Find the minimum of $MA + MB$ when $M$ is on the circle [closed]

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Let be a circle of radius $1$ and center $O$. Consider the points $A$ and $B$ such that $OA=OB$. Find $M$ on the circle such that $ MA + MB$ to be minimum. Actually I am looking for a synthetic solutions, using a construction. I tried with analytic methods but it isn't the best way and to make some inversions too.

geometry

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edited Jul 31 at 14:03

amWhy

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asked Jul 31 at 13:55

rafa

418112

closed as off-topic by amWhy, Mohammad Riazi-Kermani, John Ma, Mostafa Ayaz, max_zorn Aug 1 at 5:53

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, John Ma, Mostafa Ayaz

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  isn't it the point where perpendicular bisector of points A and B meet the circle?
  – emil
  Jul 31 at 14:03
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

Let be a circle of radius $1$ and center $O$. Consider the points $A$ and $B$ such that $OA=OB$. Find $M$ on the circle such that $ MA + MB$ to be minimum. Actually I am looking for a synthetic solutions, using a construction. I tried with analytic methods but it isn't the best way and to make some inversions too.

geometry

share|cite|improve this question

edited Jul 31 at 14:03

amWhy

189k25219431

asked Jul 31 at 13:55

rafa

418112

closed as off-topic by amWhy, Mohammad Riazi-Kermani, John Ma, Mostafa Ayaz, max_zorn Aug 1 at 5:53

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, John Ma, Mostafa Ayaz

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  isn't it the point where perpendicular bisector of points A and B meet the circle?
  – emil
  Jul 31 at 14:03
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Let be a circle of radius $1$ and center $O$. Consider the points $A$ and $B$ such that $OA=OB$. Find $M$ on the circle such that $ MA + MB$ to be minimum. Actually I am looking for a synthetic solutions, using a construction. I tried with analytic methods but it isn't the best way and to make some inversions too.

geometry

share|cite|improve this question

edited Jul 31 at 14:03

amWhy

189k25219431

asked Jul 31 at 13:55

rafa

418112

Let be a circle of radius $1$ and center $O$. Consider the points $A$ and $B$ such that $OA=OB$. Find $M$ on the circle such that $ MA + MB$ to be minimum. Actually I am looking for a synthetic solutions, using a construction. I tried with analytic methods but it isn't the best way and to make some inversions too.

geometry

share|cite|improve this question

edited Jul 31 at 14:03

amWhy

189k25219431

asked Jul 31 at 13:55

rafa

418112

share|cite|improve this question

share|cite|improve this question

edited Jul 31 at 14:03

amWhy

189k25219431

edited Jul 31 at 14:03

amWhy

189k25219431

edited Jul 31 at 14:03

amWhy

189k25219431

189k25219431

asked Jul 31 at 13:55

rafa

418112

asked Jul 31 at 13:55

rafa

418112

asked Jul 31 at 13:55

rafa

418112

418112

closed as off-topic by amWhy, Mohammad Riazi-Kermani, John Ma, Mostafa Ayaz, max_zorn Aug 1 at 5:53

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, John Ma, Mostafa Ayaz

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by amWhy, Mohammad Riazi-Kermani, John Ma, Mostafa Ayaz, max_zorn Aug 1 at 5:53

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, John Ma, Mostafa Ayaz

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  isn't it the point where perpendicular bisector of points A and B meet the circle?
  – emil
  Jul 31 at 14:03
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  isn't it the point where perpendicular bisector of points A and B meet the circle?
  – emil
  Jul 31 at 14:03
  

  
  
  
  

isn't it the point where perpendicular bisector of points A and B meet the circle?
– emil
Jul 31 at 14:03

isn't it the point where perpendicular bisector of points A and B meet the circle?
– emil
Jul 31 at 14:03

add a comment | 

3 Answers
3

active

oldest

votes

up vote
1
down vote

• $MA+MB=k$ where $k$ is a constant is an ellipse of foci $A,B$


• The big axis of the ellipse is supported by line $(AB)$


• Since $OA=OB$ then the small axis is also the perpendicular bissector of $[AB]$

For now let just consider the easy case when $[AB]$ is outside the circle (like in emil's figure).

