Line from incenter bisects side

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I got a problem recently, and have been unable to solve it.

Let $Delta ABC$ with incenter $I$ and the incircle tangent to $BC$ at $D$. Let $M$ be the midpoint of $AD$. Prove that $MI$ bisects $BC$.

I tried to prove that the midpoint of $BC$, $M$, and $I$ are collinear, used Menelaus's theorem, used this particular lemma I'd read recently, and was unable to get anywhere.

Please help.

---

The lemma:

The incircle of $Delta ABC$ is tangent to $BC, CA, AB$ at $D, E, F$ respectively. Let $M$ and $N$ be the midpoints of $BC$ and $AC$. If $K$ is the intersection of lines $BI$ and $EF$, then $BK perp KC$.

geometry contest-math geometric-transformation

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edited Jul 29 at 14:18

greedoid

26.6k93574

asked Jul 16 at 14:40

MalayTheDynamo

1,627833

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @NickD $MI$ is not parallel to $AC$. Counterexample: Equilateral triangle.
  – MalayTheDynamo
  Jul 16 at 19:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The problem is known as the degeneration of Newton's Theorem in tangential quadrilateral
  – RopuToran
  Jul 20 at 7:55
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

1

I got a problem recently, and have been unable to solve it.

Let $Delta ABC$ with incenter $I$ and the incircle tangent to $BC$ at $D$. Let $M$ be the midpoint of $AD$. Prove that $MI$ bisects $BC$.

I tried to prove that the midpoint of $BC$, $M$, and $I$ are collinear, used Menelaus's theorem, used this particular lemma I'd read recently, and was unable to get anywhere.

Please help.

---

The lemma:

The incircle of $Delta ABC$ is tangent to $BC, CA, AB$ at $D, E, F$ respectively. Let $M$ and $N$ be the midpoints of $BC$ and $AC$. If $K$ is the intersection of lines $BI$ and $EF$, then $BK perp KC$.

geometry contest-math geometric-transformation

share|cite|improve this question

edited Jul 29 at 14:18

greedoid

26.6k93574

asked Jul 16 at 14:40

MalayTheDynamo

1,627833

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @NickD $MI$ is not parallel to $AC$. Counterexample: Equilateral triangle.
  – MalayTheDynamo
  Jul 16 at 19:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The problem is known as the degeneration of Newton's Theorem in tangential quadrilateral
  – RopuToran
  Jul 20 at 7:55
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

1

up vote
0
down vote

favorite

1

1

I got a problem recently, and have been unable to solve it.

Let $Delta ABC$ with incenter $I$ and the incircle tangent to $BC$ at $D$. Let $M$ be the midpoint of $AD$. Prove that $MI$ bisects $BC$.

I tried to prove that the midpoint of $BC$, $M$, and $I$ are collinear, used Menelaus's theorem, used this particular lemma I'd read recently, and was unable to get anywhere.

Please help.

---

The lemma:

The incircle of $Delta ABC$ is tangent to $BC, CA, AB$ at $D, E, F$ respectively. Let $M$ and $N$ be the midpoints of $BC$ and $AC$. If $K$ is the intersection of lines $BI$ and $EF$, then $BK perp KC$.

geometry contest-math geometric-transformation

share|cite|improve this question

edited Jul 29 at 14:18

greedoid

26.6k93574

asked Jul 16 at 14:40

MalayTheDynamo

1,627833

I got a problem recently, and have been unable to solve it.

Let $Delta ABC$ with incenter $I$ and the incircle tangent to $BC$ at $D$. Let $M$ be the midpoint of $AD$. Prove that $MI$ bisects $BC$.

I tried to prove that the midpoint of $BC$, $M$, and $I$ are collinear, used Menelaus's theorem, used this particular lemma I'd read recently, and was unable to get anywhere.

Please help.

