Prove that a circle can be inscribed iff the given condition is satisfied
$begingroup$
I have the following question with me:
"Let $B_1$ and $C_1$ be points on the sides $AC$ and $AB$ of a triangle $ABC$. Lines $BB_1$ and $CC_1$ intersect at point $D$. Prove that a circle can be inscribed inside quadrilateral $AB_1DC_1$ if and only if the incircles of $ABD$ and $ACD$ are tangent to each other."
I start like this:
I introduce notations:
$$AB = c, AC = b, BC = a, AC_1 = x, AB_1 = z, B_1D = y, C_1D = w, BD = q, CD = p$$
To prove the above statement I basically need to prove that the statements $$x + y = z + w$$ and $$c + p = b + q$$
are equivalent. Performing a few computations involving bases I get the following equations :
$$qz(p + w) = bw(q + y)$$ and $$yc(p + w) = px(q + y)$$
How do I proceed from here? Any help is appreciated. Alternative solutions are also welcome
geometry
asked Jan 3 at 12:31
saisanjeevsaisanjeev
962212
$endgroup$
add a comment |
$begingroup$
I have the following question with me:
"Let $B_1$ and $C_1$ be points on the sides $AC$ and $AB$ of a triangle $ABC$. Lines $BB_1$ and $CC_1$ intersect at point $D$. Prove that a circle can be inscribed inside quadrilateral $AB_1DC_1$ if and only if the incircles of $ABD$ and $ACD$ are tangent to each other."
I start like this:
I introduce notations:
$$AB = c, AC = b, BC = a, AC_1 = x, AB_1 = z, B_1D = y, C_1D = w, BD = q, CD = p$$
To prove the above statement I basically need to prove that the statements $$x + y = z + w$$ and $$c + p = b + q$$
are equivalent. Performing a few computations involving bases I get the following equations :
$$qz(p + w) = bw(q + y)$$ and $$yc(p + w) = px(q + y)$$
How do I proceed from here? Any help is appreciated. Alternative solutions are also welcome
geometry
asked Jan 3 at 12:31
saisanjeevsaisanjeev
962212
$endgroup$
add a comment |
$begingroup$
I have the following question with me:
"Let $B_1$ and $C_1$ be points on the sides $AC$ and $AB$ of a triangle $ABC$. Lines $BB_1$ and $CC_1$ intersect at point $D$. Prove that a circle can be inscribed inside quadrilateral $AB_1DC_1$ if and only if the incircles of $ABD$ and $ACD$ are tangent to each other."
I start like this:
I introduce notations:
$$AB = c, AC = b, BC = a, AC_1 = x, AB_1 = z, B_1D = y, C_1D = w, BD = q, CD = p$$
To prove the above statement I basically need to prove that the statements $$x + y = z + w$$ and $$c + p = b + q$$
are equivalent. Performing a few computations involving bases I get the following equations :
$$qz(p + w) = bw(q + y)$$ and $$yc(p + w) = px(q + y)$$
How do I proceed from here? Any help is appreciated. Alternative solutions are also welcome
geometry
asked Jan 3 at 12:31
saisanjeevsaisanjeev
962212
$endgroup$
I have the following question with me:
"Let $B_1$ and $C_1$ be points on the sides $AC$ and $AB$ of a triangle $ABC$. Lines $BB_1$ and $CC_1$ intersect at point $D$. Prove that a circle can be inscribed inside quadrilateral $AB_1DC_1$ if and only if the incircles of $ABD$ and $ACD$ are tangent to each other."
I start like this:
I introduce notations:
$$AB = c, AC = b, BC = a, AC_1 = x, AB_1 = z, B_1D = y, C_1D = w, BD = q, CD = p$$
To prove the above statement I basically need to prove that the statements $$x + y = z + w$$ and $$c + p = b + q$$
are equivalent. Performing a few computations involving bases I get the following equations :
$$qz(p + w) = bw(q + y)$$ and $$yc(p + w) = px(q + y)$$
How do I proceed from here? Any help is appreciated. Alternative solutions are also welcome
geometry
geometry
asked Jan 3 at 12:31
saisanjeevsaisanjeev
962212
asked Jan 3 at 12:31
saisanjeevsaisanjeev
962212
asked Jan 3 at 12:31
saisanjeevsaisanjeev
962212
asked Jan 3 at 12:31
saisanjeevsaisanjeev
962212
asked Jan 3 at 12:31
saisanjeevsaisanjeev
962212
962212
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This is actually a problem from Bulgarian MO, third round, 1999.
For your reference I have copied the solution from "Cyclic quadrilaterals, radical axis and inscribed circles" by Charles Leytem
answered Jan 3 at 15:06
OldboyOldboy
7,7851935
$endgroup$
add a comment |
$begingroup$
For variety , here is a different solution.
Let us suppose that the incircles of $triangle$s $ABD$ and $ACD$ are not tangential . Clearly , as in the figure , there are two distinct points of tangency , $G$ and $K$
Let $AG$ = $x$ , $GK$ = $delta$ and $KD$ = $y$ . The line segments of the same color in the figure are equal.
