## Search found 36 matches

- Mon Apr 09, 2018 4:54 pm
- Forum: Geometry
- Topic: Secondary Special Camp 2011: Geometry P 4
- Replies:
**6** - Views:
**11278**

### Re: Secondary Special Camp 2011: Geometry P 4

We will have to prove that $PD.DC=RD.DQ$ $\angle{AFE}=\angle{BRQ}=\angle{BCQ}$ So, $BRCQ$ is cyclic. Then, $RD.DQ=BD.DC$ We wanna prove, $PD.DM=BD.DC$ Or, $(PB+a/2-DM)DM=(a/2-DM)(a/2+DM)$ Or, $BP.DM=a^2/4-a/2.DM$ We get $(P,D;B,C)$ is a harmonic bundle. So, $BP.CD=BD.CP$ Or, $BP(a/2+DM)=(a/2-DM)(BP+...

- Tue Apr 03, 2018 2:11 pm
- Forum: Geometry
- Topic: What is the distance?
- Replies:
**3** - Views:
**7894**

### Re: What is the distance?

I'm avoiding the calculations how I got that $PX=2R=65/2$ (By Heron's formula we got the area of the triangle $ABC$, then found out the height from on vertix to oppsite side, and then applied Brahmagupta's theorem.) Using the fact that, $CS/PS=XS/CS$ we get $CS=10$ I think you've made a typo here. ...

- Mon Apr 02, 2018 12:16 pm
- Forum: Geometry
- Topic: incircles and excircles
- Replies:
**1** - Views:
**7560**

### Re: incircles and excircles

Let $R_1$ and $P_1$ be the incenter and the C-excenter of triangle $AMC$. $R_2$ and $P_2$ be the incenter and the C-excenter of triangle $BMC$. So, we get $AP_1MR_1$ and $BP_2MR_2$ cyclic. $$\angle{AMR_1}=\angle{R_1MC}=90^o-\angle{R_2MC} \angle{R_2MC}=\angle{R_2MB} \angle{AMP_1}=90^o-\angle{R_1MC}$$...

- Mon Apr 02, 2018 12:34 am
- Forum: Geometry
- Topic: A cool Geo!
- Replies:
**1** - Views:
**7404**

### Re: A cool Geo!

$$\angle{BAP}=\angle{PAC}=\angle{PCX}$$ Again, $$\angle{PAC}=\angle{YAC}=\angle{YCA}$$ So, $$\angle{ACY}=\angle{PCX}$$ In triangle $ABC$, $CY$ is the isogonal of $CX$ So, $$\frac{AC^2}{CP^2}=\frac{AY*AX}{YP*XP}$$ Again, $$\frac{AC}{CP}=\frac{AB}{BP}$$ So, $$\frac{AB^2}{BP^2}=\frac{AY*AX}{YP*XP}$$ It...

- Fri Mar 30, 2018 10:53 pm
- Forum: Geometry
- Topic: What is the distance?
- Replies:
**3** - Views:
**7894**

### Re: What is the distance?

We use Pascal's theorem on $YXZCPB$ and get $M, N, I$ collinear where $I$ is the intersection point of $ZC$ and $BY$. In fact, $I$ is the incenter of triangle $ABC$. So, $XC=XI$ Again, $$\angle{CXY}=\angle{AXY}=\angle{CXN}=\angle{IXN}$$ And $$\angle{PCX}=\angle{NCX}=90^o$$ Ultimately we get $CNIX$ c...

- Fri Mar 30, 2018 10:07 pm
- Forum: Geometry
- Topic: What is the distance?
- Replies:
**3** - Views:
**7894**

### What is the distance?

Let $ABC$ be a triangle with $AB=26, AC=28, BC=30$. Let $X, Y, Z$ be the midpoints of arcs $BC, CA, AB$ (not containg the opposite vertics) respectively on the circumcircle of triangle $ABC$. Let $P$ be the midpoint of arc $BC$ containing point A. Suppose, lines $BP$ and $XZ$ meet at $M$, while line...

- Fri Mar 30, 2018 10:03 pm
- Forum: Geometry
- Topic: What is the distance?
- Replies:
**1** - Views:
**7053**

### What is the distance?

Let $ABC$ be a triangle with $AB=26, AC=28, BC=30$. Let $X, Y, Z$ be the midpoints of arcs $BC, CA, AB$ (not containg the opposite vertics) respectively on the circumcircle of triangle $ABC$. Let $P$ be the midpoint of arc $BC$ containing point A. Suppose, lines $BP$ and $XZ$ meet at $M$, while line...

### Hard geo?

$ABCDEF$ is a convex hexagon inscribed in a circle so that the diagonals $AD, BE$ and $CF$ are concurrent at $J$ and $$\angle{AJB}=\angle{BJC}=60^o$$. Prove that, $AJ+CJ+EJ=BJ+DJ+FJ$.

- Sat Oct 07, 2017 4:32 pm
- Forum: Asian Pacific Math Olympiad (APMO)
- Topic: APMO 2012/04
- Replies:
**3** - Views:
**6873**

### Re: APMO 2012/04

We'll have to show, $BF/CF=AB/AC$ which means $BF/CF=sin\angle{BCF}/sin\angle{CBF}=(BK*AC)/(AB*CK)$ We have to show now, $AB^2/AC^2=BK/CK$ or $AF$ is a symmedian. We get $A,O,P$ collinear Then $AEDM$ is cyclic Again, $\angle{AMC}=180-\angle{AMB}=\angle{AEF}=\angle{ABF}$ Also $\angle{AFB}=\angle{ACM}...

- Wed Oct 04, 2017 10:49 pm
- Forum: Geometry
- Topic: CGMO 2002/4
- Replies:
**2** - Views:
**8101**

### Re: CGMO 2002/4

I found this in EGMO(Euclidean geometry in mathematical Olympiad). Then it's their mistake. Not mine.