88. A nice geometric locus conditioned metrically.

by Virgil Nicula, Aug 19, 2010, 1:07 AM

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=362607
Quote:
Let $ ABC $ be a triangle with $AB=AC=b$ and $BC=a$ . Let mobile $M\in (AC)$ , $N\in (AB)$ such that

$\boxed{a^2\cdot AM\cdot AN=b^2\cdot BN\cdot CM}\ (*)$ and denote $P\in BM\cap CN$ . Find the locus of the point $P$ .

Preliminary.. Denote $AN=x$ , $AM=y$ . Thus, $BN=b-x$ , $CM=b-y$ . The relation $(*)$ becomes $\boxed{a^2xy=b^2(b-x)(b-y)}\ (1)$ .

Let $m\left(\widehat{CBM}\right)=\alpha$ , $m\left(\widehat{BCN}\right)=\beta$ . Thus, $m\left(\widehat{ABM}\right)=B-\alpha$ , $m\left(\widehat{ACN}\right)=B-\beta$ . Here are down some proofs.

Proof 1 (synthetic).Let $Q\in AC$ for which $\widehat{QBC}=\widehat{MBA}$ . Using Steiner's theorem obtain $\frac{y}{b-y}\cdot\frac {QA}{QC}\stackrel{(St.)}{=}$ $\left(\frac ba\right)^2\stackrel{(1)}{=}$ $\frac{xy}{(b-x)(b-y)}$ $\implies$ $\frac{QA}{QC}=\frac {x}{b-x}=\frac {NA}{NC}$ $\implies$

$NQ\parallel BC$ , i.e. $BCQN$ ia an isosceles trapezoid. Observe that $\widehat{NCM}\equiv$ $\widehat {NBQ}$ $\equiv\widehat{MBC}$ , i.e. $\widehat{NCM}\equiv\widehat{MBC}$ . Therefore, $m\left(\widehat {BPC}\right)=$ $m\left(\widehat{PMC}\right)+m\left(\widehat{PCM}\right)=$

$m\left(\widehat{BMC}\right)+m\left(\widehat{MBC}\right)=180^{\circ}-B$ , i.e. $\boxed {m\left(\widehat {BPC}\right)=A+B}$ (constant).

Proof 2 (metric). Apply an well-known relation to assess a ratio : $\left\|\begin{array}{c} \frac {x}{b-x}=\frac {NA}{NB}=\frac {CA}{CB}\cdot\frac {\sin\widehat{NCA}}{\sin\widehat{NCB}}=\frac ba\cdot\frac {\sin (B-\beta )}{\sin \beta}\\\\ \frac {y}{b-y}=\frac {MA}{MB}=\frac {BA}{BC}\cdot\frac {\sin\widehat{MBA}}{\sin\widehat{MBC}}=\frac ba\cdot\frac {\sin (B-\alpha )}{\sin \alpha}\end{array}\right\|$ $\bigodot\ \stackrel{(1)}{\implies}$ $\sin(B-\alpha )\sin (B-\beta )=\sin \alpha\sin\beta$ $\iff$

$\cos (\alpha -\beta )-\cos [2B-(\alpha +\beta )]=\cos (\alpha -\beta )-\cos (\alpha +\beta )$ $\iff$ $\cos [2B-(\alpha +\beta )]=\cos (\alpha +\beta )$ $\iff$ $\alpha +\beta =B$ . Thus, $\boxed {m\left(\widehat{BPC}\right)=A+B}$ (constant).

Proof 3 (trigonometric). Apply theorem of Sinus in triangles : $\left\|\begin{array}{cccc} \triangle\ ABM\ : & \frac {y}{\sin (B-\alpha )}=\frac {b}{\sin (B+\alpha )} & \implies & y=\frac {b\cdot\sin (B-\alpha )}{\sin (B+\alpha )}\\\\ \triangle\ ACN\ : & \frac {x}{\sin (B-\beta )}=\frac {b}{\sin (B+\beta )} & \implies & x=\frac {b\cdot\sin (B-\beta )}{\sin (B+\beta )}\\\\ \triangle\ BCM\ : & \frac {b-y}{\sin\alpha }=\frac {a}{\sin (B+\alpha )} & \implies & b-y=\frac {a\cdot\sin\alpha }{\sin (B+\alpha )}\\\\ \triangle\ BCN\ : & \frac {b-x}{\sin \beta }=\frac {a}{\sin (B+\beta )} & \implies & b-x=\frac {a\cdot\sin \beta }{\sin (B+\beta )}\end{array}\right\|$ .

