## Search found 217 matches

- Sun Jan 20, 2013 9:25 pm
- Forum: Higher Secondary Level
- Topic: Secondary and Higher Secondary Marathon
- Replies:
**127** - Views:
**58659**

### Re: Secondary and Higher Secondary Marathon

Problem $32$:Consider $\omega$ to be the circumcircle of $\triangle{ABC}$.$D$ is the midpoint of arc $BAC$ and $I$ is the incentre.Let $DI$ intersect $BC$ at $E$ and $\omega$ again at $F$.Let $P$ be a point on the line $AF$ such that $PE$ is parallel to $AI$.Prove that $PE$ is the bisector of $\angl...

- Sun Jan 20, 2013 9:16 pm
- Forum: Higher Secondary Level
- Topic: Secondary and Higher Secondary Marathon
- Replies:
**127** - Views:
**58659**

### Re: Secondary and Higher Secondary Marathon

This solution process can be used as a lemma.

- Sun Jan 20, 2013 9:07 pm
- Forum: Higher Secondary Level
- Topic: Secondary and Higher Secondary Marathon
- Replies:
**127** - Views:
**58659**

### Re: Secondary and Higher Secondary Marathon

Actually $\triangle{MO_{1}D}$ and $\triangle{MO_{2}C}$ are spirally symmetric.So are $\triangle{MO_{1}O_{2}}$ and

$\triangle {MDC}$.Now assuming the contrary we can show a contradiction.

$\triangle {MDC}$.Now assuming the contrary we can show a contradiction.

- Sun Jan 20, 2013 5:30 pm
- Forum: Geometry
- Topic: China National-2013-1
- Replies:
**7** - Views:
**3286**

### Re: China National-2013-1

Well,We can ensure the well definition.Let $\triangle{AEF}$ be fixed.As $AC^2=AE.AF$.So the length of $AC$ is fixed.And as $C,B,F$ and $D,B,E$ are collinear,this property forces $AC$ to be perpendicular to the angle bisector of $\angle{A}$. By restating the problem another way we can define the poin...

- Sun Jan 20, 2013 5:15 pm
- Forum: Higher Secondary Level
- Topic: Secondary and Higher Secondary Marathon
- Replies:
**127** - Views:
**58659**

### Re: Secondary and Higher Secondary Marathon

Oy Tahmid,This is my shitty solution.I find it so uncool.

The mistake in my solution is that I took the positive value of some cos when it should be negative.

The mistake in my solution is that I took the positive value of some cos when it should be negative.

- Sun Jan 20, 2013 2:13 pm
- Forum: Higher Secondary Level
- Topic: Secondary and Higher Secondary Marathon
- Replies:
**127** - Views:
**58659**

### Re: Secondary and Higher Secondary Marathon

$a=5$,$b=-2$,$c=2$ So $a+b+c=5$

I have a shitty proof.

I have a shitty proof.

- Sat Jan 19, 2013 10:00 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies:
**184** - Views:
**64044**

### Re: IMO Marathon

Pure Euclidean proof for $16.a$

All the $\triangle{AMN}$ are similar.So the area is maximum when $AM$ and $AN$ are radii of the respective circles.That is $MN$ is parallel to $BC$.

All the $\triangle{AMN}$ are similar.So the area is maximum when $AM$ and $AN$ are radii of the respective circles.That is $MN$ is parallel to $BC$.

- Sat Jan 12, 2013 10:53 pm
- Forum: Higher Secondary Level
- Topic: Trigonometric Proof
- Replies:
**1** - Views:
**1681**

### Re: Trigonometric Proof

Repeated use of the identity $Sin(2x)=2sin(x).cos(x)$

There can be no more elegant proof.

There can be no more elegant proof.

- Wed Jan 09, 2013 9:18 pm
- Forum: Combinatorics
- Topic: Count Them Differently
- Replies:
**2** - Views:
**1738**

### Re: Count Them Differently

$$\binom{k}{1}+2\binom{k}{2}=k+2\frac{k\left( k-1\right) }{2}=k^{2} $$

we get

$\sum_{k=1}^{n}k^{2}=\sum_{k=0}^{n}\binom{k}{1}+2\binom{k}{2}% =\sum_{i=1}^{n}\binom{k}{1}+2\sum_{k=1}^{n}\binom{k}{2} \\ &=&\binom{n+1}{2}+2\binom{n+1}{3} \\ &=&\frac{n\left( n+1\right) \left( 2n+1\right) }{6}$

we get

$\sum_{k=1}^{n}k^{2}=\sum_{k=0}^{n}\binom{k}{1}+2\binom{k}{2}% =\sum_{i=1}^{n}\binom{k}{1}+2\sum_{k=1}^{n}\binom{k}{2} \\ &=&\binom{n+1}{2}+2\binom{n+1}{3} \\ &=&\frac{n\left( n+1\right) \left( 2n+1\right) }{6}$

- Wed Jan 09, 2013 8:23 pm
- Forum: Secondary Level
- Topic: Cyclic Quad
- Replies:
**5** - Views:
**2186**

### Re: Cyclic Quad

W.l.o.g assume $\angle{A}>\angle{C}$

and $\angle{ABD}>\angle{BDC}$

$\Longrightarrow$ $AD>BC$

Let $AB$ and $CD$ meet at $X$

$AC>BD$ implies

$\angle{ADC}<\angle{BAD}$

so $AX>DX$

as $\frac{AB}{CD}=\frac{AX}{DX}$

$AB>CD$

So $(AB-CD)(AD-BC)>0$

and $\angle{ABD}>\angle{BDC}$

$\Longrightarrow$ $AD>BC$

Let $AB$ and $CD$ meet at $X$

$AC>BD$ implies

$\angle{ADC}<\angle{BAD}$

so $AX>DX$

as $\frac{AB}{CD}=\frac{AX}{DX}$

$AB>CD$

So $(AB-CD)(AD-BC)>0$