47. Interesting and difficult identities in a triangle.

by Virgil Nicula, Jun 23, 2010, 3:35 AM

PP1. Prove that in any triangle $ABC$ there are the following interesting identities :

$1.1\blacktriangleright\ \left(a^2+ab-c^2\right)(a-b)=$ $2Rbc\cdot (\sin C-\sin 2A)$

$1.2\blacktriangleright\ \left(a^2+ac-b^2\right)(a-c)=$ $2Rbc\cdot (\sin B-\sin 2A)$

$2\blacktriangleright\ \prod_{\mathrm{cyc}}\left(a^2+ac-b^2\right)-\prod_{\mathrm{cyc}}\left(a^2+ab-c^2\right)=(a+b+c)^3\cdot\prod_{\mathrm{cyc}}(a-b)$

$3\blacktriangleright\ \prod_{\mathrm{cyc}}\sin\left(B-\frac C2\right)-\prod_{\mathrm{cyc}}\sin\left(A-\frac C2\right)=\frac {2s}{R}\cdot\prod_{\mathrm{cyc}}\sin\frac {A-B}{2}$ .

PP2. Let $ M$ be a point of the side $ [BC]$ . Denote $ x = m(\widehat {BAM})$ , $ y = m(\widehat {CAM})$ . Prove that $ \boxed {\ \frac {\sin^2x}{\sin^2y} = \frac {\sin (B - x)}{\sin (C - y)}\ \Longleftrightarrow\ MB = MC\ \Longleftrightarrow\ \sin x\sin C = \sin y\sin B\ }\ .$

Proof I (ugly). Prove easily that $ \boxed {MA = MB}\ \Longleftrightarrow\ c\cdot \sin x = b\cdot\sin y\ \Longleftrightarrow\ \boxed {\ \sin x\cdot\sin C = \sin y\cdot\sin B\ }$ .

$ \boxed {\begin{array}{c} \\ \ \boxed {\sin^2x\sin (C - y) = \sin^2y\sin (B - x)} \\ \\ (1 - \cos 2x)\sin (C - y) = (1 - \cos 2y)\sin (B - x) \\ \\ \sin (B - x) - \sin (C - y) = \cos 2y\sin (B - x) - \cos 2x\sin (C - y) \\ \\ 4\sin\frac {B - C - x + y}{2}\sin A = [\sin (B - x + 2y) - \sin (x + 2y - B)] - [\sin (C - y + 2x) - \sin (2x + y - C)] \\ \\ 4\sin\frac {B - C - x + y}{2}\sin A = [\sin (B - x + 2y) - \sin (C - y + 2x)] + [\sin (x + A - C) - \sin (y + A - B)] \\ \\ 2\sin A\sin \frac {B - C - x + y}{2} = \sin \frac {B - C - 3x + 3y}{2}\cos\frac {B + C + x + y}{2} + \sin\frac {B - C + x - y}{2}\cos\frac {3A - B - C}{2} \\ \\ 2\sin A\sin\frac {B - C - x + y}{2} = \sin\frac {B - C + x - y}{2}\sin 2A \\ \\ \sin\frac {B - C - x + y}{2} = \cos A\sin\frac {B - C + x - y}{2} \\ \\ 2\sin\frac {B - C - x + y}{2} = \sin \frac {B - C + 2A + x - y}{2} + \sin\frac {B - C - 2A + x - y}{2} \\ \\ \sin\frac {B - C - x + y}{2} - \sin\frac {B - C + 2A + x - y}{2} = \sin\frac {B - C - 2A + x - y}{2} - \sin\frac {B - C - x + y}{2} \\ \\ \sin\frac {B - C - x + y - B + C - 2A - x + y}{4}\cos\frac {B - C - x + y + B - C + 2A + x - y}{4} = \sin\frac {B - C - 2A + x - y - B + C + x - y}{4}\cos\frac {B - C - 2A + x - y + B - C - x + y}{4} \\ \\ \sin\frac {A + x - y}{2}\sin C = \sin\frac {A + y - x}{2}\sin B \\ \\ \boxed {\sin x\sin C = \sin y\sin B} \\ \ \end{array}}$

Proof II (nice). Denote $ N\in AC$ , $ P\in AB$ so that $ MB = MC = MN = MP$ . Using the power of the point $ A$ w.r.t. the circle $ C\left(M,\frac a2\right)$ obtain $ c\cdot AP = b\cdot AN$ .

Apply the Sinus' theorem in the triangles $ APM$ and $ ANM\ :\ \left\{\begin{array}{c} \frac {PM}{\sin x} = \frac {AP}{\sin (B - x)} \\ \\ \frac {NM}{\sin y} = \frac {AN}{\sin (C - y)}\end{array}\right\|$ $ \implies$ $ \frac {\sin y}{\sin x} = \frac {AP}{AN}\cdot\frac {\sin (C - y)}{\sin (B - x)}$ . But $ MA = MB\Longleftrightarrow$

$c\cdot\sin x = b\cdot\sin y$ $ \Longleftrightarrow$ $ \frac {AP}{AN} = \frac bc = \frac {\sin x}{\sin y}$ . In conclusion, $ \frac {\sin y}{\sin x} = \frac {\sin x}{\sin y}\cdot\frac {\sin (C - y)}{\sin (B - x)}$ , i.e. $ \boxed {\ \frac {\sin^2x}{\sin^2y} = \frac {\sin (B - x)}{\sin (C - y)}}$ .

Remark. $\boxed{\frac {\sin^2x}{\sin^2y} = \frac {\sin (B - x)}{\sin (C - y)}}\iff$ $\frac {\sin x\sin B}{\sin y\sin C} = \frac {\sin (B - x)}{\sin (C - y)}\iff$ $\frac {\sin (B - x)}{\sin B\sin x}=\frac {\sin (C - y)}{\sin C\sin y}\iff$ $\cot B-\cot x=\cot C-\cot y\iff$

$\boxed{\cot x-\cot y=\cot B-\cot C}\iff$ $\frac {m_a^2+c^2-\frac {a^2}{4}}{2S}-\frac {m_a^2+b^2-\frac {a^2}{4}}{2S}=\frac {a^2+c^2-b^2}{4S}-\frac {a^2+b^2-c^2}{4S}\iff$ $2\left(c^2-b^2\right)=2\left(c^2-b^2\right)$ , what is true.
This post has been edited 10 times. Last edited by Virgil Nicula, Nov 23, 2015, 8:46 AM

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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 ryanbear, May 9, 2023, 6:10 AM

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    by OlympusHero, Sep 14, 2022, 4:44 AM

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    by tigerzhang, Aug 1, 2021, 12:02 AM

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    by the_mathmagician, Jan 17, 2021, 7:43 PM

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    by OlympusHero, Dec 30, 2020, 6:08 PM

  • long blog

    by MrMustache, Nov 11, 2020, 4:52 PM

  • 372554 views!

    by mrmath0720, Sep 28, 2020, 1:11 AM

  • wow... i am lost.

    369302 views!

    -piphi

    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

  • impressive :D
    awesome. 358,000 visits?????

    by OlympusHero, May 14, 2020, 8:43 PM

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