---

The lemma:

The incircle of $Delta ABC$ is tangent to $BC, CA, AB$ at $D, E, F$ respectively. Let $M$ and $N$ be the midpoints of $BC$ and $AC$. If $K$ is the intersection of lines $BI$ and $EF$, then $BK perp KC$.

geometry contest-math geometric-transformation

share|cite|improve this question

edited Jul 29 at 14:18

greedoid

26.6k93574

asked Jul 16 at 14:40

MalayTheDynamo

1,627833

share|cite|improve this question

share|cite|improve this question

edited Jul 29 at 14:18

greedoid

26.6k93574

edited Jul 29 at 14:18

greedoid

26.6k93574

edited Jul 29 at 14:18

greedoid

26.6k93574

26.6k93574

asked Jul 16 at 14:40

MalayTheDynamo

1,627833

asked Jul 16 at 14:40

MalayTheDynamo

1,627833

asked Jul 16 at 14:40

MalayTheDynamo

1,627833

1,627833

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @NickD $MI$ is not parallel to $AC$. Counterexample: Equilateral triangle.
  – MalayTheDynamo
  Jul 16 at 19:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The problem is known as the degeneration of Newton's Theorem in tangential quadrilateral
  – RopuToran
  Jul 20 at 7:55
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @NickD $MI$ is not parallel to $AC$. Counterexample: Equilateral triangle.
  – MalayTheDynamo
  Jul 16 at 19:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The problem is known as the degeneration of Newton's Theorem in tangential quadrilateral
  – RopuToran
  Jul 20 at 7:55
  

  
  
  
  

@NickD $MI$ is not parallel to $AC$. Counterexample: Equilateral triangle.
– MalayTheDynamo
Jul 16 at 19:12

@NickD $MI$ is not parallel to $AC$. Counterexample: Equilateral triangle.
– MalayTheDynamo
Jul 16 at 19:12

The problem is known as the degeneration of Newton's Theorem in tangential quadrilateral
– RopuToran
Jul 20 at 7:55

The problem is known as the degeneration of Newton's Theorem in tangential quadrilateral
– RopuToran
Jul 20 at 7:55

add a comment | 

4 Answers
4

active

oldest

votes

up vote
1
down vote



accepted

+50

Let $E$ be the reflection of $D$ with respect to $I$. Note that the tangent line $ell$ at $E$ to the incircle is parallel to $BC$. Consider a homothety centered at $A$ which transforms $ell$ to $BC$. Then it maps the incircle to the $A$-excircle, and point $E$ is mapped to the point of tangency of the excircle with $BC$, say $F$.

Since $A,E,F$ are collinear, $M$ is the midpoint of $AD$ and $I$ is the midpoint of $DE$, it follows that the midpoint of $DF$ lies on $MI$. Therefore you need to show that the midpoints of $DE$ and $BC$ coincide. In other words, the task is to prove that $BD=CE$. But this is well-known (and easy to show) that $BD = fracAB+BC-AC2 = CE$.

Let me know whether you need more details.

share|cite|improve this answer

answered Jul 19 at 15:29

timon92

3,7141724

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is embarrassing, but how is $BD=fracAB+BC-AC2=CE$?
  – MalayTheDynamo
  Jul 19 at 16:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Let the incircle be tangent to $AB$ and $AC$ at $X$, $Y$. Then $BD=BX=AB-AX=AB-AY=AB-(AC-CY)=AB-AC+CY=AB-AC+CD=AB-AC+BC-BD,$ so $BD=fracAB+BC-AC2$. You can prove the other equality in the same way.
  – timon92
  Jul 19 at 16:14
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! I'll wait till tomorrow morning in (the improbable) case any better answers come to accept and give the bounty.
  – MalayTheDynamo
  Jul 19 at 16:17
  

  
  
  
  

add a comment | 

up vote
1
down vote

Can you find the proof with the help of the figure below？

share|cite|improve this answer

answered Jul 19 at 15:55

mengdie1982

2,972216

• 
  
  
  

  
  

  
  
  
  
  
  

  
  One of the most elegant solutions!!
  – TryingHardToBecomeAGoodPrSlvr
  Jul 22 at 20:48
  

  
  
  
  

add a comment | 

up vote
0
down vote

Given $|BC|=a$, $|AC|=b$, $|AB|=c$
we can use known expressions to find that

beginalign
I&=fracaA+bB+cCa+b+c
,\
D&=tfrac12(B+C)+fracb-c2acdot(B-C)
,\
M&=tfrac12(A+D)=tfrac12 A+tfrac14(B+C)+fracb-c4acdot(B-C)
,\
E&=tfrac12(B+C)
.
endalign

If $MI$ bisects $BC$, we must always have

beginalign
I&=E(1-t)+Mt
tag1label1
endalign

for some real $tin(0,1)$.

