Search found 16 matches
- Fri Feb 17, 2017 2:21 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies: 146
- Views: 190137
Re: Geometry Marathon : Season 3
Another solution to Problem 28, i think this solu is cute sooo... We can state the problem in a new way: Let $ABC$ be a triangle and $\omega$ be its circumcircle. Let the external bisectors of $\angle A, \angle B, \angle C$ intersect $\omega$ at $A', B', C'$ respectively. If $X$ and $Y$ are two arbi...
- Tue Feb 14, 2017 1:08 am
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies: 146
- Views: 190137
Re: Geometry Marathon : Season 3
SOLUTION TO PROB 26: By some trivial angle chasing we have that $BE$ bisects $\angle DBA$. We have $AE\cap BC = M$ Now let $\odot DCB = ω$ intersect $AB$ at $D'$. Using spiral smilarity, we have $\triangle ED'A$ and $\triangle EDF$ are similar. That implies that $FD=AD=CD$. So $CF\perp AF$. We refle...
- Mon Feb 13, 2017 10:55 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies: 146
- Views: 190137
Re: Geometry Marathon : Season 3
$\text{Problem 25}$
Let $ABC$ be an acute triangle. Consider the equilateral triangle $A'UV$, with $A', U, V$ on $BC, AC, AB$ respectively, and $UV \parallel BC$. The points $B', C'$ are defined similarly. Prove that $AA', BB', CC'$ are concurrent.
Let $ABC$ be an acute triangle. Consider the equilateral triangle $A'UV$, with $A', U, V$ on $BC, AC, AB$ respectively, and $UV \parallel BC$. The points $B', C'$ are defined similarly. Prove that $AA', BB', CC'$ are concurrent.
- Mon Feb 13, 2017 3:29 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies: 146
- Views: 190137
Re: Geometry Marathon : Season 3
$\text{Problem no 24}$ Let $ABC$ be a triangle and $m$ a line which intersects $AB, BC$ and $CA$ at $D, E$ and $F$ in a way that $C$ lies between $B$ and $E$. The parallel lines from the points $A, B, C$ to the line $m$ intersects the circumcircle of $ABC$ at the points $A_1, B_1, C_1$. Prove that ...
- Mon Feb 13, 2017 3:12 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies: 146
- Views: 190137
Re: Geometry Marathon : Season 3
Firstly, we notice that $\angle CMP = \angle QMB = \angle LAC$. So, $LM$ bisects $\angle PMQ$. If $LM \cap \odot HNM = T$, then $T$ is the midpoint of arc $QAP$ of $\odot HMN$. So, $D, N, T$ are collinear. That means $AT$ extarnally bisects $\angle BAC$. That also indicates that $\odot \omega \cap \...
- Sat Jan 07, 2017 4:35 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies: 146
- Views: 190137
Re: Geometry Marathon : Season 3
$\text{Solution of problem 8:}$ Part (i): Lemma : If $(A, \ B; \ C, \ D)=-1$ and $M$ is the midpoint of $AB$, then $CM \times CD=CA \times CB$. Lemma's Proof : Consider a circle with center $M$ and radius $MA=MB$. Draw a tangent $DX$ from $D$ to $\odot (M)$. So we have $XC \perp AB$. Using similar t...