120. Two equal neighbouring angles/sides in a quadrilateral.

by Virgil Nicula, Sep 13, 2010, 12:14 AM

$$\bf\color{black}Particular\ quadrilaterals.$$
$ \mathrm{SQUARE}\subset\left\{\begin{array}{c}\mathrm{RECTANGLE}\\\\ \mathrm{RHOMBUS}\end{array}\right\|\subset\mathrm{PARALLELOGRAM}\subset\mathrm{TRAPEZIUM}\ ,$ where $ABCD$ is $:\ \left\{\begin{array}{ccccc} \mathrm{trapezium} & \Longleftrightarrow & AB\parallel CD & \vee & AD\parallel BC\\\\ \mathrm{parallelogram} & \Longleftrightarrow & AB\parallel CD & \wedge & AD\parallel BC\end{array}\right\|\ ;$

$\left\{\begin{array}{cccc} \mathrm{rectangle} & \Longleftrightarrow & \mathrm{parallelogram\ which\ has\ equally} & \mathrm{two\ neighbouring\ angles}\\\\ \mathrm{rhombus} & \Longleftrightarrow & \mathrm{parallelogram\ which\ has\ equally} & \mathrm{two\ neighbouring\ sides}\end{array}\right\|\ ;\ \mathrm{square}\ \Longleftrightarrow\ \mathrm{rectangle}\ \wedge\ \mathrm{rhombus}$ .

A cyclic convex quadrilateral $ ABCD$ $ \iff$ $ A+C=B+D\ .$ A circumscriptible/tangential convex quadrilateral $ ABCD$ $ \Longleftrightarrow$ $ AB+CD=AD+BC\ .$ From all parallelograms, only

rectangle is cyclic. From all parallelograms, only rhombus is circumscriptibly. From all trapeziums, only isosceles trapezium is cyclic. Remarks: direction "angle" $\implies$ cyclic quadrilateral,

isosceles trapezium, rectangle and square $;$ direction "side" $\implies$ circumscriptible quadrilateral, rhombus and square. See two aplications at here. See PP4 & PP5 from here and search here.
$$*****$$
PP1. Let a triangle $ABC$ and the points $ M\in (AC)$, $ N\in (AB)$ for which denote $ P\in BM\cap CN$. Prove that :

$ 1\blacktriangleright$ the quadrilateral $ ANPM$ is circumscriptible $ \Longleftrightarrow$ $ PB-PC=AB-AC$ $ \Longleftrightarrow$ $ NB+NC=MB+MC$.

Proof. Denote the incircle $w$ of $AMPN$ and its tangent points $\left\{\begin{array}{ccc} X\in AB\cap w & ; & Y\in AC\cap w\\\\ Z\in BM\cap w & ; & T\in CN\cap w\end{array}\right\|$ . Thus, $:\ NB+NC=NB+(NT+TC)=$ $(NB+NT)+TC=$

$(NB+NX)+CY=$ $BX+(CM+MY)=BZ+(CM+MZ)=$ $(BZ+MZ)+CM=MB+MC\implies$ $NB+NC=MB+MC$ . Analogously,

$PB-PC=(BZ-ZP)-(CT-PT)=(BZ-PT)-(CY-PT)=BX-CY=(AB-AX)-(AC-AY)=AB-AC\implies$ $PB-PC=AB-AC$ .

$ 2\blacktriangleright$ If for circumscriptible $ ANPM$ denote the tangent points $ E\in AC$, $ F\in AB$, $ U\in CN$, $ V\in BM$,

then the lines $ UF$, $ VE$, $ BC$ are concurrently in a point $ X$ for which $ \frac{XB}{XC}=\frac{c-x}{b-x}$, where $ x=AF=AE$.

$ 3\blacktriangleright$ Exists a circle $ w(O)$ which is tangent to the sidelines of the quadrilateral $ AMPN$ and $ O$ belongs to the exterior

of $ AMPN$ (in this case the quadrilateral $ AMPN$ is (ex)circumscriptible) $ \Longleftrightarrow$ $ AN+NP=AM+MP$.


PP2. Let $ C_{1}(O_{1})$, $ C_{2}(O_{2})$ be two exterior circles and let $ C(O)$ be a circle exterior tangent to the circles $ C_{1}$, $ C_{2}$ in the points $ A_{1}$, $ A_{2}$ respectively.

A exterior common tangent to the circles $ C_{1}$, $ C_{2}$ touches these in the points $ T_{1}$, $ T_{2}$ respectively so that the line $ O_{1}O_{2}$ separates the points $ A_{1}$, $ T_{1}$.

Prove that the quadrilateral $ A_{1}A_{2}T_{2}T_{1}$ is cyclically and the lines $ A_{1}A_{2}$, $ T_{1}T_{2}$, $ O_{1}O_{2}$ are concurrently.


PP3 (Ruben Dario). Let $\triangle ABC$ and $D\in BC$ such that $AD\perp BC$ . Suppose that $D\in (BC)$ . Denote the incircle $w$ of

$\triangle ADC$ , the point $E\in AC$ so that $BE$ is tangent to $w$ and $m\left(\widehat{BEC}\right)=2\alpha$ . Prove that $\tan\alpha +\cot A=\frac {\sin B+\cos B}{\sin A}$ .

