359. Ptolemy's theorem and generalized Ptolemy's relation.

by Virgil Nicula, Oct 23, 2012, 9:07 AM

Generalized Ptolemy's relation. Let a convex $ABCD$ . Prove that $\boxed{AC^2\cdot BD^2=AB^2\cdot CD^2+AD^2\cdot BC^2-2\cdot AB\cdot BC\cdot CD\cdot DA\cdot\cos (B+D)}\ (*)$ .

Proof. Let $\left\{\begin{array}{ccc} AB=a & ; & BC=b\\\\ CD=c & ; & DA=d\\\\ AC=e & ; & BD=f\end{array}\right\|$ . Construct outside of $ABCD$ the point $E$ so that $\triangle CDE\sim\triangle CBA$ , i.e. $\left\{\begin{array}{c} \widehat{CDE}\equiv\widehat{CBA}\\\\ \widehat{DCE}\equiv\widehat{BCA}\end{array}\right\|$ . Thus, $\frac {CD}{CB}=\frac {DE}{BA}=\frac {EC}{AC}\iff$

$\frac cb=\frac {DE}{a}=\frac {EC}{e}\implies$ $\boxed{DE=\frac {ac}{b}}$ and $EC=\frac {ce}{b}$ . Therefore, $\left\{\begin{array}{ccc} \widehat{BCA}\equiv\widehat{DCE} & \implies & \widehat{BCD}\equiv\widehat{ACE}\\\\ \frac {CD}{CB}=\frac {EC}{AC} & \implies & \frac {BC}{AC}=\frac {CD}{CE}\end{array}\right\|$ $\implies$ $\triangle BCD\sim\triangle ACE\implies$$\frac {BC}{AC}=\frac {CD}{CE}=\frac{DB}{EA}\implies$

$\frac be=\frac {c}{CE}=\frac {f}{EA}\implies$ $\boxed{AE=\frac {ef}{b}}$ and $EC=\frac {ce}{b}$ . Apply the Pythagoras' theorem in $\triangle ADE\ :\ AE^2=AD^2+DE^2-2\cdot AD\cdot DE\cdot\cos \widehat{ADE}\iff$

$\frac {e^2f^2}{b^2}=d^2+\frac {a^2c^2}{b^2}-2\cdot\frac {acd}{b}\cdot\cos \left|\begin{array}{c} B+D\\\ \mathrm{or}\\\ 360^{\circ}-(B+D)\end{array}\right|\iff$ $e^2f^2=b^2d^2+a^2c^2-2abcd\cos (B+D)$ .

Remark. Here are two remarkable particular cases :

$\blacktriangleright\ A+C=B+D$ , i.e. $ABCD$ is a cyclical quadrilateral $\implies \boxed{ef=ac+bd}$ , i.e. the Ptolemy's relation.

$\blacktriangleright\ B+D\in\left\{90^{\circ},270^{\circ}\right\}$ . In this case there is the relation $\boxed{e^2f^2=a^2c^2+b^2d^2}$ .


PP1. Let a convex cyclical $ABCDE$ with $\left\{\begin{array}{c} AB=BC=CD=a\ ;\ DE=EA=b\ ;\ b\ge a\sqrt 2\\\\ AD=z\ ;\ EB=EC=x\ ;\ AC=BD=y\end{array}\right\|$ . Determine $(x,y,z)=F(a,b,c)$ . Particular case : $a=2\ ,\ b=3$ .

