r)$ $\iff$

$2sr=a(R+r)$ $\iff$ $2S=a(R+r)$ $\iff$ $ah_a=a(R+r)$ $\iff$ $\boxed{h_a=R+r}$ . In this case for $D\in AI\cap BC$ obtain $\frac {MN}{a}=\frac {AI}{AD}$ , i.e.

$\frac {MN}{a}=\frac {b+c}{2s}$ $\iff$ $2s\cdot MN=a(b+c)$ $\iff$ $\frac {1}{MN}=\frac 1a+\frac {1}{b+c}$ .

Proof of PP2. From $\frac {LB}{c}=\frac {LC}{b}=\frac {a}{c-b}$ and $AE=s-a$ obtain $\boxed{LC=2\cdot AE}\iff$ $\frac {ab}{c-b}=2(s-a)\iff$ $ab=(c-b)(b+c-a)\iff$

$\boxed{c^2=b^2+ac}$ . Apply the Stewart's relation to the cevians $AD$ , $AF\ :\ \left\|\begin{array}{c} a\cdot AD^2+a(s-b)(s-c)=c^2(s-c)+b^2(s-b)\\\ a\cdot AF^2+c^2(s-a)=b^2s+as(s-a)\end{array}\right\|$ $\implies$

$a\cdot\left(AD^2-AF^2\right)=$ $b(c^2-ac-b^2)$ . Thus, $\boxed{AD=AF}\iff$ $ac=c^2-b^2\iff$ $\boxed{c^2=b^2+ac}$ $\iff$ $\sin^2C-\sin^2B=$ $\sin A\sin C$ $\iff$

$\sin (C-B)\sin (C+B)=\sin A\sin C$ $\iff$ $\sin (C-B)=\sin C$ $\iff$ $2C-B=180^{\circ}$ $\iff \boxed{C=90^{\circ}+\frac B2}$ .

I used the identity $s(s-a)+(s-b)(s-c)=bc$ .

Proof of PP3. Denote the point $P$ for which $PC\parallel AB$ , $PC=b$ and the line $AC$ separates $B$ , $P$ . Observe that $m(\widehat{APC})=90^{\circ}-\frac A2=$ $\frac A2-(A-90^{\circ})=$

$m(\widehat{NAC})-m(\widehat{NAM})=$ $m(\widehat {MAC})$ . Thus, $\widehat {CPA}\equiv\widehat {MAC}$ . Denote the symmetrical point $D$ of $A$ w.r.t. $M$ . Observe that $AD=2m_a$ , $C\in (DP)$ , $DC=c$ , $DP=b+c$ .

Since $\widehat {DAC}\equiv\widehat{CPA}$ obtain $DA^2=DC\cdot DP$ , i.e. $4m_a^2=c(b+c)$ $\iff$ $a^2=2b^2+c^2-bc$ . From the wejj-known relation $a^2=b^2+c^2-2bc\cdot\cos A$ obtain

$b=c(1-2\cos A)$ , i.e. $2\cos A=1-\frac bc$ .

Remark. If $b=2c$ , then $A=120^{\circ}$ and $\frac {a}{\sqrt 7}=\frac {b}{2}=\frac {c}{1}$ .
This post has been edited 36 times. Last edited by Virgil Nicula, Dec 1, 2015, 11:09 AM

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  • 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: 404395
  • Total comments: 37
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