Is there a right triangle with angles $A$, $B$, $C$ such that $A^2+B^2=C^2$?
$begingroup$
A right angle triangle with vertices $A,B,C$ ($C$ is the right angle), and the sides opposite to the vertices are $a,b,c$, respectively.
We know that this triangle (and any right angle triangle) has the following properties:
$a^2+b^2=c^2$
$a+b>c$
$a+c>b$
$b+c>a$
$A+B+C=pi$
Can we add the property that $A^2+B^2=C^2$ such that this triangle can be formed? If yes, how to find an example for such triangle, finding $A,B,C,a,b,c$?
geometry trigonometry triangle geometric-inequalities
edited Jan 15 at 8:32
Blue
48.2k870153
asked Jan 15 at 8:28
Hussain-AlqatariHussain-Alqatari
3217
$endgroup$
add a comment |
$begingroup$
A right angle triangle with vertices $A,B,C$ ($C$ is the right angle), and the sides opposite to the vertices are $a,b,c$, respectively.
We know that this triangle (and any right angle triangle) has the following properties:
$a^2+b^2=c^2$
$a+b>c$
$a+c>b$
$b+c>a$
$A+B+C=pi$
Can we add the property that $A^2+B^2=C^2$ such that this triangle can be formed? If yes, how to find an example for such triangle, finding $A,B,C,a,b,c$?
geometry trigonometry triangle geometric-inequalities
edited Jan 15 at 8:32
Blue
48.2k870153
asked Jan 15 at 8:28
Hussain-AlqatariHussain-Alqatari
3217
$endgroup$
add a comment |
$begingroup$
A right angle triangle with vertices $A,B,C$ ($C$ is the right angle), and the sides opposite to the vertices are $a,b,c$, respectively.
We know that this triangle (and any right angle triangle) has the following properties:
$a^2+b^2=c^2$
$a+b>c$
$a+c>b$
$b+c>a$
$A+B+C=pi$
Can we add the property that $A^2+B^2=C^2$ such that this triangle can be formed? If yes, how to find an example for such triangle, finding $A,B,C,a,b,c$?
geometry trigonometry triangle geometric-inequalities
edited Jan 15 at 8:32
Blue
48.2k870153
asked Jan 15 at 8:28
Hussain-AlqatariHussain-Alqatari
3217
$endgroup$
A right angle triangle with vertices $A,B,C$ ($C$ is the right angle), and the sides opposite to the vertices are $a,b,c$, respectively.
We know that this triangle (and any right angle triangle) has the following properties:
$a^2+b^2=c^2$
$a+b>c$
$a+c>b$
$b+c>a$
$A+B+C=pi$
Can we add the property that $A^2+B^2=C^2$ such that this triangle can be formed? If yes, how to find an example for such triangle, finding $A,B,C,a,b,c$?
geometry trigonometry triangle geometric-inequalities
geometry trigonometry triangle geometric-inequalities
edited Jan 15 at 8:32
Blue
48.2k870153
asked Jan 15 at 8:28
Hussain-AlqatariHussain-Alqatari
3217
edited Jan 15 at 8:32
Blue
48.2k870153
asked Jan 15 at 8:28
Hussain-AlqatariHussain-Alqatari
3217
edited Jan 15 at 8:32
Blue
48.2k870153
edited Jan 15 at 8:32
Blue
48.2k870153
edited Jan 15 at 8:32
Blue
48.2k870153
48.2k870153
asked Jan 15 at 8:28
Hussain-AlqatariHussain-Alqatari
3217
asked Jan 15 at 8:28
Hussain-AlqatariHussain-Alqatari
3217
asked Jan 15 at 8:28
Hussain-AlqatariHussain-Alqatari
3217
3217
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Because it's a right triangle, $C=frac{pi}{2}=A+B$. That means that
$$C^2=(A+B)^2=A^2+B^2+2AB > A^2+B^2$$
So no, it isn't possible. We can come close as we approach angles of $0,90^circ,90^circ$, but that's degenerate - it's not a triangle.
