Mixing
Try to find the ratio of two arcs
$begingroup$
O1 is the center of the big circle; O2 is the center of the small one.
Line(O2-P) is vertical to line(A-C).
Line(PQ):line(PB)=2:7
What's the ratio of arc(AC) and arc(CB)?
triangles circles arc-length
edited Feb 1 at 13:16
Bernard
124k741117
asked Feb 1 at 13:03
Andrew-at-TWAndrew-at-TW
1167
$endgroup$
add a comment |
$begingroup$
O1 is the center of the big circle; O2 is the center of the small one.
Line(O2-P) is vertical to line(A-C).
Line(PQ):line(PB)=2:7
What's the ratio of arc(AC) and arc(CB)?
triangles circles arc-length
edited Feb 1 at 13:16
Bernard
124k741117
asked Feb 1 at 13:03
Andrew-at-TWAndrew-at-TW
1167
$endgroup$
$begingroup$
Hint: $AC:CB=AP:PO_2$
$endgroup$
– Vasya
Feb 1 at 13:09
add a comment |
$begingroup$
O1 is the center of the big circle; O2 is the center of the small one.
Line(O2-P) is vertical to line(A-C).
Line(PQ):line(PB)=2:7
What's the ratio of arc(AC) and arc(CB)?
triangles circles arc-length
edited Feb 1 at 13:16
Bernard
124k741117
asked Feb 1 at 13:03
Andrew-at-TWAndrew-at-TW
1167
$endgroup$
O1 is the center of the big circle; O2 is the center of the small one.
Line(O2-P) is vertical to line(A-C).
Line(PQ):line(PB)=2:7
What's the ratio of arc(AC) and arc(CB)?
triangles circles arc-length
triangles circles arc-length
edited Feb 1 at 13:16
Bernard
124k741117
asked Feb 1 at 13:03
Andrew-at-TWAndrew-at-TW
1167
edited Feb 1 at 13:16
Bernard
124k741117
asked Feb 1 at 13:03
Andrew-at-TWAndrew-at-TW
1167
edited Feb 1 at 13:16
Bernard
124k741117
edited Feb 1 at 13:16
Bernard
124k741117
edited Feb 1 at 13:16
Bernard
124k741117
124k741117
asked Feb 1 at 13:03
Andrew-at-TWAndrew-at-TW
1167
asked Feb 1 at 13:03
Andrew-at-TWAndrew-at-TW
1167
asked Feb 1 at 13:03
Andrew-at-TWAndrew-at-TW
1167
1167
$begingroup$
Hint: $AC:CB=AP:PO_2$
$endgroup$
– Vasya
Feb 1 at 13:09
add a comment |
$begingroup$
Hint: $AC:CB=AP:PO_2$
$endgroup$
– Vasya
Feb 1 at 13:09
$begingroup$
Hint: $AC:CB=AP:PO_2$
$endgroup$
– Vasya
Feb 1 at 13:09
$begingroup$
Hint: $AC:CB=AP:PO_2$
$endgroup$
– Vasya
Feb 1 at 13:09
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $PQ=2$ and $PB=7$: by Pythagoras' theorem we then get $BQ=sqrt{53}$.
If $PH$ is an altitude of triangle $BPQ$, then:
$$PH=PQcdot {PBover BQ}={14oversqrt{53}}$$
and by similarity:
$$QH={4oversqrt{53}},
quad BH={49oversqrt{53}},
quad HO_2=QO_2-QH={45over 2sqrt{53}}.
$$
Triangle $APH$ is similar to $PHO_2$, hence:
$$
AH={392over45sqrt{53}},quad
AO_2=AH+HO_2={53sqrt{53}over90}
quadtext{and}quad
PA={14sqrt{53}over45}.
$$
Finally, $ABC$ is similar to $APO_2$ and
$displaystyle AB=AO_2+BO_2={49sqrt{53}over45}$, whence:
$$
BC={49oversqrt{53}}, quad AC={1372over45sqrt{53}}.
$$
It follows that:
$$
{ACover BC}={28over45},
quadtext{and}quad
{text{arc}(AC)over text{arc}(BC)}={arctan(28/45)overarctan(45/28)}.
$$
answered Feb 5 at 22:40
AretinoAretino
25.8k31545
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $PQ=2$ and $PB=7$: by Pythagoras' theorem we then get $BQ=sqrt{53}$.
