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by Virgil Nicula, Feb 28, 2011, 5:49 PM

P1 (M. O. Sanchez). Let an $A$-right $\triangle ABC$ with incircle $w=\mathbb C(I,r).$ Let $D\in BC\cap w$ and $\left\{\begin{array}{ccccc} M\in (AB) & \mathrm{so\ that} & MA=MB\ ; & Q\in MI\cap AC\\\\ N\in (AC) & \mathrm{so\ that} & NA=NC\ ; & P\in NI\cap AB\end{array}\right\|\ .$ Prove $m\left(\widehat {PDQ}\right)=135^{\circ}\ .$

Proof. Is well-known that $\boxed{A=90^{\circ}\iff bc=2s(s-a)\iff (s-b)(s-c)=s(s-a)}\ (*)\ .$ Let $\left\{\begin{array}{ccc} PB=u & \mathrm{i.e.} & PA=c-u\\\\ QC=v & \mathrm{i.e.} & QA=b-v\end{array}\right\|\ .$ Apply the particular case of the Cristea's theorem for

the incircle $I\ :\ \left\{\begin{array}{ccc} I\in MQ\ :\ \frac 1{MA}+\frac 1{QA}=\frac {2s}{bc}\implies\frac 2c+\frac 1{b-v}=\frac {2s}{bc}\implies b-v=\frac {bc}{2(s-b)} & \implies & b-v=\frac {\cancel{2(s-b)}(s-c)}{\cancel{2(s-b)}}=s-c\\\\ I\in NP\ :\ \frac 1{NA}+\frac 1{PA}=\frac {2s}{bc}\implies\frac 2b+\frac 1{c-u}=\frac {2s}{bc}\implies c-u=\frac {bc}{2(s-c)} & \implies & c-u=\frac {\cancel{2(s-c)}(s-b)}{\cancel{2(s-c)}}=s-b\end{array}\right\|\implies$ $u=v=b+c-s\implies \boxed{u=v=s-a}\ (1)\ .$

Denote $\left\{\begin{array}{ccc} m\left(\widehat{BDP}\right) & = & x\\\\ m\left(\widehat{CDQ}\right) & = & y\end{array}\right\|$ and the projections $U,$ $V$ of $P,$ $Q$ on $BC$ respectively. Thus:

$\blacktriangleright\ \triangle BPU\sim \triangle BCA\implies\frac {UP}b=\frac {UB}c=\frac {PB}a=\frac {s-a}a\implies UD=BD-BU=(s-b)-\frac {c(s-a)}a\implies$ $UD=\frac {a(s-b)-c(s-a)}a\implies \boxed{\cot x=\frac {UD}{UP}=\frac {a(s-b)}{b(s-a)}-\frac cb}\ .$

$\blacktriangleright\ \triangle CQV\sim \triangle CBA\implies \frac {VQ}c=\frac {VC}b=\frac {QC}a=\frac {s-a}a \implies VD=CD-CV=(s-c)-\frac {c(s-a)}a\implies$ $VD=\frac {a(s-c)-b(s-a)}a\implies \boxed{\cot y=\frac {VD}{VQ}=\frac {a(s-c)}{c(s-a)}-\frac bc}\ .$

Observe that $:\underline{\cot x+\cot y}=\frac a{s-a}\cdot \left(\frac {s-b}b+\frac {s-c}c\right)-\frac {a^2}{bc}=$ $\frac a{bc}\cdot \left[\frac {c(s-b)+b(s-c)}{s-a}-a\right]=$ $\frac {a\left[c(s-b)+b(s-c)-a(s-a)\right]}{bc(s-a)}=$ $\frac {a\left[2s(s-a)-2bc+a^2\right]}{bc(s-a)}=$ $\frac {a\left(a^2-bc\right)}{bc(s-a)}\implies$

$\boxed{\cot x+\cot y=\frac {a\left(a^2-bc\right)}{bc(s-a)}}\ ;\ \underline{\cot x\cot y-1}=\frac 1{bc}\cdot \left[\frac {a(s-b)}{s-a}-c\right]\cdot\left[\frac {a(s-c)}{s-a}-b\right]-1=$ $\frac {\left[a(s-b)-c(s-a)\right]\cdot\left[a(s-c)-b(s-a)\right]}{bc(s-a)^2}-1=$

$\frac {a^2(s-b)(s-c)+\cancel{bc(s-a)^2}-ab(s-a)(s-b)-ac(s-a)(s-c)-\cancel{bc(s-a)^2}}{bc(s-a)^2}=$ $\frac {a^2s\cancel{(s-a)}-a\cancel{(s-a)}[b(s-b)+c(s-c)]}{bc(s-a)\cancel{^2}}=$ $\frac a{bc}\cdot \frac {as-b(s-b)-c(s-c)}{s-a}=$

$\frac a{bc}\cdot\frac {a^2-2s(s-a)}{s-a}=$ $\frac {a\left(a^2-bc\right)}{bc(s-a)}\implies$ $\boxed{\cot x\cot y-1=\frac {a\left(a^2-bc\right)}{bc(s-a)}}\ .$ In conclusion, $\cot x+\cot y=\cot x\cot y-1\iff$ $\tan x+\tan y=1-\tan x\tan y\iff$

$\frac {\tan x+\tan y}{1-\tan x\tan y}=1\iff$ $\tan (x+y)=1\iff x+y=45^{\circ}\iff$ $m\left(\widehat {PDQ}\right)=135^{\circ}\ .$


P2 (Cristian Tello). Let an acute $\triangle ABC$ with the orthocenter $H.$ Denote $:$ the midpoint $M$ of $[BC]\ ;\ \ D\in AH\cap BC\ ;\ N\in AM$ so that $HN\perp AM\ .$ Prove that $\widehat{BCN}\equiv\widehat{CAM}\ .$

Proof. $DHNM\ -$ cyclic $\iff AH\cdot AD=AN\cdot AM\iff$ $2Rh_a\cos A=AN\cdot m_a\iff$ $bc\cdot \cos A=AN\cdot m_a\iff$ $b^2+c^2-a^2=2m_a\cdot AN\iff$ $AN=\frac {b^2+c^2-a^2}{2m_a}\iff$

$MN=m_a-\frac {b^2+c^2-a^2}{2m_a}\iff$ $4m_a\cdot MN=4m_a^2-2\left(b^2+c^2-a^2\right)=\cancel{2\left(b^2+c^2\right)}-a^2-\cancel{2\left(b^2+c^2\right)}+2a^2\iff$ $4MA\cdot MN=a^2\iff$ $MA\cdot MN=MC^2\iff$ $\widehat{BCN}\equiv\widehat{CAM}$
This post has been edited 66 times. Last edited by Virgil Nicula, Jun 19, 2017, 8:30 AM

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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 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 GoogleNebula, Apr 14, 2021, 5:25 AM

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    by samrocksnature, Apr 14, 2021, 5:16 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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  • Joined: Jun 22, 2005
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  • Total entries: 456
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