330. A triangle and an its transversal through a point.

by Virgil Nicula, Nov 30, 2011, 5:44 PM

An example. For a line $p$ which passes through the centroid $G$ of $\triangle ABC$ define the intersections

$X\in BC\cap p$ , $Y\in CA\cap p$ and $Z\in AB\cap p$ . Prove that $\frac{1}{\overline {GX}}+\frac{1}{\overline {GY}}+\frac{1}{\overline{GZ}}=0$ .

Proof. Denote the midpoint $M$ of $[BC]$ and $T\in p$ so that $AT\parallel BC$ . Apply the Menelaus' theorem to the transversal $AGM$ in the mentioned triangles :

$\left\{\begin{array}{cccc} \triangle XYC\ : & \frac {\overline{AY}}{\overline {AC}}\cdot \frac {\overline {MC}}{\overline{MX}}\cdot\frac {\overline {GX}}{\overline{GY}}=1 & \implies & \frac {\overline{GX}}{\overline{GY}}=\frac {\overline {AC}}{\overline{AY}}\cdot\frac {\overline {MX}}{\overline{MC}}\\\\ \triangle XBZ\ : & \frac {\overline{AZ}}{\overline {AB}}\cdot \frac {\overline {MB}}{\overline{MX}}\cdot\frac {\overline {GX}}{\overline{GZ}}=1 & \implies & \frac {\overline{GX}}{\overline{GZ}}=\frac {\overline {AB}}{\overline{AZ}}\cdot\frac {\overline {MX}}{\overline{MB}}\end{array}\right|$ $\bigoplus$ $\stackrel{\left(\overline{MB}+\overline {MC}=0\right)}{\implies}\ \frac {\overline{GX}}{\overline{GY}}+$ $\frac {\overline{GX}}{\overline{GZ}}=$ $2\cdot \frac {\overline{MX}}{\overline{BC}}\cdot\left(\frac {\overline {AC}}{\overline {AY}}-\frac {\overline {AB}}{\overline{AZ}}\right)=$

$2\cdot \frac {\overline{MX}}{\overline{BC}}\cdot\left(\frac {\overline {AY}+\overline{YC}}{\overline {AY}}-\frac {\overline {AZ}+\overline{ZB}}{\overline{AZ}}\right)=$ $2\cdot \frac {\overline{MX}}{\overline{BC}}\cdot\left(\frac {\overline {YC}}{\overline {AY}}-\frac {\overline {ZB}}{\overline{AZ}}\right)=$ $2\cdot \frac {\overline{MX}}{\overline{BC}}\cdot\left(\frac {\overline {XC}}{\overline {AT}}-\frac {\overline {XB}}{\overline{AT}}\right)=$ $2\cdot\frac {\overline{MX}}{\overline{BC}}\cdot\frac {\overline{BC}}{\overline {AT}}=$

$2\cdot\frac {\overline{MX}}{\overline{AT}}=$ $2\cdot\frac {\overline{GM}}{\overline{GA}}=-1\implies$ $\boxed{\ \frac{1}{\overline {GX}}+\frac{1}{\overline {GY}}+\frac{1}{\overline{GZ}}=0\ }$ .

Proposed problem. For a line $p$ which passes through the incircle $I$ of $\triangle ABC$ define the intersections

$X\in BC\cap p$ , $Y\in CA\cap p$ and $Z\in AB\cap p$ . Prove that $\frac{a}{\overline {IX}}+\frac{b}{\overline {IY}}+\frac{c}{\overline{IZ}}=0$ .


An easy extension. For a line $p$ which passes through the given point $P(x,y,z)$ , where $(x,y,z)$ are the barycentrical coordinates

w.r.t. $\triangle ABC$ define the intersections $X\in BC\cap p$ , $Y\in CA\cap p$ and $Z\in AB\cap p$ . Prove that $\frac{x}{\overline{PX}}+\frac{y}{\overline {PY}}+\frac{z}{\overline {PZ}}=0$ .

