Geometry Marathon : Season 3

For discussing Olympiad level Geometry Problems
Asif Hossain
Posts: 167
Joined: Sat Jan 02, 2021 9:28 pm

Re: Geometry Marathon : Season 3

Unread post by Asif Hossain » Tue Mar 30, 2021 8:58 am

Mehrab4226 wrote:
Mon Mar 29, 2021 10:08 pm
Asif Hossain wrote:
Mon Mar 29, 2021 10:01 pm
~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$.
What did you meant by inside of $\angle BMA$??(sry if this sound foolish)
I think he meant $\triangle BMA$
then i think the $TX=BX$ holds iff $T=B$
Hmm..Hammer...Treat everything as nail

Asif Hossain
Posts: 167
Joined: Sat Jan 02, 2021 9:28 pm

Re: Geometry Marathon : Season 3

Unread post by Asif Hossain » Tue Mar 30, 2021 9:01 am

~Aurn0b~ wrote:
Mon Mar 29, 2021 10:44 pm
Asif Hossain wrote:
Mon Mar 29, 2021 10:01 pm
~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$.
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
It said angle domain.... :?: :?: i think it is the fancier of saying $T$ is on the opposite region of $X$ respect to $AM$
Hmm..Hammer...Treat everything as nail

Dustan
Posts: 64
Joined: Mon Aug 17, 2020 10:02 pm

Re: Geometry Marathon : Season 3

Unread post by Dustan » Tue Mar 30, 2021 10:10 am

Asif Hossain wrote:
Tue Mar 30, 2021 8:58 am
Mehrab4226 wrote:
Mon Mar 29, 2021 10:08 pm
Asif Hossain wrote:
Mon Mar 29, 2021 10:01 pm

What did you meant by inside of $\angle BMA$??(sry if this sound foolish)
I think he meant $\triangle BMA$
then i think the $TX=BX$ holds iff $T=B$
i don't think so🤔

Asif Hossain
Posts: 167
Joined: Sat Jan 02, 2021 9:28 pm

Re: Geometry Marathon : Season 3

Unread post by Asif Hossain » Tue Mar 30, 2021 10:20 am

Dustan wrote:
Tue Mar 30, 2021 10:10 am
Asif Hossain wrote:
Tue Mar 30, 2021 8:58 am
Mehrab4226 wrote:
Mon Mar 29, 2021 10:08 pm

I think he meant $\triangle BMA$
then i think the $TX=BX$ holds iff $T=B$
i don't think so🤔
I am not sure but it is highly suspicious* because $B$ is kinda the farthest point...at least i couldn't find any.
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

~Aurn0b~
Posts: 42
Joined: Thu Dec 03, 2020 8:30 pm

Re: Geometry Marathon : Season 3

Unread post by ~Aurn0b~ » 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.

Asif Hossain
Posts: 167
Joined: Sat Jan 02, 2021 9:28 pm

Re: Geometry Marathon : Season 3

Unread post by Asif Hossain » Tue Apr 06, 2021 10:31 am

~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.
Solution(Do mention the source if someone solves your problem before posting the next it boosts the confidence and helps to determine his skill :) ):
We define the midpoint of $AC$ and $CB$ as $M_1$ and $M_2$ respectively. We also define $\omega \bigcap AB=\{F,G\}$ and $\omega \bigcap AB= H$
The statement in the question suffices to prove $A_2,B_2,M_1,M_2$ are concyclic since $\angle B_2 M_2 M_1 = \angle B_2 BA$ and $\angle A_2 M_1 M_2=\angle A_2 AB$
Claim: $AF=BG$
It is easy to see by the symmetry of isoceles trapezoid that $(AA_1 HB)$ and $(HA_1 FG)$ are also two isoceles trapezoids.So by SAS $\triangle AA_1 F \cong \triangle HGB$ . So, $AF=BG$ $\square$
$CA_2=CM_1 -M_1 A_2=AM_1-M_1 A_1=AA_1$ similarly $CB_2=BB_1$
So, $AF=BG \Rightarrow AF.AG=BG.BF \Rightarrow AA_1 .AC= BB_1 .BC \Rightarrow AA_1. 2CM_1=BB_1 .2CM_2 \Rightarrow CA_2 .CM_1 = CB_2 .CM_2$
So, $A_2,B_2,M_1,M_2$ are concyclic and we are done.$\square$
Screenshot from 2021-04-06 10-09-37.png
Screenshot from 2021-04-06 10-09-37.png (73.12 KiB) Viewed 294 times
Hmm..Hammer...Treat everything as nail

Asif Hossain
Posts: 167
Joined: Sat Jan 02, 2021 9:28 pm

Problem 62

Unread post by Asif Hossain » Tue Apr 06, 2021 10:18 pm

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:
Iranian National Olympiad (3rd Round) 2004
Hmm..Hammer...Treat everything as nail

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