For $k$ small the ellipse and the circle of centre $O$ have no intersection.

• When $k$ is growing the ellipse and circle will intersect, for the minimal $k$, they will be tangent to each other at a point $colorgreenM_1$ on the small axis.

You can also visualize a similar problem here:

minimal distance between two points and point on a plane

The cases where the segment is inside or intersect the circle seems more complicated.

• For instance when $O$ is the middle of the segment and it is inside the circle (case $[CD]$) then when $k$ grows the tangency will happen in $colorpurpleM_2$ on the big axis this time.




• But if the points are outside the circle (case $[EF]$) then the minimal $k$ is reached for the degenerated flat ellipse $k=EF$ and $mathbfM_4$ is just one of the two intersection points of the segment with the circle.




• The remaining cases seem even more complicated and I feel that analytic solution is the only way... For instance the point $colororangeM_3$ does not seem to have any particular property.

share|cite|improve this answer

edited Jul 31 at 21:54

answered Jul 31 at 21:12

zwim

11k527

add a comment | 

up vote
0
down vote

Say, $OA=OB=x$, also $A,B$ subtend an angle of $2theta$ on the center of the circle. Construct a perependicular bisector for points $A,B$. Point $M$ should be on the constructed perpendicular. Let $P$ be the point where $AB$ meet perpendicular bisector. So$$ AP=xsintheta$$ $$PM =|xcostheta-1| $$ $$AM=sqrtx^2-2xcostheta +1$$
$$AM+BM=2sqrtx^2-2xcostheta +1$$

P.S. When points $A, B$ lie inside of the circle, same proof can be used. Only difference is,$PM =1-xcostheta $. This is why I put modulus value for $PM.$

share|cite|improve this answer

edited Jul 31 at 15:53

answered Jul 31 at 14:13

emil

314210

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $A$ and $B$ are fixed
  – rafa
  Jul 31 at 14:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Fixed it. ( Previously I took points A and B as points on the circle)
  – emil
  Jul 31 at 14:45
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If $OA gt 1$ and line $AB$ intersects the circle with center $O$, the minimum is the straight line connecting $A$ and $B$.
  – Jens
  Jul 31 at 15:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That happens when $xcostheta=1$ in which if we substitute to the result, $AM+BM=2sqrtx^2-1$
  – emil
  Jul 31 at 15:16
  

  
  
  
  

add a comment | 

up vote
0
down vote

This solution is valid only if $A$ and $B$ lie outside the circle. If segment $AB$ intersects the circle, $M$ is obviously either intersection point.

Suppose then $AB$ doesn't intersect the circle: in that case $M$ is the nearest intersection between the circle and the perpendicular bisector of $AB$, such that the circle and $AB$ are on opposite sides with respect to the tangent $t$ at $M$.

Proof.

Let $A'$ be the reflection of $A$ about $t$: then $M$ is the midpoint of $A'B$. If $N$ is any other point on $t$, different from $M$, then $AN+BN>AM+BM$, because:
$$
AN+BN=A'N+BN>A'B=A'M+BM=AM+BM,
$$
where I applied the triangular inequality to triangle $A'BN$.
Let now $P$ be a point on the circle, different from $M$. Segment $PB$ will intersect $t$ at some point $Q$ and:
$$
AP+BP=AP+PQ+BQ>AQ+BQge AM+BM,
$$
where I applied the triangular inequality to triangle $APQ$ and used the result proved above.

share|cite|improve this answer

edited Aug 1 at 9:40

answered Jul 31 at 20:06

Aretino

21.7k21342

add a comment | 

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote

• $MA+MB=k$ where $k$ is a constant is an ellipse of foci $A,B$


• The big axis of the ellipse is supported by line $(AB)$


• Since $OA=OB$ then the small axis is also the perpendicular bissector of $[AB]$

For now let just consider the easy case when $[AB]$ is outside the circle (like in emil's figure).

For $k$ small the ellipse and the circle of centre $O$ have no intersection.

• When $k$ is growing the ellipse and circle will intersect, for the minimal $k$, they will be tangent to each other at a point $colorgreenM_1$ on the small axis.

You can also visualize a similar problem here:

minimal distance between two points and point on a plane

The cases where the segment is inside or intersect the circle seems more complicated.

