Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.2 National Olympiad Prove it JBMO

Problem:

Let $ABC$ be a triangle with $m(\widehat{BAC}) = 60^{\circ}$ . Let $D$ and $E$ be the feet of the perpendiculars from $A$ to the external angle bisectors of $\widehat{ABC}$ and $\widehat{ACB}$ , respectively. Let $O$ be the circumcenter of the triangle $ABC$ . Prove that the circumcircles of the triangles $ADE$ and $BOC$ are tangent to each other.

Solution

Solution:

Let $X$ be the intersection of the lines $BD$ and $CE$ .
We will prove that $X$ lies on the circumcircles of both triangles $ADE$ and $BOC$ and then we will prove that the centers of these circles and the point $X$ are collinear, which is enough for proving that the circles are tangent to each other.
In this proof we will use the notation $(MNP)$ to denote the circumcircle of the triangle $MNP$ .

Obviously, the quadrilateral $ADXE$ is cyclic, and the circle $(DAE)$ has $[AX]$ as diameter. (1)

Let $I$ be the incenter of triangle $ABC$ . So, the point $I$ lies on the segment $[AX]$ (2), and the quadrilateral $XBIC$ is cyclic because $IC \perp XC$ and $IB \perp XB$ . So, the circle $(BIC)$ has $[IX]$ as diameter.

Finally, $m(\widehat{BIC}) = 90^{\circ} + \frac{1}{2} m(\widehat{BAC}) = 120^{\circ}$ and $m(\widehat{BOC}) = 2 m(\widehat{BAC}) = 120^{\circ}$ .
So, the quadrilateral $BOIC$ is cyclic and the circle $(BOC)$ has $[IX]$ as diameter. (3)

(1), (2), (3) imply the conclusion.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.