103. Metrical usual relations in triangle. Applications.

by Virgil Nicula, Sep 7, 2010, 11:34 PM

PP1. Let $ABC$ be a triangle. Consider the points $X\in (BC)$ , $Y\in (CA)$ , $Z\in (AB)$ , $P\in AX\cap YZ$ . Prove easily that (for example, with areas)

$\boxed {\ \frac {XB}{XC}=\frac {AB}{AC}\cdot \frac {\sin \widehat{XAB}}{\sin\widehat{XAC}}\ (1)\ \wedge\ \ \frac {PZ}{PY}=\frac {XB}{XC}\cdot\frac {AZ}{AY}\cdot \frac {AC}{AB}\ (2)\ \wedge\ \ \frac {PX}{PA}\cdot BC=\frac {YC}{YA}\cdot BX+\frac {ZB}{ZA}\cdot XC\ (3)\ }$ .

Proof.

$1\blacktriangleright$ Apply the theorem of Sinus in the mentioned triangles: $\left\{\begin{array}{cccc} \triangle XAB\ : & \frac {XB}{\sin\widehat{XAB}} & = & \frac {XA}{\sin B}\\\\ \triangle XAC\ : & \frac {XA}{\sin C} & = & \frac {XC}{\sin\widehat{XAC}}\end{array}\right|\ \bigodot \implies$ $\ \frac {XB}{XC}=\frac {\sin C}{\sin B}\cdot \frac {\sin \widehat{XAB}}{\sin\widehat{XAC}}\implies$

$\frac {XB}{XC}=\frac {AB}{AC}\cdot \frac {\sin \widehat{XAB}}{\sin\widehat{XAC}}\ (1)$ . Otherwise (with areas). $\frac {XB}{XC}=\frac {[XAB]}{[XAC]}=\frac {AX\cdot AB\cdot\sin \widehat{XAB}}{AX\cdot AC\cdot\sin\widehat{XAC}}\implies$ $\frac {XB}{XC}=\frac {AB}{AC}\cdot \frac {\sin \widehat{XAB}}{\sin\widehat{XAC}}\ (1)$ .

$2\blacktriangleright$ Apply the relation $(1)$ in the mentioned triangles: $\left\{\begin{array}{cc} \triangle AZY\ : & \frac {PZ}{PY}=\frac {AZ}{AY}\cdot\frac {\sin\widehat{PAZ}}{\sin\widehat{PAY}}\\\\ \triangle ABC\ : & \frac {XB}{XC}=\frac {AB}{AC}\cdot\frac {\sin\widehat{XAB}}{\sin\widehat{XAC}}\end{array}\right|\implies$ $\frac {PZ}{PY}:\frac {XB}{XC}=\frac {AZ}{AY}:\frac {AB}{AC}\implies$ $\frac {PZ}{PY}=\frac {XB}{XC}\cdot\frac {AZ}{AY}\cdot \frac {AC}{AB}\ (2)$ .

$3\blacktriangleright$ First method. Denote $\left\{\begin{array}{c} U\in BC\cap YZ\\\ V\in YZ\ ,\ VA\parallel BC\end{array}\right\|$ . Thus, $\frac {PX}{PA}\cdot BC=\frac {YC}{YA}\cdot BX+\frac {ZB}{ZA}\cdot XC\iff$ $\frac {UX}{VA}\cdot BC=\frac {UC}{VA}\cdot BX+\frac {UB}{VA}\cdot XC\iff$

$\boxed{UX\cdot BC=UC\cdot BX+UB\cdot XC} \ (*)\ ,$ what is true because $UC\cdot BX+UB\cdot CX=$ $XB\cdot (UX+XC)+XC\cdot (UX-XB)=$

$UX\cdot (BX+CX)=UX\cdot BC$ . In conclusion, $\frac {PX}{PA}\cdot BC=\frac {YC}{YA}\cdot BX+\frac {ZB}{ZA}\cdot XC\ (3)$ .