Proof (particular case). Exists $0<x<\frac {\pi}{8}$ so that $m(<BAD)=4x$ and $m(<DAE)=90^{\circ}-3x$ . Thus, $\boxed{AD=2(1+2\cos 4x)=6\sin 3x}\ (*)\ \implies$

$1+$ $2(1-2\sin^22x)=3\sin x\left(3-4\sin^2x\right)\implies$ $3-16\sin^2x\left(1-\sin^2x\right)=9\sin x-12\sin^3x$ . Denote $\sin x=t$ . Therefore,

$16t^4+12t^3-16t^2-9t+3=0\iff$ $t\in \left\{-1 , \frac 14 , \frac{\sqrt 3}2 , -\frac {\sqrt 3}2\right\}$ . Since $0<x<\frac {\pi}8$ obtain only $\boxed{\sin x=\frac 14}$ . In conclusion, from the relation $(*)$ obtain that

$AD=6\sin 3x=6\sin x\left(3-4\sin^2x\right)=\frac 32\cdot\left(3-\frac 4{16}\right)=\frac {33}{8}$ .

Proof (general case). Apply the Ptolemy's theorem to $\left\{\begin{array}{ccc} ABCD\ : & \boxed{y^2=a^2+az} & (1)\\\\ ACDE\ : & \boxed{xz=ab+by} & (2)\end{array}\right\|$ . Is well-known $(*)$ the relation in $ACDE\ :\ \frac {AD}{CE}=\frac {AC\cdot AE+DE\cdot DC}{CA\cdot CD+EA\cdot ED}\implies$

$\frac zx=\frac {ab+by}{b^2+ay}\ (3)$ . Therefore, $(2)\ \wedge\ (3)\implies$ $\boxed{z^2=\frac {b^2(a+y)^2}{b^2+ay}}\ (4)\implies$ $a^2z^2=\frac {a^2b^2(a+y)^2}{b^2+ay}\stackrel{(1)}{\implies}$ $\left(y^2-a^2\right)^2=\frac {a^2b^2(a+y)^2}{b^2+ay}\implies$ $\left(y-a\right)^2=\frac {a^2b^2}{b^2+ay}\implies$

$\left(ay+b^2\right)(y-a)^2=a^2b^2\iff$ $ay^2+\left(b^2-2a^2\right)y+a\left(a^2-2b^2\right)=0\iff$ $\boxed{y=\frac {2a^2-b^2+b\sqrt{b^2+4a^2}}{2a}}$ . Prove easily that $a+y=\frac {4a^2-b^2+\sqrt{b^2+4a^2}}{2a}$ and

$x^2=b^2+ay=\frac {2a^2+b^2+b\sqrt{b^2+4a^2}}{2}\implies$ $x=\sqrt{b^2+ay}=\frac {b+\sqrt{b^2+4a^2}}{2}\implies$ $\boxed{x=\frac {b+\sqrt{b^2+4a^2}}{2}}$ and from $(4)$ obtain $z=\frac {2b(a+y)}{b+\sqrt{b^2+4a^2}}\implies$

$\boxed{z=\frac ba\cdot\frac {4a^2-b^2+b\sqrt{b^2+4a^2}}{b+\sqrt{b^2+4a^2}}}$ . Particular case. $\left\{\begin{array}{c} a=2\\\ b=3\end{array}\right\|\implies$ $\left\{\begin{array}{c} x=4\\\\ y=\frac 72\\\\ z=\frac {33}{8}\end{array}\right\|$ .

Remark. Let cyclic $ABCD\ ,\ \left\{\begin{array}{ccc} AB=a & ; & BC=b\\\\ CD=c & ; & DA=d\\\\ AC=e & ; & BD=f\end{array}\right\|$ and $I\in AC\cap BD$ . Prove easily that

$\frac {IA}{da}=\frac {IB}{ab}=\frac {IC}{bc}=\frac {ID}{cd}=$ $\frac {e}{da+bc}\stackrel{(*)}{=}\frac {f}{ab+cd}=\sqrt {\frac {-p_w(I)}{abcd}}$ and $p_w(I)$ - power of $I$ w.r.t. $w=C(O,R)$ .


PP2. Let $ABCDEFGHI$ be a regular nonagon with $X\in AD\cap CE$ and $Y\in EH\cap GI$ . Find $m\left(\widehat{EXY}\right)$ .

