angle bisector wanted, circle related

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angle bisector wanted, circle related

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Source: 2017 France JBMO TST 3.1

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#1 h Jul 20, 2020, 5:13 AM

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Let $\Gamma$ be a circle with center $O$ ¸ and $M$ a point outside the circle. Let $A$ be a point of $\Gamma$ so that the line $MA$ is tangent to the circle. Let $B$ and $C$ be two points of the circle $\Gamma$ such that $B$ belongs to the arc $AC$ . Let $H$ be the projection of $A$ on $[MO]$ and $K$ the intersection of $[MO]$ with the circle $\Gamma$ . Prove that $(BK$ is the bisector angle $\angle HBM$

This post has been edited 2 times. Last edited by parmenides51, Mar 27, 2022, 12:08 PM

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#2 h Jul 20, 2020, 5:46 AM

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parmenides51 wrote:

Let $\Gamma$ be a circle with center $O$ ¸ and $M$ a point outside the circle. Let $A$ be a point of $\Gamma$ so that the line $MA$ is tangent to the circle. Let $B$ and $C$ be two points of the circle $\Gamma$ such that $B$ belongs to the segment $[AC]$ . Let $H$ be the projection of $A$ on $[MO]$ and $K$ the intersection of $[MO]$ with the circle $\Gamma$ . Prove that $(BK$ is the bisector angle $\angle HBM$

Wait so A,B, and C lie on $\Gamma$ but $B$ lies on [AC] how is that possible?
Or you have typo?

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#3 h Jul 20, 2020, 5:52 AM

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something is not correct , this is my source, problem 1

Quote:

Fie \Gamma un cerc de centru O si M un punct exterior cercului. Fie A un punct de pe de \Gamma astfel ıncat dreapta MA sa fie tangenta la cerc. Fie B si C doua puncte ale cercului \Gamma astfelıncat B sa apartina segmentului [AC]. Fie H proiect¸ia lui A pe [MO] si K intersectia lui [MO] cu cercul \Gamma . Demonstrati ca (BK este bisectoarea unghiului HBM.

this is a Romanian translation

This post has been edited 3 times. Last edited by parmenides51, Jul 20, 2020, 6:03 AM

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#4 h Jul 20, 2020, 5:54 AM • 1 Y

Y by Mango247

Uhmmm... i don`t get the language :wacko:

:wacko:

But problem really seems interesting, try to translate it correctly please :-D

:-D

This post has been edited 1 time. Last edited by CROWmatician, Jul 20, 2020, 5:56 AM
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#5 h Jul 20, 2020, 5:59 AM

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You can read French?!?

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#6 h Jul 20, 2020, 6:02 AM

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there is nothing suprising

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#7 h Jul 20, 2020, 6:25 AM

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Well, does he live in the US?’Cause I do

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#8 h Jul 20, 2020, 7:33 AM

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Senguamar wrote:

Well, does he live in the US?’Cause I do

He wrote that his location is in Greece

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#9 h Jul 20, 2020, 8:20 AM

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I am from Greece, this is irrelevant to the problem, here we do not chat, we try to solve a few problems, and correct it if something seems not to work, do not spoil the problems / spam the posts with irrelevant topics with respect to the problem, I didn't find a French source to check the original wording, just a romanian translation of the wording

Ps. I personally speak and understand only Greek and English and use Google translate for all the rest sources

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#10 h Jul 20, 2020, 9:43 AM • 1 Y

Y by parmenides51

I guess it need to be:

problem wrote:

In $\triangle ABC$ , $O$ - circumcenter. $MA$ - tangent to $(ABC)$ . $H$ - projection of $A$ on $MO$ . Segment $MO$ intersect $(ABC)$ at $K$ .
Prove that $BK$ - angle bisector of $\angle HBM$ .

Let $MO$ cut $(ABC)$ second time at $L$ . Then $(M,H;L,K)=-1$ .
With $\angle KBL=90^\circ$ we get what we need.

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#11 h Jul 20, 2020, 10:35 PM

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In the original french text, the same problem / typo exists

Quote:

. Soit C un cercle de centre O, et M un point exterieur au cercle. Soit A un point de C tel que (MA) soit tangente au cercle. Soient B et C deux points de C tels que B appartienne au segment [AC]. Soit H le projete orthogonal de A sur [MO]. Soit K le point d’intersection de [MO] avec C. Montrer que (BK) est la bissectrice de HBM .

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#12 h Mar 27, 2022, 12:07 PM

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parmenides51 wrote:

Let $\Gamma$ be a circle with center $O$ ¸ and $M$ a point outside the circle. Let $A$ be a point of $\Gamma$ so that the line $MA$ is tangent to the circle. Let $B$ and $C$ be two points of the circle $\Gamma$ such that $B$ belongs to the arc $AC$ . Let $H$ be the projection of $A$ on $[MO]$ and $K$ the intersection of $[MO]$ with the circle $\Gamma$ . Prove that $(BK$ is the bisector angle $\angle HBM$

I have just changed the word segment into the word arc in order the problem to be correct as it had a typo,
it was validated by geogebra

This post has been edited 2 times. Last edited by parmenides51, Mar 27, 2022, 12:08 PM

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