85. Problem "slicing" : 60-24-12.

by Virgil Nicula, Aug 10, 2010, 12:05 AM

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=359982
Quote:
Let $ABC$ be a triangle with $B=60^{\circ}$ , $C=24^{\circ}$ . Denote $D\in (BC)$ so that $m\left(\widehat{BAD}\right)=12^{\circ}$ . Prove that $\boxed{\ b=c+x+y\ }$ , where $BD=x$ , $AD=y$ .
Proof 1 (trigonometric). $b=c+x+y$ $\Longleftrightarrow$ $\frac bc=1+\frac xc+\frac yc$ $\Longleftrightarrow$ $\frac {\sin 60}{\sin 24}=$ $1+\frac {\sin 12}{\sin 72}+$ $\frac {\sin 60}{\sin 72}$ $\Longleftrightarrow$ $\sin 60\sin 72=$ $\sin 24(\sin 72+\sin 12+\sin 60)$ $\Longleftrightarrow$ $\sin 60(\sin 72-\sin 24)=$ $\sin 24(\sin 72+\sin 12)$ $\Longleftrightarrow$ $2\sin 60\sin 24\cos 48=$ $2\sin 24\sin 42\cos 30$ , what is truly.
Quote:
Generalization. Let $ABC$ be a triangle with $A=90^{\circ}+\frac C4$ . Denote $D\in (BC)$ so that $m\left(\widehat{BAD}\right)=\frac C2$ . Prove that $\boxed{AC=AB+BD+DA\ \Longleftrightarrow\ 2\sin\frac B2\sin \frac {2B-C}{4}=\sin C}$ .

Proof.[/size] $BD=x$ , $AD=y$ and $\left\|\begin{array}{c} B=\beta\\\ C=2\alpha\\\ A=\beta +3\alpha\end{array}\right\|$ , where $2\beta+5\alpha=180^{\circ}$ . Thus, $b=c+x+y$ $\Longleftrightarrow$ $\frac bc=1+\frac xc+\frac yc$ $\Longleftrightarrow$ $\frac {\sin \beta}{\sin 2\alpha}=$ $1+\frac {\sin \alpha}{\sin (\alpha +\beta )}+$ $\frac {\sin \beta}{\sin (\alpha+\beta )}$ $\Longleftrightarrow$ $\sin \beta\sin (\alpha +\beta )=$ $\sin 2\alpha [\sin (\alpha +\beta )+\sin\alpha +\sin\beta ]$ $\Longleftrightarrow$ $\sin \beta [\sin (\alpha +\beta )-\sin 2\alpha ]=$ $\sin 2\alpha [\sin (\alpha +\beta )+\sin \alpha ]$ $\Longleftrightarrow$ $\sin \beta\sin\frac {\beta -\alpha}{2}\cos\frac {\beta +3\alpha}{2}=$ $\sin 2\alpha\sin\frac {\beta +2\alpha}{2}\cos\frac {\beta}{2}$ $\stackrel{(*)}{\Longleftrightarrow}$ $2\sin\frac {\beta}{2}\sin\frac {\beta -\alpha}{2}=\sin 2\alpha$ $\Longleftrightarrow$ $2\sin\frac B2\sin \frac {2B-C}{4}=\sin C$ .

$(*)\blacktriangleright$ I used $\frac {\beta +3\alpha }{2}+\frac {\beta +2\alpha}{2}=90^{\circ}$ $\implies$ $\cos\frac {\beta +3\alpha}{2}=\sin \frac {\beta +2\alpha}{2}$ .

Proof 2 (metric). Observe that $A=60^{\circ}$ $\Longleftrightarrow \boxed {b^2=a^2+c^2-ac}\ (*)$ . Denote $F\in (AB)$ , $\widehat{FCA}\equiv\widehat{FCB}$ , $G\in CF\cap AD$ and $E\in (BC)$ , $\widehat {DAB}\equiv\widehat {DAE}$ . Since $ACDF$ is cyclic obtain $\boxed {CA=CG}$ , $\boxed{AF=FD=DG}$ and $\boxed{CF=CD}$ . From here we can try to find a "slicing" proof ! Come back to be continued this proof. Apply theorem of bisector $CF$ in $\triangle\ ACB$ $\implies$ $\frac {FA}{b}=\frac {FB}{a}=\frac {c}{a+b}$ $\implies$ $\left\|\begin{array}{c} FA=\frac {bc}{a+b}\\\ FB=\frac {ac}{a+b}\end{array}\right\|$ . Apply power of $B$ w.r.t. circumcircle of $ACDF$ $\implies$ $x\cdot a=\frac {ac}{a+b}\cdot c$ $\implies$ $\boxed {x=\frac {c^2}{a+b}}\ (1)$ . $\left|\begin{array}{ccccccc} \widehat{BAE}\equiv\widehat{BCA} & \implies & BAE\sim BCA & \implies & \frac ca=\frac {AE}b=\frac {BE}c & \implies & \left|\begin{array}{c} AE=\frac {bc}a\\\\ BE=\frac {c^2}a\end{array}\right|\\\\ \widehat{CAE}\equiv\widehat{CDA} & \implies & CAE\sim CDA & \implies & \frac b{a-x}=\frac {AE}y=\frac {CE}b & \implies & \left|\begin{array}{c} AE=\frac {by}{a-x}\\\\ CE=\frac {b^2}{a-x}\end{array}\right|\end{array}\right|$

