213. Another classical "slicing" problems.

by Virgil Nicula, Jan 24, 2011, 10:23 AM

PP1. In $\triangle ABC$ with $A = 30^{\circ}$ and $B = 80^{\circ}$ consider $M$ so that $m(\angle MAC)= 10^{\circ}$ and $m(\angle MCA) = 30^{\circ}$ . Find $m\left(\widehat{BMC}\right)$ .

Proof 1 (trigonometric). Denote $x=m(\angle MBC)$ and apply the trigonometric form of the Ceva's theorem to the point $M$ in the triangle $ABC\ :$

$\sin\widehat{MCA}\cdot\sin\widehat{MAB}\cdot\sin\widehat {MBC}=\sin\widehat{MAC}\cdot\sin\widehat{MBA}\cdot\sin\widehat {MCB}=1\iff$ $\sin 30^{\circ}\sin 20^{\circ}\sin x=\sin 40^{\circ}\sin 10^{\circ}\sin (80^{\circ}-x)\iff$

$\sin 30^{\circ}\sin x=2\cos 20^{\circ}\sin 10^{\circ}\cos(10^{\circ}+x)\iff$ $\sin x=2(\sin 30^{\circ}-\sin 10^{\circ})\cos (10^{\circ}+x)\iff$ $\sin x=\cos (10^{\circ}+x)-2\sin 10^{\circ}\cos (10^{\circ}+x)\iff$

$\sin x=\cos (10^{\circ}+x)-\sin (20^{\circ}+x)+\sin x\iff$ $\cos (10^{\circ}+x)=\cos (70^{\circ}-x)\iff$ $10^{\circ}+x=70^{\circ}-x\iff$ $x=30^{\circ}\iff$ $m(\angle BMC)=110^{\circ}$ .

Proof 2. Take $D\in (BC$ so that $m\left(\widehat{ADB}\right)=60^\circ$, circumcenter $O$ of $\triangle ABD$, $Q$ - midpoint of arc $\stackrel {\frown}{BCA}$ , $P\in BO \cap AD$; $O$ is on $AC$ . Because $C$ , $P$ are on perpendicular bisector

of $BQ$ , $AQ$ respectively ($\triangle ABQ$ is equilateral) and $m\left(\widehat{DAQ}\right)=$ $m\left(\widehat{DBQ}\right)=20^\circ $ $\implies$ $ BC=CQ=PQ=AP$ and $m\left(\widehat{CQP}\right)=60^\circ$ . Hence $PC=CQ$ . But

$PC=AP \implies $ $\angle APC=10^\circ$ , $\angle BCP=80^\circ$ , $\triangle BCP$ is isosceles; $\angle ACM=30^\circ \implies \angle BCM = \frac{\angle BCP}{2}=40^\circ$ . Hence $M$ is on perpendicular bisector of $BP$ . As $AM$

is bisector of $\angle DAB=\angle PAB$ $\implies$ $ PMBA$ cyclic and $\angle BMP=180^\circ-\angle PAB=140^\circ$ $ \implies \angle CMB=\angle CMP =\frac{360^\circ-140^\circ}{2}=110^\circ$ .

Proof 3. Let $D\in AB\cap CM$ and $\left\{\begin{array}{cc} E\in (AM)\ ; & DE=DM\\\ F\in (AC)\ ; & EA=EF\end{array}\right|$ . Thus, $\triangle MDE$ and $\triangle AEF$ are isosceles, $m\left(\widehat{MDE}\right)=100^{\circ}$ , $m\left(\widehat{ADE}\right)=20^{\circ}$ , $\triangle ADE$ is isosceles,

$DE=AE=EF$ , $\triangle DEF$ is isoscels, $m\left(\widehat{MEF}\right)=20^{\circ}$ , $\triangle DEF$ is equilateral, $DE=DF=DM$ , $\triangle MDF$ is isosceles, $m\left(\widehat{MDF}\right)=40^{\circ}$ ,

$m\left(\widehat{DMF}\right)=m\left(\widehat{DFM}\right)=70^{\circ}$ . Let $G\in (BD)$ such that $m\left(\widehat{MGD}\right)=60^{\circ}$ . Denote $H\in DE\cap FG$ . $\triangle MDG$ is equilateral and $\triangle DFG$ is isosceles, $m\left(\widehat{DFG}\right)=40^{\circ}$ .