answered Jan 15 at 8:38
jmerryjmerry
6,712718
$endgroup$
add a comment |
$begingroup$
We have $C=pi/2,B=pi/2-A$, so we need to solve the quadratic equation$$A^2+(pi/2-A)^2=pi^2/4\implies2A^2=pi A$$giving $A=0,pi/2$ which is not possible.
answered Jan 15 at 8:39
Shubham JohriShubham Johri
5,082717
$endgroup$
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%2f3074189%2fis-there-a-right-triangle-with-angles-a-b-c-such-that-a2b2-c2%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
$begingroup$
Because it's a right triangle, $C=frac{pi}{2}=A+B$. That means that
$$C^2=(A+B)^2=A^2+B^2+2AB > A^2+B^2$$
So no, it isn't possible. We can come close as we approach angles of $0,90^circ,90^circ$, but that's degenerate - it's not a triangle.
answered Jan 15 at 8:38
jmerryjmerry
6,712718
$endgroup$
add a comment |
$begingroup$
Because it's a right triangle, $C=frac{pi}{2}=A+B$. That means that
$$C^2=(A+B)^2=A^2+B^2+2AB > A^2+B^2$$
So no, it isn't possible. We can come close as we approach angles of $0,90^circ,90^circ$, but that's degenerate - it's not a triangle.
answered Jan 15 at 8:38
jmerryjmerry
6,712718
$endgroup$
add a comment |
$begingroup$
Because it's a right triangle, $C=frac{pi}{2}=A+B$. That means that
$$C^2=(A+B)^2=A^2+B^2+2AB > A^2+B^2$$
So no, it isn't possible. We can come close as we approach angles of $0,90^circ,90^circ$, but that's degenerate - it's not a triangle.
answered Jan 15 at 8:38
jmerryjmerry
6,712718
$endgroup$
Because it's a right triangle, $C=frac{pi}{2}=A+B$. That means that
$$C^2=(A+B)^2=A^2+B^2+2AB > A^2+B^2$$
So no, it isn't possible. We can come close as we approach angles of $0,90^circ,90^circ$, but that's degenerate - it's not a triangle.
answered Jan 15 at 8:38
jmerryjmerry
6,712718
answered Jan 15 at 8:38
jmerryjmerry
6,712718
answered Jan 15 at 8:38
jmerryjmerry
6,712718
answered Jan 15 at 8:38
jmerryjmerry
6,712718
6,712718
add a comment |
add a comment |
$begingroup$
We have $C=pi/2,B=pi/2-A$, so we need to solve the quadratic equation$$A^2+(pi/2-A)^2=pi^2/4\implies2A^2=pi A$$giving $A=0,pi/2$ which is not possible.
answered Jan 15 at 8:39
Shubham JohriShubham Johri
5,082717
$endgroup$
add a comment |
$begingroup$
We have $C=pi/2,B=pi/2-A$, so we need to solve the quadratic equation$$A^2+(pi/2-A)^2=pi^2/4\implies2A^2=pi A$$giving $A=0,pi/2$ which is not possible.
answered Jan 15 at 8:39
Shubham JohriShubham Johri
5,082717
$endgroup$
add a comment |
$begingroup$
We have $C=pi/2,B=pi/2-A$, so we need to solve the quadratic equation$$A^2+(pi/2-A)^2=pi^2/4\implies2A^2=pi A$$giving $A=0,pi/2$ which is not possible.
answered Jan 15 at 8:39
Shubham JohriShubham Johri
5,082717
$endgroup$
We have $C=pi/2,B=pi/2-A$, so we need to solve the quadratic equation$$A^2+(pi/2-A)^2=pi^2/4\implies2A^2=pi A$$giving $A=0,pi/2$ which is not possible.
answered Jan 15 at 8:39
Shubham JohriShubham Johri
5,082717
answered Jan 15 at 8:39
Shubham JohriShubham Johri
5,082717
answered Jan 15 at 8:39
Shubham JohriShubham Johri
5,082717
answered Jan 15 at 8:39
Shubham JohriShubham Johri
5,082717
5,082717
add a comment |
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%2f3074189%2fis-there-a-right-triangle-with-angles-a-b-c-such-that-a2b2-c2%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
