Kdtyjb

Why not use SQL instead of GraphQL?

Have astronauts in space suits ever taken selfies? If so, how?

How did the USSR manage to innovate in an environment characterized by government censorship and high bureaucracy?

What does "Puller Prush Person" mean?

What does it mean to describe someone as a butt steak?

How to say job offer in Mandarin/Cantonese?

What's the output of a record cartridge playing an out-of-speed record

To string or not to string

Why "Having chlorophyll without photosynthesis is actually very dangerous" and "like living with a bomb"?

Is this a crack on the carbon frame?

Arthur Somervell: 1000 Exercises - Meaning of this notation

Prove that NP is closed under karp reduction?

Accidentally leaked the solution to an assignment, what to do now? (I'm the prof)

Can an x86 CPU running in real mode be considered to be basically an 8086 CPU?

Which models of the Boeing 737 are still in production?

How old can references or sources in a thesis be?

What are these boxed doors outside store fronts in New York?

Is a tag line useful on a cover?

The use of multiple foreign keys on same column in SQL Server

Risk of getting Chronic Wasting Disease (CWD) in the United States?

How can I make my BBEG immortal short of making them a Lich or Vampire?

How do we improve the relationship with a client software team that performs poorly and is becoming less collaborative?

Example of a continuous function that don't have a continuous extension

Can a Warlock become Neutral Good?

Finding angle with pure Geometry.

perimeter of square inscribed in the triagleTricky pure geometry proofSolid Geometry. Finding an angleGeometry- finding the measure of the angleTwo tangent circles inscribed in a rectangle (Compute the area)Geometry: Finding an angle of a trapezoidFinding angle and radius. Geometry/Trig applicationGeometry (Finding angle $x$)Foundations and Fundamental Concepts of Mathematics, Chapter Problems (1.1.3)Find the two missing angles in a quadrilateral

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 9 hours ago

Keshav SharmaKeshav Sharma

1336

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  8 hours 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 9 hours ago

Keshav SharmaKeshav Sharma

1336

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  8 hours 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 9 hours ago

Keshav SharmaKeshav Sharma

1336

$endgroup$

share|cite|improve this question

asked 9 hours ago

Keshav SharmaKeshav Sharma

1336

share|cite|improve this question

asked 9 hours ago

Keshav SharmaKeshav Sharma

1336

asked 9 hours ago

Keshav SharmaKeshav Sharma

1336

asked 9 hours ago

Keshav SharmaKeshav Sharma

1336

2 Answers
2

active

oldest

votes

3

$begingroup$

(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 7 hours ago

OldboyOldboy

9,37411138

$endgroup$

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 6 hours ago

answered 7 hours ago

Maria MazurMaria Mazur

49.9k1361124

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  ..........[+1]!
  $endgroup$
  – Dr. Mathva
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3177296%2ffinding-angle-with-pure-geometry%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

3

$begingroup$

(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 7 hours ago

OldboyOldboy

9,37411138

$endgroup$

add a comment | 

3

$begingroup$

(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 7 hours ago

OldboyOldboy

9,37411138

$endgroup$

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 7 hours ago

OldboyOldboy

9,37411138

$endgroup$

share|cite|improve this answer

answered 7 hours ago

OldboyOldboy

9,37411138

answered 7 hours ago

OldboyOldboy

9,37411138

answered 7 hours ago

OldboyOldboy

9,37411138

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 6 hours ago

answered 7 hours ago

Maria MazurMaria Mazur

49.9k1361124

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  ..........[+1]!
  $endgroup$
  – Dr. Mathva
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And I must admit that your posts always impress me! Specially when the questions are similar to Olympiad questions
  $endgroup$
  – Dr. Mathva
  6 hours 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 6 hours ago

answered 7 hours ago

Maria MazurMaria Mazur

49.9k1361124

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  ..........[+1]!
  $endgroup$
  – Dr. Mathva
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And I must admit that your posts always impress me! Specially when the questions are similar to Olympiad questions
  $endgroup$
  – Dr. Mathva
  6 hours 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 6 hours ago

answered 7 hours ago

Maria MazurMaria Mazur

49.9k1361124

$endgroup$

share|cite|improve this answer

edited 6 hours ago

answered 7 hours ago

Maria MazurMaria Mazur

49.9k1361124

answered 7 hours ago

Maria MazurMaria Mazur

49.9k1361124

answered 7 hours ago

Maria MazurMaria Mazur

49.9k1361124