Triangle Centers - Problem Solving

This wiki page shows some simple examples to solve triangle centers using simple properties like Circumcenter, Fermat point, Brocards points, Incenter, Centroid, Orthocenter, etc.

One should be able to recall definitions like

• Circumcenter: The point of which is equidistant from all the the vertices of the triangle.
• Incenter: The point which is equidistant from the sides of the triangle.
• Orthocenter: The point in which all the altitudes of the triangle intersect.
• Centroid: The point of intersection of the medians of the triangle.

An important relationship between these points is the Euler line, which states that \( O, G, H \) is a straight line and \( OG : GH = 1 : 2 \). In fact, the center of the Nine Point Circle is also the midpoint of \( OH \).

Show that if the orthocenter and the incenter of a triangle coincide, then this triangle must be equilateral.

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Consider vertex \(A\). Let these points coincide at \(P\).
Then, we know that \(AP\) is the angle bisector of \( \angle BAC \), and it is also the perpendicular to \( BC \). Thus, we obtain that \( \angle ABC = 90 ^ \circ - \angle BAP = 90 ^ \circ - \angle CAP = \angle ACB \).

Since this is true from any vertex, thus is means that all three angles are equal, and thus we have an equilateral triangle.

Let \(IA\), \(IB\), \(IC\) denote the distance between the incenter and the vertices of triangle \(A,B,C\) respectively. Prove that \(\frac{IA \cdot IB}{CA \cdot CB} + \frac{IA \cdot IC}{BA \cdot BC} + \frac{IB \cdot IC}{AB \cdot AC} = 1 \).

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Solution: (to be continued)

In triangle \(ABC\) with centroid \(G\), if \( AG=BC\), what is angle \(BGC \) in degrees?

Note: Diagram is not drawn to scale.

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