In this exploration, we tackle a rich geometry problem from the 2024 Regional Math Olympiad (RMO). Given an acute-angled isosceles triangle \( \triangle ABC \) with the circumcenter \( O \), orthocenter \( H \), and centroid \( G \), along with specific distances between them, we aim to prove that the triangle's incircle passes through the centroid \( G \).
Points \( O \), \( G \), \( H \), and the incenter \( I \) are collinear along \( AD \), where \( D \) is the foot of the altitude from \( A \) to \( BC \). This holds due to the perpendicular bisector properties in isosceles triangles.
Euler’s Line:
In any triangle, \( O \), \( G \), and \( H \) lie on Euler's line, dividing \( OH \) in the ratio 1:2. This relationship allows us to find specific segment lengths, given distances like \( HD \) and \( OD \).
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Step-by-Step Solution:
Distances Along Euler's Line: Use the given condition \( 2 \cdot HD = 23 \cdot OD \) to find segments \( OG \) and \( GH \).
Median Division by Centroid:
\( G \) divides the median \( AD \) in the ratio 2:1. With this, the distance \( AG \) can be determined.
Calculating Side Lengths:
Using known distances, the length of \( BD \) is found by applying the Pythagorean theorem in \( \triangle BOD \). Side lengths \( AB \) and \( AC \) are deduced for further calculations.
Proving the Incircle Passes Through \( G \):
Inradius Calculation:
Using the formula \( r = \frac{\text{Area}}{\text{Semiperimeter}} \), the inradius \( r \) is derived.
Centroid’s Distance Verification:
Show that \( IG \), the distance from the incenter to the centroid, matches the inradius, confirming \( G \) lies on the incircle.
Conclusion:
By proving \( IG \) equals the inradius, we show that the centroid \( G \) indeed lies on the incircle, completing the proof. This problem elegantly ties together triangle properties, collinearity, and Euler's line, demonstrating the interconnectedness of geometric points in advanced problem-solving.
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