Mixing
Intersecting Circles.
Clash Royale CLAN TAG#URR8PPP
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Let $O_1$ , $O_2$ be the centers of circles $C_1$ , $C_2$ in a plane respectively, and the circles meet at two distinct points $A$ , $B$ . Line $O_1$$A$ meets the circle $C_1$ at point $P_1$ , and line $O_2$$A$ meets the circle $C_2$ at point $P_2$. Determine the maximum number of points lying in a circle among these 6 points $A$, $B$, $O_1$ , $O_2$ , $P_1$ and $P_2$.
I drew the two circles, but both the points $P_1$ and $P_2$ coincide with the point $A$. Is this method of solving the question correct?
geometry circle
asked Jul 28 at 15:13
Math Tise
3727
add a comment |Â
up vote
1
down vote
favorite
Let $O_1$ , $O_2$ be the centers of circles $C_1$ , $C_2$ in a plane respectively, and the circles meet at two distinct points $A$ , $B$ . Line $O_1$$A$ meets the circle $C_1$ at point $P_1$ , and line $O_2$$A$ meets the circle $C_2$ at point $P_2$. Determine the maximum number of points lying in a circle among these 6 points $A$, $B$, $O_1$ , $O_2$ , $P_1$ and $P_2$.
I drew the two circles, but both the points $P_1$ and $P_2$ coincide with the point $A$. Is this method of solving the question correct?
geometry circle
asked Jul 28 at 15:13
Math Tise
3727
I think P1 is supposed to be the point opposite P1. In other words $P_1,O_1, A$ form a diameter.
– fleablood
Jul 28 at 15:25
Does the question suggest that line $O_1A$ should be extended to meet the circle $C_1$ again at $P_1$ (so that $AP_1$ is a diameter of the circle)?
– Malkin
Jul 28 at 15:25
@EMalkin I don't see how it can mean anything else. $O1$ is the center of the $C_1$ and $A$ is also on $C_1$. Any line through a center, which $O_1A$ is intersects the circle exactly twice (forming a diameter). The implication is $A$ is a different point than $P_1$ so it must be the other point. Even if $A$ could be the same as $P_1$ the question asks for the maximum number of $6$ possible points so it seems we must consider the points being different to have more of them.
– fleablood
Jul 28 at 15:29
add a comment |Â
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up vote
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favorite
Let $O_1$ , $O_2$ be the centers of circles $C_1$ , $C_2$ in a plane respectively, and the circles meet at two distinct points $A$ , $B$ . Line $O_1$$A$ meets the circle $C_1$ at point $P_1$ , and line $O_2$$A$ meets the circle $C_2$ at point $P_2$. Determine the maximum number of points lying in a circle among these 6 points $A$, $B$, $O_1$ , $O_2$ , $P_1$ and $P_2$.
I drew the two circles, but both the points $P_1$ and $P_2$ coincide with the point $A$. Is this method of solving the question correct?
geometry circle
asked Jul 28 at 15:13
Math Tise
3727
Let $O_1$ , $O_2$ be the centers of circles $C_1$ , $C_2$ in a plane respectively, and the circles meet at two distinct points $A$ , $B$ . Line $O_1$$A$ meets the circle $C_1$ at point $P_1$ , and line $O_2$$A$ meets the circle $C_2$ at point $P_2$. Determine the maximum number of points lying in a circle among these 6 points $A$, $B$, $O_1$ , $O_2$ , $P_1$ and $P_2$.
I drew the two circles, but both the points $P_1$ and $P_2$ coincide with the point $A$. Is this method of solving the question correct?
geometry circle
asked Jul 28 at 15:13
Math Tise
3727
edited Jul 29 at 2:01
asked Jul 28 at 15:13
Math Tise
3727
asked Jul 28 at 15:13
Math Tise
3727
asked Jul 28 at 15:13
Math Tise
3727
3727
I think P1 is supposed to be the point opposite P1. In other words $P_1,O_1, A$ form a diameter.
– fleablood
Jul 28 at 15:25
Does the question suggest that line $O_1A$ should be extended to meet the circle $C_1$ again at $P_1$ (so that $AP_1$ is a diameter of the circle)?
