Mixing
Side of the equilateral triangle
$begingroup$
I tried very much but since tomorrow is my exam, i cannot risk it.
The following is a geometry problem, which i have tried very much but could not grasp a solution.
I think that i require pythagoras' theorem here, but can not understand how to apply it.
I first thought of extending a line congruent to BP such that it intersects CP at a point. But that did not help me. I also have the hint that all the other angles are 60 in measure, but if P were the incentre, in that case it would not have been difficult to show to P is the subtending angle of measure 150 degrees.
Any help would be helpful!
geometry euclidean-geometry triangle circle
edited Jan 19 at 15:49
Michael Rozenberg
106k1893198
$endgroup$
|
show 1 more comment
$begingroup$
I tried very much but since tomorrow is my exam, i cannot risk it.
The following is a geometry problem, which i have tried very much but could not grasp a solution.
I think that i require pythagoras' theorem here, but can not understand how to apply it.
I first thought of extending a line congruent to BP such that it intersects CP at a point. But that did not help me. I also have the hint that all the other angles are 60 in measure, but if P were the incentre, in that case it would not have been difficult to show to P is the subtending angle of measure 150 degrees.
Any help would be helpful!
geometry euclidean-geometry triangle circle
edited Jan 19 at 15:49
Michael Rozenberg
106k1893198
$endgroup$
4
$begingroup$
Another picture on its side :-(
$endgroup$
– Lord Shark the Unknown
Jan 19 at 13:04
1
$begingroup$
How will i draw the given figure?
$endgroup$
– user636268
Jan 19 at 13:06
2
$begingroup$
To be clear: Are you specifically asking about question $19$ in the image? If so, please make an effort to type the text of the problem into your question; it's pretty short. (Questions in images aren't searchable, and they tend to be harder to read ... especially when they aren't right-side-up ;).
$endgroup$
– Blue
Jan 19 at 13:17
$begingroup$
Rotate it😑😑😑😑
$endgroup$
– user636268
Jan 19 at 13:19
$begingroup$
It is unneccessary. I constructed the diagram myself. I did not take a hint
$endgroup$
– user636268
Jan 19 at 13:19
|
show 1 more comment
$begingroup$
I tried very much but since tomorrow is my exam, i cannot risk it.
The following is a geometry problem, which i have tried very much but could not grasp a solution.
I think that i require pythagoras' theorem here, but can not understand how to apply it.
I first thought of extending a line congruent to BP such that it intersects CP at a point. But that did not help me. I also have the hint that all the other angles are 60 in measure, but if P were the incentre, in that case it would not have been difficult to show to P is the subtending angle of measure 150 degrees.
Any help would be helpful!
geometry euclidean-geometry triangle circle
edited Jan 19 at 15:49
Michael Rozenberg
106k1893198
$endgroup$
I tried very much but since tomorrow is my exam, i cannot risk it.
The following is a geometry problem, which i have tried very much but could not grasp a solution.
I think that i require pythagoras' theorem here, but can not understand how to apply it.
I first thought of extending a line congruent to BP such that it intersects CP at a point. But that did not help me. I also have the hint that all the other angles are 60 in measure, but if P were the incentre, in that case it would not have been difficult to show to P is the subtending angle of measure 150 degrees.
Any help would be helpful!
geometry euclidean-geometry triangle circle
geometry euclidean-geometry triangle circle
edited Jan 19 at 15:49
Michael Rozenberg
106k1893198
edited Jan 19 at 15:49
Michael Rozenberg
106k1893198
edited Jan 19 at 15:49
Michael Rozenberg
106k1893198
edited Jan 19 at 15:49
Michael Rozenberg
106k1893198
edited Jan 19 at 15:49
Michael Rozenberg
106k1893198
106k1893198
asked Jan 19 at 13:01
user636268
4
$begingroup$
Another picture on its side :-(
$endgroup$
– Lord Shark the Unknown
Jan 19 at 13:04
1
$begingroup$
How will i draw the given figure?
$endgroup$
– user636268
Jan 19 at 13:06
2
$begingroup$
To be clear: Are you specifically asking about question $19$ in the image? If so, please make an effort to type the text of the problem into your question; it's pretty short. (Questions in images aren't searchable, and they tend to be harder to read ... especially when they aren't right-side-up ;).
