270. Morrocan M.O. 2010 and Croatia TST 2009.

by Virgil Nicula, Apr 30, 2011, 6:01 PM

Morrocan M.O. 2010. Let $ABC$ be a triangle. The interior bisector of the angle $\angle BAC$ cuts $[BC]$ in $L$ and the circumcircle $w$ of

$\triangle ABC$ in $D$ . The perpendicular to $(AC)$ going through $D$ cuts $[AC]$ in $M$ . Find the value of $\frac{AM}{MC}$ knowing that $\frac{LB}{LC}=\frac{1}{2}$ .

Proof. Suppose w.l.o.g. $b>c$ , i.e. $M\in (AC)$ . Denote $N\in AB$ , $DN\perp AB$ . Thus, $B\in (AN)$ and $\triangle BDN\sim\triangle CDM$

because these triangles are right-angled and $BD=CD$ , $DN=DM$ . Therefore, $BN=CM=x\implies$ $\left\|\begin{array}{c} AM=b-x\\\ AN=c+x\end{array}\right\|$

and $AM=AN\implies$ $b-x=c+x\implies x=\frac {b-c}{2}$ . In conclusion, $MC=\frac {b-c}{2}$ and $AM=\frac {b+c}{2}$ $\implies$ $\boxed{\frac {MA}{MC}=\frac {b+c}{b-c}}$ .

In our particular case $\frac {LB}{LC}=\frac {AB}{AC}=\frac cb=\frac 12\ \ (b>c)$ obtain that $\boxed{\frac {MA}{MC}=\frac {b+c}{b-c}}=\frac {1+\frac cb}{1-\frac cb}=\frac {1+\frac 12}{1-\frac 12}=3\ .$

Remarks.

$\blacktriangleright$ If $\left\|\begin{array}{cc} P\in AL\ ; & BP\perp AL\\\\ N\in BC\ ; & AN\perp AL\end{array}\right\|$ , then $ABL\sim ADC\implies$ $\frac {MA}{MC}=\frac {PA}{PL}=$ $\frac {BN}{BL}=$ $\frac {\frac {ac}{|b-c||}}{\frac {ac}{b+c}}\implies$ $\frac {MA}{MC}=\frac {b+c}{|b-c|}$ . Otherwise, $Q\in BP\cap AC\implies$

$ AQ=c$ and $\widehat{CBP}\equiv\widehat{CDM}\implies$ the point $K\in BP\cap DM$ belongs to $w$ . In conclusion, $DB=DC\implies MQ=MC\implies$ $AM=\frac {b+c}{2}$ .

$\blacktriangleright\ MA=DA\cdot\cos\frac A2=$ $2R\sin\left(C+\frac A2\right)\cos\frac A2=$ $R(\sin B+\sin C)=$ $\frac {b+c}{2}$ . In conclusion, $\left\|\begin{array}{c} MA=\frac {b+c}{2}\\\ MC=\frac {|b-c|}{2}\end{array}\right\|$ and required ratio $\frac {MA}{MC}=\frac {b+c}{|b-c|}$ .

$\blacktriangleright\ \frac {MA}{MC}=\frac {DA}{DC}\cdot\frac {\sin\widehat{MDA}}{\sin\widehat{MDC}}=$ $\frac {\sin \widehat{ACD}}{\sin\widehat{CAD}}\cdot\frac {\sin\widehat{MDA}}{\sin\widehat{MDC}}=$ $\frac {\sin\left(C+\frac A2\right)}{\sin\frac A2}\cdot\frac {\cos\frac A2}{\left|\cos\left(C+\frac A2\right)\right|}=$ $\frac {\left|\tan\left(C+\frac A2\right)\right|}{\tan\frac A2}=$ $\frac {\tan\frac {B+C}{2}}{\tan\left|\frac {B-C}{2}\right|}=\frac {b+c}{|b-c|}$ .

$\blacktriangleright\ ABD\sim ALC\implies AD\cdot AL=bc\implies$ $AM=AD\cdot\cos\frac A2=\frac {bc}{AL}\cdot\cos\frac A2$ . But $AL=\frac {2bc}{b+c}\cdot\cos\frac A2\implies$ $AM=\frac {b+c}{2}$ a.s.o.

$\blacktriangleright$ If $P\in AL$ , $BP\perp AL$ , then $ABL\sim ADC\implies$ $\frac {MA}{MC}=\frac {PA}{PL}=$ $\frac {BA}{BL}\cdot\frac {\sin\widehat{PBA}}{\sin\widehat{PBL}}=$ $\frac {b+c}{a}\cdot\frac {\cos\frac A2}{\left|\cos\left(C+\frac A2\right)\right|}=$ $\frac {b+c}{a}\cdot\frac {\sin A}{\left|\sin B-\sin C\right|}=$ $\frac {b+c}{|b-c|}$ .

