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Source: BMO tst 2012-italy

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59 posts

bozzio

#1 h Jan 2, 2013, 7:24 PM • 1 Y

Y by Adventure10

ABC an acute triangle. $\omega$ is a circumference, with centre L on BC, which is tangent to AB and AC in B' and C' respectively. Suppose that the circumcircle of ABC has its centre on the smallest of the arc B'C'. Show that the circumcircle of ABC and $\omega$ meet in two distinct points.

This post has been edited 1 time. Last edited by bozzio, Jan 3, 2013, 10:00 AM

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#2 h Jan 3, 2013, 2:23 AM • 1 Y

Y by Adventure10

Suppose $X,Y$ be the projection of $L$ on $AB,BC$ now clearly $\angle{BOC}<\angle{XOY}$ now we've $\angle{BOC}=2A$ and $\angle{XOY}=\frac {\pi+A}{2}$ and hence $A<\frac {\pi}{6}$ now if they wouldn't cut each other then for any $P$ point lies on $\omega$ we would get $\angle{BPC}+A \geq \pi$ now take a special point $P$ on $\omega$ such that $OP$ is perpendicular to $BC$ . But now we've $LO=LP$ so $\angle{BOC}=\angle{BPC}$ and now $\angle{BPC}+A=\angle{BOC}+A=3A<\pi$ which is certainly a contradiction and hence done.

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bozzio

#3 h Jan 3, 2013, 10:10 AM • 1 Y

Y by Adventure10

I don't uderstand, you are calling X and Y the points which should be B' and C'? If yes why $\angle{XOY}=\frac{\pi +A}{2}$

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#4 h Jan 3, 2013, 11:18 AM • 2 Y

Y by Adventure10, Mango247

bozzio wrote:

I don't uderstand, you are calling X and Y the points which should be B' and C'? If yes why $\angle{XOY}=\frac{\pi +A}{2}$

Yes here $X=B'$ and $Y=C'$ and that is because $2\angle{XOY}+\angle{XLB}=2\angle{XOY}+\pi-A=2\pi$ so now that is true.

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bozzio

#5 h Jan 3, 2013, 1:07 PM • 1 Y

Y by Adventure10

subham1729 wrote:

Yes here $X=B'$ and $Y=C'$ and that is because $2\angle{XOY}+\angle{XLB}=2\angle{XOY}+\pi-A=2\pi$ so now that is true.

Thank you

anyway I think that the angle is $\angle{XLY}$ and not $\angle{XLB}$ . Furthermore I think that $A<\frac{\pi}{3}$ . I wish this will be very useful for my preparation to BMO tst this year

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