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Problem 1878

National Olympiad, first round

Geometry Difficulty 7.0 Prove it Hellenic Mathematical Olympiad · Greece

Let $ABC$ be an acute-angled triangle with $AB < AC < BC$ , inscribed in the circle $c(O,R)$ . The circle $c_1$ with center $A$ and radius $AC$ intersects the circle $c(O,R)$ at point $D$ and the extension of the side $CB$ at $E$ . The line $AE$ intersects the circle $c(O,R)$ at point $F$ and $G$ is the symmetric point of $E$ with respect to $B$ . Prove that the quadrilateral $FEDG$ is cyclic.

2. We consider three lines of the plane passing through point $A$ and dividing the plane in 6 sectors. At the interior of each sector there exist 5 points. We suppose that no three of the 30 points existing in the sectors are collinear. Prove that there exist at least 1000 triangles with vertices from the points of the 6 sectors which contain point $A$ either on their interior or on their sides.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Since the quadrilateral $AFBC$ is inscribed in the circle $(c)$ , we have: $\angle F_1 = \angle ACB = \angle C$ . Since triangle $AEC$ is isosceles we have $\angle E_1 = \angle ACB = \angle C$ . Therefore $\angle F_1 = \angle E_1$ , and hence the triangle $BEF$ is isosceles and hence
$BE = BF \qquad (1).$

Figure 2

We put $\angle C_1 = x$ . Then from the circle $(c_1)$ we get $E\angle AD = 2x$ , and hence
$E\angle AB + B\angle AD = 2x \qquad (2)$
Moreover from the circle $(c)$ we have:
$B\angle AD = \angle C_1 = x \qquad (3)$
From (2) and (3) we find $E\angle AB = B\angle AD = x$ , which means that $AB$ is bisector of the isosceles triangle $EAD$ . Hence it is perpendicular bisector of $ED$ , and
$BE = BD. \qquad (4)$
From (1) and (4), and from the equality $BE = BG$ , we conclude that $BE = BF = BG = BD$ , and hence the quadrilateral $FEDG$ is inscribed in a circle with center $B$ .

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