Proof. Suppose w.l.o.g. $2R=1$ . Denote $x=m\left(\widehat{EXY}\right)$ and $\left\{\begin{array}{ccc} AB=a=\sin 20^{\circ} & ; & AC=b=\sin 40^{\circ}\\\\ AD=c=\sin 60^{\circ} & ; & AE=d=\sin 80^{\circ}\end{array}\right\|$ . Therefore,

$\left\{\begin{array}{ccccc} ACDE\ : & \frac {XE}{d}=\frac {XC}{b}=\frac {CE}{b+d}=\frac {b}{b+d} & \implies & XE=\frac {bd}{b+d}\\\\ EGHI\ : & \frac {YE}{bd}=\frac {YH}{a^2}=\frac {HE}{bd+a^2}=\frac {c}{a^2+bd} & \implies & YE=\frac {bcd}{a^2+bd}\end{array}\right\|\implies$ $\frac {EX}{EY}=\frac {\frac {bd}{b+d}}{\frac {bcd}{a^2+bd}}\implies$ $\frac {EX}{EY}=\frac {a^2+bd}{c(b+d)}\implies$ $\frac {EX}{EY}=\frac {\sin^220^{\circ}+\sin40^{\circ}\sin80^{\circ}}{\sin 60^{\circ}(\sin 40^{\circ}+\sin 80^{\circ})}=$

$\frac {1-\cos40^{\circ}+\cos 40^{\circ}-\cos 120^{\circ}}{2\sqrt 3\sin 60^{\circ}\cos 20^{\circ}}\implies$ $\boxed{\frac {EX}{EY}=\frac 1{2\cos 20^{\circ}}}\ (1)$ . Theorem of Sines $\triangle XEY\ :\ \frac {EX}{EY}=\frac {\sin\widehat{EYX}}{\sin\widehat{EXY}}=$ $\frac {\sin (100^{\circ}-x)}{\sin x}\implies$ $\boxed{\frac {EX}{EY}=\frac {\cos (10^{\circ}-x)}{\sin x}}\ (2)$ .

In conclusion, $\frac {\cos (10^{\circ}-x)}{\sin x}= \frac 1{2\cos 20^{\circ}}\implies$ $2\cos 20^{\circ}\cos (10^{\circ}-x)=\sin x\implies$ $\cos (x+10^{\circ})+\cos (x-30^{\circ})=\cos(90^{\circ}-x)=$ $\cos(90^{\circ}-x)-\cos (x-30^{\circ})\implies$

$\cos (x+10^{\circ})=2\sin 30\sin (x-60^{\circ})\implies$ $\cos (x+10^{\circ})=\cos (150^{\circ}-x)\implies$ $x+10^{\circ}=150^{\circ}-x\implies$ $\boxed{x=70^{\circ}}$ .
This post has been edited 41 times. Last edited by Virgil Nicula, Nov 16, 2015, 2:00 PM

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Thanks a ton this is very nice!

by Mikazoid, Oct 26, 2012, 11:04 PM

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

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Virgil Nicula
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82. AA' , BN , CM concur in Gergonne point.
81. Sawayama and Thebault’s theorem.
80. Three concurrent perpendiculars (the Carnot's lemma).
79. A quadrilateral with many perpendicularities.
78. Jakobi's theorem.
77. Two equivalent perpendicularities.
76. A parallelogram and a circle (nice !).
75. Extended Steinbart Theorem.
74. Inegalitatea Erdos-Mordell.
73. A nice problems with a square and a perpendicularity.
+ July 2010
72. Colinearity in a triangle - H , P , M .
70. Colinearity with pole/polar - H\in PQ .
71. An angle questions (synthetic/trigonometric proofs).
69*1. Slicing problem 3.
69. Position of x1, x2 w.r.t. a given real number k.
68. Some nice applications of "pole/polar".
67. Remarkable algebraic/geometrical inequalities (1).
66. Remarkable perpendicularities in a triangle.
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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  • Total entries: 456
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