$\blacktriangleright\ \frac {b+c}{b-c}=\frac {\sin B+\sin C}{\sin B-\sin C}=$ $\frac {2\sin\frac {B+C}{2}\cos\frac {B-C}{2}}{2\sin\frac {B-C}{2}\cos\frac {B+C}{2}}=$ $\frac {\tan\frac {B+C}{2}}{\tan\frac {B-C}{2}}$ . Observe that $\left(C+\frac A2\right)+\frac {B-C}{2}=90^{\circ}\implies$ $\tan\left(C+\frac A2\right)\cdot\tan\frac {B-C}{2}=1$ .

$\blacktriangleright$ Mollweide's formulas $\frac {b+c}{a}=\frac {\cos \frac {B-C}{2}}{\sin\frac A2}$ and $\frac {b-c}{a}=\frac {\sin\frac {B-C}{2}}{\cos\frac A2}$ . These sometimes referred to in older texts as Mollweide's equations named after Karl Mollweide.


Croatia TST 2009. In $\triangle ABC$ with incenter $ I$ and $ c>b$ denote $ D\in (AB)$ , $E\in AA$ for which $ AD=AE=b$ and $ AB$ doesn't separate $ C$ , $ E\implies I\in DE$ .

Proof 1 (metric). $ X\in AI\cap DE$ and $ Y\in BC\cap AI\ \implies\ m(\angle BAE)=A+B$ , $ YD=YC=\frac {ab}{b+c}$ , $ m(\angle ADY)=C$ .

Thus, $ DY\parallel AE\ \implies\ \frac {XA}{XY}=\frac {AE}{DY}=$ $ \frac {b}{\frac {ab}{b+c}}\ \implies\ \frac {XA}{XY}=\frac {b+c}{a}$ $ \stackrel{\mathrm{Van\ Aubel}}{\ \ \implies\ \ }X: =I$ . In conclusion, $ I\in DE$ .

Proof 2 (synthetic). Denote $X\in DE\cap AI$ . Observe that $m\left(\widehat{XAC}\right)=\frac A2$ , $m\left(\widehat{CAE}\right)=B$ , $m\left(\widehat{DAE}\right)=A+B$ , $AD=AE\implies$ $m\left(\widehat{XEA}\right)=\frac C2$ ,

$AC=AE\implies$ $m\left(\widehat{AEC}\right)=90^{\circ}-\frac B2$ $\implies$ $m\left(\widehat{XEC}\right)=$ $m\left(\widehat{AEC}\right)-$ $m\left(\widehat{AEX}\right)=$ $90^{\circ}-\frac B2-\frac C2=$ $\frac A2=$ $m\left(\widehat{XAC}\right)$ $\implies$

$\widehat{XEC}\equiv\widehat{XAC}$ $\implies$ the quadrilateral $XAEC$ is cyclically $\implies$ $m\left(\widehat{ACX}\right)=m\left(\widehat{AEX}\right)=\frac C2\implies$ $X\equiv I$ $\implies$ $I\in DE$ .

Proof 3 (trigonometric). Denote $X\in AI\cap DE$ . Observe that $\frac {XD}{XE}=\frac {AD}{AE}\cdot\frac {\sin \widehat {XAD}}{\sin\widehat{XAE}}=$ $\frac {\sin\frac A2}{\sin\left(B+\frac A2\right)}=$ $\frac {2\sin \frac A2\cos\frac A2}{2\sin\left(B+\frac A2\right)\cos\frac A2}=$

$\frac {\sin A}{\sin B+\sin C}=$ $\frac {a}{b+c}$ . Thus, $\frac {XD}{a}=\frac {XE}{b+c}=\frac {DE}{2s}$ . Since $DE=2\cdot AD\cdot\sin\widehat{DAE}=$ $2b\cdot\sin\frac {A+B}{2}\implies$ $DE=2b\cdot \cos\frac C2$

obtain that $XD=\frac {ab\cdot\cos\frac C2}{s}$ . Therefore, $\frac {AX}{\sin \frac C2}=\frac {XD}{\sin\frac A2}=$ $\frac {ab\cdot\cos\frac C2}{s\cdot \sin\frac A2}\implies$ $AX=\frac {ab\cdot\sin \frac C2\cos\frac C2}{s\cdot \sin\frac A2}=$ $\frac {ab\cdot\sin C}{2s\cdot \sin\frac A2}=$ $\frac {S}{s\cdot \sin\frac A2}$

$\implies$ $AX=\frac {r}{\sin\frac A2}$ . Since $\sin\frac A2=\frac {r}{AI}$ obtain that $AX=AI$ $\iff$ $X\equiv I$ , i.e. $I\in DE$ .
This post has been edited 48 times. Last edited by Virgil Nicula, Nov 22, 2015, 8:32 AM

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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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    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

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    -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

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

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About Owner
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  • Joined: Jun 22, 2005
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  • Blog created: Apr 20, 2010
  • Total entries: 456
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