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Hand-in-Hand Teaching Guide to key in Mo Application

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Mo Application Demand Assistance

hello everyone in this lecture we're.going over 2010 Russia math Olympiad so.triangle ABC has parameter for points x.and y lie on raise a DNA C as you can.see here point x and y such that ax is.equal to a Y which is equal to 1 so.those distances are equal so we know.that this distance here a X is equal to.this segment here so you have an.isosceles triangle there segments BC and.XY they clearly intersect at this point.and so that I forget to point it out so.point M here um let's call it m so that.intersection point prove that the.perimeter of either triangle ABM or.triangle a cm is actually 2 so knowing.that the perimeter of triangle ABC is 4.and we need to check for the perimeter.ABM or let me actually first go ahead.and draw this am here um like that I.guess.ok so then it is quite plausible that.the UM it is really a BM which will be 2.here the perimeter right because a cm.would would be too large for that.purpose so therefore the statement which.is kind of bothering us this either-or.case here it is basically telling you.that if um if angle B is greater than.angle C then then it must be that the.perimeter of triangle a M is equal to 2.in the other opposite case where the.angle C is greater than M Ingleby then.it really means that the perimeter of.triangle AC M is the one which is.actually 2 so let me solve this problem.for this particular orientation where.angle B is larger than angle C therefore.I would like to show that.the perimeter of triangle ABM is.actually equal to two that's the result.that I would like to prove because we.are concerned about the parameters and.the parameter come brings to mind the.the notation s or little s is in this in.for this triangle ABC.we know the parameter is for therefore 2.s is equal to 4 or you can say s is.equal to 2 I also know occasions where.we can use the notation s it would be.related to portions of the law of the.edges between the vertices and the touch.points of the in-center or the X Center.in this particular case I will make use.of the X Center because I need to really.have this distance to so far I have a.distance of equal to 1 that's not really.useful for me so what I will do I will.go ahead and extend um this line segment.here so that I have my let me actually.go all the way here and I'll do.something similar here awesome so I have.my X circle so if that's the point X.here I'm going actually um an equal.distance beyond X so ax is 1 I want to.go all the way to let's call the point u.here um so I would say you would be.somewhere let's call it here so U is.such data so such that ax is actually.equal to AU here in a similar way I will.go beyond Y to a point Z where and so V.would be probably somewhere here that's.point V such that a Y is equal to oh.sorry for that so that's a typo a X is.equal to X u so sorry for that.and similarly a Y would be equal to Y Z.here okay so we I constructed U and V in.such a way because I know that the.perpendiculars from these to this point.and this point here will actually give.me my um my X center I sub a and I know.the point I sub a is along the internal.angle bisector a I answer okay so when I.draw this circle here um let me let me.do it like this I guess so it would be.something like this I would have a.tangency here then another one here and.that would that's kind of my X center so.let's say that the touch point is right.here um well it is okay something like.this so it's probably at this point or.let me actually fix my X center it will.be easier because okay so my X Center.should be somewhere um here I should say.okay there we go so that's my point.I sub a it's not very convincing okay.good and that's point here let me call.it point t recall a U is equal to two.notice that that's also equal to AV but.now we observe and so we have V u equals.B T because of the tangency here with.the X circle so you can see that note.that B U is equal to B T and in a.similar way and C V is equal to C T as.well but then this is kind of like a.game changer so au which is simply equal.to a B plus D T I already know is equal.to 2 so 2 is equal to AU which is equal.to a B plus B.see here in a similar way to is equal to.a V but a V is the same as a C plus C T.so therefore I can reduce the problem.here the perimeter of triangle ABM being.equal to 2 equivalently so I claim that.the problem reduces to showing that so.show that's an alternative formulation.of our problem so show that I'm a II.plus B M plus M a is equal to 2 well 2.is simply a you write but now I have.this new observation I can replace a you.with AE plus B T so I'm replacing it.with a b-plus on D T but then it.immediately implies those two are.cancelling out so therefore equivalently.what I want to show is that D M plus M a.is equal to B T here but wait a second.et itself can be split into B M plus M T.so that would further cancel things out.so therefore the problem can be.equivalently stated as and a is equal to.Mt so that's actually what the problem.wants us to compute so therefore the.problem is asking us to show that this.distance here is actually equal to this.distance if I can show ma is equal to Mt.then I'm done.ok so why would that be so here at this.moment I will introduce a new technique.here so I will consider the radical axis.of two circles you can say well I can.only you can only see one circle right.now which is the X circle here um but.then I'm also considering the circle.centered at Point a which have a radius.0 so I know this is a kind of surprising.result for a lot of you here so notice.that this set the circle centered at.a with radius zero and this X circle now.what is the radical axis of these two.circles obviously um all I do is I draw.the tangency s so the tangency of the X.circle and this circle here is you a and.V a and all I do is I locate the.midpoints which I already know are x and.y by construction recall that I.constructed u such that X would be the.midpoint or you can say ax is equal to X.U and similarly a Y is equal to Y V.therefore the radical axis of these two.circles namely the X circle here centre.that I sub a and the circle with with.Center a and radius zero radical axis is.X Y here but if that's the radical axis.and knowing that M is on that radical.axis it means that the power of point M.with respect to those two circles must.be equivalent so there are four hands so.let me write it out slide the page slide.here so we're almost done actually so I.know that the power of point at M with.respect to circle a Center that I sub a.has considered that circle Center that I.sub a we get the power of that point the.Power Point M with respect to that.excircles simply MT squared.and how about the power of point the.same point M with respect to that circle.centered at Point a with radius zero.it's simply a M a squared but I know.that those two entities must be equal to.each other because I know M is on the.radical axis so that actually proves.that after you take the square root of.both sides MT is in fact equal to M a.but that was exactly what we wanted to.show so if M T is in fact equal to M a.then the the perimeter of triangle a be.M a would be simply equal to a B be M.plus M T but the M plus M T is simply d.TB T is equal to bu therefore a B plus B.u which is simply a u is equal to two.and we're done.

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