Prove $AX$, $BY$, $CZ$ concur
1
$begingroup$
$Delta ABC$. Let $P,Q$ be two points lies in the interior of the triangle. Let $AP,BP,CP$ intersect $BC,CA,AB$ at $X_P,Y_P,Z_P$. Point $X_Q,Y_Q,Z_Q$ are defined similarly. Let $T_P$ be a point lies on $PQ$. Let $(C)$ be the conic pass through five points $Y_P,Z_P, X_Q, Y_Q, Z_Q$. Let $A',B',C'$ be the second intersection of $X_PT,Y_PY,Z_PT$ with $(C)$. Prove $AA',BB',CC'$ concur.
- What I have done:
- I proved the six points lies in conic by Carnot theorem
- Using Pascal theorem, I proved $X_QA',Y_QB',Z_QC'$ also concur at a point in $PQ$ (I called the concur point $T_Q$).
- I'm wondering do we have: $dfrac{sin{A'AB}}{sin{A'AC}}=dfrac{A'Z_Q}{A'Y_Q}.dfrac{A'Z_P}{A'Y_P}$ (I) ?. We do have (I) when $(C)$ is a circle (proved) but when (C) just a conic, does (I) still true ?
geometry trigonometry euclidean-geometry triangle conic-sections
asked Dec 9 '18 at 16:15
RopuToranRopuToran
1326
$endgroup$
add a comment |
1
$begingroup$
$Delta ABC$. Let $P,Q$ be two points lies in the interior of the triangle. Let $AP,BP,CP$ intersect $BC,CA,AB$ at $X_P,Y_P,Z_P$. Point $X_Q,Y_Q,Z_Q$ are defined similarly. Let $T_P$ be a point lies on $PQ$. Let $(C)$ be the conic pass through five points $Y_P,Z_P, X_Q, Y_Q, Z_Q$. Let $A',B',C'$ be the second intersection of $X_PT,Y_PY,Z_PT$ with $(C)$. Prove $AA',BB',CC'$ concur.
- What I have done:
- I proved the six points lies in conic by Carnot theorem
- Using Pascal theorem, I proved $X_QA',Y_QB',Z_QC'$ also concur at a point in $PQ$ (I called the concur point $T_Q$).
- I'm wondering do we have: $dfrac{sin{A'AB}}{sin{A'AC}}=dfrac{A'Z_Q}{A'Y_Q}.dfrac{A'Z_P}{A'Y_P}$ (I) ?. We do have (I) when $(C)$ is a circle (proved) but when (C) just a conic, does (I) still true ?
geometry trigonometry euclidean-geometry triangle conic-sections
asked Dec 9 '18 at 16:15
RopuToranRopuToran
1326
$endgroup$
1
$begingroup$
Since you just have to deal with a cross-ratio, you may assume without loss of generality that $(C)$ is a circle, since projective transformations preserve cross-ratios.
$endgroup$
– Jack D'Aurizio
Dec 9 '18 at 16:32
$begingroup$
So you mean in any Euclidean geometry problem involving conic, just assume the conic is a circle ?
$endgroup$
– RopuToran
Dec 10 '18 at 3:56
$begingroup$
I mean that in any purely projective geometry problem we are allowed to assume that the involved conic is a circle. And this is a purely projective problem, since it only involves concurrencies/collinearities/tangents/cross-ratios.
$endgroup$
– Jack D'Aurizio
Dec 10 '18 at 8:08
$begingroup$
How to prove it ? Thank you
$endgroup$
– RopuToran
Dec 10 '18 at 11:24
add a comment |
1
1
1
$begingroup$
$Delta ABC$. Let $P,Q$ be two points lies in the interior of the triangle. Let $AP,BP,CP$ intersect $BC,CA,AB$ at $X_P,Y_P,Z_P$. Point $X_Q,Y_Q,Z_Q$ are defined similarly. Let $T_P$ be a point lies on $PQ$. Let $(C)$ be the conic pass through five points $Y_P,Z_P, X_Q, Y_Q, Z_Q$. Let $A',B',C'$ be the second intersection of $X_PT,Y_PY,Z_PT$ with $(C)$. Prove $AA',BB',CC'$ concur.
- What I have done:
- I proved the six points lies in conic by Carnot theorem
- Using Pascal theorem, I proved $X_QA',Y_QB',Z_QC'$ also concur at a point in $PQ$ (I called the concur point $T_Q$).
- I'm wondering do we have: $dfrac{sin{A'AB}}{sin{A'AC}}=dfrac{A'Z_Q}{A'Y_Q}.dfrac{A'Z_P}{A'Y_P}$ (I) ?. We do have (I) when $(C)$ is a circle (proved) but when (C) just a conic, does (I) still true ?
geometry trigonometry euclidean-geometry triangle conic-sections
asked Dec 9 '18 at 16:15
RopuToranRopuToran
1326
$endgroup$
$Delta ABC$. Let $P,Q$ be two points lies in the interior of the triangle. Let $AP,BP,CP$ intersect $BC,CA,AB$ at $X_P,Y_P,Z_P$. Point $X_Q,Y_Q,Z_Q$ are defined similarly. Let $T_P$ be a point lies on $PQ$. Let $(C)$ be the conic pass through five points $Y_P,Z_P, X_Q, Y_Q, Z_Q$. Let $A',B',C'$ be the second intersection of $X_PT,Y_PY,Z_PT$ with $(C)$. Prove $AA',BB',CC'$ concur.
