Does there exists $6$ points in the plane such that the distance between any two of them is an integer?
Source :
A random video from Metamaths.
I think we can draw 6 points that have an integer distance from any 2, if all of them are on the same line.
Sorry, I forgot to mention, no three are collinear
Re: Geometry Marathon : Season 3
Posted: Sun Mar 28, 2021 10:16 am
by Asif Hossain
Here is solu (copied) credit goes to Tanya Khovanova(100th post of useless content )
Screenshot from 2021-03-28 10-15-39.png (20.39KiB)Viewed 9518 times
Problem 58
Posted: Sun Mar 28, 2021 12:49 pm
by Asif Hossain
$\triangle ABC$ 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 $\angle MCN$.
Re: Geometry Marathon : Season 3
Posted: Sun Mar 28, 2021 3:42 pm
by Dustan
$ACPM$ and $BCQN$ are cyclic.
$\angle PCM=\angle PAM=\frac{A}{2}$
$\angle ACN=\angle QBA=\frac{B}{2}$
So,$\angle MCN=90-\angle ACN-\angle BCM=90-\frac{1}{2}(A+B)=90-45=45$
Re: Geometry Marathon : Season 3
Posted: Sun Mar 28, 2021 9:14 pm
by Dustan
Let $ABC$ be a triangle with incentre $I$. The circle through $B$ tangent to $AI$ at $I$ meets side $AB$ again at $P$. The circle through $C$ tangent to $AI$ at $I$ meets side $AC$ again at $Q$. Prove that $PQ$ is tangent to the incircle of $ABC.$
Let $ABC$ be a triangle with incentre $I$. The circle through $B$ tangent to $AI$ at $I$ meets side $AB$ again at $P$. The circle through $C$ tangent to $AI$ at $I$ meets side $AC$ again at $Q$. Prove that $PQ$ is tangent to the incircle of $ABC.$
$\textbf{Solution 59}$
It suffices to proof that the incircle is the $A-excircle$ of $\triangle APQ$
Let $I'$ be the incircle of $\triangle APQ$, so obviously $A-I'-I$ are collinear.
Now, By alternate segment theorem, we have $\angle QIA=\angle QCI=\frac{\angle C}{2}$
similarly, $\angle PIA=\angle PBI=\frac{\angle B}{2}$
So, $\angle QIP=\angle QIA+\angle PIA=\frac{\angle C}{2}+\frac{\angle B}{2}=90^{\circ}-\frac{\angle A}{2}$
Therefore, $\angle QI'P+\angle QIP=(90^{\circ}+\frac{\angle A}{2})+(90^{\circ}-\frac{\angle A}{2})=180^{\circ}\Rightarrow Q,I',P,I$ are con-cyclic.
so we get that $\angle PQI=\angle PI'I=90^{\circ}-\frac{\angle AQP}{2}\Rightarrow \angle IQC=90^{\circ}-\frac{\angle AQP}{2}$
Hence $IQ$ is the angle-bisector of $\angle PQC$
Similarly, $IP$ is the angle-bisector of $\angle BPQ$
An thus $I$ is the $A-excenter$ of $\triangle APQ$, as the incircle of $ABC$ is tangent to the extension of $AP$ and $AQ$, we can say that the incircle of $ABC$ is indeed the excircle of $APQ$, and so it must be tangent to $PQ.\blacksquare$
Re: Geometry Marathon : Season 3
Posted: Sun Mar 28, 2021 11:21 pm
by ~Aurn0b~
$\textbf{Problem 60}$
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$.
What did you meant by inside of $\angle BMA$??(sry if this sound foolish)
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$.
What did you meant by inside of $\angle BMA$??(sry if this sound foolish)
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$.
What did you meant by inside of $\angle BMA$??(sry if this sound foolish)
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 to
)