Lithuania TS 2009

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cashaz

#1 h Dec 29, 2009, 9:39 AM • 2 Y

Y by Adventure10, Mango247

In triangle ABC, angle A=30, points S and I are ABC circumcentre and incentre respectively. Points D and E are on sides BA and CA respectively such that BD=CE=BC. Prove that SI is perpendicular and equal to DE.

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cashaz

#2 h Jan 1, 2010, 6:27 PM • 2 Y

Y by Adventure10, Mango247

so if there is anyone who can help me and post the solution?

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7054 posts

#3 h Jan 1, 2010, 7:53 PM • 2 Y

Y by Adventure10, Mango247

cashaz wrote:

Let $ABC$ be a triangle with the circumcentre $O$ and the incentre $I$ . Suppose that exist $D\in (AB)$

and $E\in (AC)$ so that $BD=CE=BC$ . Prove that $\|\ \begin{array}{cc} 1. & OI\perp DE\\\ 2. & A=30^{\circ}\stackrel{?!}{\Longrightarrow} DE=OI\end{array}$ .

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4394 posts

#4 h Jan 1, 2010, 8:34 PM • 1 Y

Y by Adventure10

1) $DE \perp OI$ is easy to prove, and already old; I think the easiest way is to prove $DI^2-EI^2=DO^2-EO^2=a\cdot (c-b)$ , by drawing the perpendiculars from $I$ and $O$ onto $AB$ and $AC$ .
2) It's easy as well to see that $\angle DIE = 45^\circ$ and $\angle DOE = 135^\circ$ , hence $O$ is the orthocenter of $\triangle DIE$ , done.

Best regards,
sunken rock

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4149 posts

#5 h Jan 1, 2010, 10:24 PM • 2 Y

Y by Adventure10, Mango247

There's a more general result coming from this configuration.

Proposition. Let $\triangle ABC$ be a scalene triangle and $M,N$ are two points on $AC, AB,$ such that $BN = CM = BC.$ Then $MN$ is perpendicular to $OI,$ where $O , I$ are the circumcenter and incenter of $\triangle ABC$ and $\frac {_{MN}}{^{BC}} = \frac {_{OI}}{^{R}}.$

Proof. In the isosceles triangles $\triangle OAC$ and $\triangle OAB,$ we have

$R^2 - OM^2 = AM \cdot MC , \ R^2 - ON^2 = AN \cdot BN$

$\Longrightarrow ON^2 - OM^2 = AM \cdot MC - AN \cdot BN$

$\Longrightarrow ON^2 - OM^2= (CA - BC)BC - (AB - BC)BC = BC(CA - AB)$

Since $\triangle MIC$ and $\triangle MIB$ are isosceles, we have $IM = IB,$ $IN = IC.$ Then

$IN^2 - IM^2 = IC^2 - IB^2 = BC(CA - AB)$

$\Longrightarrow ON^2 - OM^2 = IN^2 - IM^2 \Longrightarrow IO \perp MN$

On the other hand, $\angle NBM = \angle OCI = 90^{\circ} - \frac {_1}{^2}\angle C - \angle A.$ If $L \equiv{} CI \cap BM,$ then it follows that $\angle NMB$ and $\angle OIC$ are supplementary, which implies that $\triangle IOC$ and $\triangle MNB$ are pseudo-similar, thus

$\frac {MN}{OI} = \frac {BN}{OC} \Longrightarrow \frac {MN}{BC} = \frac {OI}{R}.$

Note that $\angle A = 30^{\circ} \Longleftrightarrow BC = R \Longleftrightarrow DE = OI$ (the proposed problem).

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27 posts

cashaz

#6 h Jan 4, 2010, 9:56 AM • 2 Y

Y by Adventure10, Mango247

Quote:

sunken rock Re: Lithuania TS 2009
Nice solution

1) is easy to prove, and already old; I think the easiest way is to prove , by drawing the perpendiculars from and onto and .
2) It's easy as well to see that and , hence is the orthocenter of , done.