• For instance when $O$ is the middle of the segment and it is inside the circle (case $[CD]$) then when $k$ grows the tangency will happen in $colorpurpleM_2$ on the big axis this time.




• But if the points are outside the circle (case $[EF]$) then the minimal $k$ is reached for the degenerated flat ellipse $k=EF$ and $mathbfM_4$ is just one of the two intersection points of the segment with the circle.




• The remaining cases seem even more complicated and I feel that analytic solution is the only way... For instance the point $colororangeM_3$ does not seem to have any particular property.

share|cite|improve this answer

edited Jul 31 at 21:54

answered Jul 31 at 21:12

zwim

11k527

add a comment | 

up vote
1
down vote

• $MA+MB=k$ where $k$ is a constant is an ellipse of foci $A,B$


• The big axis of the ellipse is supported by line $(AB)$


• Since $OA=OB$ then the small axis is also the perpendicular bissector of $[AB]$

For now let just consider the easy case when $[AB]$ is outside the circle (like in emil's figure).

For $k$ small the ellipse and the circle of centre $O$ have no intersection.

• When $k$ is growing the ellipse and circle will intersect, for the minimal $k$, they will be tangent to each other at a point $colorgreenM_1$ on the small axis.

You can also visualize a similar problem here:

minimal distance between two points and point on a plane

The cases where the segment is inside or intersect the circle seems more complicated.

• For instance when $O$ is the middle of the segment and it is inside the circle (case $[CD]$) then when $k$ grows the tangency will happen in $colorpurpleM_2$ on the big axis this time.




• But if the points are outside the circle (case $[EF]$) then the minimal $k$ is reached for the degenerated flat ellipse $k=EF$ and $mathbfM_4$ is just one of the two intersection points of the segment with the circle.




• The remaining cases seem even more complicated and I feel that analytic solution is the only way... For instance the point $colororangeM_3$ does not seem to have any particular property.

share|cite|improve this answer

edited Jul 31 at 21:54

answered Jul 31 at 21:12

zwim

11k527

add a comment | 

up vote
1
down vote

up vote
1
down vote

• $MA+MB=k$ where $k$ is a constant is an ellipse of foci $A,B$


• The big axis of the ellipse is supported by line $(AB)$


• Since $OA=OB$ then the small axis is also the perpendicular bissector of $[AB]$

For now let just consider the easy case when $[AB]$ is outside the circle (like in emil's figure).

For $k$ small the ellipse and the circle of centre $O$ have no intersection.

• When $k$ is growing the ellipse and circle will intersect, for the minimal $k$, they will be tangent to each other at a point $colorgreenM_1$ on the small axis.

You can also visualize a similar problem here:

minimal distance between two points and point on a plane

The cases where the segment is inside or intersect the circle seems more complicated.

• For instance when $O$ is the middle of the segment and it is inside the circle (case $[CD]$) then when $k$ grows the tangency will happen in $colorpurpleM_2$ on the big axis this time.




• But if the points are outside the circle (case $[EF]$) then the minimal $k$ is reached for the degenerated flat ellipse $k=EF$ and $mathbfM_4$ is just one of the two intersection points of the segment with the circle.




• The remaining cases seem even more complicated and I feel that analytic solution is the only way... For instance the point $colororangeM_3$ does not seem to have any particular property.

share|cite|improve this answer

edited Jul 31 at 21:54

answered Jul 31 at 21:12

zwim

11k527

• $MA+MB=k$ where $k$ is a constant is an ellipse of foci $A,B$


• The big axis of the ellipse is supported by line $(AB)$


• Since $OA=OB$ then the small axis is also the perpendicular bissector of $[AB]$

For now let just consider the easy case when $[AB]$ is outside the circle (like in emil's figure).

For $k$ small the ellipse and the circle of centre $O$ have no intersection.

• When $k$ is growing the ellipse and circle will intersect, for the minimal $k$, they will be tangent to each other at a point $colorgreenM_1$ on the small axis.

You can also visualize a similar problem here:

minimal distance between two points and point on a plane

The cases where the segment is inside or intersect the circle seems more complicated.