And indeed, eqref1 has one real solution

beginalign
t&=frac2aa+b+c
.
endalign

Since $a,b,c$ are the sides of $triangle ABC$,
beginalign
a&<b+c
,\
2a&<a+b+c
,\
texthence, quad
tin(0,1)
.
endalign

share|cite|improve this answer

answered Jul 19 at 16:36

g.kov

5,5321717

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What do you mean when you write $=textsomething$?
  – MalayTheDynamo
  Jul 19 at 16:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ah, typo. I meant $I=$something. How do you equate a point with an expression?
  – MalayTheDynamo
  Jul 19 at 17:09
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @MalayTheDynamo: We consider the points $A,B,C,D,E,M,I$ either as vectors in $2D$ or as complex numbers, and $a,b,c,t$ as real numbers.
  – g.kov
  Jul 19 at 17:29
  

  
  
  
  

add a comment | 

up vote
0
down vote

In unnormalized barycentric coordinates,
$$I = (a, b, c), \
D = (0, a + b - c, a - b + c), \
M = (2a, a + b - c, a - b + c), \
E = (0, 1, 1).$$
Then the problem reduces to verifying the identity
$$beginvmatrix
0 & 1 & 1 \
a & b & c \
2 a & a + b - c & a - b + c
endvmatrix = 0.$$

share|cite|improve this answer

edited Jul 20 at 2:37

answered Jul 16 at 20:31

Maxim

2,115113

add a comment | 

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

+50

Let $E$ be the reflection of $D$ with respect to $I$. Note that the tangent line $ell$ at $E$ to the incircle is parallel to $BC$. Consider a homothety centered at $A$ which transforms $ell$ to $BC$. Then it maps the incircle to the $A$-excircle, and point $E$ is mapped to the point of tangency of the excircle with $BC$, say $F$.

Since $A,E,F$ are collinear, $M$ is the midpoint of $AD$ and $I$ is the midpoint of $DE$, it follows that the midpoint of $DF$ lies on $MI$. Therefore you need to show that the midpoints of $DE$ and $BC$ coincide. In other words, the task is to prove that $BD=CE$. But this is well-known (and easy to show) that $BD = fracAB+BC-AC2 = CE$.

Let me know whether you need more details.

share|cite|improve this answer

answered Jul 19 at 15:29

timon92

3,7141724

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is embarrassing, but how is $BD=fracAB+BC-AC2=CE$?
  – MalayTheDynamo
  Jul 19 at 16:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Let the incircle be tangent to $AB$ and $AC$ at $X$, $Y$. Then $BD=BX=AB-AX=AB-AY=AB-(AC-CY)=AB-AC+CY=AB-AC+CD=AB-AC+BC-BD,$ so $BD=fracAB+BC-AC2$. You can prove the other equality in the same way.
  – timon92
  Jul 19 at 16:14
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! I'll wait till tomorrow morning in (the improbable) case any better answers come to accept and give the bounty.
  – MalayTheDynamo
  Jul 19 at 16:17
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

+50

Let $E$ be the reflection of $D$ with respect to $I$. Note that the tangent line $ell$ at $E$ to the incircle is parallel to $BC$. Consider a homothety centered at $A$ which transforms $ell$ to $BC$. Then it maps the incircle to the $A$-excircle, and point $E$ is mapped to the point of tangency of the excircle with $BC$, say $F$.

Since $A,E,F$ are collinear, $M$ is the midpoint of $AD$ and $I$ is the midpoint of $DE$, it follows that the midpoint of $DF$ lies on $MI$. Therefore you need to show that the midpoints of $DE$ and $BC$ coincide. In other words, the task is to prove that $BD=CE$. But this is well-known (and easy to show) that $BD = fracAB+BC-AC2 = CE$.

