National BDMO 2017 : Junior 8
In $\bigtriangleup ABC$, the perpendicular bisector of $AB$ and $AC$ meet at $O$. $AO$ meets $BC$ at $D$. Now, $OD$ = $BD$ = $\dfrac {1}{3}BC$. Find the angles of $\bigtriangleup ABC$.
The study of mathematics, like the Nile, begins in minuteness but ends in magnificence.
- Charles Caleb Colton
- Charles Caleb Colton
- Thamim Zahin
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Re: National BDMO 2017 : Junior 8
Are there anyone who solve this in the exam time? Did you @dshasan?
I think we judge talent wrong. What do we see as talent? I think I have made the same mistake myself. We judge talent by the trophies on their showcases, the flamboyance the supremacy. We don't see things like determination, courage, discipline, temperament.
- Thamim Zahin
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Re: National BDMO 2017 : Junior 8
We know that O is the circumcenter. So, $OA=OB=OC$. And $DB=DO$. So, $ \angle DBO=\angle DOB=a$.
Take D' such that $D'C= \frac {1}{3} BC$. So, $DO=DB=DD'=D'C$.
$OB=OC$. So, $\angle OBC= \angle OCB=a$. So by $SAS$ congruence $\triangle OBD \cong \triangle OCD'$.
So, $OD=OD'$, And, $\triangle ODD'$ is equilateral. And $\angle ODD'=2a$. So, $a=30^o$. The rest are calculation.
So, $\angle BOA = 180^o-a=150^o$. So, $\angle OAB=\angle OBA= 15^o$.
And now, $\angle COA = 180^o-3a=90^o$. So, $\angle OAC=\angle OCA= 45^o$.
So, $\angle ABC=30^o+15^o= 45^o$. $\angle BCA= 30^o+45^o=75^o$.
And, $\angle CAB= 45^o+15^o= 60^o$.
Answer $(\angle A,\angle B,\angle C)=(60^o,45^o,75^o)$
Take D' such that $D'C= \frac {1}{3} BC$. So, $DO=DB=DD'=D'C$.
$OB=OC$. So, $\angle OBC= \angle OCB=a$. So by $SAS$ congruence $\triangle OBD \cong \triangle OCD'$.
So, $OD=OD'$, And, $\triangle ODD'$ is equilateral. And $\angle ODD'=2a$. So, $a=30^o$. The rest are calculation.
So, $\angle BOA = 180^o-a=150^o$. So, $\angle OAB=\angle OBA= 15^o$.
And now, $\angle COA = 180^o-3a=90^o$. So, $\angle OAC=\angle OCA= 45^o$.
So, $\angle ABC=30^o+15^o= 45^o$. $\angle BCA= 30^o+45^o=75^o$.
And, $\angle CAB= 45^o+15^o= 60^o$.
Answer $(\angle A,\angle B,\angle C)=(60^o,45^o,75^o)$
I think we judge talent wrong. What do we see as talent? I think I have made the same mistake myself. We judge talent by the trophies on their showcases, the flamboyance the supremacy. We don't see things like determination, courage, discipline, temperament.
Re: National BDMO 2017 : Junior 8
The study of mathematics, like the Nile, begins in minuteness but ends in magnificence.
- Charles Caleb Colton
- Charles Caleb Colton
- Thamim Zahin
- Posts:98
- Joined:Wed Aug 03, 2016 5:42 pm
Re: National BDMO 2017 : Junior 8
I am actually kicking me for not trying 8,9 in the exam. They both are easy.
I think we judge talent wrong. What do we see as talent? I think I have made the same mistake myself. We judge talent by the trophies on their showcases, the flamboyance the supremacy. We don't see things like determination, courage, discipline, temperament.
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Re: National BDMO 2017 : Junior 8
Don't kick yourself. You have 1 year more to be the COC or COO.
- ahmedittihad
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Re: National BDMO 2017 : Junior 8
How does one actually kick himself?
Frankly, my dear, I don't give a damn.
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Re: National BDMO 2017 : Junior 8
You are right,Thamim...I also feel like kicking myself for not paying much attention to this problem as I was working on other problems...or else I could have attained a bit more than being 2nd runners up......Now,This problem does need only a bit of algebraic calculations..We may let angle BAO to be y,then we may ensure that both AOB And AOC are isosceles triangles.Then by angle chasing,find angle BOD=angle OBD=angle OCD.Then by the rest angle calculations,derive an equation which follows 3y+5y+4y=180,find y =15..That fulfills our requirements of 60-75-45 angled triangle...
- ahmedittihad
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Re: National BDMO 2017 : Junior 8
This is a valid question.ahmedittihad wrote:How does one actually kick himself?
Frankly, my dear, I don't give a damn.