Dhaka 2014,Junior,P9

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tanmoy
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Dhaka 2014,Junior,P9

Unread post by tanmoy » Sat Dec 06, 2014 3:44 pm

In square $ABCD$,the length of its sides is $5$.$E$,$F$ are two points on $AB$ and $AD$ in such a way so that $\angle ECF=45^{\circ}$.Find the maximum value of the perimeter of $\Delta AEF$.
"Questions we can't answer are far better than answers we can't question"

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Raiyan Jamil
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Re: Dhaka 2014,Junior,P9

Unread post by Raiyan Jamil » Sat Dec 13, 2014 11:17 am

I think the ans is 10.
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tanmoy
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Location:Rangpur,Bangladesh

Re: Dhaka 2014,Junior,P9

Unread post by tanmoy » Sun Dec 14, 2014 3:02 pm

Raiyan Jamil wrote:I think the ans is 10.
Post the full solution.I have also got the answer $10$.But I couldn't prove that this is the maximum value of the perimeter of $\Delta AEF$.
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Raiyan Jamil
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Re: Dhaka 2014,Junior,P9

Unread post by Raiyan Jamil » Tue Dec 16, 2014 7:12 pm

The perimeter of $\Delta AEF$ is AE+EF+FA. Now, if we try to draw the figure, you will discover that, the smallest perimeter of $\Delta AEF$ will be at that time when AE=AF and $\angle AFE=\angle AEF$ =45 degrees.And it's perimeter will be 8.53553....... Then, if you keep the point C in the centre and move the $\triangle CEF$ clockwise or anticlockwise, you can observe that the area of $\triangle CEF$ will be increasing. And then at one point, the point E and F will be on B and A and the area of $\triangle CEF$ will be (5*5)/2=12.5. At that point, you can observe that, the $\triangle AEF$ will have no area. But it's perimeter will be 5+0+5(AE+EF+FA)=10. By this, you can tell that the highest perimeter will be 10.
But I also have a doubt that if this solution will be suitable and acceptable in the math olympiad. :roll:
A smile is the best way to get through a tough situation, even if it's a fake smile.

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