$m\left(\widehat{EFG}\right)=100^{\circ}$ , $m\left(\widehat{DHG}\right)=20^{\circ}$ , $\triangle DGH$ is isosceles, $GH=DG=MG$ , $\triangle MGH$ is isosceles, ∠$m\left(\widehat{GMB}\right)=10^{\circ}$ , $m\left(\widehat{DMB}\right)=70^{\circ}$ , $m\left(\widehat{BMC}\right)=110^{\circ}$ .


PP2. Let $ABC$ be an $A$-isosceles triangle with $A=80^{\circ}$ . Let $P$ be an inner point of $\triangle ABC$ so that $\left\{\begin{array}{c} m\left(\widehat{PBC}\right)=10^{\circ}\\\ m\left(\widehat{PCB}\right)=30^{\circ}\end{array}\right|$ . Find the value of the angle $\widehat {PAB}$ .

Proof 1 (trigonometric). Denote $m\left(\widehat {PAB}\right)=x$ . Thus, $m\left(\widehat {PAC}\right)=80^{\circ}-x$ and apply the trigonometric form of the Menelaus' theorem:

$\sin 40^{\circ}\sin 30^{\circ}\sin \left(80^{\circ}-x\right)=\sin x\sin 10^{\circ}\sin 20^{\circ}\iff$ $\cos 20^{\circ}\sin \left(80^{\circ}-x\right)=\sin x\sin 10^{\circ}\iff$ $\cos 20^{\circ}\cos \left(10^{\circ}+x\right)=\sin x\sin 10^{\circ}\iff$

$\cos (30^{\circ}+x)+\cos (x-10^{\circ})=\cos (x-10^{\circ})-\cos (x+10^{\circ})\iff$ $\cos (30^{\circ}+x)=\cos \left(170^{\circ}-x\right)\iff$ $30^{\circ}+x=170^{\circ}-x\iff$ $x=70^{\circ}$ .

Proof 2 (synthetic). Let $P$ be the circumcenter of $\Delta ABC$. See that $\Delta BPC$ is $P$-isosceles, $40^\circ-100^\circ-40^\circ$ , also $\Delta CPO$ is $P$-isosceles, $20^\circ-140^\circ-20^\circ$ .

Take $O'$ the symmetrical of $O$ w.r.t. perpendicular bisector of $BC$ . Then $\Delta OPO'$ is equilateral, so $BCOO'$ is an isosceles trapezoid with $CO=OO'=O'B$ . Therefore,

$\angle CBO'=\angle OCB=20^\circ$ and $BO$ is the bisector of $\angle CBO'$ . Consequently $\angle ABO=70^\circ$ . Hence $\Delta ABO$ is isosceles with $\angle BAO=40^\circ$ and $\angle BOA=70^\circ$ .

Proof 3 (metric). $a=2b\sin 40^{\circ}$ . Apply theorem of Sinus in $\triangle BPC\ :\ \frac {PB}{\sin \widehat{BCP}}=$ $\frac {BC}{\sin\widehat{BPC}}\iff$ $\frac {PB}{\sin 30^{\circ}}=$ $\frac {a}{\sin 140^{\circ}}\iff$

$PB=\frac {a}{2\sin 40^{\circ}}\iff$ $PB=b\iff$ the triangle $ABP$ is $B$-isosceles with $m\left(\widehat{ABP}\right)=40^{\circ}\iff$ $m\left(\widehat{BAP}\right)=70^{\circ}$ .

Proof 4 (synthetic). Denote the midpoint $M$ of $[BC]$ and the projection $S$ of $B$ on $PC$ . Observe that the triangle $BMS$ is equilateral, i.e. $BS=SM=MB=MC$

and $\triangle BSP\equiv\triangle BMA$ because $BS=BM$ and $\widehat{BPS}\equiv\widehat{BAM}$ . In conclusion, $BP=BA$ , i.e. $m\left(\widehat{BAP}\right)=70^{\circ}$ .


