0
$\begingroup$

Source: Regional Math Olympiad of BD.

In the given diagram, both $\angle ABC$ and $\angle DBE$ are right-angled triangles. $B$ is the right angle for both (according to the diagram). They have hypotenuses $AC$ and $DE$ of the same length and $F$ is the middle point of both the hypotenuse. $DE$ is perpendicular to $BC$. Area of $\triangle ABC$ is $\sqrt 3$ times of $DEF$. What is the value of angle $\angle BDE$?

Here is a likelier figure of the contextual problem:

At first, I thought $DEF$ meant $DE+EF$. Therefore, $DE$ + $\frac{1}{2} DE$ = $\frac {3}{2} DE$. As it is mentioned in the above post that $AC= DE$, so $\frac {3}{2} DE$ = $\frac{3}{2} AC$.

Let us denote the angle $\angle ACB$ = $\theta$

From another condition, we can write that:

$\frac{1}{2} AB × BC$ = $\frac {3 \sqrt 3}{2} AC$.

$AB×BC$ = $3 \sqrt 3 AC$

And then, dividing with $BC$ to both side, we get:

$AB$ = $\frac {3 \sqrt 3 C}{BC}$

$AB$ = $3 \sqrt 3 \sec\theta$ (by denoting $\angle ACB$ as $\theta$)

Here is the problem. I failed to show any relation of $\theta$ with $AB$. I don't know whether I'm approaching towards the right direction or not. If not so, then please help me find my mistake.

Kindly pardon my error. Thanks in advance.

And if you correct my mistake, then I'll very much glad.

$\endgroup$
  • $\begingroup$ Where is triangle DEF? $\endgroup$ – Aqua Jan 27 at 11:33
  • $\begingroup$ @greedoid That's so why I'm little bit confused and disturbed. I can see that $D, E and F$ are collinear. I didn't understand the fact. This problem was appeared in a math contest. So I don't know so much about it. But there can be another way but I ain't so sure about it. $\endgroup$ – Anirban Niloy Jan 27 at 11:37
2
$\begingroup$

I replace the alleged $\triangle DEF$ by $\triangle DEB$.

Let $|FG|=:h$, $|BG|=:s$, and let $r$ be the radius of the circle with center $F$ going through $A$, $B$, E$, C$, and $D$. Then ${\rm area}(ABC)=\sqrt{3}\>{\rm area}(DEB)$ implies $2hs=\sqrt{3}rs$, so that together with $h^2+s^2=r^2$ we obtain $$\angle(BFG)=\arctan{s\over h}={1\over\sqrt{3}}=30^\circ\ .$$ This implies $\angle(BDG)=15^\circ$.

$\endgroup$
  • $\begingroup$ I was thinking so but not was sured. Maybe this could be. Thank you for your effort. I will check it out and later I'll inform you. But can't a triangle be $n$ times of a side of a distinctive triangle where $n$ is described as positive integer number? Then somehow we might have got the point that the area of $\triangle ABC$ is $\sqrt 3$ times of length $DEF$. $\endgroup$ – Anirban Niloy Jan 27 at 14:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.