If $PH$ is an altitude of triangle $BPQ$, then:
$$PH=PQcdot {PBover BQ}={14oversqrt{53}}$$
and by similarity:
$$QH={4oversqrt{53}},
quad BH={49oversqrt{53}},
quad HO_2=QO_2-QH={45over 2sqrt{53}}.
$$
Triangle $APH$ is similar to $PHO_2$, hence:
$$
AH={392over45sqrt{53}},quad
AO_2=AH+HO_2={53sqrt{53}over90}
quadtext{and}quad
PA={14sqrt{53}over45}.
$$
Finally, $ABC$ is similar to $APO_2$ and
$displaystyle AB=AO_2+BO_2={49sqrt{53}over45}$, whence:
$$
BC={49oversqrt{53}}, quad AC={1372over45sqrt{53}}.
$$
It follows that:
$$
{ACover BC}={28over45},
quadtext{and}quad
{text{arc}(AC)over text{arc}(BC)}={arctan(28/45)overarctan(45/28)}.
$$
answered Feb 5 at 22:40
AretinoAretino
25.8k31545
$endgroup$
add a comment |
$begingroup$
Let $PQ=2$ and $PB=7$: by Pythagoras' theorem we then get $BQ=sqrt{53}$.
If $PH$ is an altitude of triangle $BPQ$, then:
$$PH=PQcdot {PBover BQ}={14oversqrt{53}}$$
and by similarity:
$$QH={4oversqrt{53}},
quad BH={49oversqrt{53}},
quad HO_2=QO_2-QH={45over 2sqrt{53}}.
$$
Triangle $APH$ is similar to $PHO_2$, hence:
$$
AH={392over45sqrt{53}},quad
AO_2=AH+HO_2={53sqrt{53}over90}
quadtext{and}quad
PA={14sqrt{53}over45}.
$$
Finally, $ABC$ is similar to $APO_2$ and
$displaystyle AB=AO_2+BO_2={49sqrt{53}over45}$, whence:
$$
BC={49oversqrt{53}}, quad AC={1372over45sqrt{53}}.
$$
It follows that:
$$
{ACover BC}={28over45},
quadtext{and}quad
{text{arc}(AC)over text{arc}(BC)}={arctan(28/45)overarctan(45/28)}.
$$
answered Feb 5 at 22:40
AretinoAretino
25.8k31545
$endgroup$
add a comment |
$begingroup$
Let $PQ=2$ and $PB=7$: by Pythagoras' theorem we then get $BQ=sqrt{53}$.
If $PH$ is an altitude of triangle $BPQ$, then:
$$PH=PQcdot {PBover BQ}={14oversqrt{53}}$$
and by similarity:
$$QH={4oversqrt{53}},
quad BH={49oversqrt{53}},
quad HO_2=QO_2-QH={45over 2sqrt{53}}.
$$
Triangle $APH$ is similar to $PHO_2$, hence:
$$
AH={392over45sqrt{53}},quad
AO_2=AH+HO_2={53sqrt{53}over90}
quadtext{and}quad
PA={14sqrt{53}over45}.
$$
Finally, $ABC$ is similar to $APO_2$ and
$displaystyle AB=AO_2+BO_2={49sqrt{53}over45}$, whence:
$$
BC={49oversqrt{53}}, quad AC={1372over45sqrt{53}}.
$$
It follows that:
$$
{ACover BC}={28over45},
quadtext{and}quad
{text{arc}(AC)over text{arc}(BC)}={arctan(28/45)overarctan(45/28)}.
$$
answered Feb 5 at 22:40
AretinoAretino
25.8k31545
$endgroup$
Let $PQ=2$ and $PB=7$: by Pythagoras' theorem we then get $BQ=sqrt{53}$.
If $PH$ is an altitude of triangle $BPQ$, then:
$$PH=PQcdot {PBover BQ}={14oversqrt{53}}$$
and by similarity:
$$QH={4oversqrt{53}},
quad BH={49oversqrt{53}},
quad HO_2=QO_2-QH={45over 2sqrt{53}}.
$$
Triangle $APH$ is similar to $PHO_2$, hence:
$$
AH={392over45sqrt{53}},quad
AO_2=AH+HO_2={53sqrt{53}over90}
quadtext{and}quad
PA={14sqrt{53}over45}.
$$
Finally, $ABC$ is similar to $APO_2$ and
$displaystyle AB=AO_2+BO_2={49sqrt{53}over45}$, whence:
$$
BC={49oversqrt{53}}, quad AC={1372over45sqrt{53}}.
$$
It follows that:
$$
{ACover BC}={28over45},
quadtext{and}quad
{text{arc}(AC)over text{arc}(BC)}={arctan(28/45)overarctan(45/28)}.
$$
answered Feb 5 at 22:40
AretinoAretino
25.8k31545
edited Feb 6 at 20:58
answered Feb 5 at 22:40
AretinoAretino
25.8k31545
answered Feb 5 at 22:40
AretinoAretino
25.8k31545
answered Feb 5 at 22:40
AretinoAretino
25.8k31545
25.8k31545
add a comment |
add a comment |
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$begingroup$
Hint: $AC:CB=AP:PO_2$
$endgroup$
– Vasya
Feb 1 at 13:09