Still not interested. $SIGH$

btw,Happy Birthday.

## Search found 217 matches

- Thu Jan 24, 2013 6:18 pm
- Forum: Secondary Level
- Topic: Solve this congruence equation
- Replies:
**2** - Views:
**1525**

- Wed Jan 23, 2013 1:59 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies:
**184** - Views:
**62189**

### Re: IMO Marathon

I propose a new rule that anyone who posts a solution of a problem must certainly post another problem.Or we'll have to wait for someone's grace. Problem $\boxed{22}$:Let $ABCD$ be a parallelogram.A variable line $l$ passing through the point $A$ intersects the rays $BC$ and $CD$ at points $X$ and $...

- Wed Jan 23, 2013 1:29 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies:
**184** - Views:
**62189**

### Re: IMO Marathon

I Love Ratio. :mrgreen: $\frac{SC}{SB}=\frac{SA}{SC}$ $\Longrightarrow \frac{SP}{SB}=\frac{SA}{SP}$ $\triangle{SPB} \sim \triangle{SAP}$ and $\triangle{SCB} \sim \triangle{SAC}$ $\frac{AP}{BP}=\frac{AS}{PS}=\frac{AS}{CS}=\frac{AC}{BC}$ from alternate segment theorem, $\frac{MK}{MP}=\frac{AC}{AP}$ an...

- Wed Jan 23, 2013 10:03 am
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies:
**184** - Views:
**62189**

### Re: IMO Marathon

I wish every geometry problems were as easy and as beautiful as this.

- Wed Jan 23, 2013 9:56 am
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies:
**184** - Views:
**62189**

### Re: IMO Marathon

**Problem 21:**Let $ABC$ be a triangle with $P$ in its interior(with $BC \neq AC$).The lines $AP,BP,CP$ meet $\odot{ABC}$ again at $K,L,M$.The tangent line at $C$ intersects $AB$ at $S$.Show that from $SC=SP$ it follows that $MK=ML$

Source: IMO 2010-Problem: 4

- Tue Jan 22, 2013 10:13 pm
- Forum: Higher Secondary Level
- Topic: Secondary and Higher Secondary Marathon
- Replies:
**127** - Views:
**57096**

### Re: Secondary and Higher Secondary Marathon

Problem $\boxed{34}$:$ABCD$ is a cyclic quadrilateral.$E$ and $F$ are variable points on sides $AB$ and $CD$ respectively such that $\displaystyle \frac{AE}{BE}=\frac{CF}{DF}$. $\; P$ is a point on the segment $EF$ such that $\displaystyle \frac{EP}{PF}=\frac{AB}{CD}$.Show that $\displaystyle \frac{...

- Tue Jan 22, 2013 8:10 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies:
**184** - Views:
**62189**

### Re: IMO Marathon

Solution $\boxed {19}$:Same as Tahmid.

Solution $\boxed {20}$:Actually $APIE$ is concyclic .Which implies $\angle {A}$=$90$

Problem $19$ seemed harder to me than problem $20$.

Solution $\boxed {20}$:Actually $APIE$ is concyclic .Which implies $\angle {A}$=$90$

Problem $19$ seemed harder to me than problem $20$.

- Tue Jan 22, 2013 1:21 pm
- Forum: Higher Secondary Level
- Topic: Secondary and Higher Secondary Marathon
- Replies:
**127** - Views:
**57096**

### Re: Secondary and Higher Secondary Marathon

If we draw a circle centering at $A$ with radius $AC$ as $CD^2=AD.BD$ it holds that $D$ lies on the radical axis of the two circles.And $DK$ is the radical axis.And we now that the radical axis is perpendicular to the line joining the centres of the concerning circles.The result follows.

- Tue Jan 22, 2013 12:59 pm
- Forum: Higher Secondary Level
- Topic: Secondary and Higher Secondary Marathon
- Replies:
**127** - Views:
**57096**

### Re: Secondary and Higher Secondary Marathon

Well,Adib notice that $AK=AC$ he said.

- Mon Jan 21, 2013 12:07 am
- Forum: Higher Secondary Level
- Topic: Secondary and Higher Secondary Marathon
- Replies:
**127** - Views:
**57096**

### Re: Secondary and Higher Secondary Marathon

Well,this is problem 3 of geomtry problem set of BDMC 2012.