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
1667
$endgroup$
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
1667
$endgroup$
$begingroup$
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
1667
$endgroup$
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
1667
asked Apr 6 at 18:09
Keshav SharmaKeshav Sharma
1667
asked Apr 6 at 18:09
Keshav SharmaKeshav Sharma
1667
asked Apr 6 at 18:09
Keshav SharmaKeshav Sharma
1667
asked Apr 6 at 18:09
Keshav SharmaKeshav Sharma
1667
1667
$begingroup$
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
9,42911138
$endgroup$
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
50k1361125
$endgroup$
$begingroup$
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
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 |
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
});
}
});
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
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
Required, but never shown
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
9,42911138
$endgroup$
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:
$$angle ECF=frac12angle BCG+frac 12angle GCD=45^circ$$
answered Apr 6 at 19:46
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:
$$angle ECF=frac12angle BCG+frac 12angle GCD=45^circ$$
answered Apr 6 at 19:46
OldboyOldboy
9,42911138
$endgroup$
(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
9,42911138
answered Apr 6 at 19:46
OldboyOldboy
9,42911138
answered Apr 6 at 19:46
OldboyOldboy
9,42911138
answered Apr 6 at 19:46
OldboyOldboy
9,42911138
9,42911138
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
50k1361125
$endgroup$
$begingroup$
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
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 $$angle QCP = angle ECP = 45^{circ}$$
answered Apr 6 at 20:01
Maria MazurMaria Mazur
50k1361125
$endgroup$
$begingroup$
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
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 $$angle QCP = angle ECP = 45^{circ}$$
answered Apr 6 at 20:01
Maria MazurMaria Mazur
50k1361125
$endgroup$
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
50k1361125
edited 2 days ago
answered Apr 6 at 20:01
Maria MazurMaria Mazur
50k1361125
answered Apr 6 at 20:01
Maria MazurMaria Mazur
50k1361125
answered Apr 6 at 20:01
Maria MazurMaria Mazur
50k1361125
50k1361125
$begingroup$
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
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 |
$begingroup$
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
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
$begingroup$
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
2 days ago
$begingroup$
..........[+1]!
$endgroup$
– Dr. Mathva
2 days ago
1
1
$begingroup$
@Dr.Mathva Also thank you for not voting for close down...
$endgroup$
– Maria Mazur
2 days ago
$begingroup$
@Dr.Mathva Also thank you for not voting for close down...
$endgroup$
– Maria Mazur
2 days ago
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
2 days ago
$begingroup$
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
$begingroup$
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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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
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
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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