Sorry, I forgot to mention, no three are collinearMehrab4226 wrote: ↑Sun Mar 21, 2021 10:36 pmI think we can draw 6 points that have an integer distance from any 2, if all of them are on the same line.Anindya Biswas wrote: ↑Sun Mar 21, 2021 9:02 pmDoes there exists $6$ points in the plane such that the distance between any two of them is an integer?
Source :
Geometry Marathon : Season 3
- Anindya Biswas
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"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
— John von Neumann
— John von Neumann
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Re: Geometry Marathon : Season 3
Here is solu (copied) credit goes to Tanya Khovanova(100th post of useless content )
Hmm..Hammer...Treat everything as nail
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Problem 58
$\triangle ABC$ has a right triangle at $C$ the internal bisectors of $\angle BAC$ and $\angle ABC$ meet $BC$ and $CA$ at $P$ and $Q$ respectively.
The points $M$ and $N$ are the feet of perpendiculars from $P$ and $Q$ to $AB$. find $\angle MCN$.
The points $M$ and $N$ are the feet of perpendiculars from $P$ and $Q$ to $AB$. find $\angle MCN$.
Hmm..Hammer...Treat everything as nail
Re: Geometry Marathon : Season 3
$\textbf{Solution 59}$
Re: Geometry Marathon : Season 3
$\textbf{Problem 60}$
Denote by $ M$ midpoint of side $ BC$ in an isosceles triangle $ \triangle ABC$ with $ AC = AB$. Take a point $ X$ on a smaller arc $MA$ of circumcircle of triangle $ \triangle ABM$. Denote by $ T$ point inside of angle $ BMA$ such that $ \angle TMX = 90$ and $ TX = BX$.
Prove that $ \angle MTB - \angle CTM$ does not depend on choice of $ X$.
Denote by $ M$ midpoint of side $ BC$ in an isosceles triangle $ \triangle ABC$ with $ AC = AB$. Take a point $ X$ on a smaller arc $MA$ of circumcircle of triangle $ \triangle ABM$. Denote by $ T$ point inside of angle $ BMA$ such that $ \angle TMX = 90$ and $ TX = BX$.
Prove that $ \angle MTB - \angle CTM$ does not depend on choice of $ X$.
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Re: Geometry Marathon : Season 3
What did you meant by inside of $\angle BMA$??(sry if this sound foolish)~Aurn0b~ wrote: ↑Sun Mar 28, 2021 11:21 pm$\textbf{Problem 60}$
Denote by $ M$ midpoint of side $ BC$ in an isosceles triangle $ \triangle ABC$ with $ AC = AB$. Take a point $ X$ on a smaller arc $MA$ of circumcircle of triangle $ \triangle ABM$. Denote by $ T$ point inside of angle $ BMA$ such that $ \angle TMX = 90$ and $ TX = BX$.
Prove that $ \angle MTB - \angle CTM$ does not depend on choice of $ X$.
Hmm..Hammer...Treat everything as nail
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Re: Geometry Marathon : Season 3
I think he meant $\triangle BMA$Asif Hossain wrote: ↑Mon Mar 29, 2021 10:01 pmWhat did you meant by inside of $\angle BMA$??(sry if this sound foolish)~Aurn0b~ wrote: ↑Sun Mar 28, 2021 11:21 pm$\textbf{Problem 60}$
Denote by $ M$ midpoint of side $ BC$ in an isosceles triangle $ \triangle ABC$ with $ AC = AB$. Take a point $ X$ on a smaller arc $MA$ of circumcircle of triangle $ \triangle ABM$. Denote by $ T$ point inside of angle $ BMA$ such that $ \angle TMX = 90$ and $ TX = BX$.
Prove that $ \angle MTB - \angle CTM$ does not depend on choice of $ X$.
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré
-Henri Poincaré
Re: Geometry Marathon : Season 3
I think it kinda means TM is inside the angle BMA(That is what i assumed when i tried to solve it, i also never heard anything like it, but that's what the problem states). It's from 2007 IMOSL u can see the question if u want toAsif Hossain wrote: ↑Mon Mar 29, 2021 10:01 pmWhat did you meant by inside of $\angle BMA$??(sry if this sound foolish)~Aurn0b~ wrote: ↑Sun Mar 28, 2021 11:21 pm$\textbf{Problem 60}$
Denote by $ M$ midpoint of side $ BC$ in an isosceles triangle $ \triangle ABC$ with $ AC = AB$. Take a point $ X$ on a smaller arc $MA$ of circumcircle of triangle $ \triangle ABM$. Denote by $ T$ point inside of angle $ BMA$ such that $ \angle TMX = 90$ and $ TX = BX$.
Prove that $ \angle MTB - \angle CTM$ does not depend on choice of $ X$.