Best regards,
sunken rock

can you explain me how you get DIE=45 and all yout second part of the solution.

Thnx luis, very interesting solution.

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MJ GEO

#7 h Jan 4, 2010, 10:22 AM • 2 Y

Y by Adventure10, Mango247

its interesting and easy to prove that $OI=R_ADE$

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1312 posts

#8 h Jan 4, 2010, 12:23 PM • 2 Y

Y by Adventure10, Mango247

In $\triangle ABC$ , $\angle A = 30^\circ$ , the points $O$ and $I$ are the circumcentre and incentre of $\triangle ABC$ respectively. Points $D$ and $E$ are on the sides $BA$ and $CA$ respectively such that $BD = CE = BC$ . Prove that $OI$ is perpendicular and equal to $DE$ .
(I'm used to $O$ instead of $S$

)

I have a solution using trigonometry:
Let $\angle DBC = \beta, \angle ECB = \gamma$ , then $\beta + \gamma = 150^{\circ}$ . We easily find $\angle OBI = 60^\circ - \frac {\beta}{2} = \frac {\gamma}{2} - 15^\circ = \angle DCE$ . Let $X$ be the point on $DC$ , such that $BI = CX$ . Then $\triangle CXE$ is congruent to $\triangle BIO$ . Let $\angle ECD = u$ . Then
$IO^2 = EX^2 = CX^2 + EC^2 - 2 CX \cdot EC \cdot \cos u$ , $DE^2 = DC^2 + EC^2 - 2DC \cdot EC \cdot \cos u$ .
So $IO^2 = DE^2 \iff$
$BI^2 + EC^2 - 2BI \cdot EC \cdot \cos u = DC^2 + EC^2 - 2DC \cdot EC \cdot \cos u \iff$
$(BI - DC)(BI + DC - 2EC \cdot \cos u) = 0$
I will prove $BI + DC = 2EC \cdot \cos u$ , which implies $IO = DE$ .
Let $BD = BC = CE = BO = OC = t$ .
From the law of sines relation we get: $BI = \frac {\sin \frac {\gamma}{2}}{\sin 75^\circ} t$ , since $\angle BCI = \frac {\gamma}{2}$ , and $\angle BIC = 105^\circ$ .
Using the law of sines once again: $CD = \frac {\sin \beta}{\sin 90^\circ - \frac {\beta}{2}} t = 2\cos \left ( \frac {\beta}{2} \right )t$ .
Letting $x = \frac {\beta}{2}$ , and using $\beta + \gamma = 150^\circ$ , we see that $BI + DC = 2EC \cdot \cos u \iff$
$\frac {\sin 75^\circ - x}{\sin 75^\circ} + 2 \cos x = 2 \cos 60^\circ - x$ .
But using the relations $\sin a + b = \sin a \cos b + \sin b \cos a$ , $\cos a + b = \cos a \cos b - \sin a \sin b$ , and $\cot 75^\circ = 2 - \sqrt {3}$ it is easy to see that both sides equal $\sin x + \sqrt {3}\cos x$ .
Hence $DE = IO$ . But then we also have $DE = EX$ . Let $v = \angle EDC = angle EDX = \angle DXE = 180^\circ - \angle EXC = 180^\circ - \angle BID$ .
$EC$ can be obtained by rotating and moving $BC$ with an angle of $ 180^\circ - \ECB = 180^\circ - \gamma$ in positive direction. $DE$ can be obtained by rotating, moving, and scaling $CE$ with an angle of $180^\circ - \angle DEC = \angle EDC + \angle DCE$ , so $DE$ can be obtained by rotationg, moving, and scaling $BC$ with an angle of $90^\circ + \left ( v + \frac {\beta}{2} \right )$
Similarily $OI$ can be obtained by rotating, moving and scaling $BC$ with an angle of $v + \frac {\beta}{2}$ . It follows that $ED$ can be obtained by rotating, moving, and scaling $OI$ with and angle of $90^\circ$ , and hence $ED \perp OI$ . And we are done

