BdMO National 2012: Junior 5

Discussion on Bangladesh Mathematical Olympiad (BdMO) National
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Zzzz
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BdMO National 2012: Junior 5

Unread post by Zzzz » Sun Feb 12, 2012 9:11 am

Problem 5:
$ABC$ is a right triangle with hypotenuse $AC$. $D$ is the midpoint of $AC$. $E$ is a point on the extension of $BD$. The perpendicular drawn on $BC$ from $E$ intersects $AC$ at $F$ and $BC$ at $G.$ (a) Prove that, if $DEF$ is an equilateral triangle then $\angle ACB = 30^0$. (b) Prove that, if $\angle ACB = 30^0$ then $DEF$ is an equilateral triangle.
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Arafat
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Re: BdMO National 2012: Junior 5

Unread post by Arafat » Fri Feb 24, 2012 1:26 am

I could solve the 1st part only....can anyone explain the another one.....

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Shafin
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Re: BdMO National 2012: Junior 5

Unread post by Shafin » Thu Jan 10, 2013 4:42 pm

Arafat wrote:I could solve the 1st part only....can anyone explain the another one.....
Here is my ans ( I'm not sure wheter i'm correct ) :?:
2nd Part

$\angle ACB = 30^{\circ}$
$\angle ABC = 90^{\circ}$ ( Since it is a right angled triangle )
$\angle BAC + 30^{\circ} + 90^{\circ} = 180^{\circ}$
$\angle BAC= 60^{\circ}$
$\angle ABD= \angle ABC - \angle CBD\; \; (\angle CBD = \angle BCD=30^{\circ}) =90^{\circ}-30^{\circ}=60^{\circ}$
$\angle BDA+\angle DAB+\angle ABD= 180^{\circ}$
$\angle BDA = 60^{\circ}$
$\angle BDA =\angle EDF= 60^{\circ}$

$\angle FGC + \angle FCG + \angle GFC = 180^{\circ}$
$90^{\circ} + 30^{\circ} + \angle GFC = 180^{\circ}$
$\angle GFC = 60^{\circ}$
$\angle GFC = \angle DFE=60^{\circ}$
$\angle EDF +\angle DFE + \angle DEF = 180^{\circ}$
$60^{\circ} + 60^{\circ} + \angle DEF = 180^{\circ}$
$\angle DEF= 60^{\circ}$

All three angles are equal so $\triangle DEF$ is isosceles { Solved }
I hope I'm right. :D

Anyone plz check my ans
Last edited by Phlembac Adib Hasan on Sat Jan 12, 2013 3:01 pm, edited 1 time in total.
Reason: Latexed

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Fahim Shahriar
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Re: BdMO National 2012: Junior 5

Unread post by Fahim Shahriar » Sun Jan 13, 2013 12:32 am

I've smaller solution.


2nd Part

$\angle DFE=60°$

In $\triangle CBD$, $\displaystyle \frac {BD}{\sin 30°} = \frac {CD}{\sin CBD}$

In $\triangle ABD$, $\displaystyle \frac {BD}{\sin 60°} = \frac {AD}{\sin (90°-CBD)}$

From the 2 equations, we get $\angle CBD = 30°$. Then $\angle DEF = 60°$.

$\angle DFE = \angle DEF = \angle EDF = 60°$

So $\triangle DEF$ is equilateral.
Name: Fahim Shahriar Shakkhor
Notre Dame College

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atiab jobayer
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Re: BdMO National 2012: Junior 5

Unread post by atiab jobayer » Fri Feb 07, 2014 11:12 pm

Arafat wrote:I could solve the 1st part only....can anyone explain the another one.....
Brother please show your ans of 1st part :|
math champion

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Fahim Shahriar
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Re: BdMO National 2012: Junior 5

Unread post by Fahim Shahriar » Sat Feb 08, 2014 6:56 pm

atiab jobayer wrote:
Arafat wrote:I could solve the 1st part only....can anyone explain the another one.....
Brother please show your ans of 1st part :|
1st part
ACB 30.jpg
ACB 30.jpg (13.7 KiB) Viewed 1319 times
In equilateral triangle $\triangle DEF$,
$\angle DFE = \angle DEF = \angle EDF = 60^\circ$
$\angle DFE =\angle CFG = 60^\circ$
$\angle ACB = 90^\circ - \angle CFG = 90^\circ - 60^\circ = 30^\circ$

[Proved]
Name: Fahim Shahriar Shakkhor
Notre Dame College

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