## Search found 48 matches

- Sat Apr 22, 2017 7:04 pm
- Forum: Geometry
- Topic: USA(J)MO 2017 #3
- Replies:
**6** - Views:
**9547**

### Re: USA(J)MO 2017 #3

For those who loves synthetic geometry Throughout the proof signed area will be used. Lemma : Let $ABC$ be an equilateral triangle, and point $P$ on its circumcircle. Let $PB$ and $AC$ intersect at $E$, and $PC$ and $AB$ intersect at $F$.Then $ {[EPF]}={[ABPC]}$ Proof: Let the tangent to $(ABC)$ at ...

- Sun Apr 02, 2017 11:34 pm
- Forum: Secondary Level
- Topic: Find the angle
- Replies:
**2** - Views:
**1411**

### Re: Find the angle

Let $O$ be the circumcenter of $\triangle ABC$, Then $\angle ABC =180^{\circ}-30^{\circ} -70^{\circ}=80^{\circ} ,\angle OAC =90^{\circ}-\angle ABC=10^{\circ}.$ So $O \in AM$ .$\angle OBC=\angle ABC -\angle ABO =80^{\circ}-20^{\circ}=60^{\circ} $ .As $OB=OC$ so , $\triangle OBC$ is equilateral .$\ang...

- Mon Feb 27, 2017 9:13 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**110** - Views:
**52510**

### Re: Geometry Marathon : Season 3

**Problem 38 :**

In $\triangle ABC$ let the angle bisector of $\angle BAC$ meet $BC$ at $A_o$. Define $B_o,C_o$ similarly.Prove that

the circumcircle of $\triangle A_oB_oC_o$ goes though the Feuerbach point of $\triangle ABC$.

- Mon Feb 27, 2017 7:57 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**110** - Views:
**52510**

### Re: Geometry Marathon : Season 3

Solution of problem 36: Let $S$ be the midpoint of arc $AC$ (containing $B$) & $Q$ be the midpoint of arc $AC$ (not containing $B$).$R$ be the reflection of point $P$ wrt $XY$.Now $M$ be the midpoint of $AC$ & $K$ be the orthocenter of $\triangle SAC $.$N$ be the midpoint of $BH$. $K$ is the reflec...

- Sun Feb 26, 2017 3:37 pm
- Forum: Geometry
- Topic: IGO 2016 Elementary/2
- Replies:
**5** - Views:
**2115**

### Re: IGO 2016 Elementary/2

As, $ \angle YXC = \angle CPY + \angle PCA$. So ,$ \angle YXC = \angle XYC \Rightarrow \angle CPY + \angle PCA = \angle XYC \Rightarrow \widehat{AP} + \widehat{CY} = \widehat{PC}$

- Sat Feb 25, 2017 6:02 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**110** - Views:
**52510**

### Re: Geometry Marathon : Season 3

**Problem 34:**

Let $O$ & $I$ denote the circumcenter & incenter of $\triangle ABC$ respectively.Prove that The reflections of the

$OI$ line in the sides of the intouch triangle of $\triangle ABC$ concur at the Feuerbach point of $\triangle ABC$.

- Thu Feb 23, 2017 10:22 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**110** - Views:
**52510**

### Re: Geometry Marathon : Season 3

Solution of problem 33: Let $I_a$ denote the excenter opposite to $A$ .The $A$- excircle touches $BC$ at $P$ .Let $A_0D \cap OI = J$ & $AI \cap BC=K$.Let the perpendiculer from $O$ to $BC$ meet $AI$ & $A_0D$ at $M,N$ respectively. Lemma : $A_0 ,D ,I_a$ are collinear. Proof :Let $I_aD$ meet $AH$ at ...

- Tue Feb 21, 2017 4:10 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**110** - Views:
**52510**

### Re: Geometry Marathon : Season 3

Problem 32 : In triangle $ABC$ with incenter $I$ and circumcenter $O$, let $A',B',C'$ be the points of tangency of its circumcircle with its $A,B,C$-mixtilinear incircles, respectively. Let $\omega_A$ be the circle through $A'$ that is tangent to $AI$ at $I$, and define $\omega_B, \omega_C$ similar...

- Tue Feb 21, 2017 4:00 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**110** - Views:
**52510**

### Re: Geometry Marathon : Season 3

Solution of problem 31: Let $D,E,F $ be the midpoints of the arc $BC$ (not containg $A$) ,arc $CA$ (not containg $B$),arc $AB$ (not containg $C$) respectively. Let $H_a,H_b,H_c$ be the orthocenters of $\triangle IBC ,\triangle ICA ,\triangle IAB$ respectively .$M ,N ,P$ be the midpoints of $BC,CA,A...

- Sat Feb 18, 2017 1:51 am
- Forum: Geometry
- Topic: IMO Shortlist 2010 G7
- Replies:
**1** - Views:
**1382**

### Re: IMO Shortlist 2010 G7

Lemma : Let two circular arcs $\alpha $ & $\beta$ connect pionts $A,B$. If two circle $ \varpi_1$ & $ \varpi_2$ are tangent to $\alpha $ & $\beta$ ,then their external center of simillitude lies in $AB$. Proof : Let an external common tangent $l$ of $\alpha $ & $\beta$ meet $AB$ at piont $M$.Then t...