Let me know whether you need more details.

share|cite|improve this answer

answered Jul 19 at 15:29

timon92

3,7141724

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is embarrassing, but how is $BD=fracAB+BC-AC2=CE$?
  – MalayTheDynamo
  Jul 19 at 16:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Let the incircle be tangent to $AB$ and $AC$ at $X$, $Y$. Then $BD=BX=AB-AX=AB-AY=AB-(AC-CY)=AB-AC+CY=AB-AC+CD=AB-AC+BC-BD,$ so $BD=fracAB+BC-AC2$. You can prove the other equality in the same way.
  – timon92
  Jul 19 at 16:14
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! I'll wait till tomorrow morning in (the improbable) case any better answers come to accept and give the bounty.
  – MalayTheDynamo
  Jul 19 at 16:17
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

+50

up vote
1
down vote



accepted

+50

+50

Let $E$ be the reflection of $D$ with respect to $I$. Note that the tangent line $ell$ at $E$ to the incircle is parallel to $BC$. Consider a homothety centered at $A$ which transforms $ell$ to $BC$. Then it maps the incircle to the $A$-excircle, and point $E$ is mapped to the point of tangency of the excircle with $BC$, say $F$.

Since $A,E,F$ are collinear, $M$ is the midpoint of $AD$ and $I$ is the midpoint of $DE$, it follows that the midpoint of $DF$ lies on $MI$. Therefore you need to show that the midpoints of $DE$ and $BC$ coincide. In other words, the task is to prove that $BD=CE$. But this is well-known (and easy to show) that $BD = fracAB+BC-AC2 = CE$.

Let me know whether you need more details.

share|cite|improve this answer

answered Jul 19 at 15:29

timon92

3,7141724

Let $E$ be the reflection of $D$ with respect to $I$. Note that the tangent line $ell$ at $E$ to the incircle is parallel to $BC$. Consider a homothety centered at $A$ which transforms $ell$ to $BC$. Then it maps the incircle to the $A$-excircle, and point $E$ is mapped to the point of tangency of the excircle with $BC$, say $F$.

Since $A,E,F$ are collinear, $M$ is the midpoint of $AD$ and $I$ is the midpoint of $DE$, it follows that the midpoint of $DF$ lies on $MI$. Therefore you need to show that the midpoints of $DE$ and $BC$ coincide. In other words, the task is to prove that $BD=CE$. But this is well-known (and easy to show) that $BD = fracAB+BC-AC2 = CE$.

Let me know whether you need more details.

share|cite|improve this answer

answered Jul 19 at 15:29

timon92

3,7141724

share|cite|improve this answer

share|cite|improve this answer

answered Jul 19 at 15:29

timon92

3,7141724

answered Jul 19 at 15:29

timon92

3,7141724

answered Jul 19 at 15:29

timon92

3,7141724

3,7141724

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is embarrassing, but how is $BD=fracAB+BC-AC2=CE$?
  – MalayTheDynamo
  Jul 19 at 16:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Let the incircle be tangent to $AB$ and $AC$ at $X$, $Y$. Then $BD=BX=AB-AX=AB-AY=AB-(AC-CY)=AB-AC+CY=AB-AC+CD=AB-AC+BC-BD,$ so $BD=fracAB+BC-AC2$. You can prove the other equality in the same way.
  – timon92
  Jul 19 at 16:14
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! I'll wait till tomorrow morning in (the improbable) case any better answers come to accept and give the bounty.
  – MalayTheDynamo
  Jul 19 at 16:17
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is embarrassing, but how is $BD=fracAB+BC-AC2=CE$?
  – MalayTheDynamo
  Jul 19 at 16:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Let the incircle be tangent to $AB$ and $AC$ at $X$, $Y$. Then $BD=BX=AB-AX=AB-AY=AB-(AC-CY)=AB-AC+CY=AB-AC+CD=AB-AC+BC-BD,$ so $BD=fracAB+BC-AC2$. You can prove the other equality in the same way.
  – timon92
  Jul 19 at 16:14
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! I'll wait till tomorrow morning in (the improbable) case any better answers come to accept and give the bounty.
  – MalayTheDynamo
  Jul 19 at 16:17
  

  
  
  
  

This is embarrassing, but how is $BD=fracAB+BC-AC2=CE$?
– MalayTheDynamo
Jul 19 at 16:09

This is embarrassing, but how is $BD=fracAB+BC-AC2=CE$?
– MalayTheDynamo
Jul 19 at 16:09

