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BdMO Online Forum • View topic - BDMO REGIONAL 2015

BDMO REGIONAL 2015

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BDMO REGIONAL 2015

Post Number:#1  Unread postby Math Mad Muggle » Mon Feb 06, 2017 3:32 pm

Hey, guys .Help me to solve this.....
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Math Mad Muggle
 
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Re: BDMO REGIONAL 2015

Post Number:#2  Unread postby Absur Khan Siam » Tue Feb 07, 2017 3:52 pm

Let's draw a line such that it goes through $E$ and
perpendicular to $AB$ and $CD$.It intersects $AB$
at $F$ and $CD$ at $G$.We also can say that $BF = CG , AF = DG$.
We can form four equations:
$DE^2 = EG^2 + DG^2...(i)$
$CE^2 = EG^2 + CG^2...(ii)$
$AE^2 = EF^2 + AF^2...(iii)$
$BE^2 = EF^2 + BF^2...(iv)$
$(iv) -(iii) \Rightarrow BE^2 - AE^2 = BF^2 - AF^2 \Rightarrow 20 = BF^2 - AF^2$
$ = CG^2 - DG^2$
$(i) - (ii) \Rightarrow DE^2 = DG^2 - CG^2 + CE^2 = -20 + 25 = 5$
$\therefore DE = \sqrt{5}$ :D
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid
Absur Khan Siam
 
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Re: BDMO REGIONAL 2015

Post Number:#3  Unread postby Absur Khan Siam » Wed Mar 08, 2017 10:59 pm

A simpler solution:
Draw two diagonals $AC$ and $BD$ and let them intersect in $O$.
$AO = BO = CO = DO$
Thus,in $\triangle AEC$ , $EO$ is the median.
Appolonius' theorem implies that $AE^2 + CE^2 = 2(AO^2 + EO^2)$
Similarly , from $\triangle BED$ , $DE^2 + BE^2 = 2(BO^2 + EO^2) = 2(AO^2 + EO^2)$

Thus , we can say ,$AE^2 + CE^2 = BE^2 + DE^2 \Rightarrow DE^2 = AE^2 + CE^2 - BE^2 = 5$

$\therefore DE = \sqrt{5}$ :D
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid
Absur Khan Siam
 
Posts: 53
Joined: Tue Dec 08, 2015 4:25 pm
Location: Bashaboo , Dhaka


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