Mixing
Two triangles $triangle ABC$ and $triangle BDE$ are placed in a way that $B$ is the right-angle of both. Find...
$begingroup$
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:
$AB$ = $frac {3 sqrt 3 C}{BC}$
$AB$ = $3 sqrt 3 sectheta$ (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.
geometry proof-verification contest-math triangles alternative-proof
edited Feb 4 at 6:47
the_fox
2,90031538
asked Jan 27 at 11:23
Anirban NiloyAnirban Niloy
8311218
$endgroup$
add a comment |
$begingroup$
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:
$AB$ = $frac {3 sqrt 3 C}{BC}$
$AB$ = $3 sqrt 3 sectheta$ (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.
geometry proof-verification contest-math triangles alternative-proof
edited Feb 4 at 6:47
the_fox
2,90031538
asked Jan 27 at 11:23
Anirban NiloyAnirban Niloy
8311218
$endgroup$
$begingroup$
Where is triangle DEF?
$endgroup$
– Maria Mazur
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
add a comment |
$begingroup$
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:
$AB$ = $frac {3 sqrt 3 C}{BC}$
$AB$ = $3 sqrt 3 sectheta$ (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.
geometry proof-verification contest-math triangles alternative-proof
edited Feb 4 at 6:47
the_fox
2,90031538
asked Jan 27 at 11:23
Anirban NiloyAnirban Niloy
8311218
$endgroup$
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:
$AB$ = $frac {3 sqrt 3 C}{BC}$
$AB$ = $3 sqrt 3 sectheta$ (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.
geometry proof-verification contest-math triangles alternative-proof
geometry proof-verification contest-math triangles alternative-proof
edited Feb 4 at 6:47
the_fox
2,90031538
asked Jan 27 at 11:23
Anirban NiloyAnirban Niloy
8311218
edited Feb 4 at 6:47
the_fox
2,90031538
asked Jan 27 at 11:23
Anirban NiloyAnirban Niloy
8311218
edited Feb 4 at 6:47
the_fox
2,90031538
edited Feb 4 at 6:47
the_fox
2,90031538
edited Feb 4 at 6:47
the_fox
2,90031538
2,90031538
asked Jan 27 at 11:23
Anirban NiloyAnirban Niloy
8311218
asked Jan 27 at 11:23
Anirban NiloyAnirban Niloy
8311218
asked Jan 27 at 11:23
Anirban NiloyAnirban Niloy
8311218
8311218
$begingroup$
Where is triangle DEF?
$endgroup$
– Maria Mazur
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
add a comment |
$begingroup$
Where is triangle DEF?
$endgroup$
– Maria Mazur
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
$begingroup$
Where is triangle DEF?
$endgroup$
– Maria Mazur
Jan 27 at 11:33
$begingroup$
Where is triangle DEF?
$endgroup$
– Maria Mazur
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
$begingroup$
@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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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
$$angle(BFG)=arctan{sover h}={1oversqrt{3}}=30^circ .$$
This implies $angle(BDG)=15^circ$.
answered Jan 27 at 14:20
Christian BlatterChristian Blatter
175k8115327
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
votes
active
oldest
votes
$begingroup$
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
$$angle(BFG)=arctan{sover h}={1oversqrt{3}}=30^circ .$$
This implies $angle(BDG)=15^circ$.
answered Jan 27 at 14:20
Christian BlatterChristian Blatter
175k8115327
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
$$angle(BFG)=arctan{sover h}={1oversqrt{3}}=30^circ .$$
This implies $angle(BDG)=15^circ$.
answered Jan 27 at 14:20
Christian BlatterChristian Blatter
175k8115327
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
$$angle(BFG)=arctan{sover h}={1oversqrt{3}}=30^circ .$$
This implies $angle(BDG)=15^circ$.
answered Jan 27 at 14:20
Christian BlatterChristian Blatter
175k8115327
$endgroup$
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
$$angle(BFG)=arctan{sover h}={1oversqrt{3}}=30^circ .$$
This implies $angle(BDG)=15^circ$.
answered Jan 27 at 14:20
Christian BlatterChristian Blatter
175k8115327
answered Jan 27 at 14:20
Christian BlatterChristian Blatter
175k8115327
answered Jan 27 at 14:20
Christian BlatterChristian Blatter
175k8115327
answered Jan 27 at 14:20
Christian BlatterChristian Blatter
175k8115327
175k8115327
$begingroup$
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
add a comment |
$begingroup$
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
$begingroup$
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
$begingroup$
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
add a comment |
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$begingroup$
Where is triangle DEF?
$endgroup$
– Maria Mazur
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