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4394 posts

#9 h Jan 4, 2010, 5:03 PM • 2 Y

Y by Adventure10, Mango247

To expand little bit my compressed solution:
Due to symmetry the following angle equalities hold:
$\angle BID = \angle BIC = \angle CIE = 90^\circ + \frac{1}{2} \cdot m(\angle A)= 105^\circ$ $\implies \angle DIE = 360^\circ - 3 \cdot 105^\circ = 45^\circ$ .
$\angle BOC = 60^\circ$ , as $\triangle BOC$ is equilateral; $\angle OBD + \angle OCE = 180^\circ - \angle A - \angle OBC - \angle OCB = 30^\circ$ . The triangles $OBD$ and $OCE$ are isosceles, hence $\angle BOD = 90^\circ - \frac{1}{2} \cdot \angle OBD$ , $\angle COE = 90^\circ - \frac{1}{2} \cdot \angle OCD$ , so $\angle BOD + \angle COE = 180^\circ - \frac {\angle OBD + \angle OCE}{2} = 165^\circ$ , consequently $\angle DOE = 360^\circ - (\angle BOD + \angle COE + \angle BOC ) = 135^\circ$ . Under the circumstances $O$ is the orthocenter of $\triangle DIE$ ( $O$ is placed on the altitude from $I$ and $\angle DIE + \angle DOE = 180^\circ$ , hence it's true, as claimed).

I hope to be all clear now!

Best regards,
sunken rock

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27 posts

cashaz

#10 h Jan 5, 2010, 9:49 AM • 2 Y

Y by Adventure10, Mango247

Quote:

I hope to be all clear now!

yes its all clear now, thanks a lot

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behdad.math.math

255 posts

behdad.math.math

#11 h Jan 6, 2010, 8:26 PM • 2 Y

Y by Adventure10, Mango247

$\triangle ABC$ , $D$ is on $AC$ and $E$ is on $BC$ such that $AD=AB=BE$ . $O$ , $I$ are circumcenter and incenter of $\triangle ABC$ respectively.prove that
1) $OI \bot DE$
2)the radius of circumcircle of $\triangle CDE$ is equal to $OI$ .

solution 1)
$M$ is the midpoint of arc $BC$ , $K$ is the midpoint of $BC$ , $AM$ intersect $BD$ at $N$ , $R$ is the radius of circumcircle of $\triangle ABC$ .
$MI$ = $2R$ $\sin\frac{A}{2}$ .
$BD$ = $2BN$ = $2AB\sin\frac{A}{2}$ .
Now it is easy to compute that $\frac{BD}{MI}$ = $\frac{BE}{OM}$ (1)
$\triangle EDB$ and $\triangle OIM$ : $OM \bot BE$ , $IM \bot DB$ (2)
(1),(2) $\Longrightarrow$ $\triangle IMO$ $\sim$ $\triangle DBE$
Now $\triangle IMO$ and $\triangle DBE$ are similar $\Longrightarrow$ $IM \bot BE$ , $BE \bot OM$ $\Longrightarrow$ $OI \bot DE$
solution 2)
$\triangle IMO$ $\sim$ $\triangle DBE$ $\longrightarrow$ $\frac{DE}{OI}$ = $\frac{BE}{OM}$ = $\frac{AB}{R}$
$\frac{AB}{R}$ = $2\sinh$ $\Longrightarrow$ $\frac{DE}{OI}$ = $2\sinh$ . radius of circumcircle $CDE$ =$ R^'$.
$ \frac{DE}{R^'}$= $2\sinh$ , hence $\frac{DE}{OI}$ = $ \frac{DE}{R^'}$ $\Longrightarrow$ $ OI=R^'$

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9796 posts

jayme

#12 h Apr 30, 2011, 11:39 AM • 2 Y

Y by Adventure10, Mango247

Dear Mathlinkers,
an article concerning some results about “two equal segments on two sides of a triangle” has been put on my website
http://perso.orange.fr/jl.ayme vol. 6 Deux segments égaux sur deux côtés d'un triangle p. 23
You can use Google translator
Sincerely
Jean-Louis

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