## Search found 138 matches

- Sun Nov 26, 2017 11:16 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**110** - Views:
**5608**

### Re: Geometry Marathon : Season 3

$\text{Problem 50:}$ Let $\vartriangle ABC$ be a triangle, $O$ be the circumcenter of $\vartriangle ABC$, $N$ be the center of nine point circle of $\vartriangle ABC$ and $X$ be the midpoint of the line segment $ON$. Let $A',B',C'$ be the midpoints of the line segments $BC,CA,AB$, respectively. Let ...

- Sun Nov 26, 2017 10:48 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**110** - Views:
**5608**

### Re: Geometry Marathon : Season 3

$\text{Solution to Problem 49:}$ Let $BH \cap AC=E, CH \cap AB=F, AP \cap BC=X$. Now, by miquel's theorem on circle $BCEF, M,H,P$ are collinear. Also by radical axis theorem on circle $ABC, APFHE, BCEF$, we get $X,F,E$ are collinear. Now, we get that $H$ is the orthocentre of triangle $AXM$ so $XH \...

- Sat Nov 25, 2017 11:59 am
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**110** - Views:
**5608**

### Re: Geometry Marathon : Season 3

$\text{Problem 48:}$

Prove that for a scalene triangle $ABC$, one can never find two pairs of isogonal conjugates ${P,P'}$ and ${Q,Q'}$ such that $P,P',Q,Q'$ are all collinear and given that they're not self isogonal conjugates.

Prove that for a scalene triangle $ABC$, one can never find two pairs of isogonal conjugates ${P,P'}$ and ${Q,Q'}$ such that $P,P',Q,Q'$ are all collinear and given that they're not self isogonal conjugates.

- Sat Nov 25, 2017 11:56 am
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**110** - Views:
**5608**

### Re: Geometry Marathon : Season 3

$\text{Solution to Problem 47:}$

Let $AC \cap BD=X$, $AC \cap EF=Y$ and $BD \cap EF=Z$. Now $P,M,N$ are collinear by newton line. Then, $XM.XZ=XB.XD=XA.XC=XN.XY \Rightarrow MNYZ$ is cyclic and $PE^2=PY.PZ=PN.PM$.

Let $AC \cap BD=X$, $AC \cap EF=Y$ and $BD \cap EF=Z$. Now $P,M,N$ are collinear by newton line. Then, $XM.XZ=XB.XD=XA.XC=XN.XY \Rightarrow MNYZ$ is cyclic and $PE^2=PY.PZ=PN.PM$.

- Sat Oct 21, 2017 4:20 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**110** - Views:
**5608**

### Re: Geometry Marathon : Season 3

$\text{Problem 45}$ Let $ABC$ be a triangle with orthocentre $H$ and circumcircle $\omega$ centered at $O$. Let $M_a,M_b,M_c$ be the midpoints of $BC,CA,AB$. Lines $AM_a,BM_b,CM_c$ meet $\omega$ again at $P_a,P_b,P_c$. Rays $M_aH,M_bH,M_cH$ intersect $\omega$ at $Q_a,Q_b,Q_c$. Prove that $P_aQ_a,P_b...

- Sat Oct 21, 2017 3:58 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**110** - Views:
**5608**

### Re: Geometry Marathon : Season 3

Let $BH,CH$ meet $(O)$ at $X,Y$. We get $CXNH,BYMH$ are similar rhombuses. Let the perpendicular bisectors of $HM,HN$ meet at $K$ and perpendicular bisectors of $BY,CX$ meet at $O$. Let perpendicular bisectors of $BY,CX$ meet $HM,HN$ at $P,Q$ repectively. Its enough to prove that $\frac{HP}{HM/2}= \...

- Sun Oct 01, 2017 2:59 pm
- Forum: Geometry
- Topic: CGMO 2002/4
- Replies:
**2** - Views:
**246**

### Re: CGMO 2002/4

Correction: $G$ lies on $T_1$, not $T_2$.

Solution: $I$ be a point on line $AF$ such that it is the reflection of $D$ under $AB$. We get $HEBI$ is cyclic. So, $AH \times AI=AE \times AB=AF^2$ from which we can get the result.

Solution: $I$ be a point on line $AF$ such that it is the reflection of $D$ under $AB$. We get $HEBI$ is cyclic. So, $AH \times AI=AE \times AB=AF^2$ from which we can get the result.

- Tue Aug 08, 2017 12:58 am
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**110** - Views:
**5608**

### Re: Geometry Marathon : Season 3

Problem 43: An equilateral pentagon $AMNPQ$ is inscribed in triangle $ABC$ such that $M\in\overline{AB}$, $Q\in\overline{AC}$, and $N,P\in\overline{BC}$. Let $S$ be the intersection of $ \overleftrightarrow{MN}$ and $ \overleftrightarrow{PQ}$. Denote by $\ell$ the angle bisector of $\angle MSQ$. Pr...

- Tue Aug 08, 2017 12:37 am
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**110** - Views:
**5608**

### Re: Geometry Marathon : Season 3

Solution to Problem 42: $a)$ $P$ is just the reflection of $F$ under $AC$ and $Q$ is the reflection of $E$ under $AB$. $AFDCP$ and $AEDBQ$ cyclic. The rest is trivial by taking homothety with centre $A$ and ratio 2. $b)$ Let $H_1$ be the reflection of $H$ under $A$. We can find that the perpendicul...

- Sat Apr 29, 2017 11:01 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**110** - Views:
**5608**

### Re: Geometry Marathon : Season 3

$\text{Problem 40:}$ Let the Nagel point of triangle $ABC$ be $N$. We draw lines from $B$ and $C$ to $N$ so that these lines intersect sides $AC$ and $AB$ in $D$ and $E$ respectively. $M$ and $T$ are midpoints of segments $BE$ and $CD$ respectively. $P$ is the second intersection point of circumcirc...