• For instance when $O$ is the middle of the segment and it is inside the circle (case $[CD]$) then when $k$ grows the tangency will happen in $colorpurpleM_2$ on the big axis this time.




• But if the points are outside the circle (case $[EF]$) then the minimal $k$ is reached for the degenerated flat ellipse $k=EF$ and $mathbfM_4$ is just one of the two intersection points of the segment with the circle.




• The remaining cases seem even more complicated and I feel that analytic solution is the only way... For instance the point $colororangeM_3$ does not seem to have any particular property.

share|cite|improve this answer

edited Jul 31 at 21:54

answered Jul 31 at 21:12

zwim

11k527

share|cite|improve this answer

share|cite|improve this answer

edited Jul 31 at 21:54

edited Jul 31 at 21:54

edited Jul 31 at 21:54

answered Jul 31 at 21:12

zwim

11k527

answered Jul 31 at 21:12

zwim

11k527

answered Jul 31 at 21:12

zwim

11k527

11k527

add a comment | 

add a comment | 

up vote
0
down vote

Say, $OA=OB=x$, also $A,B$ subtend an angle of $2theta$ on the center of the circle. Construct a perependicular bisector for points $A,B$. Point $M$ should be on the constructed perpendicular. Let $P$ be the point where $AB$ meet perpendicular bisector. So$$ AP=xsintheta$$ $$PM =|xcostheta-1| $$ $$AM=sqrtx^2-2xcostheta +1$$
$$AM+BM=2sqrtx^2-2xcostheta +1$$

P.S. When points $A, B$ lie inside of the circle, same proof can be used. Only difference is,$PM =1-xcostheta $. This is why I put modulus value for $PM.$

share|cite|improve this answer

edited Jul 31 at 15:53

answered Jul 31 at 14:13

emil

314210

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $A$ and $B$ are fixed
  – rafa
  Jul 31 at 14:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Fixed it. ( Previously I took points A and B as points on the circle)
  – emil
  Jul 31 at 14:45
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If $OA gt 1$ and line $AB$ intersects the circle with center $O$, the minimum is the straight line connecting $A$ and $B$.
  – Jens
  Jul 31 at 15:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That happens when $xcostheta=1$ in which if we substitute to the result, $AM+BM=2sqrtx^2-1$
  – emil
  Jul 31 at 15:16
  

  
  
  
  

add a comment | 

up vote
0
down vote

Say, $OA=OB=x$, also $A,B$ subtend an angle of $2theta$ on the center of the circle. Construct a perependicular bisector for points $A,B$. Point $M$ should be on the constructed perpendicular. Let $P$ be the point where $AB$ meet perpendicular bisector. So$$ AP=xsintheta$$ $$PM =|xcostheta-1| $$ $$AM=sqrtx^2-2xcostheta +1$$
$$AM+BM=2sqrtx^2-2xcostheta +1$$

P.S. When points $A, B$ lie inside of the circle, same proof can be used. Only difference is,$PM =1-xcostheta $. This is why I put modulus value for $PM.$

share|cite|improve this answer

edited Jul 31 at 15:53

answered Jul 31 at 14:13

emil

314210

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $A$ and $B$ are fixed
  – rafa
  Jul 31 at 14:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Fixed it. ( Previously I took points A and B as points on the circle)
  – emil
  Jul 31 at 14:45
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If $OA gt 1$ and line $AB$ intersects the circle with center $O$, the minimum is the straight line connecting $A$ and $B$.
  – Jens
  Jul 31 at 15:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That happens when $xcostheta=1$ in which if we substitute to the result, $AM+BM=2sqrtx^2-1$
  – emil
  Jul 31 at 15:16
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

Say, $OA=OB=x$, also $A,B$ subtend an angle of $2theta$ on the center of the circle. Construct a perependicular bisector for points $A,B$. Point $M$ should be on the constructed perpendicular. Let $P$ be the point where $AB$ meet perpendicular bisector. So$$ AP=xsintheta$$ $$PM =|xcostheta-1| $$ $$AM=sqrtx^2-2xcostheta +1$$
$$AM+BM=2sqrtx^2-2xcostheta +1$$

