Finding angle with pure Geometry.
$begingroup$
Each side of a square ABCD has a length of 1 unit. Points P and Q belong to AB and DA respectively. The perimeter of triangle APQ is 2 units. What will be the angle of PCQ.
I was able to do this with simple trignometry and found it to be 45 degrees but the book I am reading doesn't yet talked about trignometry or similarity of triangle so I want rather pure geometric proof which doesn't include trigonometry or similarity of triangle concepts.
geometry euclidean-geometry
asked Apr 6 at 18:09
Keshav SharmaKeshav Sharma
1847
$endgroup$
add a comment |
$begingroup$
Each side of a square ABCD has a length of 1 unit. Points P and Q belong to AB and DA respectively. The perimeter of triangle APQ is 2 units. What will be the angle of PCQ.
I was able to do this with simple trignometry and found it to be 45 degrees but the book I am reading doesn't yet talked about trignometry or similarity of triangle so I want rather pure geometric proof which doesn't include trigonometry or similarity of triangle concepts.
geometry euclidean-geometry
asked Apr 6 at 18:09
Keshav SharmaKeshav Sharma
1847
$endgroup$
$begingroup$
Let $AQ=x,AP=y.$ Then you can find that $2x+2y=xy+2,$ which has infinitely many solutions for $0 < x,y leq 1.$
$endgroup$
– Dbchatto67
Apr 6 at 18:56
add a comment |
$begingroup$
Each side of a square ABCD has a length of 1 unit. Points P and Q belong to AB and DA respectively. The perimeter of triangle APQ is 2 units. What will be the angle of PCQ.
I was able to do this with simple trignometry and found it to be 45 degrees but the book I am reading doesn't yet talked about trignometry or similarity of triangle so I want rather pure geometric proof which doesn't include trigonometry or similarity of triangle concepts.
geometry euclidean-geometry
asked Apr 6 at 18:09
Keshav SharmaKeshav Sharma
1847
$endgroup$
Each side of a square ABCD has a length of 1 unit. Points P and Q belong to AB and DA respectively. The perimeter of triangle APQ is 2 units. What will be the angle of PCQ.
I was able to do this with simple trignometry and found it to be 45 degrees but the book I am reading doesn't yet talked about trignometry or similarity of triangle so I want rather pure geometric proof which doesn't include trigonometry or similarity of triangle concepts.
geometry euclidean-geometry
geometry euclidean-geometry
asked Apr 6 at 18:09
Keshav SharmaKeshav Sharma
1847
asked Apr 6 at 18:09
Keshav SharmaKeshav Sharma
1847
asked Apr 6 at 18:09
Keshav SharmaKeshav Sharma
1847
asked Apr 6 at 18:09
Keshav SharmaKeshav Sharma
1847
asked Apr 6 at 18:09
Keshav SharmaKeshav Sharma
1847
1847
$begingroup$
Let $AQ=x,AP=y.$ Then you can find that $2x+2y=xy+2,$ which has infinitely many solutions for $0 < x,y leq 1.$
$endgroup$
– Dbchatto67
Apr 6 at 18:56
add a comment |
$begingroup$
Let $AQ=x,AP=y.$ Then you can find that $2x+2y=xy+2,$ which has infinitely many solutions for $0 < x,y leq 1.$
$endgroup$
– Dbchatto67
Apr 6 at 18:56
$begingroup$
Let $AQ=x,AP=y.$ Then you can find that $2x+2y=xy+2,$ which has infinitely many solutions for $0 < x,y leq 1.$
$endgroup$
– Dbchatto67
Apr 6 at 18:56
$begingroup$
Let $AQ=x,AP=y.$ Then you can find that $2x+2y=xy+2,$ which has infinitely many solutions for $0 < x,y leq 1.$
$endgroup$
– Dbchatto67
Apr 6 at 18:56
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
(I apologize for replacing points $PQ$ with $EF$, I'm too tired to draw the picture again :)
Draw circle with center $E$ and radius $EB$ and circle with center $F$ and radius $FD$. These circles intersect segment $EF$ at points $G',G''$ These points are identical! Why?
Because perimeter of triangle $AEF$ is 2 which is $AB+AD=AE+EB+AF+FD=AE+EG'+AF+FG''$. It means that $EG'+FG''=EF$ and therefore $G'equiv G''equiv Gin EF$.
So these two circles touch at point $G$ as shown in the picture. Power of point $C$ with respect to both circles is equal $(CB=CD)$ and therefore it has to be on the radical axis of the circles. These circles touch at point $G$ and their radical axis is defined by the common tangent at point $G$. Becuase of that $CG$ must be tangent to both circles and, at the same time, $CGbot EF$.
