Romanian National Olympiad 2018 - Grade 9 - problem 1

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Romanian National Olympiad 2018 - Grade 9 - problem 1

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Source: Romania NMO - 2018

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381 posts

#1 h Apr 12, 2018, 2:39 PM • 2 Y

Y by Adventure10, Mango247

Prove that if in a triangle the orthocenter, the centroid and the incenter are collinear, then the triangle is isosceles.

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#2 h Apr 12, 2018, 2:41 PM • 2 Y

Y by Adventure10, Mango247

Edmonds, Allan L.; Hajja, Mowaffaq; Martini, Horst (2008), "Orthocentric simplices and biregularity", Results in Mathematics, 52 (1-2): 41–50, doi:10.1007/s00025-008-0294-4, MR 2430410, "It is well known that the incenter of a Euclidean triangle lies on its Euler line connecting the centroid and the circumcenter if and only if the triangle is isosceles."

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#3 h Aug 6, 2018, 9:44 AM • 3 Y

Y by APWLGeometry, Amin_Hashemi-1379, Adventure10

See here: https://artofproblemsolving.com/community/q1h1621342p10147701

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Taco12

#4 h Jan 3, 2023, 12:24 PM

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Solution 1 (Barycentric Coordinates). Use the sidelength form of $H$ . The determinant of $G, H, I$ simplifies as $(a - b) (a - c) (b - c) (a + b + c)^2 = 0 \rightarrow a=c \text{ OR } b=c \text{ OR } a=b,$ which finishes.

Solution 2 (Complex Numbers). Set $(ABC)$ as the unit circle and $A=a^2, B=b^2, C=c^2$ . By Euler Line it suffices to show $H,I,O$ collinear which means we must show $\frac{I}{H} \in \mathbb{R}$ . We have $H=a^2+b^2+c^2, I=-(ab+bc+ac)$ and expanding the conjugate and setting it equal gives the desired result upon some algebra.

This post has been edited 1 time. Last edited by Taco12, Jan 3, 2023, 12:24 PM

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Myfavoruitepenguin

#5 h Mar 31, 2025, 12:42 PM

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How did you reach to that equation using Barycentric coordinates?

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