P.S. When points $A, B$ lie inside of the circle, same proof can be used. Only difference is,$PM =1-xcostheta $. This is why I put modulus value for $PM.$

share|cite|improve this answer

edited Jul 31 at 15:53

answered Jul 31 at 14:13

emil

314210

Say, $OA=OB=x$, also $A,B$ subtend an angle of $2theta$ on the center of the circle. Construct a perependicular bisector for points $A,B$. Point $M$ should be on the constructed perpendicular. Let $P$ be the point where $AB$ meet perpendicular bisector. So$$ AP=xsintheta$$ $$PM =|xcostheta-1| $$ $$AM=sqrtx^2-2xcostheta +1$$
$$AM+BM=2sqrtx^2-2xcostheta +1$$

P.S. When points $A, B$ lie inside of the circle, same proof can be used. Only difference is,$PM =1-xcostheta $. This is why I put modulus value for $PM.$

share|cite|improve this answer

edited Jul 31 at 15:53

answered Jul 31 at 14:13

emil

314210

share|cite|improve this answer

share|cite|improve this answer

edited Jul 31 at 15:53

edited Jul 31 at 15:53

edited Jul 31 at 15:53

answered Jul 31 at 14:13

emil

314210

answered Jul 31 at 14:13

emil

314210

answered Jul 31 at 14:13

emil

314210

314210

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $A$ and $B$ are fixed
  – rafa
  Jul 31 at 14:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Fixed it. ( Previously I took points A and B as points on the circle)
  – emil
  Jul 31 at 14:45
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If $OA gt 1$ and line $AB$ intersects the circle with center $O$, the minimum is the straight line connecting $A$ and $B$.
  – Jens
  Jul 31 at 15:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That happens when $xcostheta=1$ in which if we substitute to the result, $AM+BM=2sqrtx^2-1$
  – emil
  Jul 31 at 15:16
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $A$ and $B$ are fixed
  – rafa
  Jul 31 at 14:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Fixed it. ( Previously I took points A and B as points on the circle)
  – emil
  Jul 31 at 14:45
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If $OA gt 1$ and line $AB$ intersects the circle with center $O$, the minimum is the straight line connecting $A$ and $B$.
  – Jens
  Jul 31 at 15:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That happens when $xcostheta=1$ in which if we substitute to the result, $AM+BM=2sqrtx^2-1$
  – emil
  Jul 31 at 15:16
  

  
  
  
  

$A$ and $B$ are fixed
– rafa
Jul 31 at 14:20

$A$ and $B$ are fixed
– rafa
Jul 31 at 14:20

Fixed it. ( Previously I took points A and B as points on the circle)
– emil
Jul 31 at 14:45

Fixed it. ( Previously I took points A and B as points on the circle)
– emil
Jul 31 at 14:45

If $OA gt 1$ and line $AB$ intersects the circle with center $O$, the minimum is the straight line connecting $A$ and $B$.
– Jens
Jul 31 at 15:04

If $OA gt 1$ and line $AB$ intersects the circle with center $O$, the minimum is the straight line connecting $A$ and $B$.
– Jens
Jul 31 at 15:04

That happens when $xcostheta=1$ in which if we substitute to the result, $AM+BM=2sqrtx^2-1$
– emil
Jul 31 at 15:16

That happens when $xcostheta=1$ in which if we substitute to the result, $AM+BM=2sqrtx^2-1$
– emil
Jul 31 at 15:16

add a comment | 

up vote
0
down vote

This solution is valid only if $A$ and $B$ lie outside the circle. If segment $AB$ intersects the circle, $M$ is obviously either intersection point.

Suppose then $AB$ doesn't intersect the circle: in that case $M$ is the nearest intersection between the circle and the perpendicular bisector of $AB$, such that the circle and $AB$ are on opposite sides with respect to the tangent $t$ at $M$.

Proof.

