Search found 217 matches
- Sun Jan 20, 2013 9:25 pm
- Forum: Higher Secondary Level
- Topic: Secondary and Higher Secondary Marathon
- Replies: 128
- Views: 301462
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: 128
- Views: 301462
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: 128
- Views: 301462
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: 4936
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: 128
- Views: 301462
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: 128
- Views: 301462
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: 114841
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: 2747
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: 2922
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: 3535
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 $(AB-CD)(AD-BC)>0$