## How to use LaTeX

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nabid72019
Posts: 2
Joined: Wed Jul 22, 2020 1:06 pm

### Re: How to use LaTeX

I have written a Latex document with bangla writing. But the problem is য ফলা gets really weird.

wa-sickofmath
Posts: 4
Joined: Sat Apr 10, 2021 9:56 pm

### BdMO National 2021 Junior P8

$ABCD$ একটি বর্গ যার বাহুর দৈর্ঘ্য $8$। $E$ আর $F$ বিন্দু যথাক্রমে $\overrightarrow{DA}$ এবং $\overrightarrow{DC}$ রশ্মির উপরে এমনভাবে আছে যেন $E$,$B$ আর $F$ বিন্দুগুলো সমরেখ এবং $EF=19$। $\frac{1}{DE}+\frac{1}{DF}$ এর মানকে $\frac{a}{b}$ আকারে লেখা যায় যেখানে $a$ আর $b$ সহমৌলিক ধনাত্বক পূর্ণসংখ্যা। $\left(a+10b\right)$ - এর মান কত $?$

Let $ABCD$ be a square with side length $8$. Points $E$ and $F$ are on ray $\overrightarrow{DA}$ and $\overrightarrow{DC}$ respectively such that $E$,$B$ and $F$ are collinear and $EF=19$. The value of $\frac{1}{DE}+\frac{1}{DF}$ can be expressed as $\frac{a}{b}$, where $a$ and $b$ are relatively prime positive integers. What is the value of $\left(a+10b\right) ?$

wa-sickofmath
Posts: 4
Joined: Sat Apr 10, 2021 9:56 pm

### BdMO National 2021 Junior Problem 6

$ABCD$ একটা সমদ্বিবাহু ট্রাপিজিয়াম যেন $AD=BC$ , $AB=5$ আর $CD=10$। $E$ এমন একটা বিন্দু যেন $AE\perp EC$ এবং $BC=EC$ । $AE$ -এর দৈর্ঘ্যকে $a\sqrt{b}$আকারে লেখা যায় যেখানে $a$ আর $b$ পূর্ণসংখ্যা এবং $b$ , $1$ বাদে অন্য কোনো পূর্ণবর্গ সংখ্যা দিয়ে বিভাজ্য না। $(b-a)$ -এর মান বের করো।

$ABCD$ be an isosceles trapezium such that, $AD=BC$ , $AB=5$ and $CD=10$ . 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 $(b-a)$

wa-sickofmath
Posts: 4
Joined: Sat Apr 10, 2021 9:56 pm

### Re: BdMO National 2021 Junior Problem 6

For our advantage, let us draw the point $E$ on the other side of $DC$ line where points $A$ and $B$ lie. Suppose, $AE$ intersects line $DC$ at point $H$. Now, draw a perpendicular on line $DC$ from the point $A$ and let it intersect $DC$ at $F$. Again, draw a perpendicular on line $DC$ but now from the point $B$, and let it intersect at point $G$.
Now, we know that both $DF$ and $GC$ are both $2.5$ because $FG$ is $5$ and $\triangle ADF$ and $\triangle BCG$ are congruent which refers to the fact that $DF$ and $GC$ both are equal to each other.

It's all Pythagoras now.
$AD^2 = (2.5)^2 + AF^2$ and $AC^2 = (2.5+5)^2 + AF^2$

We know that, $EC = BC$ $or$ $EC = AD$ so,

$AC^2 = EC^2 + AE^2$
$or,$ $AC^2 = AD^2 + AE^2$
$or,$ $(7.5)^2 + AF^2 = {(2.5)^2 + AF^2} + AE^2$
$or,$ $(7.5)^2 + AF^2 - (2.5)^2 - AF^2 = AE^2$
$or,$ $(7.5)^2 - (2.5)^2 = AE^2$
$or,$ $(7.5 + 2.5)(7.5 - 2.5) = AE^2$
$or,$ $10\times5 = AE^2$
$or,$ $AE^2 = 50$
$or,$ $AE = \surd50$
$or,$ $AE = 5\surd2$
So, $(b-a) = 5-2 = 3$