- What I have done:
- I proved the six points lies in conic by Carnot theorem
- Using Pascal theorem, I proved $X_QA',Y_QB',Z_QC'$ also concur at a point in $PQ$ (I called the concur point $T_Q$).
- I'm wondering do we have: $dfrac{sin{A'AB}}{sin{A'AC}}=dfrac{A'Z_Q}{A'Y_Q}.dfrac{A'Z_P}{A'Y_P}$ (I) ?. We do have (I) when $(C)$ is a circle (proved) but when (C) just a conic, does (I) still true ?
geometry trigonometry euclidean-geometry triangle conic-sections
geometry trigonometry euclidean-geometry triangle conic-sections
asked Dec 9 '18 at 16:15
RopuToranRopuToran
1326
asked Dec 9 '18 at 16:15
RopuToranRopuToran
1326
edited Dec 9 '18 at 19:07
user593746
asked Dec 9 '18 at 16:15
RopuToranRopuToran
1326
asked Dec 9 '18 at 16:15
RopuToranRopuToran
1326
asked Dec 9 '18 at 16:15
RopuToranRopuToran
1326
1326
1
$begingroup$
Since you just have to deal with a cross-ratio, you may assume without loss of generality that $(C)$ is a circle, since projective transformations preserve cross-ratios.
$endgroup$
– Jack D'Aurizio
Dec 9 '18 at 16:32
$begingroup$
So you mean in any Euclidean geometry problem involving conic, just assume the conic is a circle ?
$endgroup$
– RopuToran
Dec 10 '18 at 3:56
$begingroup$
I mean that in any purely projective geometry problem we are allowed to assume that the involved conic is a circle. And this is a purely projective problem, since it only involves concurrencies/collinearities/tangents/cross-ratios.
$endgroup$
– Jack D'Aurizio
Dec 10 '18 at 8:08
$begingroup$
How to prove it ? Thank you
$endgroup$
– RopuToran
Dec 10 '18 at 11:24
add a comment |
1
$begingroup$
Since you just have to deal with a cross-ratio, you may assume without loss of generality that $(C)$ is a circle, since projective transformations preserve cross-ratios.
$endgroup$
– Jack D'Aurizio
Dec 9 '18 at 16:32
$begingroup$
So you mean in any Euclidean geometry problem involving conic, just assume the conic is a circle ?
$endgroup$
– RopuToran
Dec 10 '18 at 3:56
$begingroup$
I mean that in any purely projective geometry problem we are allowed to assume that the involved conic is a circle. And this is a purely projective problem, since it only involves concurrencies/collinearities/tangents/cross-ratios.
$endgroup$
– Jack D'Aurizio
Dec 10 '18 at 8:08
$begingroup$
How to prove it ? Thank you
$endgroup$
– RopuToran
Dec 10 '18 at 11:24
1
1
$begingroup$
Since you just have to deal with a cross-ratio, you may assume without loss of generality that $(C)$ is a circle, since projective transformations preserve cross-ratios.
$endgroup$
– Jack D'Aurizio
Dec 9 '18 at 16:32
$begingroup$
Since you just have to deal with a cross-ratio, you may assume without loss of generality that $(C)$ is a circle, since projective transformations preserve cross-ratios.
$endgroup$
– Jack D'Aurizio
Dec 9 '18 at 16:32
$begingroup$
So you mean in any Euclidean geometry problem involving conic, just assume the conic is a circle ?
$endgroup$
– RopuToran
Dec 10 '18 at 3:56
$begingroup$
So you mean in any Euclidean geometry problem involving conic, just assume the conic is a circle ?
$endgroup$
– RopuToran
Dec 10 '18 at 3:56
$begingroup$
I mean that in any purely projective geometry problem we are allowed to assume that the involved conic is a circle. And this is a purely projective problem, since it only involves concurrencies/collinearities/tangents/cross-ratios.
$endgroup$
– Jack D'Aurizio
Dec 10 '18 at 8:08
$begingroup$
I mean that in any purely projective geometry problem we are allowed to assume that the involved conic is a circle. And this is a purely projective problem, since it only involves concurrencies/collinearities/tangents/cross-ratios.
$endgroup$
– Jack D'Aurizio
Dec 10 '18 at 8:08
$begingroup$
How to prove it ? Thank you
$endgroup$
– RopuToran
Dec 10 '18 at 11:24
$begingroup$
How to prove it ? Thank you
$endgroup$
– RopuToran
Dec 10 '18 at 11:24
add a comment |
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1
$begingroup$
Since you just have to deal with a cross-ratio, you may assume without loss of generality that $(C)$ is a circle, since projective transformations preserve cross-ratios.
$endgroup$
– Jack D'Aurizio
Dec 9 '18 at 16:32
$begingroup$
So you mean in any Euclidean geometry problem involving conic, just assume the conic is a circle ?
$endgroup$
– RopuToran
Dec 10 '18 at 3:56
$begingroup$
I mean that in any purely projective geometry problem we are allowed to assume that the involved conic is a circle. And this is a purely projective problem, since it only involves concurrencies/collinearities/tangents/cross-ratios.
$endgroup$
– Jack D'Aurizio
Dec 10 '18 at 8:08
$begingroup$
How to prove it ? Thank you
$endgroup$
– RopuToran
Dec 10 '18 at 11:24