IMO-2001-G1

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Tahmid Hasan
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IMO-2001-G1

Unread post by Tahmid Hasan » Sat Jun 02, 2012 2:04 pm

Let $A_1$ be the center of the square inscribed in acute triangle $ABC$ with two vertices of the square on side $BC$.Thus one of the two remaining vertices of the square is on side $AB$ and the other is on $AC$.Points $B_1,C_1$ are defined in a similar way for inscribed squares with two vertices on sides $AC$ and $AB$, respectively.Prove that lines $AA_1,BB_1,CC_1$ are concurrent.
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Tahmid Hasan
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Location: Khulna,Bangladesh.

Re: IMO-2001-G1

Unread post by Tahmid Hasan » Sat Jun 02, 2012 2:06 pm

I used a tricky idea.
Erect a square on $BC$ with side length $BC$ out side the triangle,let it's centre be $A_2$,use homothety to prove $A,A_1,A_2$ are collinear.Hence the problem diverts into proving $AA_2,BB_2,CC_2$ are concurrent which is a trivial transformation problem. :)
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SANZEED
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Re: IMO-2001-G1

Unread post by SANZEED » Wed Jun 06, 2012 11:43 pm

My idea uses ratios:
Let $MNPQ$ be the square where $M\in AB,N\in AC$. Let $AA_{1}$ bisect $MN,PQ$ at $X,Y$ respectively.Now it is easy to show that $\triangle MXA_{1},\triangle PYA_{1}$ are congruent. Similarly $\triangle NXA_{1},\triangle QYA_{1}$ are congruent. Now,$\frac {BY}{CY}=\frac{MX}{NX}=\frac{BY+YP}{CY+YQ}=\frac{BQ+QP}{CP+PQ}=\frac{MN cotB+MN}{MN cotC+MN}=\frac{cotB+1}{cotC+1}$. Derive similar expressions for other cevians and use Ceva's theorem.
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RC.
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Re: IMO-2001-G1

Unread post by RC. » Mon Jul 10, 2017 3:12 pm

I however have an another idea. The first user had told some fractions of it but then he told to do a question... Which is generally solved using trigonometry or cevians. Here is

Let X on AB and Y on AC be the vertices of square on side BC. Draw AA_1.Join XA_1 and YA_1, let *P* be a point on AA_1 such that BP || XA_1 then CP || XA_1 . Similarly define the points *Q* and *R* with respect to vertices B and C and lying on BB_1 and CC_1 respectively. Now, the game is starting.... Draw a circle centered at Q passing through A and C similarly draw a circle centered at R and passing through A and B. Let the two circles centered at Q and R intersect at some point D other than the point A.
Angle chasing tells that PBDC is cyclic. Therefore angle PCB = PDB = 45 nad ADB = 135 so *ADP* must for a straight line but AD is also the common chord of circles centered at R and Q. Therefore, AP perpendicular RQ.
Similarly, BQ perpendicular PR and CR perpendicular PQ.
Therefore, AP,BQ,RC are altitudes of DELTA PQR. Therefore, they should be concurrent.
One can back up the first part
using homothety. You can see that the transformation which I have done is actually a homothety centered at the vertices.

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