then i think the $TX=BX$ holds iff $T=B$Mehrab4226 wrote: ↑Mon Mar 29, 2021 10:08 pmI 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$.
Geometry Marathon : Season 3

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Re: Geometry Marathon : Season 3
Hmm..Hammer...Treat everything as nail

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Re: Geometry Marathon : Season 3
It said angle domain.... i think it is the fancier of saying $T$ is on the opposite region of $X$ respect to $AM$~Aurn0b~ wrote: ↑Mon Mar 29, 2021 10:44 pmI 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$.
Hmm..Hammer...Treat everything as nail
Re: Geometry Marathon : Season 3
i don't think soAsif Hossain wrote: ↑Tue Mar 30, 2021 8:58 amthen i think the $TX=BX$ holds iff $T=B$Mehrab4226 wrote: ↑Mon Mar 29, 2021 10:08 pmI think he meant $\triangle BMA$Asif Hossain wrote: ↑Mon Mar 29, 2021 10:01 pm
What did you meant by inside of $\angle BMA$??(sry if this sound foolish)

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Re: Geometry Marathon : Season 3
I am not sure but it is highly suspicious* because $B$ is kinda the farthest point...at least i couldn't find any.Dustan wrote: ↑Tue Mar 30, 2021 10:10 ami don't think so
You may try to prove it Set $M(0,0),B(a,0),A(0,b),T(x',y')$ and $X(m,n)$ where $m$ and $n$ satisfies $(m \frac{a}{2})^2 +(n \frac{b}{2})^2 =\frac{\sqrt{a^2+b^2}}{2}$ and $0<n<b$ then noticing $0<x' <a$ and $0<y'<b$
Hmm..Hammer...Treat everything as nail
Re: Geometry Marathon : Season 3
$\textbf{Problem 61}$
Let $ABCD$ be a isosceles trapezoid with $AB\parallel CD$ and $ \Omega $ is a circle passing through $A,B,C,D$. Let $ \omega $ be the circle passing through $C,D$ and intersecting with $CA,CB$ at $A_1$, $B_1$ respectively. $A_2$ and $B_2$ are the points symmetric to $A_1$ and $B_1$ respectively, with respect to the midpoints of $CA$ and $CB$. Prove that the points $A,B,A_2,B_2$ are concyclic.
Let $ABCD$ be a isosceles trapezoid with $AB\parallel CD$ and $ \Omega $ is a circle passing through $A,B,C,D$. Let $ \omega $ be the circle passing through $C,D$ and intersecting with $CA,CB$ at $A_1$, $B_1$ respectively. $A_2$ and $B_2$ are the points symmetric to $A_1$ and $B_1$ respectively, with respect to the midpoints of $CA$ and $CB$. Prove that the points $A,B,A_2,B_2$ are concyclic.

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Re: Geometry Marathon : Season 3
Solution(Do mention the source if someone solves your problem before posting the next it boosts the confidence and helps to determine his skill ):~Aurn0b~ wrote: ↑Mon Apr 05, 2021 7:36 pm$\textbf{Problem 61}$
Let $ABCD$ be a isosceles trapezoid with $AB\parallel CD$ and $ \Omega $ is a circle passing through $A,B,C,D$. Let $ \omega $ be the circle passing through $C,D$ and intersecting with $CA,CB$ at $A_1$, $B_1$ respectively. $A_2$ and $B_2$ are the points symmetric to $A_1$ and $B_1$ respectively, with respect to the midpoints of $CA$ and $CB$. Prove that the points $A,B,A_2,B_2$ are concyclic.
Hmm..Hammer...Treat everything as nail

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Problem 62
Problem 62
Incircle of $\triangle ABC$ with center $I$ touches $AB,AC$ at $P,Q$. $BI,CI$ intersect with $PQ$ at $K,L$. Prove that circumcircle of $ILK$ is tangent to incircle of $ABC$ if and only if $AB+AC=3BC$.
Source:
Incircle of $\triangle ABC$ with center $I$ touches $AB,AC$ at $P,Q$. $BI,CI$ intersect with $PQ$ at $K,L$. Prove that circumcircle of $ILK$ is tangent to incircle of $ABC$ if and only if $AB+AC=3BC$.
Source:
Hmm..Hammer...Treat everything as nail