– Malkin
Jul 28 at 15:25
@EMalkin I don't see how it can mean anything else. $O1$ is the center of the $C_1$ and $A$ is also on $C_1$. Any line through a center, which $O_1A$ is intersects the circle exactly twice (forming a diameter). The implication is $A$ is a different point than $P_1$ so it must be the other point. Even if $A$ could be the same as $P_1$ the question asks for the maximum number of $6$ possible points so it seems we must consider the points being different to have more of them.
– fleablood
Jul 28 at 15:29
add a comment |Â
I think P1 is supposed to be the point opposite P1. In other words $P_1,O_1, A$ form a diameter.
– fleablood
Jul 28 at 15:25
Does the question suggest that line $O_1A$ should be extended to meet the circle $C_1$ again at $P_1$ (so that $AP_1$ is a diameter of the circle)?
– Malkin
Jul 28 at 15:25
@EMalkin I don't see how it can mean anything else. $O1$ is the center of the $C_1$ and $A$ is also on $C_1$. Any line through a center, which $O_1A$ is intersects the circle exactly twice (forming a diameter). The implication is $A$ is a different point than $P_1$ so it must be the other point. Even if $A$ could be the same as $P_1$ the question asks for the maximum number of $6$ possible points so it seems we must consider the points being different to have more of them.
– fleablood
Jul 28 at 15:29
I think P1 is supposed to be the point opposite P1. In other words $P_1,O_1, A$ form a diameter.
– fleablood
Jul 28 at 15:25
I think P1 is supposed to be the point opposite P1. In other words $P_1,O_1, A$ form a diameter.
– fleablood
Jul 28 at 15:25
Does the question suggest that line $O_1A$ should be extended to meet the circle $C_1$ again at $P_1$ (so that $AP_1$ is a diameter of the circle)?
– Malkin
Jul 28 at 15:25
Does the question suggest that line $O_1A$ should be extended to meet the circle $C_1$ again at $P_1$ (so that $AP_1$ is a diameter of the circle)?
– Malkin
Jul 28 at 15:25
@EMalkin I don't see how it can mean anything else. $O1$ is the center of the $C_1$ and $A$ is also on $C_1$. Any line through a center, which $O_1A$ is intersects the circle exactly twice (forming a diameter). The implication is $A$ is a different point than $P_1$ so it must be the other point. Even if $A$ could be the same as $P_1$ the question asks for the maximum number of $6$ possible points so it seems we must consider the points being different to have more of them.
– fleablood
Jul 28 at 15:29
@EMalkin I don't see how it can mean anything else. $O1$ is the center of the $C_1$ and $A$ is also on $C_1$. Any line through a center, which $O_1A$ is intersects the circle exactly twice (forming a diameter). The implication is $A$ is a different point than $P_1$ so it must be the other point. Even if $A$ could be the same as $P_1$ the question asks for the maximum number of $6$ possible points so it seems we must consider the points being different to have more of them.
– fleablood
Jul 28 at 15:29
add a comment |Â
2 Answers
2
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up vote
1
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Analysis & Solution
Notice that $angle ABP_1+angle ABP_2=90^o+90^o=180^o$. Hence, $P_1, B, P_2$ are collinear. Thus, we have three groups of collinear points, which are $(A, O_1, P_1), (A, O_2, P_2)$ and $(P_1, B,P_2).$
On one hand，we may claim that, no matter what $5$ points we pick among the given $6$ points, or we pick all of them, there necessarily exists at least one group of collinear points, which is not concylic. For this reason, we obtain the fact,there exists no more than $4$ concyclic points among the given $6$ points.
One the other hand, we may pick $4$ points such that they are concylic, for example, we want $O_1, O_2, P_1, P_2$ are concylic. For this purpose, we only need $c_1$ and $c_2$ are equal circles. (Can you prove this?)
Now, we may conclude, the maximum number of points lying in a circle among the six is $4.$
answered Jul 28 at 17:00
mengdie1982
2,827216
add a comment |Â
up vote
0
down vote
If the correction as described in the comments is taken to be true, then you can prove that $P_1, B$ and $P_2$ are collinear. (Sketch it out and first prove that $ABP_1$ is a right angle.)
