BDMO REGIONAL 2015
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Please don't post problems (by starting a topic) in the "X: Solved" forums. Those forums are only for showcasing the problems for the convenience of the users. You can always post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.

 Posts: 29
 Joined: Mon Jan 23, 2017 10:32 am
 Location: Rajshahi,Bangladesh
BDMO REGIONAL 2015
Hey, guys .Help me to solve this.....
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 Posts: 53
 Joined: Tue Dec 08, 2015 4:25 pm
 Location: Bashaboo , Dhaka
Re: BDMO REGIONAL 2015
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$
$(i)  (ii) \Rightarrow DE^2 = DG^2  CG^2 + CE^2 = 20 + 25 = 5$
$\therefore DE = \sqrt{5}$
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$
$(i)  (ii) \Rightarrow DE^2 = DG^2  CG^2 + CE^2 = 20 + 25 = 5$
$\therefore DE = \sqrt{5}$
"(To Ptolemy I) There is no 'royal road' to geometry."  Euclid

 Posts: 53
 Joined: Tue Dec 08, 2015 4:25 pm
 Location: Bashaboo , Dhaka
Re: BDMO REGIONAL 2015
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 ,$AE^2 + CE^2 = BE^2 + DE^2 \Rightarrow DE^2 = AE^2 + CE^2  BE^2 = 5$
$\therefore DE = \sqrt{5}$
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 ,$AE^2 + CE^2 = BE^2 + DE^2 \Rightarrow DE^2 = AE^2 + CE^2  BE^2 = 5$
$\therefore DE = \sqrt{5}$
"(To Ptolemy I) There is no 'royal road' to geometry."  Euclid