310. Some properties of the tangential quadrilateral.

by Virgil Nicula, Aug 16, 2011, 1:25 PM

PP1. Let convex $ABCD$ with inscribed circle $w$ so that $\left\{\begin{array}{ccccc} E\in AB\cap w & ; & F\in BC\cap w & ; & Y\in HF\cap EG\\\\ G\in CD\cap w & ; & H\in DA\cap w & ; & X\in EF\cap HG\end{array}\right\|$ . Prove that $Y\in AC\cap BD$ , $X\in AC$ and $\frac {YA}{YC}=\frac {XA}{XC}$ .

Proof. Let $\left\{\begin{array}{c} Y_1=EG\cap AC\\\\ Y_2\in FH\cap AC\end{array}\right|$ . Apply the Sinus' theorem $:\ \left\{\begin{array}{ccc} \triangle Y_1CG\ : & \frac {Y_1C}{\sin \widehat{Y_1GC}}=\frac {GC}{\sin\widehat{GY_1C}}\\\\ \triangle Y_1AE\ : & \frac {Y_1A}{\sin \widehat{Y_1EA}}=\frac {EA}{\sin\widehat{EY_1A}}\end{array}\right|$ . Since $\left\{\begin{array}{c} \widehat{GY_1C}\equiv\widehat{EY_1A}\\\\ m\left(\widehat{Y_1GC}\right)+m\left(\widehat{Y_1EA}\right)=180^{\circ}\end{array}\right|$ obtain that $\frac {Y_1A}{Y_1C}=\frac {EA}{GC}$ .

Apply the Sines' theorem $:\ \left\{\begin{array}{ccc} \triangle Y_2CF\ : & \frac {Y_2C}{\sin \widehat{Y_2FC}}=\frac {FC}{\sin\widehat{FY_2C}}\\\\ \triangle Y_2AH\ : & \frac {Y_2A}{\sin \widehat{Y_2HA}}=\frac {HA}{\sin\widehat{HY_2A}}\end{array}\right|$ . Since $\left\{\begin{array}{c} \widehat{FY_2C}\equiv\widehat{HY_2A}\\\\ m\left(\widehat{Y_2FC}\right)+m\left(\widehat{Y_2HA}\right)=180^{\circ}\end{array}\right|$ obtain that $\frac {Y_2A}{Y_2C}=\frac {HA}{FC}$ . Since $\left\{\begin{array}{c} CG=CF\\\\ AE=AH\end{array}\right|$ obtain that

$\frac {Y_1A}{Y_1C}=\frac {EA}{GC}=$ $\frac {HA}{FC}=\frac {Y_2A}{Y_2C}$ , i.e. $\frac {Y_1A}{Y_1C}=\frac {Y_2A}{Y_2C}$ $\implies$ $Y_1\equiv Y_2\equiv Y$ and $\frac {YA}{YC}=\frac {HA}{FC}$ . Hence the diagonal $AC$ pass through $Y\in GE\cap FH$ . Prove analogously that and the

diagonal $BD$ pass through same $Y$ . Hence $Y\in AC\cap BD\cap GE\cap FH$ (the Newton's point). Let $X_1\in AC\cap HG$ and $X_2\in AC\cap EF$ . Apply the Menelaus' theorem to

$:\ \left\{\begin{array}{cccc} \overline {X_1GH}/\triangle ADC\ : & \frac {X_1C}{X_1A}\cdot\frac {HA}{HD}\cdot\frac {GD}{GC}=1 & \implies & \frac {X_1A}{X_1C}=\frac {HA}{GC}\\\\ \overline {X_2FE}/\triangle ABC\ : & \frac {X_2C}{X_2A}\cdot\frac {EA}{EB}\cdot\frac {FB}{FC}=1 & \implies & \frac {X_2A}{X_2C}=\frac {EA}{FC}\end{array}\right|$ . Since $\left\{\begin{array}{c} CG=CF\\\\ AE=AH\end{array}\right|$ obtain that $\frac {X_1A}{X_1C}=\frac {HA}{GC}=$ $\frac {EA}{FC}=\frac {X_2A}{X_2C}$ , i.e. $\frac {X_1A}{X_1C}=\frac {X_2A}{X_2C}$ $\implies$

$X_1\equiv X_2\equiv X$ and $\frac {XA}{XC}=\frac {HA}{GC}$ , i.e. $HG\cap EF\cap AC\ne\emptyset$ . Hence $\frac {XA}{XC}=\frac {YA}{YC}$ . See here.


