IMO 2012: Day 2 Problem 5

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Moon
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IMO 2012: Day 2 Problem 5

Unread post by Moon » Thu Jul 12, 2012 1:02 am

Let $ABC$ be a triangle with $\angle {ACB}=90^0$ and let $D$ be the foot of the altitude from $C$. Let $X$ be a point in the interior of the segment $CD$. Let $K$ be the point on the segment $AX$ such that $BK=BC$. Similarly, let $L$ be the point on the segment $BX$ such that $AL=AC$. Let $M$ be the point of intersection of $AL$ and $BK$.

Show that $MK=ML$.
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Re: IMO 2012: Day 2 Problem 5

Unread post by SANZEED » Thu Jul 12, 2012 1:24 am

Can anyone please post an image? I am having a problem. :oops:
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Re: IMO 2012: Day 2 Problem 5

Unread post by SANZEED » Thu Jul 12, 2012 3:48 am

I think I did it too,but latexing is the problem because my net connection is slow. It seems that to show that $MK$ and $ML$ touches the same circle and we need angle chasing vitally.However,I am a bit confused about what I have done.I'm trying to latex my solution as early as possible. :oops:
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Re: IMO 2012: Day 2 Problem 5

Unread post by Tahmid Hasan » Thu Jul 12, 2012 6:57 pm

First we construct three circles:the circumcircles of $\triangle ABC(\Gamma),$,The circle($\alpha$) with centre $A$ and radius $AC$ and the circle($\beta$) with centre $B$ and radius $BC$.
Note that the centre of $\Gamma$ lies on the midpoint of $AB$,so the three circles are co-axial with radical axis $CD$.
Let $AX \cap \Gamma =Y \neq A,BX \cap \Gamma =Z \neq B,AY \cap BZ =P$.
so $\angle AYB =\angle AZB=90^{\circ}$,so $X$ must be the orthocenter of $\triangle ABP$ implying $P$ lies on the radical axis.
let us denote $PWR(P)$ by the power of $P$ wrt $\alpha,\beta,\Gamma$.[which are equal]
From similar triangles $ABC,ACD$ we get $AC^2=AD*AB=AL^2 \Rightarrow \angle ALD = \angle ABL$
similarly $\angle BKD = \angle BAK$
now $\angle APD=\angle ABZ=\angle ALZ$ implying $ADLP$ is cyclic.
so $AL \bot PL$.....(1) meaning $PL^2=PWR(P)$.
similarly we get $BK \bot PK$.....(2) meaning $PK^2=PWR(P)$
hence $PK=PL$.....(3)
so $\triangle MKP \cong MLP$[using (1),(2),(3)]
proving $MK=ML$
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Re: IMO 2012: Day 2 Problem 5

Unread post by Tahmid Hasan » Thu Jul 12, 2012 6:59 pm

Although I used a reconstructed version pf SANZEED's hint,I didn't need to use it.It is to be noted that I also posted the same solution in AOPS.
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Re: IMO 2012: Day 2 Problem 5

Unread post by SANZEED » Thu Jul 12, 2012 7:32 pm

Well,at last I afford to post it (Gracious Sanzeed!). Here it is:
$MK$ and $ML$ are both tangents to a circle.let $BX$ meets the circle with center $A$ and radius $AC$ at J And $AX$ meets circle with center $B$ and radius $BC$ at $I$. Now,since $CD$ is perpendicular on the joining segments of the centers $A,B$,clearly it is the common cord and the radical axis.So,we deduce that $IX\times XK=CX\times XZ=JX\times XL$,where $Z$ is the second point of intersection of the two circles.Easily we can find that $JKLI$ is cyclic .Let us name this that have these four points on itself circle $\tau$.Easily we can find that $BK$ is tangent to $\tau$ (because we have $BC^{2}=Bl\times BJ$ and we have $BC =BK$,so $BK^{2}=BL\times BJ$). With a similar work we can see that $AL$ is tangent to the $\tau$ too.So $MK$ and $ML$ are both tangents to the circle $\tau$. So $MK=ML$.
P.S.I think the idea here is to note the equal segments,tangents and some right angles. At least I noticed that and used the circles and cyclic quad and the tangents.And I think it is enough to work with power of points. ;)
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