Angle in Circle

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Source: 2004 IrMO Paper 2 Problem 2

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#1 h Dec 28, 2017, 7:14 PM • 1 Y

Y by Adventure10

$A$ and $B$ are distinct points on a circle $T$ . $C$ is a point distinct from $B$ such that $|AB|=|AC|$ , and such that $BC$ is tangent to $T$ at $B$ . Suppose that the bisector of $\angle ABC$ meets $AC$ at a point $D$ inside $T$ . Show that $\angle ABC>72^\circ$ .

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#2 h Dec 29, 2017, 10:10 PM • 3 Y

Y by laegolas, Adventure10, Mango247

Let O be the center of circle $T$ , and let X be the intersection of AC and the angle bisector of $ABC$ .

Let $\alpha=\angle XBC=\angle XBA$ . Since $AB=AC$ , $\angle ACB=\angle ABC=2\alpha$ so $\angle AXB=3\alpha$

Since BC is a tangent, $\angle OBC=90\implies \angle AOB=180-2\angle OBA=180-2(90-2\alpha)=4\alpha$
In order for $X$ to lie inside circle $T$ , $\angle AXB$ must be greater than half of $360-\angle AOB$ (this is essentially the $180^\circ\le \angle AOB <360^\circ$ )

In other words, $3\alpha>\frac{360-4\alpha}{2}\implies \alpha>36$

Since $\alpha$ is half of $\angle ABC$ , this means $\angle ABC>72^\circ$

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