$\left\|\begin{array}{ccccc} \frac {bc}{a}=\frac {by}{a-x} & \implies & cx+ay=ac & \implies & \boxed {\frac xa+\frac yc=1}\\\\ a=BE+EC & \implies & a=\frac {c^2}{a}+\frac {b^2}{a-x} & \stackrel {(*)}{\implies} & \boxed {x=\frac {ac(a-2c)}{a^2-c^2}}\ (2)\end{array}\right\|$ $\stackrel {(*)}{\implies}\boxed {y=\frac {b^2c}{a^2-c^2}}$ $\stackrel {(*)}{\implies}\boxed {x+y=\frac {c(2a-c)}{a+c}}$ . From relations $(1)$ , $(2)$ obtain $\frac {c^2}{a+b}=\frac {ac(a-2c)}{a^2-c^2}$ $\implies$ $a^3+c^3+a^2(b-3c)-2abc=0$ $\stackrel{(*)}{\implies}$ $b^2(a+c)+a^2(b-3c)-2abc=0$ $\implies$ $(a+c)\cdot\underline b^2+a(a-2c)\cdot\underline b-3a^2c=0$ $\implies$ $[(a+c)b-3ac](a+b)=0$ $\implies$ $b(a+c)=3ac$ , i.e. $\boxed {\frac 1a+\frac 1c=\frac 3b}$ . Thus $x+y=\frac {3ac-c(a+c)}{a+c}=$ $\frac {b(a+c)-c(a+c)}{a+c}$ $\implies$ $x+y=b-c$ .
Quote:
Proposed problem. In $\triangle ABC$ with $B=60^{\circ}$ denote length $l_b$ of the interior $B$-bisector. Prove that $\boxed{C=24^{\circ}\Longleftrightarrow\frac 1a+\frac 1c=\frac 3b\Longleftrightarrow b=l_b\cdot\sqrt 3}$ .
Proof.
$\blacktriangleright$ I"ll use the well-known relation - the length of the interior $B$-bisector : $l_b=\frac {2ac\cdot\cos\frac B2}{a+c}=$ $\frac {2\sqrt{acs(s-b)}}{a+c}\ (1)$ , where $2s=a+b+c$ . Thus, $\boxed {\frac 3b=\frac 1a+\frac 1c}$ $\Longleftrightarrow$ $b=3\cdot \frac {ac}{a+c}$ $\stackrel{(1)}{\Longleftrightarrow}$ $b=3\cdot \frac {l_a}{2\cos\frac B2}$ $\Longleftrightarrow$ $\boxed {b=l_b\sqrt 3}$ .
$\blacktriangleright$ I"ll use the well-known relation $\sin 54^{\circ}=\cos 36^{\circ}=\frac{1+\sqrt 5}{4}\ \ (2)$ . Therefore, $\boxed {\frac 3b=\frac 1a+\frac 1c}$ $\Longleftrightarrow$ $b(a+c)=3ac$ $\Longleftrightarrow$ $\sin 60\cdot [\sin (60+C)+\sin C]=$ $3\sin (60+C)\sin C$ $\Longleftrightarrow$ $\frac {\sqrt 3}{2}\cdot 2\sin (30+C)\cos 30=$ $\frac 32\cdot [\cos 60-\cos (60+2C)]$ $\Longleftrightarrow$ $\sin (30+C)=\frac 12-\cos (60+2C)$ . Denote $\sin (30+C)=t$ . Thus our relation becomes $4t^2-2t-1=0$ $\Longleftrightarrow$ $t=\frac {1+\sqrt 5}{4}$ $\stackrel{(2)}{\Longleftrightarrow}$ $\sin (30+C)=\sin 54^{\circ}$ $\Longleftrightarrow$ $\boxed {C=24^{\circ}}$ .
$\blacktriangleright$ (otherwise) I'll use $B=60^{\circ}\Longleftrightarrow a^2+c^2-ac=b^2\ (3)$ . Therefore, $\boxed {b=l_a\sqrt 3}$ $\Longleftrightarrow$ $b^2=3\cdot l_a^2$ $\stackrel{(1)}{\Longleftrightarrow}$ $b^2=3\cdot \frac {ac\left[(a+c)^2-b^2\right]}{(a+c)^2}$ $\Longleftrightarrow$ $b^2(a+c)^2=3ac\left[(a+c)^2-b^2\right]$ $\stackrel{(3)}{\Longleftrightarrow}$ $b^2(a+c)^2=3ac\left[\left(a^2+c^2+2ac\right)-\left(a^2+c^2-ac\right)\right]$ $\Longleftrightarrow$ $b^2(a+c)^2=3ac\cdot 2ac$ $\Longleftrightarrow$ $\boxed {b(a+c)=3ac}$ .
Quote:
Proposed problem. Let $ABC$ be a triangle with $A\le 90^{\circ}$ . Prove that if exists $k>0$ so that $\frac 1b+\frac 1c=\frac ka$ , then exists the chain $\frac 1b+\frac 1c=\frac ka$ $\Longleftrightarrow$ $l_a=\frac {2\cos\frac A2}{k}\cdot a$ $\Longleftrightarrow$ $\cos \frac {B-C}{2}=\frac {\sin\frac A2}{k}\cdot \left[1+\cos A+\sqrt {k^2+(1+\cos A)^2}\right]$ .
Remark. For $A=60^{\circ}$ and $k=3$ obtain upper proposed problem or from here.
This post has been edited 27 times. Last edited by Virgil Nicula, Nov 23, 2015, 2:35 PM

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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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