41. ONM 2010 Calarasi Problema 3.

by Virgil Nicula, Jun 1, 2010, 2:51 PM

Quote:
In triunghiul $ABC$ punctul $M$ este mijlocul laturii $[BC]$ si $C = 15^{\circ}$ .

Stiind ca $m( \angle{AMB}) = 45^{\circ}$, determinati masura unghiului $\angle BAC$.
Desi problema este propusa la nivelul clasei a VI - a voi oferi mai multe soutii pentru aceasta frumoasa problema "slicing". Voi incepe cu

metodele metrice si/sau trigonometrice si sfarsind cu cele sintetice, de fapt singurele de interes estetic intr-o problema de tip "slicing".

Metoda "killer" 0 (metrica). Aplicam teorema sinusurilor in $\triangle AMC$ $\Longleftrightarrow$ $\frac {CM}{CA}=\frac {\sin \widehat {CAM}}{\sin\widehat {CMA}}$ $\Longleftrightarrow$ $\frac {\frac a2}{b}=\frac {\sin 30^{\circ}}{\sin 45^{\circ}}$ $\Longleftrightarrow$ $a=b\sqrt 2$ $\Longleftrightarrow$ $CA^2=CM\cdot CB$ $\Longleftrightarrow$

$\triangle CAM\sim \triangle CBA$ , adica dreapta $CA$ este tangenta in $A$ la cercul circumscris triunghiului $ABM$ $\Longleftrightarrow$ $\widehat {CBA}\equiv\widehat {CAM}$ $\Longleftrightarrow$ $B=30^{\circ}$ .

Observatie. $\triangle CAM\sim \triangle CBA$ $\Longrightarrow$ $\frac {b}{a}=\frac {m_a}{c}=\frac {a}{2b}=\frac {\sqrt 2}{2}$ , adica $c=m_a\sqrt 2$ . Teorema Pitagora generalizata in $\triangle ABM$ $\Longrightarrow$ $c^2=m_a^2+\frac {a^2}{4}-am_a\frac {\sqrt 2}{2}$ $\Longrightarrow$

$a^2-2ac-2c^2=0$ $\Longrightarrow$ $a=c\left(1+\sqrt 3\right)$ . In concluzie $\boxed {\ \frac {a}{2}=\frac {b}{\sqrt 2}=\frac {c}{\sqrt 3-1}\ }$ de unde rezulta cu usurinta masura unghiului $\angle ABC .$ .

Metoda 1 (metrica). Teorema Sinusurilor in $\triangle AMC$ $\Longrightarrow$ $\frac {b}{\sin 45}=\frac {a}{2\sin 30}$ $\Longrightarrow$ $a=b\sqrt 2\ \ (1)$ . Teorema medianei in $\triangle ABC$ $\Longrightarrow$ $4m_a^2=2\left(b^2+c^2\right)-a^2\stackrel{(1)}{\ =\ }$

$2\left(b^2+c^2\right)-2b^2$ $\Longrightarrow$ $c=m_a\sqrt 2\ \ (2)$ . Teorema Sinusurilor in $\triangle ABM$ $\Longrightarrow$ $\frac {m_a}{\sin B}=\frac {c}{\sin 45}\stackrel{(2)}{\ \Longrightarrow\ }\frac {m_a}{\sin B}=\frac {m_a\sqrt 2}{\frac {\sqrt 2}{2}}$ $\Longrightarrow$ $\sin B=\frac 12$ $\Longrightarrow$ $B=30^{\circ}$ .

Metoda 2. Aplicam o relatie cunoscuta cevienei $AM$ in $\triangle ABC$ : $\frac {MB}{MC}=$ $\frac {AB}{AC}\cdot\frac {\sin \widehat {MAB}}{\sin\widehat {MAC}}$ $\Longrightarrow$ $1=$ $\frac {\sin 15^{\circ}}{\sin B}\cdot \frac {\sin (B+45^{\circ})}{\sin 30^{\circ}}$ $\Longrightarrow$ $\sin (B+45^{\circ})=$ $2\sin B\cos 15^{\circ}$ $\Longrightarrow$

$\sin\left(B+45^{\circ}\right)=$ $\sin \left(B+15^{\circ}\right)+\sin\left(B-15^{\circ}\right)$ $\Longleftrightarrow$ $\sin (B+45^{\circ})-\sin\left(B-15^{\circ}\right)=$ $\sin (B+15^{\circ})$ $\Longrightarrow$ $2\sin 30^{\circ}\cos \left(B+15^{\circ}\right)=\sin \left(B+15^{\circ}\right)$ $\Longleftrightarrow$

$\tan\left(B+15^{\circ}\right)=1$ $\Longleftrightarrow$ $B=30^{\circ}$ . Daca in $\triangle ABC$ mijlocul $M$ pt. $[BC]$ are proprietatea ca $m(\widehat {MAC})=2\cdot m(\widehat {MCA})=2x$ , atunci $\tan (B+x)=2\sin 2x$ .

==============================================================================

Este binecunoscuta urmatoarea problema "slicing" intr-un triunghi :
Quote:
Lema $1$. In $\triangle ABC$ cu $B=30^{\circ}$ si $C=15^{\circ}$ avem $m(\widehat {MAC})=30^{\circ}$ , unde punctul $M$ este mijlocul laturii $[BC]$ .

Demonstratie. Fie $C(O)$ circumscris $\triangle ABC$ . Se observa ca $OB\perp OC$ , $\triangle AOC$ este echilateral si $AM$ este axa de simetrie pentru $\triangle AOC$ . In concluzie $m(\widehat{ MAC})=30^{\circ}$ .