PP2 (Generalization). Let the incircle $w=C(I,r)$ of the cyclical convex quadrilateral $ABCD$ and $\left|\begin{array}{cc} M\in w\cap AB\ ; & N\in w\cap BC\\\\ P\in w\cap CD\ ; & Q\in w\cap DA\end{array}\right|$ . For $S\in \overarc[]{MQ}$ denote

$\left\{\begin{array}{ccc} X\in AB\ ,\ SX\perp AB & ; & SX=x\\\\ Y\in BC\ ,\ SY\perp BC & ; & SY=y\end{array}\right\|$ and $\left\{\begin{array}{ccc} Z\in CD\ ,\ SZ\perp CD & ; & SZ=z\\\\ T\in DA\ , \ ST\perp DA & ; & ST=t\end{array}\right\|$ . Prove that $\boxed{\sqrt{yz}\cos\frac A2=\sqrt {zt}\cos\frac B2+\sqrt{xt}\cos \frac C2+\sqrt {xy}\cos\frac D2}$ ..

Proof. Let $\left\{\begin{array}{c} U\in AB\cap CD\\\\ V\in AD\cap BC\end{array}\right\|$ . Thus, $\left\{\begin{array}{cccc} MN=2r\cos\frac B2\ ; & NP=2r\cos \frac C2\ ; & PQ=2r\cos \frac D2\\\\ QM=2r\cos\frac A2\ ; & MP=2r\cos\frac U2\ ; & NQ=2r\cos\frac V2\end{array}\right\|\implies$ $\frac {MN}{\cos\frac B2}=\frac {NP}{\cos\frac C2}=\frac {PQ}{\cos\frac D2}=\frac {QM}{\cos \frac A2}=\frac {MP}{\cos \frac U2}=\frac {NQ}{\cos\frac V2}\ (1)$ .

Apply upper lemma to the trapezoids $:\ \left\{\begin{array}{cc} SXMI\ : & SM^2=2xr\\\\ SYNI\ : & SN^2=2ry\\\\ SZPI\ : & SP^2=2rz\\\\ STQI\ : & SQ^2=2rt\end{array}\right\|\implies$ $\frac {SM}{\sqrt x}=\frac {SN}{\sqrt y}=\frac {SP}{\sqrt z}=\frac {SQ}{\sqrt t}=\sqrt {2r}\ (2)$ . I"ll use the proportions $(1)$ and $(2)$ after the application of the

Ptolemy's theorem to $:$ $\left\{\begin{array}{ccccc} SMNP\ : & SN\cdot MP=SM\cdot NP+SP\cdot MN & \implies & \sqrt y\cos \frac U2=\sqrt x\cos\frac C2+\sqrt z\cos\frac B2 & (3)\\\\ SMNQ\ : & SN\cdot MQ=SM\cdot NQ+SQ\cdot MN & \implies & \sqrt y\cos \frac A2=\sqrt x\cos\frac V2+\sqrt t\cos\frac B2 & (4)\\\\ SMPQ\ : & SP\cdot MQ=SM\cdot PQ+SQ\cdot MP & \implies & \sqrt z\cos\frac A2=\sqrt x\cos\frac D2+\sqrt t\cos \frac U2 & (5)\\\\ SNPQ\ : & SP\cdot QN=SQ\cdot NP+SN\cdot PQ & \implies & \sqrt z\cos \frac V2=\sqrt t\cos\frac C2+\sqrt y\cos\frac D2 & (6)\end{array}\right\|$ . Eliminate $\cos\frac U2$ between $(3)$ , $(5)$ and