After proving this, you can reason that a maximum of four of the six points can lie on a circle.
answered Jul 28 at 15:39
Malkin
1,236421
angle $ABP_1$ is an angle in a semicircle, so it is a right angle. ... same reasoning for angle $ABP_2$ , I believe ...
– Math Tise
Jul 29 at 2:00
That's correct! Easy peasy.
– Malkin
Jul 29 at 17:54
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Analysis & Solution
Notice that $angle ABP_1+angle ABP_2=90^o+90^o=180^o$. Hence, $P_1, B, P_2$ are collinear. Thus, we have three groups of collinear points, which are $(A, O_1, P_1), (A, O_2, P_2)$ and $(P_1, B,P_2).$
On one hand，we may claim that, no matter what $5$ points we pick among the given $6$ points, or we pick all of them, there necessarily exists at least one group of collinear points, which is not concylic. For this reason, we obtain the fact,there exists no more than $4$ concyclic points among the given $6$ points.
One the other hand, we may pick $4$ points such that they are concylic, for example, we want $O_1, O_2, P_1, P_2$ are concylic. For this purpose, we only need $c_1$ and $c_2$ are equal circles. (Can you prove this?)
Now, we may conclude, the maximum number of points lying in a circle among the six is $4.$
answered Jul 28 at 17:00
mengdie1982
2,827216
add a comment |Â
up vote
1
down vote
Analysis & Solution
Notice that $angle ABP_1+angle ABP_2=90^o+90^o=180^o$. Hence, $P_1, B, P_2$ are collinear. Thus, we have three groups of collinear points, which are $(A, O_1, P_1), (A, O_2, P_2)$ and $(P_1, B,P_2).$
On one hand，we may claim that, no matter what $5$ points we pick among the given $6$ points, or we pick all of them, there necessarily exists at least one group of collinear points, which is not concylic. For this reason, we obtain the fact,there exists no more than $4$ concyclic points among the given $6$ points.
One the other hand, we may pick $4$ points such that they are concylic, for example, we want $O_1, O_2, P_1, P_2$ are concylic. For this purpose, we only need $c_1$ and $c_2$ are equal circles. (Can you prove this?)
Now, we may conclude, the maximum number of points lying in a circle among the six is $4.$
answered Jul 28 at 17:00
mengdie1982
2,827216
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Analysis & Solution
Notice that $angle ABP_1+angle ABP_2=90^o+90^o=180^o$. Hence, $P_1, B, P_2$ are collinear. Thus, we have three groups of collinear points, which are $(A, O_1, P_1), (A, O_2, P_2)$ and $(P_1, B,P_2).$
On one hand，we may claim that, no matter what $5$ points we pick among the given $6$ points, or we pick all of them, there necessarily exists at least one group of collinear points, which is not concylic. For this reason, we obtain the fact,there exists no more than $4$ concyclic points among the given $6$ points.
One the other hand, we may pick $4$ points such that they are concylic, for example, we want $O_1, O_2, P_1, P_2$ are concylic. For this purpose, we only need $c_1$ and $c_2$ are equal circles. (Can you prove this?)
Now, we may conclude, the maximum number of points lying in a circle among the six is $4.$
answered Jul 28 at 17:00
mengdie1982
2,827216
Analysis & Solution
Notice that $angle ABP_1+angle ABP_2=90^o+90^o=180^o$. Hence, $P_1, B, P_2$ are collinear. Thus, we have three groups of collinear points, which are $(A, O_1, P_1), (A, O_2, P_2)$ and $(P_1, B,P_2).$
On one hand，we may claim that, no matter what $5$ points we pick among the given $6$ points, or we pick all of them, there necessarily exists at least one group of collinear points, which is not concylic. For this reason, we obtain the fact,there exists no more than $4$ concyclic points among the given $6$ points.
One the other hand, we may pick $4$ points such that they are concylic, for example, we want $O_1, O_2, P_1, P_2$ are concylic. For this purpose, we only need $c_1$ and $c_2$ are equal circles. (Can you prove this?)
