MathNugget
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- Feb 1, 2024
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If ABC is a triangle, O is the intersection of the perpendicular bisectors, and H is the orthocenter (intersection of heights / altitudes) not so sure these are the proper words. (I attached a pdf with what I could find so far. it's part of a bigger exercise, but I am not following up the part where it proves what I want).
This is what I'd like to see how it's proven: [imath]\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} = \overrightarrow{OH}[/imath] . Proving that [imath]\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} = \overrightarrow{ON} \Longrightarrow N = H[/imath] doesn't really do it.
I can see that, alternatively, it's enough to prove [imath]2 \overrightarrow{OU} = \overrightarrow{AH}[/imath], where U is the middle of BC.
I found this result is Sylvester 's triangle problem. Still cannot fully understand if the proof is complete, or if maybe there's a way to prove those things without starting with the conclusion.
This is what I'd like to see how it's proven: [imath]\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} = \overrightarrow{OH}[/imath] . Proving that [imath]\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} = \overrightarrow{ON} \Longrightarrow N = H[/imath] doesn't really do it.
I can see that, alternatively, it's enough to prove [imath]2 \overrightarrow{OU} = \overrightarrow{AH}[/imath], where U is the middle of BC.
I found this result is Sylvester 's triangle problem. Still cannot fully understand if the proof is complete, or if maybe there's a way to prove those things without starting with the conclusion.
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