185. The Durrande's relation.

by Virgil Nicula, Dec 8, 2010, 2:36 PM

Quote:
Let $ABCD$ be a convex quadrilateral with the circumcircle $w=C(O,R)$ and the incircle $C(I,r)$ . Denote

$\left\|\begin{array}{c} AB=b\ ,\ BC=b\ ,\ CD=c\\\ DA=d\ ,\ AC=e\ ,\ BD=f\end{array}\right\|$ and $a+b+c+d=s\ ,\ S=[ABCD]$ (area). Prove that :

$1.\blacktriangleright\ \ S=sr=\sqrt {abcd}\ ,\ ef=2r\left(r+\sqrt {4R^2+r^2}\right)$ .

$2.\blacktriangleright\ \boxed{\ OI^2=R^2+r^2-r\sqrt {4R^2+r^2}\ }$ (Durrande's relation).

$3.\blacktriangleright$ If $M\ ,\ N$ are the midpoints of $[AC]\ ,\ [BD]$ respectively, then $I\in MN\ ,\ \frac {IM}{IN}=\frac ef$ and $\frac 1e+\frac 1f=\frac {s}{4Rr}$ .

Proof. Denote $P\in AC\cap BD$ . I"ll use the well-known relations $a+c=b+d$ (Pithot's relation), $ac+bd=ef$ (Ptolemy's relation) and

$\frac {PA}{da}=\frac {PB}{ab}=$ $\frac {PC}{bc}=\frac {PD}{cd}=$ $\frac {e}{ad+bc}=\frac {f}{ab+cd}=$ $\frac {e+f}{s^2}=\sqrt {\frac {ef}{(ad+bc)(ab+cd)}}=$ $\sqrt {\frac {-p_w(P)}{abcd}}$ , where $p_w(P)=PO^2-R^2$ (power).

Obtain easily that $S=sr=\sqrt {abcd}\ ,\ 4RS=\sqrt {(ab+cd)(ac+bd)(ad+bc)}$ and $4Rr(e+f)=sef$ , i.e. $\frac 1e+\frac 1f=\frac {s}{4Rr}$ . It is well-known

or prove easily that $I\in MN$ . Therefore, $IM\cdot [NBC]+IN\cdot [MBC]=MN\cdot [IBC]$ $\implies$ $IM\cdot [BCD]+IN\cdot [ABC]=MN\cdot br$ $\implies$

$IM\cdot bc\cdot \sin C+IN\cdot ab\cdot\sin B=2\cdot MN\cdot br$ $\implies$ $IM\cdot c\cdot\sin C+IN\cdot a\cdot\sin B=2\cdot MN\cdot r$ $\implies$

$IM\cdot c\cdot \frac {2S}{ad+bc}+IN\cdot a\cdot\frac {2S}{ab+cd}=2\cdot MN\cdot \frac Ss$ $\implies$ $IM\cdot \frac {c}{ad+bc}+IN\cdot\frac {a}{ab+cd}=MN\cdot\frac 1s$ $\implies$

$IM\cdot \left(\frac {c}{ad+bc}-\frac 1s\right)=IN\cdot\left(\frac 1s-\frac {a}{ab+cd}\right)$ $\implies$ $IM\cdot (ab+cd)=IN\cdot (ad+bc)$ $\implies$ $\boxed {\ \frac {IM}{IN}=\frac ef\ }$ , i.e. $\frac {IM}{e}=\frac {IN}{f}=\frac {MN}{e+f}$ .

$\blacktriangleright$ Apply the Stewart's relation in $\triangle MON$ for the cevian $IO\ :\ OM^2\cdot IN+ON^2\cdot IM=OI^2\cdot MN+IM\cdot IN\cdot MN$ . Using the relations

$OM^2=R^2-\frac 14\cdot e^2$ and $ON^2=R^2-\frac 14\cdot f^2$ obtain that $\left(R^2-\frac 14\cdot e^2\right)\cdot\frac {f}{e+f}+\left(R^2-\frac 14\cdot f^2\right)\cdot \frac {e}{e+f}=OI^2+IM\cdot IN$ $\implies$ $R^2-\frac 14\cdot ef=OI^2+IM\cdot IN$ $\implies$

$4\cdot \left(R^2-OI^2\right)=ef+4ef\cdot \left(\frac {MN}{e+f}\right)^2$ . Using the Euler's relation $4\cdot MN^2=a^2+b^2+c^2+d^2-e^2-f^2$ obtain that $4\cdot MN^2=(a+c)^2+(b+d)^2-(e+f)^2$ $\implies$

$4\cdot MN^2=2s^2-(e+f)^2$ . In conclusion, $4\left(R^2-OI^2\right)=ef+2s^2\cdot\frac {ef}{(e+f)^2}-ef$ , i.e. $R^2-OI^2=\frac {s^2ef}{2(e+f)^2}$ . Thus, the power $\rho_w(I)=\frac {s^2ef}{2(e+f)^2}$ But

$e+f=2R\cdot (\sin A+\sin B)=2R\left(\frac {2S}{ad+bc}+\frac {2S}{ab+cd}\right)=$ $\frac {4RSs^2(e+f)^2}{s^4ef}=$ $\frac {4Rr(e+f)^2}{sef}$ . So $e+f=\frac {sef}{4Rr}$ , i.e. $\frac 1e+\frac 1f=\frac {s}{4Rr}$ . Thus, $R^2-OI^2=\frac {s^2ef}{2(e+f)^2}=$

$\frac {8R^2r^2s^2ef}{s^2e^2f^2}=$ $\frac {8R^2r^2}{ef}$ . From relation $\frac {ef}{2r}=\sqrt {4R^2+r^2}+r$ obtain $R^2-OI^2=\frac {4R^2r}{r+\sqrt {4R^2+r^2}}=$ $r\left(\sqrt {4R^2+r^2}-r\right)$ . Thus, $OI^2=R^2+r^2-r\sqrt {4R^2+r^2}$ .
This post has been edited 29 times. Last edited by Virgil Nicula, Nov 26, 2015, 6:43 PM

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What does "I'll come back soon" mean? Will you be busy these days, taking a vacation from the blog for the holidays? No more math blog posts for a period of time? :(

by Thomas_, Dec 8, 2010, 6:44 PM

Own problems or extensions/generalizations of some problems which was posted here.

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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

  • 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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About Owner
  • Posts: 7054
  • Joined: Jun 22, 2005
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  • Blog created: Apr 20, 2010
  • Total entries: 456
  • Total visits: 404395
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