The rest is trivial: you can easily show that triangles $FCD$ and $FCG$ are congruent. The same is true for triangles $ECG$ and $ECB$. Because of that:
$$angle ECF=frac12angle BCG+frac 12angle GCD=45^circ$$
answered Apr 6 at 19:46
OldboyOldboy
9,55911138
$endgroup$
add a comment |
$begingroup$
Rotate $Q$ for $90^{circ}$ around $C$ in to new point $E$. Then $P,B,E$ are collinear and $PE = PQ$. So triangles $CQP$ and $CEP$ are congruent by (sss), so $$angle QCP = angle ECP = 45^{circ}$$
answered Apr 6 at 20:01
Maria MazurMaria Mazur
50.5k1361126
$endgroup$
$begingroup$
You're right... And how can $PB=PQ$? Isn't that a particular case?
$endgroup$
– Dr. Mathva
Apr 6 at 20:47
1
$begingroup$
..........[+1]!
$endgroup$
– Dr. Mathva
Apr 6 at 21:23
1
$begingroup$
@Dr.Mathva Also thank you for not voting for close down...
$endgroup$
– Maria Mazur
Apr 6 at 21:24
1
$begingroup$
I don't really understand why that question should be closed or downvoted (which is the reason why I upvoted and voted not to close it)... It sometimes happens that good questions are voted to be closed...
$endgroup$
– Dr. Mathva
Apr 6 at 21:27
$begingroup$
And I must admit that your posts always impress me! Specially when the questions are similar to Olympiad questions
$endgroup$
– Dr. Mathva
Apr 6 at 21:30
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
(I apologize for replacing points $PQ$ with $EF$, I'm too tired to draw the picture again :)
Draw circle with center $E$ and radius $EB$ and circle with center $F$ and radius $FD$. These circles intersect segment $EF$ at points $G',G''$ These points are identical! Why?
Because perimeter of triangle $AEF$ is 2 which is $AB+AD=AE+EB+AF+FD=AE+EG'+AF+FG''$. It means that $EG'+FG''=EF$ and therefore $G'equiv G''equiv Gin EF$.
So these two circles touch at point $G$ as shown in the picture. Power of point $C$ with respect to both circles is equal $(CB=CD)$ and therefore it has to be on the radical axis of the circles. These circles touch at point $G$ and their radical axis is defined by the common tangent at point $G$. Becuase of that $CG$ must be tangent to both circles and, at the same time, $CGbot EF$.
The rest is trivial: you can easily show that triangles $FCD$ and $FCG$ are congruent. The same is true for triangles $ECG$ and $ECB$. Because of that:
$$angle ECF=frac12angle BCG+frac 12angle GCD=45^circ$$
answered Apr 6 at 19:46
OldboyOldboy
9,55911138
$endgroup$
add a comment |
$begingroup$
(I apologize for replacing points $PQ$ with $EF$, I'm too tired to draw the picture again :)
Draw circle with center $E$ and radius $EB$ and circle with center $F$ and radius $FD$. These circles intersect segment $EF$ at points $G',G''$ These points are identical! Why?
Because perimeter of triangle $AEF$ is 2 which is $AB+AD=AE+EB+AF+FD=AE+EG'+AF+FG''$. It means that $EG'+FG''=EF$ and therefore $G'equiv G''equiv Gin EF$.
So these two circles touch at point $G$ as shown in the picture. Power of point $C$ with respect to both circles is equal $(CB=CD)$ and therefore it has to be on the radical axis of the circles. These circles touch at point $G$ and their radical axis is defined by the common tangent at point $G$. Becuase of that $CG$ must be tangent to both circles and, at the same time, $CGbot EF$.
The rest is trivial: you can easily show that triangles $FCD$ and $FCG$ are congruent. The same is true for triangles $ECG$ and $ECB$. Because of that:
$$angle ECF=frac12angle BCG+frac 12angle GCD=45^circ$$
answered Apr 6 at 19:46
OldboyOldboy
9,55911138
$endgroup$
add a comment |
$begingroup$
(I apologize for replacing points $PQ$ with $EF$, I'm too tired to draw the picture again :)
Draw circle with center $E$ and radius $EB$ and circle with center $F$ and radius $FD$. These circles intersect segment $EF$ at points $G',G''$ These points are identical! Why?
Because perimeter of triangle $AEF$ is 2 which is $AB+AD=AE+EB+AF+FD=AE+EG'+AF+FG''$. It means that $EG'+FG''=EF$ and therefore $G'equiv G''equiv Gin EF$.