Now consider quadrilaterals $ABDC$ , $AB_1DC_1$ . Let us assume that a circle can be circumscribed by $AB_1DC_1$ . Join points of tangency to form segments $K’L’$ , $M’N’$ . Join $JL$ , $FH$ . The respective dotted segments are parallel , by Thales’ Theorem.
Let $DL’ = a $ , $L’B_1 = b $ , $M’A = c$ and $K’C_1 = d $ . Let $C_1J = q_1$ and $B_1H = q_2 $.
Using properties of tangents and isosceles $triangle$s, we have:- $$ AJ = x + delta = c + d - q_1$$
$$ AH = x = c + b - q_2 $$
From these , we get $$ delta = d - b - q_1 + q_2 $$ Call this equation (1). Also , we have :-
$$LL’ = a + y = d - q_1 $$ $$ FN’= a+ delta + y = b - q_2 $$ From these , we get :- $$ delta = b-d+q_1-q_2$$ Call this equation (2) . Clearly , from (1) and (2) , we get $delta = 0 $ , a contradiction if the points are distinct . QED
Note:- This is a case in which $H$ and $J$ lie inside the quadrilateral $AB_1DC_1$ , while $F$ and $L$ lie outside it . There exist other cases , which can be proven similarly
answered Jan 3 at 18:29
RahuboyRahuboy
48011
$endgroup$
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%2f3060520%2fprove-that-a-circle-can-be-inscribed-iff-the-given-condition-is-satisfied%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
$begingroup$
This is actually a problem from Bulgarian MO, third round, 1999.
For your reference I have copied the solution from "Cyclic quadrilaterals, radical axis and inscribed circles" by Charles Leytem
answered Jan 3 at 15:06
OldboyOldboy
7,7851935
$endgroup$
add a comment |
$begingroup$
This is actually a problem from Bulgarian MO, third round, 1999.
For your reference I have copied the solution from "Cyclic quadrilaterals, radical axis and inscribed circles" by Charles Leytem
answered Jan 3 at 15:06
OldboyOldboy
7,7851935
$endgroup$
add a comment |
$begingroup$
This is actually a problem from Bulgarian MO, third round, 1999.
For your reference I have copied the solution from "Cyclic quadrilaterals, radical axis and inscribed circles" by Charles Leytem
answered Jan 3 at 15:06
OldboyOldboy
7,7851935
$endgroup$
This is actually a problem from Bulgarian MO, third round, 1999.
For your reference I have copied the solution from "Cyclic quadrilaterals, radical axis and inscribed circles" by Charles Leytem
answered Jan 3 at 15:06
OldboyOldboy
7,7851935
answered Jan 3 at 15:06
OldboyOldboy
7,7851935
answered Jan 3 at 15:06
OldboyOldboy
7,7851935
answered Jan 3 at 15:06
OldboyOldboy
7,7851935
7,7851935
add a comment |
add a comment |
$begingroup$
For variety , here is a different solution.
Let us suppose that the incircles of $triangle$s $ABD$ and $ACD$ are not tangential . Clearly , as in the figure , there are two distinct points of tangency , $G$ and $K$
Let $AG$ = $x$ , $GK$ = $delta$ and $KD$ = $y$ . The line segments of the same color in the figure are equal.
Now consider quadrilaterals $ABDC$ , $AB_1DC_1$ . Let us assume that a circle can be circumscribed by $AB_1DC_1$ . Join points of tangency to form segments $K’L’$ , $M’N’$ . Join $JL$ , $FH$ . The respective dotted segments are parallel , by Thales’ Theorem.
Let $DL’ = a $ , $L’B_1 = b $ , $M’A = c$ and $K’C_1 = d $ . Let $C_1J = q_1$ and $B_1H = q_2 $.
Using properties of tangents and isosceles $triangle$s, we have:- $$ AJ = x + delta = c + d - q_1$$
$$ AH = x = c + b - q_2 $$
From these , we get $$ delta = d - b - q_1 + q_2 $$ Call this equation (1). Also , we have :-
$$LL’ = a + y = d - q_1 $$ $$ FN’= a+ delta + y = b - q_2 $$ From these , we get :- $$ delta = b-d+q_1-q_2$$ Call this equation (2) . Clearly , from (1) and (2) , we get $delta = 0 $ , a contradiction if the points are distinct . QED
Note:- This is a case in which $H$ and $J$ lie inside the quadrilateral $AB_1DC_1$ , while $F$ and $L$ lie outside it . There exist other cases , which can be proven similarly
answered Jan 3 at 18:29
RahuboyRahuboy
48011
$endgroup$
add a comment |
$begingroup$
For variety , here is a different solution.
Let us suppose that the incircles of $triangle$s $ABD$ and $ACD$ are not tangential . Clearly , as in the figure , there are two distinct points of tangency , $G$ and $K$
Let $AG$ = $x$ , $GK$ = $delta$ and $KD$ = $y$ . The line segments of the same color in the figure are equal.