Now, we may conclude, the maximum number of points lying in a circle among the six is $4.$
answered Jul 28 at 17:00
mengdie1982
2,827216
answered Jul 28 at 17:00
mengdie1982
2,827216
answered Jul 28 at 17:00
mengdie1982
2,827216
answered Jul 28 at 17:00
mengdie1982
2,827216
2,827216
add a comment |Â
add a comment |Â
up vote
0
down vote
If the correction as described in the comments is taken to be true, then you can prove that $P_1, B$ and $P_2$ are collinear. (Sketch it out and first prove that $ABP_1$ is a right angle.)
After proving this, you can reason that a maximum of four of the six points can lie on a circle.
answered Jul 28 at 15:39
Malkin
1,236421
angle $ABP_1$ is an angle in a semicircle, so it is a right angle. ... same reasoning for angle $ABP_2$ , I believe ...
– Math Tise
Jul 29 at 2:00
That's correct! Easy peasy.
– Malkin
Jul 29 at 17:54
add a comment |Â
up vote
0
down vote
If the correction as described in the comments is taken to be true, then you can prove that $P_1, B$ and $P_2$ are collinear. (Sketch it out and first prove that $ABP_1$ is a right angle.)
After proving this, you can reason that a maximum of four of the six points can lie on a circle.
answered Jul 28 at 15:39
Malkin
1,236421
angle $ABP_1$ is an angle in a semicircle, so it is a right angle. ... same reasoning for angle $ABP_2$ , I believe ...
– Math Tise
Jul 29 at 2:00
That's correct! Easy peasy.
– Malkin
Jul 29 at 17:54
add a comment |Â
up vote
0
down vote
up vote
0
down vote
If the correction as described in the comments is taken to be true, then you can prove that $P_1, B$ and $P_2$ are collinear. (Sketch it out and first prove that $ABP_1$ is a right angle.)
After proving this, you can reason that a maximum of four of the six points can lie on a circle.
answered Jul 28 at 15:39
Malkin
1,236421
If the correction as described in the comments is taken to be true, then you can prove that $P_1, B$ and $P_2$ are collinear. (Sketch it out and first prove that $ABP_1$ is a right angle.)
After proving this, you can reason that a maximum of four of the six points can lie on a circle.
answered Jul 28 at 15:39
Malkin
1,236421
edited Jul 28 at 17:08
answered Jul 28 at 15:39
Malkin
1,236421
answered Jul 28 at 15:39
Malkin
1,236421
answered Jul 28 at 15:39
Malkin
1,236421
1,236421
angle $ABP_1$ is an angle in a semicircle, so it is a right angle. ... same reasoning for angle $ABP_2$ , I believe ...
– Math Tise
Jul 29 at 2:00
That's correct! Easy peasy.
– Malkin
Jul 29 at 17:54
add a comment |Â
angle $ABP_1$ is an angle in a semicircle, so it is a right angle. ... same reasoning for angle $ABP_2$ , I believe ...
– Math Tise
Jul 29 at 2:00
That's correct! Easy peasy.
– Malkin
Jul 29 at 17:54
angle $ABP_1$ is an angle in a semicircle, so it is a right angle. ... same reasoning for angle $ABP_2$ , I believe ...
– Math Tise
Jul 29 at 2:00
angle $ABP_1$ is an angle in a semicircle, so it is a right angle. ... same reasoning for angle $ABP_2$ , I believe ...
– Math Tise
Jul 29 at 2:00
That's correct! Easy peasy.
– Malkin
Jul 29 at 17:54
That's correct! Easy peasy.
– Malkin
Jul 29 at 17:54
add a comment |Â
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I think P1 is supposed to be the point opposite P1. In other words $P_1,O_1, A$ form a diameter.
– fleablood
Jul 28 at 15:25
Does the question suggest that line $O_1A$ should be extended to meet the circle $C_1$ again at $P_1$ (so that $AP_1$ is a diameter of the circle)?
– Malkin
Jul 28 at 15:25
@EMalkin I don't see how it can mean anything else. $O1$ is the center of the $C_1$ and $A$ is also on $C_1$. Any line through a center, which $O_1A$ is intersects the circle exactly twice (forming a diameter). The implication is $A$ is a different point than $P_1$ so it must be the other point. Even if $A$ could be the same as $P_1$ the question asks for the maximum number of $6$ possible points so it seems we must consider the points being different to have more of them.
– fleablood
Jul 28 at 15:29