So these two circles touch at point $G$ as shown in the picture. Power of point $C$ with respect to both circles is equal $(CB=CD)$ and therefore it has to be on the radical axis of the circles. These circles touch at point $G$ and their radical axis is defined by the common tangent at point $G$. Becuase of that $CG$ must be tangent to both circles and, at the same time, $CGbot EF$.
The rest is trivial: you can easily show that triangles $FCD$ and $FCG$ are congruent. The same is true for triangles $ECG$ and $ECB$. Because of that:
$$angle ECF=frac12angle BCG+frac 12angle GCD=45^circ$$
answered Apr 6 at 19:46
OldboyOldboy
9,55911138
$endgroup$
(I apologize for replacing points $PQ$ with $EF$, I'm too tired to draw the picture again :)
Draw circle with center $E$ and radius $EB$ and circle with center $F$ and radius $FD$. These circles intersect segment $EF$ at points $G',G''$ These points are identical! Why?
Because perimeter of triangle $AEF$ is 2 which is $AB+AD=AE+EB+AF+FD=AE+EG'+AF+FG''$. It means that $EG'+FG''=EF$ and therefore $G'equiv G''equiv Gin EF$.
So these two circles touch at point $G$ as shown in the picture. Power of point $C$ with respect to both circles is equal $(CB=CD)$ and therefore it has to be on the radical axis of the circles. These circles touch at point $G$ and their radical axis is defined by the common tangent at point $G$. Becuase of that $CG$ must be tangent to both circles and, at the same time, $CGbot EF$.
The rest is trivial: you can easily show that triangles $FCD$ and $FCG$ are congruent. The same is true for triangles $ECG$ and $ECB$. Because of that:
$$angle ECF=frac12angle BCG+frac 12angle GCD=45^circ$$
answered Apr 6 at 19:46
OldboyOldboy
9,55911138
answered Apr 6 at 19:46
OldboyOldboy
9,55911138
answered Apr 6 at 19:46
OldboyOldboy
9,55911138
answered Apr 6 at 19:46
OldboyOldboy
9,55911138
9,55911138
add a comment |
add a comment |
$begingroup$
Rotate $Q$ for $90^{circ}$ around $C$ in to new point $E$. Then $P,B,E$ are collinear and $PE = PQ$. So triangles $CQP$ and $CEP$ are congruent by (sss), so $$angle QCP = angle ECP = 45^{circ}$$
answered Apr 6 at 20:01
Maria MazurMaria Mazur
50.5k1361126
$endgroup$
$begingroup$
You're right... And how can $PB=PQ$? Isn't that a particular case?
$endgroup$
– Dr. Mathva
Apr 6 at 20:47
1
$begingroup$
..........[+1]!
$endgroup$
– Dr. Mathva
Apr 6 at 21:23
1
$begingroup$
@Dr.Mathva Also thank you for not voting for close down...
$endgroup$
– Maria Mazur
Apr 6 at 21:24
1
$begingroup$
I don't really understand why that question should be closed or downvoted (which is the reason why I upvoted and voted not to close it)... It sometimes happens that good questions are voted to be closed...
$endgroup$
– Dr. Mathva
Apr 6 at 21:27
$begingroup$
And I must admit that your posts always impress me! Specially when the questions are similar to Olympiad questions
$endgroup$
– Dr. Mathva
Apr 6 at 21:30
add a comment |
$begingroup$
Rotate $Q$ for $90^{circ}$ around $C$ in to new point $E$. Then $P,B,E$ are collinear and $PE = PQ$. So triangles $CQP$ and $CEP$ are congruent by (sss), so $$angle QCP = angle ECP = 45^{circ}$$
answered Apr 6 at 20:01
Maria MazurMaria Mazur
50.5k1361126
$endgroup$
$begingroup$
You're right... And how can $PB=PQ$? Isn't that a particular case?
$endgroup$
– Dr. Mathva
Apr 6 at 20:47
1
$begingroup$
..........[+1]!
$endgroup$
– Dr. Mathva
Apr 6 at 21:23
1
$begingroup$
@Dr.Mathva Also thank you for not voting for close down...
$endgroup$
– Maria Mazur
Apr 6 at 21:24
1
$begingroup$
I don't really understand why that question should be closed or downvoted (which is the reason why I upvoted and voted not to close it)... It sometimes happens that good questions are voted to be closed...