Now consider quadrilaterals $ABDC$ , $AB_1DC_1$ . Let us assume that a circle can be circumscribed by $AB_1DC_1$ . Join points of tangency to form segments $K’L’$ , $M’N’$ . Join $JL$ , $FH$ . The respective dotted segments are parallel , by Thales’ Theorem.
Let $DL’ = a $ , $L’B_1 = b $ , $M’A = c$ and $K’C_1 = d $ . Let $C_1J = q_1$ and $B_1H = q_2 $.
Using properties of tangents and isosceles $triangle$s, we have:- $$ AJ = x + delta = c + d - q_1$$
$$ AH = x = c + b - q_2 $$
From these , we get $$ delta = d - b - q_1 + q_2 $$ Call this equation (1). Also , we have :-
$$LL’ = a + y = d - q_1 $$ $$ FN’= a+ delta + y = b - q_2 $$ From these , we get :- $$ delta = b-d+q_1-q_2$$ Call this equation (2) . Clearly , from (1) and (2) , we get $delta = 0 $ , a contradiction if the points are distinct . QED
Note:- This is a case in which $H$ and $J$ lie inside the quadrilateral $AB_1DC_1$ , while $F$ and $L$ lie outside it . There exist other cases , which can be proven similarly
answered Jan 3 at 18:29
RahuboyRahuboy
48011
$endgroup$
add a comment |
$begingroup$
For variety , here is a different solution.
Let us suppose that the incircles of $triangle$s $ABD$ and $ACD$ are not tangential . Clearly , as in the figure , there are two distinct points of tangency , $G$ and $K$
Let $AG$ = $x$ , $GK$ = $delta$ and $KD$ = $y$ . The line segments of the same color in the figure are equal.
Now consider quadrilaterals $ABDC$ , $AB_1DC_1$ . Let us assume that a circle can be circumscribed by $AB_1DC_1$ . Join points of tangency to form segments $K’L’$ , $M’N’$ . Join $JL$ , $FH$ . The respective dotted segments are parallel , by Thales’ Theorem.
Let $DL’ = a $ , $L’B_1 = b $ , $M’A = c$ and $K’C_1 = d $ . Let $C_1J = q_1$ and $B_1H = q_2 $.
Using properties of tangents and isosceles $triangle$s, we have:- $$ AJ = x + delta = c + d - q_1$$
$$ AH = x = c + b - q_2 $$
From these , we get $$ delta = d - b - q_1 + q_2 $$ Call this equation (1). Also , we have :-
$$LL’ = a + y = d - q_1 $$ $$ FN’= a+ delta + y = b - q_2 $$ From these , we get :- $$ delta = b-d+q_1-q_2$$ Call this equation (2) . Clearly , from (1) and (2) , we get $delta = 0 $ , a contradiction if the points are distinct . QED
Note:- This is a case in which $H$ and $J$ lie inside the quadrilateral $AB_1DC_1$ , while $F$ and $L$ lie outside it . There exist other cases , which can be proven similarly
answered Jan 3 at 18:29
RahuboyRahuboy
48011
$endgroup$
For variety , here is a different solution.
Let us suppose that the incircles of $triangle$s $ABD$ and $ACD$ are not tangential . Clearly , as in the figure , there are two distinct points of tangency , $G$ and $K$
Let $AG$ = $x$ , $GK$ = $delta$ and $KD$ = $y$ . The line segments of the same color in the figure are equal.
Now consider quadrilaterals $ABDC$ , $AB_1DC_1$ . Let us assume that a circle can be circumscribed by $AB_1DC_1$ . Join points of tangency to form segments $K’L’$ , $M’N’$ . Join $JL$ , $FH$ . The respective dotted segments are parallel , by Thales’ Theorem.
Let $DL’ = a $ , $L’B_1 = b $ , $M’A = c$ and $K’C_1 = d $ . Let $C_1J = q_1$ and $B_1H = q_2 $.
Using properties of tangents and isosceles $triangle$s, we have:- $$ AJ = x + delta = c + d - q_1$$
$$ AH = x = c + b - q_2 $$
From these , we get $$ delta = d - b - q_1 + q_2 $$ Call this equation (1). Also , we have :-
$$LL’ = a + y = d - q_1 $$ $$ FN’= a+ delta + y = b - q_2 $$ From these , we get :- $$ delta = b-d+q_1-q_2$$ Call this equation (2) . Clearly , from (1) and (2) , we get $delta = 0 $ , a contradiction if the points are distinct . QED
Note:- This is a case in which $H$ and $J$ lie inside the quadrilateral $AB_1DC_1$ , while $F$ and $L$ lie outside it . There exist other cases , which can be proven similarly
answered Jan 3 at 18:29
RahuboyRahuboy
48011
answered Jan 3 at 18:29
RahuboyRahuboy
48011
answered Jan 3 at 18:29
RahuboyRahuboy
48011
answered Jan 3 at 18:29
RahuboyRahuboy
48011
48011
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%2f3060520%2fprove-that-a-circle-can-be-inscribed-iff-the-given-condition-is-satisfied%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