Let $A'$ be the reflection of $A$ about $t$: then $M$ is the midpoint of $A'B$. If $N$ is any other point on $t$, different from $M$, then $AN+BN>AM+BM$, because:
$$
AN+BN=A'N+BN>A'B=A'M+BM=AM+BM,
$$
where I applied the triangular inequality to triangle $A'BN$.
Let now $P$ be a point on the circle, different from $M$. Segment $PB$ will intersect $t$ at some point $Q$ and:
$$
AP+BP=AP+PQ+BQ>AQ+BQge AM+BM,
$$
where I applied the triangular inequality to triangle $APQ$ and used the result proved above.

share|cite|improve this answer

edited Aug 1 at 9:40

answered Jul 31 at 20:06

Aretino

21.7k21342

add a comment | 

up vote
0
down vote

This solution is valid only if $A$ and $B$ lie outside the circle. If segment $AB$ intersects the circle, $M$ is obviously either intersection point.

Suppose then $AB$ doesn't intersect the circle: in that case $M$ is the nearest intersection between the circle and the perpendicular bisector of $AB$, such that the circle and $AB$ are on opposite sides with respect to the tangent $t$ at $M$.

Proof.

Let $A'$ be the reflection of $A$ about $t$: then $M$ is the midpoint of $A'B$. If $N$ is any other point on $t$, different from $M$, then $AN+BN>AM+BM$, because:
$$
AN+BN=A'N+BN>A'B=A'M+BM=AM+BM,
$$
where I applied the triangular inequality to triangle $A'BN$.
Let now $P$ be a point on the circle, different from $M$. Segment $PB$ will intersect $t$ at some point $Q$ and:
$$
AP+BP=AP+PQ+BQ>AQ+BQge AM+BM,
$$
where I applied the triangular inequality to triangle $APQ$ and used the result proved above.

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edited Aug 1 at 9:40

answered Jul 31 at 20:06

Aretino

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This solution is valid only if $A$ and $B$ lie outside the circle. If segment $AB$ intersects the circle, $M$ is obviously either intersection point.

Suppose then $AB$ doesn't intersect the circle: in that case $M$ is the nearest intersection between the circle and the perpendicular bisector of $AB$, such that the circle and $AB$ are on opposite sides with respect to the tangent $t$ at $M$.

Proof.

Let $A'$ be the reflection of $A$ about $t$: then $M$ is the midpoint of $A'B$. If $N$ is any other point on $t$, different from $M$, then $AN+BN>AM+BM$, because:
$$
AN+BN=A'N+BN>A'B=A'M+BM=AM+BM,
$$
where I applied the triangular inequality to triangle $A'BN$.
Let now $P$ be a point on the circle, different from $M$. Segment $PB$ will intersect $t$ at some point $Q$ and:
$$
AP+BP=AP+PQ+BQ>AQ+BQge AM+BM,
$$
where I applied the triangular inequality to triangle $APQ$ and used the result proved above.

share|cite|improve this answer

edited Aug 1 at 9:40

answered Jul 31 at 20:06

Aretino

21.7k21342

This solution is valid only if $A$ and $B$ lie outside the circle. If segment $AB$ intersects the circle, $M$ is obviously either intersection point.

Suppose then $AB$ doesn't intersect the circle: in that case $M$ is the nearest intersection between the circle and the perpendicular bisector of $AB$, such that the circle and $AB$ are on opposite sides with respect to the tangent $t$ at $M$.

Proof.

Let $A'$ be the reflection of $A$ about $t$: then $M$ is the midpoint of $A'B$. If $N$ is any other point on $t$, different from $M$, then $AN+BN>AM+BM$, because:
$$
AN+BN=A'N+BN>A'B=A'M+BM=AM+BM,
$$
where I applied the triangular inequality to triangle $A'BN$.
Let now $P$ be a point on the circle, different from $M$. Segment $PB$ will intersect $t$ at some point $Q$ and:
$$
AP+BP=AP+PQ+BQ>AQ+BQge AM+BM,
$$
where I applied the triangular inequality to triangle $APQ$ and used the result proved above.

share|cite|improve this answer

edited Aug 1 at 9:40

answered Jul 31 at 20:06

Aretino

21.7k21342

share|cite|improve this answer

share|cite|improve this answer

edited Aug 1 at 9:40

edited Aug 1 at 9:40

edited Aug 1 at 9:40

answered Jul 31 at 20:06

Aretino

21.7k21342

answered Jul 31 at 20:06

Aretino

21.7k21342

answered Jul 31 at 20:06

Aretino

21.7k21342

21.7k21342

add a comment | 

add a comment |