$endgroup$
– Dr. Mathva
Apr 6 at 21:27
$begingroup$
And I must admit that your posts always impress me! Specially when the questions are similar to Olympiad questions
$endgroup$
– Dr. Mathva
Apr 6 at 21:30
add a comment |
$begingroup$
Rotate $Q$ for $90^{circ}$ around $C$ in to new point $E$. Then $P,B,E$ are collinear and $PE = PQ$. So triangles $CQP$ and $CEP$ are congruent by (sss), so $$angle QCP = angle ECP = 45^{circ}$$
answered Apr 6 at 20:01
Maria MazurMaria Mazur
50.5k1361126
$endgroup$
Rotate $Q$ for $90^{circ}$ around $C$ in to new point $E$. Then $P,B,E$ are collinear and $PE = PQ$. So triangles $CQP$ and $CEP$ are congruent by (sss), so $$angle QCP = angle ECP = 45^{circ}$$
answered Apr 6 at 20:01
Maria MazurMaria Mazur
50.5k1361126
edited Apr 6 at 21:18
answered Apr 6 at 20:01
Maria MazurMaria Mazur
50.5k1361126
answered Apr 6 at 20:01
Maria MazurMaria Mazur
50.5k1361126
answered Apr 6 at 20:01
Maria MazurMaria Mazur
50.5k1361126
50.5k1361126
$begingroup$
You're right... And how can $PB=PQ$? Isn't that a particular case?
$endgroup$
– Dr. Mathva
Apr 6 at 20:47
1
$begingroup$
..........[+1]!
$endgroup$
– Dr. Mathva
Apr 6 at 21:23
1
$begingroup$
@Dr.Mathva Also thank you for not voting for close down...
$endgroup$
– Maria Mazur
Apr 6 at 21:24
1
$begingroup$
I don't really understand why that question should be closed or downvoted (which is the reason why I upvoted and voted not to close it)... It sometimes happens that good questions are voted to be closed...
$endgroup$
– Dr. Mathva
Apr 6 at 21:27
$begingroup$
And I must admit that your posts always impress me! Specially when the questions are similar to Olympiad questions
$endgroup$
– Dr. Mathva
Apr 6 at 21:30
add a comment |
$begingroup$
You're right... And how can $PB=PQ$? Isn't that a particular case?
$endgroup$
– Dr. Mathva
Apr 6 at 20:47
1
$begingroup$
..........[+1]!
$endgroup$
– Dr. Mathva
Apr 6 at 21:23
1
$begingroup$
@Dr.Mathva Also thank you for not voting for close down...
$endgroup$
– Maria Mazur
Apr 6 at 21:24
1
$begingroup$
I don't really understand why that question should be closed or downvoted (which is the reason why I upvoted and voted not to close it)... It sometimes happens that good questions are voted to be closed...
$endgroup$
– Dr. Mathva
Apr 6 at 21:27
$begingroup$
And I must admit that your posts always impress me! Specially when the questions are similar to Olympiad questions
$endgroup$
– Dr. Mathva
Apr 6 at 21:30
$begingroup$
You're right... And how can $PB=PQ$? Isn't that a particular case?
$endgroup$
– Dr. Mathva
Apr 6 at 20:47
$begingroup$
You're right... And how can $PB=PQ$? Isn't that a particular case?
$endgroup$
– Dr. Mathva
Apr 6 at 20:47
1
1
$begingroup$
..........[+1]!
$endgroup$
– Dr. Mathva
Apr 6 at 21:23
$begingroup$
..........[+1]!
$endgroup$
– Dr. Mathva
Apr 6 at 21:23
1
1
$begingroup$
@Dr.Mathva Also thank you for not voting for close down...
$endgroup$
– Maria Mazur
Apr 6 at 21:24
$begingroup$
@Dr.Mathva Also thank you for not voting for close down...
$endgroup$
– Maria Mazur
Apr 6 at 21:24
1
1
$begingroup$
I don't really understand why that question should be closed or downvoted (which is the reason why I upvoted and voted not to close it)... It sometimes happens that good questions are voted to be closed...
$endgroup$
– Dr. Mathva
Apr 6 at 21:27
$begingroup$
I don't really understand why that question should be closed or downvoted (which is the reason why I upvoted and voted not to close it)... It sometimes happens that good questions are voted to be closed...
$endgroup$
– Dr. Mathva
Apr 6 at 21:27
$begingroup$
And I must admit that your posts always impress me! Specially when the questions are similar to Olympiad questions
$endgroup$
– Dr. Mathva
Apr 6 at 21:30
$begingroup$
And I must admit that your posts always impress me! Specially when the questions are similar to Olympiad questions
$endgroup$
– Dr. Mathva
Apr 6 at 21:30
add a comment |
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$begingroup$
Let $AQ=x,AP=y.$ Then you can find that $2x+2y=xy+2,$ which has infinitely many solutions for $0 < x,y leq 1.$
$endgroup$
– Dbchatto67
Apr 6 at 18:56