incenter, circumcenter symmetric in isosceles (Oral Moscow Team MO 2006.1.8B6)

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incenter, circumcenter symmetric in isosceles (Oral Moscow Team MO 2006.1.8B6)

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#1 h Dec 2, 2020, 6:33 PM • 1 Y

Y by Mango247

Find the angles of an isosceles triangle with the centers of the inscribed and circumscribed circle symmetric wrt the base.

Remarks

This post has been edited 1 time. Last edited by parmenides51, Dec 2, 2020, 6:33 PM

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#2 h Dec 2, 2020, 6:45 PM

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Well maybe with symmetric they meant homothetic? I'm not sure of it though

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#3 h Dec 2, 2020, 6:49 PM • 1 Y

Y by Mango247

any noncongruent 2 circles are considered homothetic, so 'the base'' term seams irrelevant

This post has been edited 1 time. Last edited by parmenides51, Dec 2, 2020, 7:50 PM
Reason: added the word ''noncongruent'' , to be correct

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#4 h Dec 2, 2020, 6:56 PM

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Maybe the incenter is a reflection of the circumcenter over the base? That's how I interpreted it.

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#5 h Dec 2, 2020, 7:04 PM

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Well it is simpler than I thought

So notice that we need to find the $\angle ABC$ in the given situation, that the centers coincide.

It is just simply counting in two ways,
Let the center be $O$ .
$\angle BOC=90-\tfrac{A}{2}=2A$ Solving this we get $A=36$ so $\angle ABC=72$

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#6 h Dec 2, 2020, 7:08 PM

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parmenides51 wrote:

any 2 circles are considered homothetic, so 'the base'' term seams irrelevant

Not just any two circles are homothetic! The incircle and excircle being homothetic is a really special scenario. Generally, the incircle and the excircle are always homothetic. Not the incircle and the circumcircle.

This post has been edited 1 time. Last edited by Repzram06, Dec 2, 2020, 7:16 PM
Reason: fixed errors

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#7 h Dec 2, 2020, 7:19 PM

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All noncongruent circles are homothetic. You can find the center of homothety by finding the intersection of the common external tangents.

This post has been edited 2 times. Last edited by natmath, Dec 2, 2020, 7:27 PM

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#8 h Dec 2, 2020, 7:26 PM

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@above yeah I agree to that but in case of the incircle and circumcircle there is no common tangents. So they cannot be homothetic right?

Repzram06 wrote:

parmenides51 wrote:

any 2 circles are considered homothetic, so 'the base'' term seams irrelevant

Not just any two circles are homothetic! The incircle and excircle being homothetic is a really special scenario. Generally, the incircle and the circumcircle are always homothetic. Not the incircle and the circumcircle.

I got what mistake I was doing here. The fact that this is related to homothety is actually comletely wrong.

However in my opinion, the proof is correct. This could be a two marker in the PRMOs

This post has been edited 1 time. Last edited by Repzram06, Dec 2, 2020, 7:34 PM
Reason: Circumcircles

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#9 h Dec 2, 2020, 7:28 PM

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Repzram06 wrote:

Well it is simpler than I thought

So notice that we need to find the $\angle ABC$ in the given situation, that the centers coincide.

It is just simply counting in two ways,
Let the center be $O$ .
$\angle BOC=90-\tfrac{A}{2}=2A$ Solving this we get $A=36$ so $\angle ABC=72$

I think you are making a mistake. $\Delta ABC$ should not be an acute triangle because the circumcenter and incenter must be on different sides of the triangle.

my solution

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#10 h Dec 2, 2020, 7:31 PM

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Repzram06 wrote:

@above yeah I agree to that but in case of the incircle and excircle there is no common tangents. So they cannot be homothetic right?

Aren't the sides of the triangle external common tangents of the incircle and excircle.

Also why are you talking about the excircle? The problem only contains information on the incircle and circumcircle?

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#11 h Dec 2, 2020, 7:33 PM

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natmath wrote:

Repzram06 wrote:

@above yeah I agree to that but in case of the incircle and excircle there is no common tangents. So they cannot be homothetic right?

Aren't the sides of the triangle external common tangents of the incircle and excircle.

Also why are you talking about the excircle? The problem only contains information on the incircle and circumcircle?

Ugh sorry about that I meant the circumcircle.

Also @above could you please explain what does one mean by symmetric in the problem? I think my errors are due to not knowing their definition.

This post has been edited 1 time. Last edited by Repzram06, Dec 2, 2020, 7:36 PM
Reason: Additionals

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#12 h Dec 2, 2020, 7:53 PM

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@above I was going off of parmenides51 remarks because I thought that was the only feasible interpretation.

I assumed that the reflecting $I$ about the base brought $I$ to $O$ . Where $I$ and $O$ are the incenter and circumcenter respectively.

So if you look here, you will see that the perpendicular bisector of $OI$ is the same as line $BC$ . Hence the reflection of $O$ about $BC$ is $I$ and vice versa

This post has been edited 1 time. Last edited by natmath, Dec 2, 2020, 7:59 PM

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#13 h Dec 2, 2020, 8:05 PM

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I thought that it means that the centers are coinciding

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#14 h Dec 2, 2020, 8:07 PM

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If the circumcenter and incenter coincide, then isn't it an equilateral triangle?

I think symmetric, in this case, means like a reflection symmetry.

This post has been edited 1 time. Last edited by natmath, Dec 2, 2020, 8:07 PM

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