Proof. Denote $M\in AP\cap BC$ and $T\in p$ so that $AT\parallel BC$ . Apply the Menelaus' theorem to the transversal $APM$ in the mentioned triangles :

$\left\{\begin{array}{cccc} \triangle XYC\ : & \frac {\overline{AY}}{\overline {AC}}\cdot \frac {\overline {MC}}{\overline{MX}}\cdot\frac {\overline {PX}}{\overline{PY}}=1 & \implies & \frac {\overline{PX}}{\overline{PY}}=\frac {\overline {AC}}{\overline{AY}}\cdot\frac {\overline {MX}}{\overline{MC}}\\\\ \triangle XBZ\ : & \frac {\overline{AZ}}{\overline {AB}}\cdot \frac {\overline {MB}}{\overline{MX}}\cdot\frac {\overline {PX}}{\overline{PZ}}=1 & \implies & \frac {\overline{PX}}{\overline{PZ}}=\frac {\overline {AB}}{\overline{AZ}}\cdot\frac {\overline {MX}}{\overline{MB}}\end{array}\right|$ $\bigoplus$ $\implies\ \overline {MC}\cdot\frac {\overline{PX}}{\overline{PY}}-$ $\overline{MB}\cdot \frac {\overline{PX}}{\overline{PZ}}=$ $\overline{MX}\cdot\left(\frac {\overline {AC}}{\overline {AY}}-\frac {\overline {AB}}{\overline{AZ}}\right)=$

$\overline{MX}\cdot\left(\frac {\overline {AY}+\overline{YC}}{\overline {AY}}-\frac {\overline {AZ}+\overline{ZB}}{\overline{AZ}}\right)=$ $\overline{MX}\cdot\left(\frac {\overline {YC}}{\overline {AY}}-\frac {\overline {ZB}}{\overline{AZ}}\right)=$ $\overline{MX}\cdot\left(\frac {\overline {XC}}{\overline {AT}}-\frac {\overline {XB}}{\overline{AT}}\right)=$ $\overline{MX}\cdot\frac {\overline{BC}}{\overline {AT}}=$ $\overline{BC}\cdot\frac {\overline{MX}}{\overline{AT}}=$ $\overline{BC}\cdot\frac {\overline{PM}}{\overline{PA}}\implies$

$\overline {MC}\cdot\frac {\overline{PX}}{\overline{PY}}+$ $\overline{BM}\cdot \frac {\overline{PX}}{\overline{PZ}}+\overline{BC}\cdot\frac {\overline{PM}}{\overline{AP}}=0\implies$ $\boxed{\ \frac {\overline {PM}}{\overline{AP}}\cdot \frac{1}{\overline{PX}}+\frac {\overline {MC}}{\overline{BC}}\cdot \frac{1}{\overline {PY}}+\frac {\overline{BM}}{\overline{BC}}\cdot \frac{1}{\overline {PZ}}=0\ }\ (*)$ .

Suppose that $P$ has the barycentrical coordinates $(x,y,z)$ w.r.t. the triangle $ABC$ and denote $\left\{\begin{array}{c} U\in BP\cap CA\\\ V\in CP\cap AB\end{array}\right\|$ . Apply the Menelaus' theorem :
$\left\{\begin{array}{cccc} \overline{BPU}/\triangle AMC\ : & \frac{\overline {BM}}{\overline {BC}}\cdot\frac {\overline {UC}}{\overline {UA}}\cdot\frac {\overline {PA}}{\overline {PM}}=1 & \implies & \frac {\overline {BM}}{\overline {BC}}=\frac zx\cdot\frac {\overline{PM}}{\overline{AP}}\\\\ \overline{CPV}/\triangle AMB\ : & \frac{\overline {CM}}{\overline {CB}}\cdot\frac {\overline {VB}}{\overline {VA}}\cdot\frac {\overline {PA}}{\overline {PM}}=1 & \implies & \frac {\overline {MC}}{\overline {BC}}=\frac yx\cdot\frac {\overline{PM}}{\overline{AP}}\end{array}\right\|$ . Thus, the relation $(*)$ becomes $\frac {x}{\overline {PX}}+\frac {y}{\overline {PY}}+\frac {z}{\overline {PZ}}=0$ . See and here.
This post has been edited 34 times. Last edited by Virgil Nicula, Nov 19, 2015, 4:13 PM

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18. O problema a lui Toshio Seimiya.
17. O inegalitate "tare" intr-un triunghi.
16. Length of the (interior) angle bisector (10 proofs).
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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    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

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

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    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

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    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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  • Joined: Jun 22, 2005
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  • Total entries: 456
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