Let the incircle be tangent to $AB$ and $AC$ at $X$, $Y$. Then $BD=BX=AB-AX=AB-AY=AB-(AC-CY)=AB-AC+CY=AB-AC+CD=AB-AC+BC-BD,$ so $BD=fracAB+BC-AC2$. You can prove the other equality in the same way.
– timon92
Jul 19 at 16:14

Let the incircle be tangent to $AB$ and $AC$ at $X$, $Y$. Then $BD=BX=AB-AX=AB-AY=AB-(AC-CY)=AB-AC+CY=AB-AC+CD=AB-AC+BC-BD,$ so $BD=fracAB+BC-AC2$. You can prove the other equality in the same way.
– timon92
Jul 19 at 16:14

Thanks! I'll wait till tomorrow morning in (the improbable) case any better answers come to accept and give the bounty.
– MalayTheDynamo
Jul 19 at 16:17

Thanks! I'll wait till tomorrow morning in (the improbable) case any better answers come to accept and give the bounty.
– MalayTheDynamo
Jul 19 at 16:17

add a comment | 

up vote
1
down vote

Can you find the proof with the help of the figure below？

share|cite|improve this answer

answered Jul 19 at 15:55

mengdie1982

2,972216

• 
  
  
  

  
  

  
  
  
  
  
  

  
  One of the most elegant solutions!!
  – TryingHardToBecomeAGoodPrSlvr
  Jul 22 at 20:48
  

  
  
  
  

add a comment | 

up vote
1
down vote

Can you find the proof with the help of the figure below？

share|cite|improve this answer

answered Jul 19 at 15:55

mengdie1982

2,972216

• 
  
  
  

  
  

  
  
  
  
  
  

  
  One of the most elegant solutions!!
  – TryingHardToBecomeAGoodPrSlvr
  Jul 22 at 20:48
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

Can you find the proof with the help of the figure below？

share|cite|improve this answer

answered Jul 19 at 15:55

mengdie1982

2,972216

Can you find the proof with the help of the figure below？

share|cite|improve this answer

answered Jul 19 at 15:55

mengdie1982

2,972216

share|cite|improve this answer

share|cite|improve this answer

answered Jul 19 at 15:55

mengdie1982

2,972216

answered Jul 19 at 15:55

mengdie1982

2,972216

answered Jul 19 at 15:55

mengdie1982

2,972216

2,972216

• 
  
  
  

  
  

  
  
  
  
  
  

  
  One of the most elegant solutions!!
  – TryingHardToBecomeAGoodPrSlvr
  Jul 22 at 20:48
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  One of the most elegant solutions!!
  – TryingHardToBecomeAGoodPrSlvr
  Jul 22 at 20:48
  

  
  
  
  

One of the most elegant solutions!!
– TryingHardToBecomeAGoodPrSlvr
Jul 22 at 20:48

One of the most elegant solutions!!
– TryingHardToBecomeAGoodPrSlvr
Jul 22 at 20:48

add a comment | 

up vote
0
down vote

Given $|BC|=a$, $|AC|=b$, $|AB|=c$
we can use known expressions to find that

beginalign
I&=fracaA+bB+cCa+b+c
,\
D&=tfrac12(B+C)+fracb-c2acdot(B-C)
,\
M&=tfrac12(A+D)=tfrac12 A+tfrac14(B+C)+fracb-c4acdot(B-C)
,\
E&=tfrac12(B+C)
.
endalign

If $MI$ bisects $BC$, we must always have

beginalign
I&=E(1-t)+Mt
tag1label1
endalign

for some real $tin(0,1)$.

