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Finding angle with pure Geometry.

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4

1

$begingroup$

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

share|cite|improve this question

asked 2 days ago

Keshav SharmaKeshav Sharma

1457

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  2 days ago
  
  
  

  
  
  
  

add a comment | 

4

1

$begingroup$

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

share|cite|improve this question

asked 2 days ago

Keshav SharmaKeshav Sharma

1457

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  2 days ago
  
  
  

  
  
  
  

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

share|cite|improve this question

asked 2 days ago

Keshav SharmaKeshav Sharma

1457

$endgroup$

share|cite|improve this question

asked 2 days ago

Keshav SharmaKeshav Sharma

1457

share|cite|improve this question

asked 2 days ago

Keshav SharmaKeshav Sharma

1457

asked 2 days ago

Keshav SharmaKeshav Sharma

1457

asked 2 days ago

Keshav SharmaKeshav Sharma

1457

2 Answers
2

active

oldest

votes

5

$begingroup$

(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:

$$angle ECF=frac12angle BCG+frac 12angle GCD=45^circ$$

share|cite|improve this answer

answered 2 days ago

OldboyOldboy

9,42911138

$endgroup$

add a comment | 

2

$begingroup$

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 $$angle QCP = angle ECP = 45^{circ}$$

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Maria MazurMaria Mazur

50.1k1361125

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're right... And how can $PB=PQ$? Isn't that a particular case?
  $endgroup$
  – Dr. Mathva
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  ..........[+1]!
  $endgroup$
  – Dr. Mathva
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Dr.Mathva Also thank you for not voting for close down...
  $endgroup$
  – Maria Mazur
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  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
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And I must admit that your posts always impress me! Specially when the questions are similar to Olympiad questions
  $endgroup$
  – Dr. Mathva
  2 days ago
  

  
  
  
  

add a comment | 

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5

$begingroup$

(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:

$$angle ECF=frac12angle BCG+frac 12angle GCD=45^circ$$

share|cite|improve this answer

answered 2 days ago

OldboyOldboy

9,42911138

$endgroup$

add a comment | 

5

$begingroup$

(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:

$$angle ECF=frac12angle BCG+frac 12angle GCD=45^circ$$

share|cite|improve this answer

answered 2 days ago

OldboyOldboy

9,42911138

$endgroup$

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:

$$angle ECF=frac12angle BCG+frac 12angle GCD=45^circ$$

share|cite|improve this answer

answered 2 days ago

OldboyOldboy

9,42911138

$endgroup$

share|cite|improve this answer

answered 2 days ago

OldboyOldboy

9,42911138

answered 2 days ago

OldboyOldboy

9,42911138

answered 2 days ago

OldboyOldboy

9,42911138

2

$begingroup$

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 $$angle QCP = angle ECP = 45^{circ}$$

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Maria MazurMaria Mazur

50.1k1361125

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're right... And how can $PB=PQ$? Isn't that a particular case?
  $endgroup$
  – Dr. Mathva
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  ..........[+1]!
  $endgroup$
  – Dr. Mathva
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Dr.Mathva Also thank you for not voting for close down...
  $endgroup$
  – Maria Mazur
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  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
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And I must admit that your posts always impress me! Specially when the questions are similar to Olympiad questions
  $endgroup$
  – Dr. Mathva
  2 days ago
  

  
  
  
  

add a comment | 

2

$begingroup$

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 $$angle QCP = angle ECP = 45^{circ}$$

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Maria MazurMaria Mazur

50.1k1361125

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're right... And how can $PB=PQ$? Isn't that a particular case?
  $endgroup$
  – Dr. Mathva
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  ..........[+1]!
  $endgroup$
  – Dr. Mathva
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Dr.Mathva Also thank you for not voting for close down...
  $endgroup$
  – Maria Mazur
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  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
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And I must admit that your posts always impress me! Specially when the questions are similar to Olympiad questions
  $endgroup$
  – Dr. Mathva
  2 days ago
  

  
  
  
  

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 $$angle QCP = angle ECP = 45^{circ}$$

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Maria MazurMaria Mazur

50.1k1361125

$endgroup$

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Maria MazurMaria Mazur

50.1k1361125

answered 2 days ago

Maria MazurMaria Mazur

50.1k1361125

answered 2 days ago

Maria MazurMaria Mazur

50.1k1361125