PP3. Let an $A$-isosceles $\triangle ABC$ with $A=20^{\circ}$ . Denote $\left\{\begin{array}{cc} D\in (AB)\ ; & m\left(\widehat{ACD}\right)=20^{\circ}\\\\ E\in (AC)\ ; & m\left(\widehat{ABE}\right)=10^{\circ}\end{array}\right|$ . Find $m\left(\widehat{DEB}\right)$ .

Proof 1 (trigonometric). Denote $m\left(\widehat{DEB}\right)=x$ and apply the trigonometric form of the Ceva's therem in the quadrilateral $BDEC\ :$

\[\boxed{\begin{array}{c} \sin (50^{\circ}+x)\sin 10^{\circ}\sin 60^{\circ}\sin 30^{\circ}=\sin x\sin 40^{\circ}\sin 70^{\circ}\sin 20^{\circ}\\\\ \cos (40^{\circ}-x)\sin 10^{\circ}\sin 60^{\circ}=2\sin x\sin 40^{\circ}\cos 20^{\circ}\sin 20^{\circ}\\\\ \cos (40^{\circ}-x)\sin 10^{\circ}\sin 60^{\circ}=\sin x\sin^240^{\circ}\\\\ \cos (40^{\circ}-x)(\cos 50^{\circ}-\cos 70^{\circ})=\sin x(1-\cos 80^{\circ})\\\\ \cos (40^{\circ}-x)(\sin 40^{\circ}-\sin 20^{\circ})=\sin x (1-\sin 10^{\circ})\\\\ \sin (80^{\circ}-x)+\sin x-\sin (60^{\circ}-x)+\sin (20^{\circ}-x)=2\sin x-\cos(x-10^{\circ})+\cos (x+10^{\circ})\\\\ -\sin (60^{\circ}-x)+\sin (20^{\circ}-x)=\sin x-\cos (x-10^{\circ})\\\\ \sin (60^{\circ}-x)+\sin x=\sin (20^{\circ}-x)+\cos (x-10^{\circ})\\\\ 2\sin 30^{\circ}\cos (30^{\circ}-x)=\sin (20^{\circ}-x)+\cos (x-10^{\circ})\\\\ \cos (30^{\circ}-x)-\cos (x-10^{\circ})=\sin (20^{\circ}-x)\\\\ -2\sin 10^{\circ}\sin (20^{\circ}-x)=\sin (20^{\circ}-x)\\\\ \sin (20^{\circ}-x)(1+2\sin 10^{\circ})=0\\\\ \sin (20^{\circ}-x)=0\\\\ x=20^{\circ}\\\\ m\left(\widehat{DEB}\right)=20^{\circ}\end{array}}\]
Proof 2 (synthetic). Denote intersection $N$ of $CD$ with bisector of $\widehat {BAC}$ . Prove easily that $BCN$ is an equilateral triangle, $ABNE$ is an isosceles trapezoid with $NE\parallel AB$ ,

$CNE$ is a $N$-isosceles triangle and $ANE$ is an $E$-isosceles triangle. In conclusion, $BC=BN=NC=NE=AE\implies$ $\boxed{AE=BC}\ (*)$ . Denote $F\in (AB)$ ,

$G\in (AC)$ so that $CB=CF=CG$ . Prove easily that $BCF$ is a $C$-isosceles triangle, $CFG$ is an equilateral triangle, $DFG$ is a $F$-isosceles triangle. Since $AE=BC\implies$

$AE=CG$ and by symmetry w.r.t. the midpoint of $[AC]$ obtain that $D$ belongs to the bisector of $[AC]$ , i.e. $DEG$ is $D$-isosceles. In conclusion, $m\left(\widehat{DEB}\right)=20^{\circ}$ .
This post has been edited 47 times. Last edited by Virgil Nicula, Nov 26, 2015, 6:18 PM

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87. An interesting vectorial identity.
86. A very nice maxmin.
85. Problem "slicing" : 60-24-12.
84. IRAN national math olympiad - 2010
83. A metrical characterization of concurrence.
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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    by MrMustache, Nov 11, 2020, 4:52 PM

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    by piphi, Jun 10, 2020, 11:44 PM

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  • Total entries: 456
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