Observatie. Problema propusa este o reciproca a acestei leme, in sensul ca de data aceasta se cunoaste $m(\widehat{ MAC})=30^{\circ}$ si se cere a se arata ca $B=30^{\circ}$ . :!:

Metoda a 3 -a ("slicing"). Notam mijlocul $N$ al laturii $[AC]$ . Aplicam lema $(1)$ in $\triangle AMC$ si rezulta $m(\widehat{CMN})=30^{\circ}$ . Din $MN\parallel AB$ rezulta $B=30^{\circ}$ .

Metoda 4 ("slicing"). Construim triunghiul $BCD$ astfel ca dreapta $BC$ separa punctele $A$ , $D$ si $m(\angle CBD)=30^{\circ}$ si $m(\angle BCD)=15^{\circ}$ . Aplicam

lema $(1)$ in $\triangle BDC$ si rezulta $m(\widehat{MDC})=30^{\circ}$ . Deoarece triunghiurile $AMC$ si $DMC$ sunt simetrice fata de dreapta $BC$ rezulta ca $B=30^{\circ}$ .

Metoda 5 ("slicing"). Cercul circumscris $w$ al triunghiului $ABM$ taie (eventual) a doua oara dreapta $AC$ in punctul $D$ . Daca $D\equiv A$ , atunci dreapta $AC$ este tangenta cercului $w$ in

punctul $A$ si in acest caz este limpede ca $B=m(\angle MAC)=30^{\circ}$ . Daca $D\in (AC)$ , atunci $30^{\circ}=m(\angle MAD)=m(\angle MBD)$ si aplicand lema $(1)$ in triunghiul $BDC$ rezulta

$30^{\circ}=m(\angle MDC)>m(\angle MAC)=30^{\circ}$ , ceea ce este absurd. Daca $A\in (DC)$ , atunci $30^{\circ}=m(\angle MAC)=m(\angle MBD)$ si aplicand lema $(1)$ in triunghiul $BDC$ rezulta

$30^{\circ}=m(\angle MDC)<m(\angle MAC)=30^{\circ}$ , ceea ce este absurd.
Quote:
Lema $2$ (own). In interiorul triunghiului $A$-dreptunghic si isoscel $ABC$ consideram punctul $M$ pentru care $m(\widehat {ABM})=$ $m(\widehat {BCM})=15^{\circ}$ . Atunci $MA=MB$ si $MC=CA$ .

Observatie. Construim patratul $ABDC$ si se observa ca se reproduce o alta problema cunoscuta de "slicing" : "Fie patratul

$ABDC$ si un punct interior $M$ pentru care $m(\angle ABM)=$ $m(\angle BAM)=15^{\circ}$ . Sa se arate ca triunghiul $MCD$ este echilateral".

Metoda 6 ("slicing"). Construim triunghiul $D$-dreptunghic isoscel $ACD$ astfel ca dreapta $BC$ sa separe punctele $A$ , $D$ . Aplicand lema $(2)$ mentionata rezulta $MA=MD$

si $MC=CD$ . Consideram mijlocul $N$ al segmentului $[AD]$ . Insa $MA=MD$ $\Longrightarrow$ $MN\perp AD$ $\Longrightarrow$ $MN\parallel CD$ si deoarece punctele $M$ , $N$ sunt mijloacele segmentelor

$[BC]$ , $[AD]$ respectiv rezulta $AB\parallel CD$ , adica $m(\widehat {ABC})=m(\widehat {DCB})$ $\Longrightarrow$ $B=30^{\circ}$ .

Metoda 7 ("slicing"). Construim triunghiul echilateral $ACD$ astfel ca dreapta $BC$ sa separe punctele $A$ , $D$ . Asadar $DA=DC=AC$ $\Longrightarrow$ $m(\angle MAD)=30^{\circ}$ , $M(\angle BCD)=45^{\circ}$

si semidreapta $[AM$ este bisectoarea $\angle CAD$ $\Longrightarrow$ $AM\perp CD$ si punctul $N\in AM\cap CD$ este mijlocul laturii $[CD]$ $\Longrightarrow$ segmentul $[MN]$ este linie mijlocie in triunghiul $BCD$ $\Longrightarrow$

$MN\parallel BD$ $\Longrightarrow$ $m(\angle CBD)=45^{\circ}$ si $m(\angle ADB)=30^{\circ}$ $\Longrightarrow$ triunghiul $BDC$ este $A$-dreptunghic isoscel $\Longrightarrow$ $DB=DC=DA$ $\Longrightarrow$ $DB=DA$ $\Longrightarrow$ triunghiul $ADB$ este

$A$-isoscel cu $m(\angle ADB)=30^{\circ}$ $\Longrightarrow$ $m(\angle BAD)=75^{\circ}$ $\Longrightarrow$ $m(\angle BAC)=135^{\circ}$ .

Metoda 8 ("slicing"). Construim triunghiul echilateral $BCD$ astfel ca dreapta $BC$ sa nu separe punctele $A$ , $D$ . Se observa ca $DM\perp BC$ si $m(\angle MDC)=$ $m(\angle MAC)=30^{\circ}$ ceea

ce inseamna ca patrulaterul $ADCM$ este inscriptibil. Rezulta $AD\perp AC$ , $m(\angle MDA)=$ $m(\angle MCA)=$ $m(\angle ADB)=15^{\circ}$ . Asadar semidreptele $[DA$ si $[MA$ sunt bisectoarele

unghiurilor $\angle BMD$ si $\angle BDM$ respectiv ceea ce inseamna ca $A$ este centrul cercului inscris in triunghiul $BDM$ si semidreapta $[BA$ este bisectoarea unghiului $\angle DBM$ , adica $B=30^{\circ}$ .
This post has been edited 55 times. Last edited by Virgil Nicula, Nov 27, 2015, 8:10 AM

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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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