eliminate $\cos\frac V2$ between $(4)$ , $(6) \ :\ \left\{\begin{array}{c} \sqrt{yz}\cos\frac A2=\sqrt {xy}\cos\frac D2+\sqrt t\left(\sqrt x\cos\frac C2+\sqrt z\cos\frac B2\right)\\\\ \sqrt {yz}\cos \frac A2=\sqrt x\left(\sqrt t\cos\frac C2+\sqrt y\cos\frac D2\right)+\sqrt {tz}\cos\frac B2\end{array}\right\|\implies$ $\boxed{\ \sqrt{yz}\cos\frac A2=\sqrt {zt}\cos\frac B2+\sqrt{xt}\cos \frac C2+\sqrt {xy}\cos\frac D2\ }\ (*)$ .

If eliminate $\cos\frac A2$ between $(4)$ , $(5)$ or eliminate $\cos\frac C2$ between $(3)$ , $(6)$ , then obtain same relation $:\ \boxed{\ \sqrt{yt}\cos\frac U2-\sqrt {xz}\cos\frac V2=\sqrt{zt}\cos \frac B2-\sqrt {xy}\cos\frac D2\ }$ .

If eliminate $\cos\frac B2$ between $(3)$ , $(4)$ or eliminate $\cos\frac D2$ between $(5)$ , $(6)$ , then obtain same relation $:\ \boxed{\ \sqrt{yt}\cos\frac U2+\sqrt {xz}\cos\frac V2=\sqrt{xt}\cos \frac C2+\sqrt {yz}\cos\frac A2\ }$ .

Remark. If $D\in (CA)$ , then $z=t$ and the relation $(*)$ becomes the proposed problem PP7 from here.[/size]
This post has been edited 31 times. Last edited by Virgil Nicula, Nov 20, 2015, 10:19 AM

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57. Opinii si comentarii relativ la inv. rom. II
56. A extremum problem with areas.
55. Two important circles - mixtilinear incircle/excircle.
54. Cinci grade de libertate.
53. An remarkable concurrency.
52. Harmonical division.
51*1. JBTST-3/2010 Aplicatii la proprietatea fluturelui.
51. A nice problem with perpendicularity from Jaime.
50. IMAC 2010 seniors, the first day.
49. The midpoint of a cevian in an acute triangle.
+ June 2010
48. An inequality in a triangle : h_a+max{b,c} ...
47. Interesting and difficult identities in a triangle.
46. Outside of a quadrilateral.
45. \cos B+\cos C=1 <==> nice property.
44. Relatia IA=IO sau IA=IM intr-un triunghi ABC etc.
43. A "slicing" problem with a parameter.
42. IMAC 2010 Problema 2.
41. ONM 2010 Calarasi Problema 3.
+ May 2010
40. Lema lui Haruki.
39. A nice and remarkable problem with Brocard's point.
38. Minimum area of triangle touching with semicircle.
37. Own extension of the Steiner-Lehmus' problem.
36. ONM (juniori) - 2010 Iasi, problema 4.
35. Nice problem from Arqady (Eduard_Tur).
34. IMO Shortlist 2002 geometry problem 7.
+ April 2010
32. Linkuri diverse.
31.Some nice metrical problems.
29. Inegalitate integrala de la Kunny.
28. A fost odată ...
27.Problem of Darij Grinberg (I - centroid of triangle DEF).
26. Outside of a triangle.
25. Geometric equations.
24. A fine and cool inequalities.
23. A very nice exponential inequality (own - from CRUX).
22. Some interesting limits (sequence/function).
21 bis. Some problems with complex numbers.
21. Probleme de geometrie (Yetty & I).
20. O ineg.cu suma de AG/AI .
19. Own proposed problems in CRUX Mathematicorum - Canada.
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 MrMustache, Nov 11, 2020, 4:52 PM

  • 372554 views!

    by mrmath0720, Sep 28, 2020, 1:11 AM

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    -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

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    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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