249. An usual and nice metrical relations in quadrilateral.

by Virgil Nicula, Mar 9, 2011, 10:43 AM

Lemma 1. Let a convex $ABCD$ with $P\in AC\cap BD$ . Then there is the relation $\frac {PA}{PC}=\frac {\cot\widehat{CAB}+\cot\widehat{CAD}}{\cot\widehat{ACB}+\cot\widehat {ACD}}$ .

Proof. Let $\begin{array}{ccc} AB=a & ; & BC=b\\\ CD=c & ; & DA=d\end{array}\ \ \wedge\ \ \begin{array}{ccc} m\left(\widehat{CAB}\right)=x & ; & m\left(\widehat{CAD}\right)=y\\\ m\left(\widehat{ACB}\right)=z & ; & m\left(\widehat{ACD}\right)=t\end{array}$ . Observe that

$\frac {PA}{PC}=\frac {[ABD]}{[CBD]}=\frac ab\cdot\frac dc\cdot\frac {\sin A}{\sin C}=$ $\frac {\sin z}{\sin x}\cdot\frac {\sin t}{\sin y}\cdot\frac {\sin (x+y)}{\sin (z+t)}=$ $\frac {\sin (x+y)}{\sin x\sin y}\cdot\frac {\sin z\sin t}{\sin (z+t)}=$ $\frac {\cot x+\cot y}{\cot z+\cot t}$ .

Lemma 2. Let a convex $ABDC$ with $R\in AC\cap BD$ . Then there is the relation $\frac {RA}{RC}=\left|\frac {\cot\widehat{CAB}-\cot\widehat{CAD}}{\cot\widehat{ACB}-\cot\widehat {ACD}}\right|$ .

Proof. Let $\left|\begin{array}{ccc} AB=a & ; & BC=b\\\\ CD=c & ; & DA=d\end{array}\right|\ \ \wedge\ \ \left|\begin{array}{ccc} m\left(\widehat{CAB}\right)=x & ; & m\left(\widehat{CAD}\right)=y\\\\ m\left(\widehat{ACB}\right)=z & ; & m\left(\widehat{ACD}\right)=t\end{array}\right|$ . Observe that

$\frac {RA}{RC}=\frac {[ABD]}{[CBD]}=\frac ab\cdot\frac dc\cdot\left|\frac {\sin \widehat{BAD}}{\sin \widehat{BCD}}\right|=$ $\left|\frac {\sin z}{\sin x}\cdot\frac {\sin t}{\sin y}\cdot\frac {\sin (x-y)}{\sin (z-t)}\right|=$ $\left|\frac {\sin (x-y)}{\sin x\sin y}\cdot\frac {\sin z\sin t}{\sin (z-t)}\right|=$ $\left|\frac {\cot x-\cot y}{\cot z-\cot t}\right|$ .

Lemma 3. Prove that in any convex $ABCD$ there is the relation $\frac{\sin\widehat{CAB}}{\sin\widehat{CAD}} =\frac{\sin\widehat{CBA}}{\sin\widehat{CDA}}\cdot\frac{\sin\widehat{CDB}}{\sin\widehat{CBD}} \ (*)$ .

Proof. Let $X\in AB$ so that $CX\perp AB$ and $Y\in AD$ so that $CY\perp AD$ . Then from the ratio of the relations $\left\{\begin{array}{c} CB\cdot\sin\widehat{CBA}=CX=CA\cdot\sin\widehat{CAB}\\\\ CD\cdot\sin\widehat{CDA}=CY=CA\cdot\sin\widehat{CAD}\end{array}\right\|$ obtain the relation $(*)$ .


Application. Let $X$, $Y$, $Z$ be three interior points w.r.t. $\triangle ABC$ such that $\widehat{XBC}\equiv \widehat{ABZ}$ , $\widehat{YCA}\equiv\widehat{BCX}$ and $\widehat{ZAB}\equiv\widehat{CAY}$ . Let $X^{\prime}$, $Y^{\prime}$, $Z^{\prime}$ be three

exterior points w.r.t. $\triangle ABC$ such that the sidelines $BC$ , $CA$ , $AB$ separates the pairs of points $\{X,X'\}$ , $\{Y,Y'\}$ , $\{Z,Z'\}$ respectively and $\widehat{X'BC}\equiv \widehat{ABZ'}$ ,

$\widehat{Y'CA}\equiv\widehat{BCX'}$ and $\widehat{Z'AB}\equiv\widehat{CAY'}$ . Denote $K\in X'X\cap BC$ , $L\in Y'Y\cap CA$ , $M\in Z'Z\cap AB$ . Prove that $AK\cap BL\cap CM\ne\emptyset$

(Source : Hans Walser: Ein Schnittpunktsatz, Praxis der Mathematik 2/1991 pp. 70-71).


Proof. Denote $\left\|\begin{array}{c} m\left(\widehat{XBC}\right)=m\left(\widehat{ABZ}\right)=y\\\ m\left(\widehat{YCA}\right)=m\left(\widehat{BCX}\right)=z\\\ m\left(\widehat{ZAB}\right)=m\left(\widehat{CAY}\right)=x\end{array}\right\|\ \ \wedge\ \ \left\|\begin{array}{c} m\left(\widehat{X'BC}\right)=m\left(\widehat{ABZ'}\right)=y'\\\ m\left(\widehat{Y'CA}\right)=m\left(\widehat{BCX'}\right)=z'\\\ m\left(\widehat{Z'AB}\right)=m\left(\widehat{CAY'}\right)=x'\end{array}\right\|$ . Apply upper lemma 1 to the convex

quadrilaterals : $\left\{\begin{array}{ccc} BXCX' & \implies & \frac {KB}{KC}=\frac {\cot y+\cot y'}{\cot z+\cot z'}\\\\ CYAY' & \implies & \frac {LC}{LA}=\frac {\cot z+\cot z'}{\cot x+\cot x'}\\\\ AZBZ' & \implies & \frac {MA}{MB}=\frac {\cot x+\cot x'}{\cot y+\cot y'}\end{array}\right\|\ \bigodot\ \implies$ $\frac {KB}{KC}\cdot\frac {LC}{LA}\cdot\frac {MA}{MB}=1\implies AK\cap BL\cap CM\ne\emptyset$ .

Remark. Pairs of the rays $\{[AY,[AZ\}$ and $\{[AY',[AZ'\}$ are $A$-isogonally. Analogously pairs of the rays $\{[BZ,[BX\}$

and $\{[BZ',[BX'\}$ are $B$-isogonally and pairs of the rays $\{[CX,[CY\}$ and $\{[CX',[CY'\}$ are $C$-isogonally

Particular case. If $X\equiv Y\equiv Z\equiv I$ , where $I$ is incenter $\triangle ABC$ and $BCX'$ , $CAY'$ , $ ABZ'$ are equilateral triangles, then obtain this problem.
This post has been edited 31 times. Last edited by Virgil Nicula, Nov 22, 2015, 12:18 PM

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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

  • this css is sus

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

  • 391345 views moment

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

  • We need virgil nicula to return to aops, this blog is top 10 all time.

    by OlympusHero, Sep 14, 2022, 4:44 AM

  • :omighty: blog

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