India TST 2009 DAY 1:PROBLEM 1

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India TST 2009 DAY 1:PROBLEM 1

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#1 h May 14, 2009, 10:44 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Let $ABC$ be a triangle with $\angle A = 60^{\circ}$ .Prove that if $T$ is point of contact of Incircle And Nine-Point Circle, Then $AT = r$ ,
$r$ being inradius.

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Ahiles

#2 h May 14, 2009, 12:19 PM • 2 Y

Y by Adventure10, Mango247

1. Points $B,O,I,H,C$ are conciclic

Proof

2. $AH = OH$

Proof

3. $OI = IH$

Proof

4. $T$ is midpoint of $AI$

Proof

Therefore, $AT = TI = r$ .

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#3 h May 16, 2009, 8:56 AM • 2 Y

Y by Adventure10, Mango247

my solution:
Prove that N lies on AI. so A,T,I are colliner.AI=2r,Done.

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#4 h Jun 16, 2009, 5:34 PM • 4 Y

Y by Adventure10, Mango247, and 2 other users

Another solution

Let $T'$ be the midpoint of $AI$ . Drop perpendicular $IX$ to $AB$ . right triangle $AIX$ has $\angle A = 30^0$ So $AI = 2IX = 2r$ . So $AT' = r$ .
Let $D,E,F$ be the respective midpoints of sides $BC,CA,AB$ of triangle $ABC$

It is sufficient to show $T = T'$ .
Consider the homothety with ratio $\frac {1}{2}$ centered at $A$ .
$B$ gets mapped to $F$ , $C$ to $E$ and $I$ to $T'$ .
Hence $\angle FT'E = \angle BIC = 90^0 + \frac {1}{2}\angle A = 120^0$ .
And $\angle EDF = \angle A = 60^0$ . So $\angle EDF + \angle FT'E = 180^0$
Hence $DFT'E$ concyclic. So $T'$ lies on the circumcircle of $\Delta DEF$ which is the medial triangle of $\Delta ABC$ .
This triangle's circumcircle is the nine-point circle of $\Delta ABC$ .
Hence $T'$ belongs to the incircle as well as nine-point circle of $ABC$ .
But they touch (at an unique point) by feurbach's theorem. Hence $T = T'$
QED

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r1234

#5 h Jan 8, 2011, 6:26 AM • 1 Y

Y by Adventure10

this problem is quite easy. we denote the nine point centre by N , incenter by I, circumcenter by O and orthocenter by H. As the incircle touches the nine point circle at T we get N,I,T collinear.Now applying <A=60 we get AH=R and AI=2r and hence we get that the triangle AOH is isosceles. Hence AN is perpendicular bisector of OH( since N bisects OH).Hence AN is the angular bisector of OAH and by angle chasing we get AN is the angular bisector of BAC. Hence A,T,I,N are all collinear.Now AI=2r and IT=r. Hence AT=r

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