Mixing
Geometrical problem in Newton's “Principia”.
$begingroup$
Let VQPA be the circumference
of the circle, S the given
point toward which the force
tends as to its center, P the body
revolving in the circumference, Q
the place to which it will move
next, and PRZ the tangent of
the circle at the previous place.
Through point S draw chord PV;
and when the diameter VA of
the circle has been drawn, join
AP; and to SP drop perpendicular
QT, which when produced meets the tangent PR at Z; and finally through
point Q draw LR parallel to SP and meeting both the circle at L and the
tangent PZ at R. Then because the triangles ZQR, ZTP, and VPA are
similar, RP2 (that is, QR x RL) will be to QT2 as AV2 to PV2.
My question is how can we prove the similarity of triangles ZQR, ZTP, and VPA?
geometry euclidean-geometry
asked Jan 6 at 17:43
Vaggelis KyrilasVaggelis Kyrilas
385
$endgroup$
add a comment |
$begingroup$
Let VQPA be the circumference
of the circle, S the given
point toward which the force
tends as to its center, P the body
revolving in the circumference, Q
the place to which it will move
next, and PRZ the tangent of
the circle at the previous place.
Through point S draw chord PV;
and when the diameter VA of
the circle has been drawn, join
AP; and to SP drop perpendicular
QT, which when produced meets the tangent PR at Z; and finally through
point Q draw LR parallel to SP and meeting both the circle at L and the
tangent PZ at R. Then because the triangles ZQR, ZTP, and VPA are
similar, RP2 (that is, QR x RL) will be to QT2 as AV2 to PV2.
My question is how can we prove the similarity of triangles ZQR, ZTP, and VPA?
geometry euclidean-geometry
asked Jan 6 at 17:43
Vaggelis KyrilasVaggelis Kyrilas
385
$endgroup$
1
$begingroup$
$triangle ZQRsim triangle ZTP$ because $overline{QR}paralleloverline{TP}$. For the other similarity, first note that $angle VPA$ is a right angle (via Thales' Theorem). Also, if we introduce $O$ as the center of the circle (aka, the midpoint of $overline{VA}$), we can do a little angle-chasing to show $$angle ZPT=90^circ−angle OPT=angle OPA = angle A$$ Thus, $triangle ZTPsimtriangle VPA$ by Angle-Angle Similarity.
$endgroup$
– Blue
Jan 6 at 18:44
add a comment |
$begingroup$
Let VQPA be the circumference
of the circle, S the given
point toward which the force
tends as to its center, P the body
revolving in the circumference, Q
the place to which it will move
next, and PRZ the tangent of
the circle at the previous place.
Through point S draw chord PV;
and when the diameter VA of
the circle has been drawn, join
AP; and to SP drop perpendicular
QT, which when produced meets the tangent PR at Z; and finally through
point Q draw LR parallel to SP and meeting both the circle at L and the
tangent PZ at R. Then because the triangles ZQR, ZTP, and VPA are
similar, RP2 (that is, QR x RL) will be to QT2 as AV2 to PV2.
My question is how can we prove the similarity of triangles ZQR, ZTP, and VPA?
geometry euclidean-geometry
asked Jan 6 at 17:43
Vaggelis KyrilasVaggelis Kyrilas
385
$endgroup$
Let VQPA be the circumference
of the circle, S the given
point toward which the force
tends as to its center, P the body
revolving in the circumference, Q
the place to which it will move
next, and PRZ the tangent of
the circle at the previous place.
Through point S draw chord PV;
and when the diameter VA of
the circle has been drawn, join
AP; and to SP drop perpendicular
QT, which when produced meets the tangent PR at Z; and finally through
point Q draw LR parallel to SP and meeting both the circle at L and the
tangent PZ at R. Then because the triangles ZQR, ZTP, and VPA are
similar, RP2 (that is, QR x RL) will be to QT2 as AV2 to PV2.
My question is how can we prove the similarity of triangles ZQR, ZTP, and VPA?
geometry euclidean-geometry
geometry euclidean-geometry
asked Jan 6 at 17:43
Vaggelis KyrilasVaggelis Kyrilas
385
asked Jan 6 at 17:43
Vaggelis KyrilasVaggelis Kyrilas
385
asked Jan 6 at 17:43
Vaggelis KyrilasVaggelis Kyrilas
385
asked Jan 6 at 17:43
Vaggelis KyrilasVaggelis Kyrilas
385
asked Jan 6 at 17:43
Vaggelis KyrilasVaggelis Kyrilas
385
385
1
$begingroup$
$triangle ZQRsim triangle ZTP$ because $overline{QR}paralleloverline{TP}$. For the other similarity, first note that $angle VPA$ is a right angle (via Thales' Theorem). Also, if we introduce $O$ as the center of the circle (aka, the midpoint of $overline{VA}$), we can do a little angle-chasing to show $$angle ZPT=90^circ−angle OPT=angle OPA = angle A$$ Thus, $triangle ZTPsimtriangle VPA$ by Angle-Angle Similarity.
