Problem #4

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[INMO-1995]: In an acute-angled traingle $ABC$, $\angle A = 30^0$, $H$ is the orthocentre, and $M$ is the midpoint of $BC$. On the line $HM$, take a point $T$ such that $HM = MT$. Show that $AT = 2BC$.

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Solution:

Since $M$ is the midpoint of $HT$ and $M$ is also the midpoint of $BC => BC$ and $HT$ bisect each other. This implies that $BHCT$ is a parallelogram; which makes $BH$ parallel to $TC$. Since $BH \perp AC => TC \perp AC$. Similarly, $TB \perp AB$.

The quadrilateral $ACTB$ becomes a cyclic quadrilateral because $\angle ABT + \angle ACT = 180^o$. Furthermore, since $\angle ABT = \angle ACT = 90^o, AT$ is the diameter of the circumcircle of $ACTB$.

Say we call the midpoint of $AT$ as $O$. Since $AT$ is the diameter, $O$ becomes the centre of the circumcircle of $ACTB$. Since, $\angle BAC = 30^o => \angle BOC = 60^o$. Since $OB = OC$ (= radius of circumcircle), $\triangle OBC$ becomes an equilateral triangle. Thus, the length of $BC$ is same as the radius of the circumcircle which is half the diameter $AT$.