The relation $(1)$ becomes $\frac {a^2b^2\sin (B-\alpha )\sin (B-\beta )}{\sin (B+\alpha )\sin (B+\beta )}=\frac {b^2a^2\sin \alpha \sin\beta }{\sin (B+\alpha )\sin (B+\beta )}$ $\iff$ $\sin (B-\alpha )\sin (B-\beta )=\sin\alpha\sin\beta$ $\iff$

$\cos (\alpha -\beta )-\cos [2B-(\alpha +\beta )]=\cos (\alpha -\beta )-\cos (\alpha +\beta )$ $\iff$ $\cos [2B-(\alpha +\beta )]=\cos (\alpha +\beta )$ $\iff$ $\alpha +\beta =B$ .

Thus, $\boxed {m\left(\widehat{BPC}\right)=A+B}$ (constant).

Proof 4 (metric). Apply an well-known area relation $[PBN]\cdot [PCM]=[PMN]\cdot [BPC]$ . Thus $PB\cdot (b-x)\cdot \sin (B-\alpha )\cdot PC\cdot (b-y)\cdot\sin (C-\beta )=$

$\frac {a^2\sin \alpha\sin\beta}{\sin (\alpha +\beta )}\cdot PN\cdot PM\cdot \sin (\alpha +\beta )$ $\iff$ $\frac {\sin (B-\alpha )\sin (B-\beta )}{\sin\alpha\sin \beta}=\frac {a^2}{(b-x)(b-y)}\cdot\frac {PM\cdot PN}{PB\cdot PC}\ (2)$ . Apply the Menelaus' theorem to $:$

$\left\|\begin{array}{cc} \overline {CPN}/ \triangle ABM \ : & \frac {b-y}{b}\cdot \frac {x}{b-x}\cdot \frac {PB}{PM}=1\\\\ \overline{BPM}/ \triangle ACN\ : & \frac {b-x}{b}\cdot\frac {y}{b-y}\cdot\frac {PC}{PN}=1\end{array}\right\|$ $\bigodot\implies$ $\frac {xy}{b^2}\cdot\frac {PB\cdot PC}{PM\cdot PN}=1\ (3)$ . Therefore, $\left\|\begin{array}{cc} (1) & a^2xy=b^2(b-x)(b-y)\\\\ (2) & \frac {\sin (B-\alpha )\sin (B-\beta )}{\sin\alpha\sin \beta}=\frac {a^2}{(b-x)(b-y)}\cdot\frac {PM\cdot PN}{PB\cdot PC}\\\\ (3) & 1=\frac {xy}{b^2}\cdot\frac {PB\cdot PC}{PM\cdot PN}\end{array}\right\|\ \bigodot $ $\implies$

$\sin (B-\alpha )\sin (B-\beta)=\sin \alpha\sin\beta$ $\iff$ $\cos (\alpha -\beta )-\cos [2B-(\alpha +\beta )]=\cos (\alpha -\beta )-\cos (\alpha +\beta )$ $\iff$

$\cos [2B-(\alpha +\beta )]=\cos (\alpha +\beta )$ $\iff$ $\alpha +\beta =B$ . Thus, $\boxed {m\left(\widehat{BPC}\right)=A+B}$ (constant).

Conclusion. Construct the rombus $ABDC$ . Observe that $D$ is $A$-exincenter $I_a$ and $m\left(\widehat{BDC}\right)=B$ . In conclusion,

the point $P$ belongs to the circle with diameter $[II_a]$ (where $I$ is incenter) so that the sideline $BC$ separates $P$ , $I_a$.

Very nice your proof without words (see lower), Sunken Rock ! Thank you. Here is what means $\alpha +\beta =B$ $\Longleftrightarrow$ $AR=CQ\ \wedge\ AQ=BR$ ,

i.e. $AR\parallel BQ\ \wedge\ AQ\parallel CR$ . Observe that $\left\|\begin{array}{c} \frac {x}{b-x}=\frac {b\cdot AR}{a\cdot BR}\\\\ \frac {y}{b-y}=\frac {b\cdot AQ}{a\cdot CQ}\end{array}\right\|\ \bigodot\ \implies\ \frac {xy}{(b-x)(b-y)}=\left(\frac ba\right)^2$ , i.e. $a^2xy=b^2(b-x)(b-y)$ .
This post has been edited 32 times. Last edited by Virgil Nicula, Nov 27, 2015, 7:00 AM

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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 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

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

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About Owner
  • Posts: 7054
  • Joined: Jun 22, 2005
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  • Blog created: Apr 20, 2010
  • Total entries: 456
  • Total visits: 404396
  • Total comments: 37
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