And indeed, eqref1 has one real solution

beginalign
t&=frac2aa+b+c
.
endalign

Since $a,b,c$ are the sides of $triangle ABC$,
beginalign
a&<b+c
,\
2a&<a+b+c
,\
texthence, quad
tin(0,1)
.
endalign

share|cite|improve this answer

answered Jul 19 at 16:36

g.kov

5,5321717

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What do you mean when you write $=textsomething$?
  – MalayTheDynamo
  Jul 19 at 16:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ah, typo. I meant $I=$something. How do you equate a point with an expression?
  – MalayTheDynamo
  Jul 19 at 17:09
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @MalayTheDynamo: We consider the points $A,B,C,D,E,M,I$ either as vectors in $2D$ or as complex numbers, and $a,b,c,t$ as real numbers.
  – g.kov
  Jul 19 at 17:29
  

  
  
  
  

add a comment | 

up vote
0
down vote

Given $|BC|=a$, $|AC|=b$, $|AB|=c$
we can use known expressions to find that

beginalign
I&=fracaA+bB+cCa+b+c
,\
D&=tfrac12(B+C)+fracb-c2acdot(B-C)
,\
M&=tfrac12(A+D)=tfrac12 A+tfrac14(B+C)+fracb-c4acdot(B-C)
,\
E&=tfrac12(B+C)
.
endalign

If $MI$ bisects $BC$, we must always have

beginalign
I&=E(1-t)+Mt
tag1label1
endalign

for some real $tin(0,1)$.

And indeed, eqref1 has one real solution

beginalign
t&=frac2aa+b+c
.
endalign

Since $a,b,c$ are the sides of $triangle ABC$,
beginalign
a&<b+c
,\
2a&<a+b+c
,\
texthence, quad
tin(0,1)
.
endalign

share|cite|improve this answer

answered Jul 19 at 16:36

g.kov

5,5321717

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What do you mean when you write $=textsomething$?
  – MalayTheDynamo
  Jul 19 at 16:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ah, typo. I meant $I=$something. How do you equate a point with an expression?
  – MalayTheDynamo
  Jul 19 at 17:09
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @MalayTheDynamo: We consider the points $A,B,C,D,E,M,I$ either as vectors in $2D$ or as complex numbers, and $a,b,c,t$ as real numbers.
  – g.kov
  Jul 19 at 17:29
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

Given $|BC|=a$, $|AC|=b$, $|AB|=c$
we can use known expressions to find that

beginalign
I&=fracaA+bB+cCa+b+c
,\
D&=tfrac12(B+C)+fracb-c2acdot(B-C)
,\
M&=tfrac12(A+D)=tfrac12 A+tfrac14(B+C)+fracb-c4acdot(B-C)
,\
E&=tfrac12(B+C)
.
endalign

If $MI$ bisects $BC$, we must always have

beginalign
I&=E(1-t)+Mt
tag1label1
endalign

for some real $tin(0,1)$.

And indeed, eqref1 has one real solution

beginalign
t&=frac2aa+b+c
.
endalign

Since $a,b,c$ are the sides of $triangle ABC$,
beginalign
a&<b+c
,\
2a&<a+b+c
,\
texthence, quad
tin(0,1)
.
endalign

share|cite|improve this answer

answered Jul 19 at 16:36

g.kov

5,5321717

Given $|BC|=a$, $|AC|=b$, $|AB|=c$
we can use known expressions to find that

beginalign
I&=fracaA+bB+cCa+b+c
,\
D&=tfrac12(B+C)+fracb-c2acdot(B-C)
,\
M&=tfrac12(A+D)=tfrac12 A+tfrac14(B+C)+fracb-c4acdot(B-C)
,\
E&=tfrac12(B+C)
.
endalign

If $MI$ bisects $BC$, we must always have

beginalign
I&=E(1-t)+Mt
tag1label1
endalign

for some real $tin(0,1)$.

And indeed, eqref1 has one real solution

beginalign
t&=frac2aa+b+c
.
endalign

Since $a,b,c$ are the sides of $triangle ABC$,
beginalign
a&<b+c
,\
2a&<a+b+c
,\
texthence, quad
tin(0,1)
.
endalign

