KHULNA 2013,Junior,P10
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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.
In trapezium $ABCD$,$AD||BC$,$AD < BC$ and nonparallel sides are equal.Perpendicular drawn at $A$ on $AB$ meets $BC$ at $F$ where $BF:FC=3:2$.Perpendicular from $A$on $BC$ meets $BC$ at $E$.$BC = 10AE$,$BE < EF$,the ratio $EF:BC$ can be expressed as $\frac{a\sqrt{a}+b}{c}$.Find the LCM of $a,b,c$.
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- Raiyan Jamil
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Re: KHULNA 2013,Junior,P10
I think the ans is 240.
Let AE be x. So, BC= 10x , BF=6x , FC=4x . We take any point Z as the midpoint of BF. So, BZ=FZ=3x. We join A and Z. So, AZ=3x. And therefore, $EZ^2=(3x)^2-(x)^2$. And EZ= $\sqrt{8x^2}$ or, $\sqrt{8}x$.
Now, EF=EZ+FZ or, $\sqrt{8}x + 3x$. So, the ratio EF:BC can be expressed as $\frac{x\sqrt{8}+3x}{10x}$
Here, we can discover that the value of x is 8. Or, the value of a is 8, b is 3*8=24 and c is 10*8=80.So, the lcm of a,b,c will be the lcm of 8,24 and 80.That is 240.
Let AE be x. So, BC= 10x , BF=6x , FC=4x . We take any point Z as the midpoint of BF. So, BZ=FZ=3x. We join A and Z. So, AZ=3x. And therefore, $EZ^2=(3x)^2-(x)^2$. And EZ= $\sqrt{8x^2}$ or, $\sqrt{8}x$.
Now, EF=EZ+FZ or, $\sqrt{8}x + 3x$. So, the ratio EF:BC can be expressed as $\frac{x\sqrt{8}+3x}{10x}$
Here, we can discover that the value of x is 8. Or, the value of a is 8, b is 3*8=24 and c is 10*8=80.So, the lcm of a,b,c will be the lcm of 8,24 and 80.That is 240.
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Re: KHULNA 2013,Junior,P10
$\frac{x\sqrt{8}+3x}{10x}=\frac{x(\sqrt{8}+3)}{10x}=\frac{\sqrt{8}+3}{10}$=$\frac{2\sqrt{2}+3}{10}$.$\therefore$ LCM of $a,b,c$ is $30$.
"Questions we can't answer are far better than answers we can't question"
- Raiyan Jamil
- Posts:138
- Joined:Fri Mar 29, 2013 3:49 pm
Re: KHULNA 2013,Junior,P10
Sorry, I actually did it by not taking it in the least form. This solution of yours is correct. Thanks for that.
A smile is the best way to get through a tough situation, even if it's a fake smile.
Re: KHULNA 2013,Junior,P10
Also thanks to you for posting the solution
"Questions we can't answer are far better than answers we can't question"