## BdMO National 2012: Junior 5

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

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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Joined: Tue Feb 21, 2012 2:11 pm

### Re: BdMO National 2012: Junior 5

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

Shafin
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Joined: Thu Dec 27, 2012 3:05 pm

### Re: BdMO National 2012: Junior 5

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. 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

Fahim Shahriar
Posts: 138
Joined: Sun Dec 18, 2011 12:53 pm

### Re: BdMO National 2012: Junior 5

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

atiab jobayer
Posts: 23
Joined: Thu Dec 19, 2013 7:40 pm

### Re: BdMO National 2012: Junior 5

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

Fahim Shahriar
Posts: 138
Joined: Sun Dec 18, 2011 12:53 pm

### Re: BdMO National 2012: Junior 5

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 (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