Proof. Let $F\in BE\cap AD$ and apply property $(1)$ of PP1 to tangential $CDFE\ :\ \frac {DA+DB}{AB}=\frac {EA+EB}{AB}$ $\iff$ $\sin B+\cos B=\frac {\sin (2\alpha -A)+\sin A}{\sin 2\alpha}=$

$\frac {2\sin\alpha\cos (A-\alpha )}{2\sin \alpha\cos\alpha }=$ $\frac {\cos (A-\alpha )}{\cos\alpha }=$ $\cos A+\sin A\tan\alpha\implies$ $\sin B+\cos B=\cos A+\sin A\tan\alpha$ $\implies$ $\boxed{\tan\alpha +\cot A=\frac {\sin B+\cos B}{\sin A}}\ (*)$ .

Particular case. If $A=90^{\circ}$ (degrees), then the relation $(*)$ becomes $\boxed{\tan\alpha =\cos C+\sin C}$ .


PP4 (Ruben Dario). Let an $A$-right $\triangle ABC$ and $D\in BC$ so that $AD\perp BC$ . Denote the incircles $\left\{\begin{array}{ccc} w_c & \mathrm{of} & \triangle ADC\\\\ w_b & \mathrm{of} & \triangle ADB\end{array}\right\|$ , the point

$E\in AC$ so that $BE$ is tangent to $w_c$ , the point $F\in AB$ so that $CF$ is tangent to $w_b$ and $\left\{\begin{array}{c} m\left(\widehat{ABE}\right)=x\\\ m\left(\widehat{ACF}\right)=y\end{array}\right\|$ . Prove that $x=y$ .

Proof. Let $\left\{\begin{array}{ccc} X\in BE\cap AD & ; & m\left(\widehat{BEC}\right)=2\alpha\\\\ Y\in CF\cap AD & ; & m\left(\widehat{BFC}\right)=2\beta\end{array}\right\|$ . Apply the particular case of the previous problem to the tangential quadrilaterals $:$

$\left\{\begin{array}{cc} CDXE\ : & \tan\alpha =\cos C+\sin C\\\\ BDYF\ : & \tan\beta =\cos B+\sin B\end{array}\right\|\ \stackrel{A=90^{\circ}}{\implies}\ \tan\alpha =$ $\tan\beta\implies \alpha =\beta$ . In conclusion, $x=y$ because $\left\{\begin{array}{ccc} x+90^{\circ}=2\alpha\\\ y+90^{\circ}=2\beta\end{array}\right\|$ .


Some properties of the circumscribed quadrilateral.

Let $ABCD$ be a convex quadrilateral which is circumscribed to the circle $w=C(I,r)$. Denote $a=AB$ , $b=BC$ , $c=CD$ , $d=DA$ ,

$e=AC$ , $f=BD$ , the midpoints $M$, $N$ of $AC$, $BD$ respectively, $E\in AC\cap BD$ , $F\in AB\cap CD$ , $G\in AD\cap BC$ , the tangentpoints

$X\in AB$ , $Y\in BC$ , $Z\in CD$ , $T\in DA$ with incircle $w$ , $U\in XT\cap YZ$, $V\in XY\cap ZT$ and $S=[ABCD]$ , $2p=a+b+c+d$. Then:

$1.$ The quadrilateral $ABCD$ is circumscribed $\Longleftrightarrow a+c=b+d$ (Pithot).

$2.\ E\in XZ\cap YT\ ;\ I\in MN$ (Newton's line).

$3.1.\ U\in BD\ \wedge\ (D,E,B,U)$ is a harmonic division.

$3.2.\ V\in AC\ \wedge\ (A,E,C,V)$ is a harmonic division.

$4.$ The quadrilateral $IEUV$ is orthocentric, i.e. $UI\perp EV$ and $VI\perp EU$ .


$5.$ If the quadrilateral $ABCD$ is circumscribed and inscribed in the circle $e=C(O,R)$ , then:

$5.1.\ ef=2r\left(r+\sqrt{4R^{2}+r^{2}}\right)\ ;\ OI^{2}=R^{2}+r^{2}-r\sqrt{4R^{2}+r^{2}}$ (Durrande's relation).

$5.2.\ \frac{IM}{IN}=\frac{e}{f}\ ;\ \frac{1}{e}+\frac{1}{f}=\frac{p}{4Rr}$ .

$5.3.$ The power of the point $E$ w.r.t the circle $e$ is $p_{e}(E)=\left(\frac{ef}{4R}\right)^{2}$ .

$5.4.$ The power of the incenter $I$ w.r.t. the circumcircle $e$ is $p_{e}(I)=\frac{8R^{2}r^{2}}{ef}$ .

Remark. Maybe now the problem from here seems more easily.
This post has been edited 76 times. Last edited by Virgil Nicula, Oct 18, 2016, 9:08 AM

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65. Parallel to BC through I .
64. Problems of the type "instant" (at most one minute).
63. A nice problem with radical center (livetolove212).
62. Nice and easy problem with a right triangle.
61. IMO Shortlist 2009 (very nice problem and its proof).
60. Mediteranean M.C. 2010 Problem 3.
59. IMO 2010, problem 4.
58. A cevian and two incircles.
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!

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    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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