BdMO National 2021 Secondary P4

Discussion on Bangladesh Mathematical Olympiad (BdMO) National
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Pro_GRMR
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BdMO National 2021 Secondary P4

Unread post by Pro_GRMR » Sun Apr 11, 2021 1:37 pm

$ABCD$ be an isosceles trapezium such that $AD=BC$,$AB=6$ and $CD=8$. A point $E$ on the plane is such that $AE\perp EC$ and $BC=EC$. The length of $AE$ can be expressed as $a\sqrt{b}$ where $a$ and $b$ are integers and $b$ is not divisible by any square number other than $1$. Find the value of $(a-b)$.
"When you change the way you look at things, the things you look at change." - Max Planck

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Pro_GRMR
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Re: BdMO National 2021 Secondary P4

Unread post by Pro_GRMR » Sun Apr 11, 2021 3:58 pm

Pro_GRMR wrote:
Sun Apr 11, 2021 1:37 pm
$ABCD$ be an isosceles trapezium such that $AD=BC$,$AB=6$ and $CD=8$. A point $E$ on the plane is such that $AE\perp EC$ and $BC=EC$. The length of $AE$ can be expressed as $a\sqrt{b}$ where $a$ and $b$ are integers and $b$ is not divisible by any square number other than $1$. Find the value of $(a-b)$.
We take a line $CD= 8$. We can just make a circle $g$ with a radius of $7$ from center $D$. We can create point $E$ on $CD$ such that $ED=1$. We draw a perpendicular from $E$ that intersects the circle $g$ at point $A$. We can get point $B$ in a similar way.
Screenshot 2021-04-11 134855.png
Screenshot 2021-04-11 134855.png (20.24KiB)Viewed 1767 times
Notice This trapezium satisfies all such constraints.
Now, we can easily get $AE = \sqrt{AD^2+DE^2} =\sqrt{7^2-1^2}=\sqrt{48}=4\sqrt{3}$

Comparing this to $a\sqrt{b}$, we get $a=4, b=3$. And so, $a-b=4-3=\boxed{1}$


Last bumped by Pro_GRMR on Sun Apr 11, 2021 3:58 pm.
"When you change the way you look at things, the things you look at change." - Max Planck

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