Second method. Denote $U\in YZ\cap BC$ . Apply the Menelaus' theorem to the transversals $:\ \left\{\begin{array}{cccc} \overline{PZU}/\triangle ABX\ : & \frac {UB}{UX}\cdot\frac {PX}{PA}\cdot\frac {ZA}{ZB}=1 & \implies & \frac {ZB}{ZA}=\frac {UB}{UX}\cdot \frac {PX}{PA}\\\\ \overline{PYU}/\triangle ACX\ : & \frac {UX}{UC}\cdot\frac {YC}{YA}\cdot\frac {PA}{PX}=1 & \implies & \frac {YC}{YA}=\frac {UC}{UX}\cdot\frac {PX}{PA}\end{array}\right\|$ and

$\left|\begin{array}{c} \odot\ XC\\\\ \odot\ XB\end{array}\right\|$ $\bigoplus$ $\implies$ $\frac {YC}{YA}\cdot BX+\frac {ZB}{ZA}\cdot XC=$ $\frac {PX}{PA}\cdot \frac {UB\cdot XC+UC\cdot XB}{UX}\ \stackrel{(*)}{=}\ \frac {PX}{PA}$ $\cdot \frac {BC\cdot UX}{UX}=$ $\frac {PX}{PA}\cdot BC$ $\implies$ $\frac {PX}{PA}\cdot BC=\frac {YC}{YA}\cdot BX+\frac {ZB}{ZA}\cdot XC\ (3)$ .

Remark. For third relation can use an equivalent form : $\boxed {\ \frac {AX}{AP}\cdot BC=\frac {AC}{AY}\cdot BX+\frac {AB}{AZ}\cdot XC\ }$ .


Application 1. Let $ABC$ be a triangle, let $D\in BC$ , $E\in CA$ , $F\in AB$ so that $AD\cap BE\cap CF\ne\emptyset$

and let $P\in EF$ , $Q\in FD$ , $R\in DE$ so that $DP\cap EQ\cap FR\ne\emptyset$ . Prove that $AP\cap BQ\cap CR\ne\emptyset$ .

Proof. Denote $\left\{\begin{array}{c} X\in BC\cap AP\\\ Y\in CA\cap BQ\\\ Z\in AB\cap CR\end{array}\right\|$ . Apply the Ceva's theorem to the concurrent lines $\left\{\begin{array}{ccc} AD\cap BE\cap CF\ne\emptyset & \implies & \frac {DB}{DC}\cdot\frac {EC}{EA}\cdot\frac {FA}{FB}=1\\\\ DP\cap EQ\cap FR\ne\emptyset & \implies & \frac {PE}{PF}\cdot\frac {QF}{QD}\cdot\frac {RD}{RE}=1\end{array}\right\|\ (*)$ .

Apply upper relation $(2)\ :\ \left\{\begin{array}{c} \frac {XB}{XC}=\frac {PF}{PE}\cdot\frac {AF}{AE}\cdot\frac {AC}{AB}\\\\ \frac {YC}{YA}=\frac {QD}{QF}\cdot\frac {BD}{BF}\cdot\frac {BA}{BC}\\\\ \frac {ZA}{ZB}=\frac {RE}{RD}\cdot\frac {CE}{CD}\cdot\frac {CB}{CA}\end{array}\right\|\bigodot\ \stackrel{(*)}{\implies}\ \frac {XB}{XC}$ $\cdot\frac {YC}{YA}\cdot\frac {ZA}{ZB}=1\implies$ $AX\cap BY\cap CZ\ne\emptyset\implies$ $AP\cap BQ\cap CR\ne\emptyset$ .
This post has been edited 35 times. Last edited by Virgil Nicula, Nov 26, 2015, 9:40 PM

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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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  • orzzzzzzzzz

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

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

  • 391345 views moment

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

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    by OlympusHero, Sep 14, 2022, 4:44 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

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About Owner
  • Posts: 7054
  • Joined: Jun 22, 2005
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  • Blog created: Apr 20, 2010
  • Total entries: 456
  • Total visits: 404398
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