Search found 65 matches
- Sun Feb 26, 2017 3:21 pm
- Forum: Geometry
- Topic: IGO 2016 Elementary/2
- Replies: 5
- Views: 4589
Re: IGO 2016 Elementary/2
How we can write $\angle YXC = \angle XYC \Rightarrow \widehat{AP} + \widehat{CY} = \widehat{PC}$?
- Thu Feb 23, 2017 1:21 pm
- Forum: Geometry
- Topic: IGO 2016 Elementary/2
- Replies: 5
- Views: 4589
Re: IGO 2016 Elementary/2
Here is the official solution :
We know that $CX = CY$ therefore:
$\angle YXC = \angle XYC \Rightarrow \widehat{AP} + \widehat{CY} = \widehat{PC}$ [But how?]
Also we have $AB = CY$ therefore $\widehat{AP} + \widehat{CY} = \widehat{AP} + \widehat{AB} = \widehat{PB}$
So $PB = PC$.
We know that $CX = CY$ therefore:
$\angle YXC = \angle XYC \Rightarrow \widehat{AP} + \widehat{CY} = \widehat{PC}$ [But how?]
Also we have $AB = CY$ therefore $\widehat{AP} + \widehat{CY} = \widehat{AP} + \widehat{AB} = \widehat{PB}$
So $PB = PC$.
- Thu Feb 23, 2017 1:11 pm
- Forum: Geometry
- Topic: IGO 2016 Elementary/2
- Replies: 5
- Views: 4589
Re: IGO 2016 Elementary/2
Let , $\angle CPY = \angle CBY = \angle p$ $\angle PBC = \angle PYC = \angle q$ $CX = CY \Rightarrow \angle CYX = \angle CXY = \angle q$ $AB = CY , BY = BY , \angle BAY = \angle BCY$ So,$\triangle ABY \cong \triangle BCY$ Thus, $\angle CBY = \angle AYB = \angle ACB = \angle p$ From $\triangle PXC$, ...
- Mon Feb 20, 2017 2:46 pm
- Forum: Geometry
- Topic: Looking for non-trig solution
- Replies: 3
- Views: 3337
Re: Looking for non-trig solution
Thanks to @Thanic nur samin .Can anyone give a synthetic solution?
- Mon Feb 20, 2017 12:38 pm
- Forum: Geometry
- Topic: Looking for non-trig solution
- Replies: 3
- Views: 3337
Re: Looking for non-trig solution
$AB = 4 \Rightarrow OA = OB = 2$. $BC = OB\sin30 = 1$ , $OC = \sqrt{2^2 - 1^2} = \sqrt{3}$ , $CD = 2\sqrt{3}$ and $BD = \sqrt{13}$ Let , $\angle EAO = \alpha$ From the sine rule , $\frac{AC}{\sin AOC} = \frac{OC}{\sin\alpha} \Rightarrow \sin\alpha = \frac{\sqrt{3}}{2\sqrt{13}}$ From the cosine rule,...
- Mon Feb 20, 2017 12:19 pm
- Forum: Geometry
- Topic: Looking for non-trig solution
- Replies: 3
- Views: 3337
Looking for non-trig solution
In a circle , $AB = 4$ is the diameter and $O$ is the centre.On the diameter that makes an angle of $30^{\circ}$ with $AB$ at the centre, two points $C$ and $D$ are chosen that $OC = OD$ and $\angle BCO = 90^{\circ}$.The perpendicular on $AB$ through $O$ meets $AC$ at $E$ and $BD$ at $F$.If $EF = \f...
- Sat Feb 18, 2017 10:57 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO NATIONAL JUNIOR 2012/09
- Replies: 1
- Views: 2218
Re: BDMO NATIONAL JUNIOR 2012/09
It's confusing to use $R$ for both a vertex of $PQRS$ and the cirum-radius of $\triangle ABC$.So,we will use $r$ for the cirum-radius of $\triangle ABC$. For the proof , we will prove two things : $(i)\frac{AS}{PS} = \frac{c}{a}$ $(ii)\frac{PS}{SB} = \frac{b}{2r}$ $(i)$ By some angle chasing,$\angle...
- Tue Feb 14, 2017 10:37 pm
- Forum: Geometry
- Topic: A smart geo
- Replies: 1
- Views: 2792
Re: A smart geo
The problem statement is incomplete. The radius/ diameter of the smaller circle should be given.$AB$ can be determined in respect of the smaller circle's radius. Let the radii of the bigger and smaller circle are respectively $O_1$ and $O_2$ and the touching point of the two circle is $T$.Using the ...
- Sat Feb 11, 2017 10:59 pm
- Forum: Junior Level
- Topic: geomerty
- Replies: 10
- Views: 8323
Re: geomerty
Let's extend AC and define two points $F$ and $E$ such that $CE = BC$ and
$CD = CF$.
By some angle chasing,
$\angle DEF = \angle EDF \Rightarrow DF = EF$
$\angle DFA = \angle FAD \Rightarrow DF = AD$
$\therefore AD = EF$
Now ,
$AD + CD = EF + CF = CE = BC = 2017$
$CD = CF$.
By some angle chasing,
$\angle DEF = \angle EDF \Rightarrow DF = EF$
$\angle DFA = \angle FAD \Rightarrow DF = AD$
$\therefore AD = EF$
Now ,
$AD + CD = EF + CF = CE = BC = 2017$
- Tue Feb 07, 2017 8:38 pm
- Forum: Divisional Math Olympiad
- Topic: Geometry
- Replies: 10
- Views: 7225
Re: Geometry
The answer will be $\binom{2013}{4}$