# Problem #4

[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 $$AT = 2BC$$.

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 $$AT$$.