$endgroup$
– Blue
Jan 6 at 18:44
add a comment |
1
$begingroup$
$triangle ZQRsim triangle ZTP$ because $overline{QR}paralleloverline{TP}$. For the other similarity, first note that $angle VPA$ is a right angle (via Thales' Theorem). Also, if we introduce $O$ as the center of the circle (aka, the midpoint of $overline{VA}$), we can do a little angle-chasing to show $$angle ZPT=90^circ−angle OPT=angle OPA = angle A$$ Thus, $triangle ZTPsimtriangle VPA$ by Angle-Angle Similarity.
$endgroup$
– Blue
Jan 6 at 18:44
1
1
$begingroup$
$triangle ZQRsim triangle ZTP$ because $overline{QR}paralleloverline{TP}$. For the other similarity, first note that $angle VPA$ is a right angle (via Thales' Theorem). Also, if we introduce $O$ as the center of the circle (aka, the midpoint of $overline{VA}$), we can do a little angle-chasing to show $$angle ZPT=90^circ−angle OPT=angle OPA = angle A$$ Thus, $triangle ZTPsimtriangle VPA$ by Angle-Angle Similarity.
$endgroup$
– Blue
Jan 6 at 18:44
$begingroup$
$triangle ZQRsim triangle ZTP$ because $overline{QR}paralleloverline{TP}$. For the other similarity, first note that $angle VPA$ is a right angle (via Thales' Theorem). Also, if we introduce $O$ as the center of the circle (aka, the midpoint of $overline{VA}$), we can do a little angle-chasing to show $$angle ZPT=90^circ−angle OPT=angle OPA = angle A$$ Thus, $triangle ZTPsimtriangle VPA$ by Angle-Angle Similarity.
$endgroup$
– Blue
Jan 6 at 18:44
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Accepting the comment of @Blue on the similarity of triangles $ZQR$ and $ZTP$, then without further construction, since $PZ$ is tangent at $P$, in right triangles $ZTP$ and $VPA$, $angle ZPT=angle VAP$ (Euclid, Elements III, 32). Therefore$$triangle ZTPsimtriangle VPA$$
answered Jan 15 at 22:27
Edward PorcellaEdward Porcella
1,4311511
$endgroup$
add a comment |
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$begingroup$
Accepting the comment of @Blue on the similarity of triangles $ZQR$ and $ZTP$, then without further construction, since $PZ$ is tangent at $P$, in right triangles $ZTP$ and $VPA$, $angle ZPT=angle VAP$ (Euclid, Elements III, 32). Therefore$$triangle ZTPsimtriangle VPA$$
answered Jan 15 at 22:27
Edward PorcellaEdward Porcella
1,4311511
$endgroup$
add a comment |
$begingroup$
Accepting the comment of @Blue on the similarity of triangles $ZQR$ and $ZTP$, then without further construction, since $PZ$ is tangent at $P$, in right triangles $ZTP$ and $VPA$, $angle ZPT=angle VAP$ (Euclid, Elements III, 32). Therefore$$triangle ZTPsimtriangle VPA$$
answered Jan 15 at 22:27
Edward PorcellaEdward Porcella
1,4311511
$endgroup$
add a comment |
$begingroup$
Accepting the comment of @Blue on the similarity of triangles $ZQR$ and $ZTP$, then without further construction, since $PZ$ is tangent at $P$, in right triangles $ZTP$ and $VPA$, $angle ZPT=angle VAP$ (Euclid, Elements III, 32). Therefore$$triangle ZTPsimtriangle VPA$$
answered Jan 15 at 22:27
Edward PorcellaEdward Porcella
1,4311511
$endgroup$
Accepting the comment of @Blue on the similarity of triangles $ZQR$ and $ZTP$, then without further construction, since $PZ$ is tangent at $P$, in right triangles $ZTP$ and $VPA$, $angle ZPT=angle VAP$ (Euclid, Elements III, 32). Therefore$$triangle ZTPsimtriangle VPA$$
answered Jan 15 at 22:27
Edward PorcellaEdward Porcella
1,4311511
answered Jan 15 at 22:27
Edward PorcellaEdward Porcella
1,4311511
answered Jan 15 at 22:27
Edward PorcellaEdward Porcella
1,4311511
answered Jan 15 at 22:27
Edward PorcellaEdward Porcella
1,4311511
1,4311511
add a comment |
add a comment |
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$begingroup$
$triangle ZQRsim triangle ZTP$ because $overline{QR}paralleloverline{TP}$. For the other similarity, first note that $angle VPA$ is a right angle (via Thales' Theorem). Also, if we introduce $O$ as the center of the circle (aka, the midpoint of $overline{VA}$), we can do a little angle-chasing to show $$angle ZPT=90^circ−angle OPT=angle OPA = angle A$$ Thus, $triangle ZTPsimtriangle VPA$ by Angle-Angle Similarity.
$endgroup$
– Blue
Jan 6 at 18:44