$endgroup$
– Blue
Jan 19 at 13:17
$begingroup$
Rotate it😑😑😑😑
$endgroup$
– user636268
Jan 19 at 13:19
$begingroup$
It is unneccessary. I constructed the diagram myself. I did not take a hint
$endgroup$
– user636268
Jan 19 at 13:19
|
show 1 more comment
4
$begingroup$
Another picture on its side :-(
$endgroup$
– Lord Shark the Unknown
Jan 19 at 13:04
1
$begingroup$
How will i draw the given figure?
$endgroup$
– user636268
Jan 19 at 13:06
2
$begingroup$
To be clear: Are you specifically asking about question $19$ in the image? If so, please make an effort to type the text of the problem into your question; it's pretty short. (Questions in images aren't searchable, and they tend to be harder to read ... especially when they aren't right-side-up ;).
$endgroup$
– Blue
Jan 19 at 13:17
$begingroup$
Rotate it😑😑😑😑
$endgroup$
– user636268
Jan 19 at 13:19
$begingroup$
It is unneccessary. I constructed the diagram myself. I did not take a hint
$endgroup$
– user636268
Jan 19 at 13:19
4
4
$begingroup$
Another picture on its side :-(
$endgroup$
– Lord Shark the Unknown
Jan 19 at 13:04
$begingroup$
Another picture on its side :-(
$endgroup$
– Lord Shark the Unknown
Jan 19 at 13:04
1
1
$begingroup$
How will i draw the given figure?
$endgroup$
– user636268
Jan 19 at 13:06
$begingroup$
How will i draw the given figure?
$endgroup$
– user636268
Jan 19 at 13:06
2
2
$begingroup$
To be clear: Are you specifically asking about question $19$ in the image? If so, please make an effort to type the text of the problem into your question; it's pretty short. (Questions in images aren't searchable, and they tend to be harder to read ... especially when they aren't right-side-up ;).
$endgroup$
– Blue
Jan 19 at 13:17
$begingroup$
To be clear: Are you specifically asking about question $19$ in the image? If so, please make an effort to type the text of the problem into your question; it's pretty short. (Questions in images aren't searchable, and they tend to be harder to read ... especially when they aren't right-side-up ;).
$endgroup$
– Blue
Jan 19 at 13:17
$begingroup$
Rotate it😑😑😑😑
$endgroup$
– user636268
Jan 19 at 13:19
$begingroup$
Rotate it😑😑😑😑
$endgroup$
– user636268
Jan 19 at 13:19
$begingroup$
It is unneccessary. I constructed the diagram myself. I did not take a hint
$endgroup$
– user636268
Jan 19 at 13:19
$begingroup$
It is unneccessary. I constructed the diagram myself. I did not take a hint
$endgroup$
– user636268
Jan 19 at 13:19
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Let in $Delta ABC$ vertexes by the rotation similar to the clock.
The first problem.
Let $Kin AM$ such that $BM=KM$.
Thus, since
$$measuredangle BMK=measuredangle BMA=measuredangle BCA=60^{circ},$$
we obtain that $Delta BMK$ is equilateral triangle.
Now, take $Delta BAK$ and rotate around point $B$ by $60^{circ}.$
We obtain that $A$ goes to $C$ and $K$ goes to $M$,
which says $AK=MC$ and
$$BM+MC=MK+AK=AM$$ and we are done!
The second problem.
Take the rotation around $C$ by $60^{circ}.$
We see that $B$ goes to $A$ and let $M$ goes to $M'$.
Thus, $BM=AM'$ and $MC=MM'$ and we got $Delta AMM'$ with needed properties.
answered Jan 19 at 15:48
Michael RozenbergMichael Rozenberg
106k1893198
$endgroup$
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 20 at 22:45
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let in $Delta ABC$ vertexes by the rotation similar to the clock.
The first problem.
Let $Kin AM$ such that $BM=KM$.
Thus, since
$$measuredangle BMK=measuredangle BMA=measuredangle BCA=60^{circ},$$
we obtain that $Delta BMK$ is equilateral triangle.
Now, take $Delta BAK$ and rotate around point $B$ by $60^{circ}.$
We obtain that $A$ goes to $C$ and $K$ goes to $M$,
which says $AK=MC$ and
$$BM+MC=MK+AK=AM$$ and we are done!
The second problem.
Take the rotation around $C$ by $60^{circ}.$
We see that $B$ goes to $A$ and let $M$ goes to $M'$.