share|cite|improve this answer

answered Jul 19 at 16:36

g.kov

5,5321717

share|cite|improve this answer

share|cite|improve this answer

answered Jul 19 at 16:36

g.kov

5,5321717

answered Jul 19 at 16:36

g.kov

5,5321717

answered Jul 19 at 16:36

g.kov

5,5321717

5,5321717

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What do you mean when you write $=textsomething$?
  – MalayTheDynamo
  Jul 19 at 16:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ah, typo. I meant $I=$something. How do you equate a point with an expression?
  – MalayTheDynamo
  Jul 19 at 17:09
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @MalayTheDynamo: We consider the points $A,B,C,D,E,M,I$ either as vectors in $2D$ or as complex numbers, and $a,b,c,t$ as real numbers.
  – g.kov
  Jul 19 at 17:29
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What do you mean when you write $=textsomething$?
  – MalayTheDynamo
  Jul 19 at 16:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ah, typo. I meant $I=$something. How do you equate a point with an expression?
  – MalayTheDynamo
  Jul 19 at 17:09
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @MalayTheDynamo: We consider the points $A,B,C,D,E,M,I$ either as vectors in $2D$ or as complex numbers, and $a,b,c,t$ as real numbers.
  – g.kov
  Jul 19 at 17:29
  

  
  
  
  

What do you mean when you write $=textsomething$?
– MalayTheDynamo
Jul 19 at 16:38

What do you mean when you write $=textsomething$?
– MalayTheDynamo
Jul 19 at 16:38

Ah, typo. I meant $I=$something. How do you equate a point with an expression?
– MalayTheDynamo
Jul 19 at 17:09

Ah, typo. I meant $I=$something. How do you equate a point with an expression?
– MalayTheDynamo
Jul 19 at 17:09

1

1

@MalayTheDynamo: We consider the points $A,B,C,D,E,M,I$ either as vectors in $2D$ or as complex numbers, and $a,b,c,t$ as real numbers.
– g.kov
Jul 19 at 17:29

@MalayTheDynamo: We consider the points $A,B,C,D,E,M,I$ either as vectors in $2D$ or as complex numbers, and $a,b,c,t$ as real numbers.
– g.kov
Jul 19 at 17:29

add a comment | 

up vote
0
down vote

In unnormalized barycentric coordinates,
$$I = (a, b, c), \
D = (0, a + b - c, a - b + c), \
M = (2a, a + b - c, a - b + c), \
E = (0, 1, 1).$$
Then the problem reduces to verifying the identity
$$beginvmatrix
0 & 1 & 1 \
a & b & c \
2 a & a + b - c & a - b + c
endvmatrix = 0.$$

share|cite|improve this answer

edited Jul 20 at 2:37

answered Jul 16 at 20:31

Maxim

2,115113

add a comment | 

up vote
0
down vote

In unnormalized barycentric coordinates,
$$I = (a, b, c), \
D = (0, a + b - c, a - b + c), \
M = (2a, a + b - c, a - b + c), \
E = (0, 1, 1).$$
Then the problem reduces to verifying the identity
$$beginvmatrix
0 & 1 & 1 \
a & b & c \
2 a & a + b - c & a - b + c
endvmatrix = 0.$$

share|cite|improve this answer

edited Jul 20 at 2:37

answered Jul 16 at 20:31

Maxim

2,115113

add a comment | 

up vote
0
down vote

up vote
0
down vote

In unnormalized barycentric coordinates,
$$I = (a, b, c), \
D = (0, a + b - c, a - b + c), \
M = (2a, a + b - c, a - b + c), \
E = (0, 1, 1).$$
Then the problem reduces to verifying the identity
$$beginvmatrix
0 & 1 & 1 \
a & b & c \
2 a & a + b - c & a - b + c
endvmatrix = 0.$$

share|cite|improve this answer

edited Jul 20 at 2:37

answered Jul 16 at 20:31

Maxim

2,115113

In unnormalized barycentric coordinates,
$$I = (a, b, c), \
D = (0, a + b - c, a - b + c), \
M = (2a, a + b - c, a - b + c), \
E = (0, 1, 1).$$
Then the problem reduces to verifying the identity
$$beginvmatrix
0 & 1 & 1 \
a & b & c \
2 a & a + b - c & a - b + c
endvmatrix = 0.$$

share|cite|improve this answer

edited Jul 20 at 2:37

answered Jul 16 at 20:31

Maxim

2,115113

share|cite|improve this answer

share|cite|improve this answer

edited Jul 20 at 2:37

edited Jul 20 at 2:37

edited Jul 20 at 2:37

answered Jul 16 at 20:31

Maxim

2,115113

answered Jul 16 at 20:31

Maxim

2,115113

answered Jul 16 at 20:31

Maxim

2,115113

2,115113

add a comment | 

add a comment | 

 

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