Thus, $BM=AM'$ and $MC=MM'$ and we got $Delta AMM'$ with needed properties.
answered Jan 19 at 15:48
Michael RozenbergMichael Rozenberg
106k1893198
$endgroup$
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 20 at 22:45
add a comment |
$begingroup$
Let in $Delta ABC$ vertexes by the rotation similar to the clock.
The first problem.
Let $Kin AM$ such that $BM=KM$.
Thus, since
$$measuredangle BMK=measuredangle BMA=measuredangle BCA=60^{circ},$$
we obtain that $Delta BMK$ is equilateral triangle.
Now, take $Delta BAK$ and rotate around point $B$ by $60^{circ}.$
We obtain that $A$ goes to $C$ and $K$ goes to $M$,
which says $AK=MC$ and
$$BM+MC=MK+AK=AM$$ and we are done!
The second problem.
Take the rotation around $C$ by $60^{circ}.$
We see that $B$ goes to $A$ and let $M$ goes to $M'$.
Thus, $BM=AM'$ and $MC=MM'$ and we got $Delta AMM'$ with needed properties.
answered Jan 19 at 15:48
Michael RozenbergMichael Rozenberg
106k1893198
$endgroup$
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 20 at 22:45
add a comment |
$begingroup$
Let in $Delta ABC$ vertexes by the rotation similar to the clock.
The first problem.
Let $Kin AM$ such that $BM=KM$.
Thus, since
$$measuredangle BMK=measuredangle BMA=measuredangle BCA=60^{circ},$$
we obtain that $Delta BMK$ is equilateral triangle.
Now, take $Delta BAK$ and rotate around point $B$ by $60^{circ}.$
We obtain that $A$ goes to $C$ and $K$ goes to $M$,
which says $AK=MC$ and
$$BM+MC=MK+AK=AM$$ and we are done!
The second problem.
Take the rotation around $C$ by $60^{circ}.$
We see that $B$ goes to $A$ and let $M$ goes to $M'$.
Thus, $BM=AM'$ and $MC=MM'$ and we got $Delta AMM'$ with needed properties.
answered Jan 19 at 15:48
Michael RozenbergMichael Rozenberg
106k1893198
$endgroup$
Let in $Delta ABC$ vertexes by the rotation similar to the clock.
The first problem.
Let $Kin AM$ such that $BM=KM$.
Thus, since
$$measuredangle BMK=measuredangle BMA=measuredangle BCA=60^{circ},$$
we obtain that $Delta BMK$ is equilateral triangle.
Now, take $Delta BAK$ and rotate around point $B$ by $60^{circ}.$
We obtain that $A$ goes to $C$ and $K$ goes to $M$,
which says $AK=MC$ and
$$BM+MC=MK+AK=AM$$ and we are done!
The second problem.
Take the rotation around $C$ by $60^{circ}.$
We see that $B$ goes to $A$ and let $M$ goes to $M'$.
Thus, $BM=AM'$ and $MC=MM'$ and we got $Delta AMM'$ with needed properties.
answered Jan 19 at 15:48
Michael RozenbergMichael Rozenberg
106k1893198
edited Jan 19 at 16:03
answered Jan 19 at 15:48
Michael RozenbergMichael Rozenberg
106k1893198
answered Jan 19 at 15:48
Michael RozenbergMichael Rozenberg
106k1893198
answered Jan 19 at 15:48
Michael RozenbergMichael Rozenberg
106k1893198
106k1893198
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 20 at 22:45
add a comment |
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 20 at 22:45
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 20 at 22:45
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 20 at 22:45
add a comment |
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4
$begingroup$
Another picture on its side :-(
$endgroup$
– Lord Shark the Unknown
Jan 19 at 13:04
1
$begingroup$
How will i draw the given figure?
$endgroup$
– user636268
Jan 19 at 13:06
2
$begingroup$
To be clear: Are you specifically asking about question $19$ in the image? If so, please make an effort to type the text of the problem into your question; it's pretty short. (Questions in images aren't searchable, and they tend to be harder to read ... especially when they aren't right-side-up ;).
$endgroup$
– Blue
Jan 19 at 13:17
$begingroup$
Rotate it😑😑😑😑
$endgroup$
– user636268
Jan 19 at 13:19
$begingroup$
It is unneccessary. I constructed the diagram myself. I did not take a hint
$endgroup$
– user636268
Jan 19 at 13:19