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ABC is a triangle. D, E and F are the respective middles of segments [AB], [AC] and [BC].

A line d passing through A intersects with (DE) in G, a line d' passing through C intersects with (EF) in H.

Under what condition are lines (AG) and (CH) parallel?

 

You can get to an hypothesis, then prove it, using the Cartesian coordinate system \((\vec{ED},\vec{EF})\)of origin E.

Good luck!

Nota Bene: The problem was originally in French, I had to translate it by myself. I apologize if the translation is wrong or if there are mistakes in notations or something. If you cannot do the exercise because of this, say it in the comments, I'll do my best to fix it. Thanks.

EinsteinJr  Nov 20, 2015

Best Answer 

 #2
avatar+18715 
+20

ABC is a triangle. D, E and F are the respective middles of segments [AB], [AC] and [BC].

A line d passing through A intersects with (DE) in G, a line d' passing through C intersects with (EF) in H.

Under what condition are lines (AG) and (CH) parallel?

You can get to an hypothesis, then prove it, using the Cartesian coordinate system of origin E. Good luck!

Nota Bene: The problem was originally in French, I had to translate it by myself. I apologize if the translation is wrong or if there are mistakes in notations or something. If you cannot do the exercise because of this, say it in the comments, I'll do my best to fix it. Thanks.

 

I. Ths sides of the triangle ABC are: a = (BC), b = (AC) , c = (AB)

 

II.  (DE) \(\parallel\) (BC) and (EF) \(\parallel\) (AB).

Proof:

The angles of the triangle are: A = BAC,  B = CBA, C = ACB

1. Cos-Rule:

Let x = (DE) then: \(\begin{array}{rcll} (1): & \quad x^2 &=& (\frac{b}{2})^2 + (\frac{c}{2})^2 - 2\frac{b}{2}\frac{c}{2}\cdot\cos{(A)} \qquad | \qquad \cdot 4\\ & 4x^2 &=& b^2 +c^2 - 2bc\cdot\cos{(A)} \\ (2): & a^2 &=& b^2 +c^2 - 2bc\cdot\cos{(A)} \\ \hline \text{compare:} & 4x^2 &=& a^2 \\ \text{we find:} & x &=& \frac{a}{2}\\ \text{or} & (DE) &=& \frac{(BC)}{2} \end{array}\)

 

2. Sin-Rule:

Let \(\epsilon\) = angle ADE then: \(\begin{array}{rcll} (1): & \quad \frac{\sin{(\epsilon)} } {\frac{b}{2} } &=& \frac{\sin{(A)} } { \frac{a}{2} } \\ & \quad \frac{\sin{(\epsilon)} } { b } &=& \frac{\sin{(A)} } { a } \\ (2): & \quad \frac{\sin{(B)} } { b } &=& \frac{\sin{(A)} } { a } \\ \hline \text{compare:} & \sin{(\epsilon)} &=& \sin{(B)}\\ \text{we find:} & \epsilon &=& B\\ \end{array}\)

 

so we have (DE) \(\parallel\) (BC)

\(\begin{array}{rcll} \text{permute and we have: } \quad (EF) &=& \frac{(AB)}{2} = \frac{(c)}{2} \\ \text{and angle } CFE &=& B\\ \text{so we have also } \quad \mathbf{(FE) \parallel (AB)} \end{array}\)

 

III.

\(\begin{array}{rcll} \vec{a} &=& \vec{B} - \vec{C}\\ \vec{c} &=& \vec{B} - \vec{A}\\ \vec{GE} = \vec{G} - \vec{E} &=& \lambda \cdot \vec{a} \\ \vec{HE} = \vec{H} - \vec{E} &=& \mu \cdot \vec{c} \\ \vec{AE} = \vec{A} - \vec{E} &=& \frac{ \vec{a} }{2} - \frac{ \vec{c} }{2} \\ \vec{CE} = \vec{C} - \vec{E} &=& \frac{ \vec{c} }{2} - \frac{ \vec{a} }{2} \\ \hline \vec{d} &=& \vec{GE} - \vec{AE}\\ \vec{d'} &=& \vec{HE} - \vec{CE}\\ \hline \vec{d} = \lambda \cdot \vec{a} - \frac{ \vec{a} }{2} + \frac{ \vec{c} }{2} \\ \vec{d'} = \mu \cdot \vec{c} - \frac{ \vec{c} }{2} + \frac{ \vec{a} }{2} \\ \mathbf{If } \quad |\vec{d} \times \vec{d'} | = 0 \quad \text{ then } \quad \vec{d} \parallel \vec{d'}\\ \hline | \vec{d} \times \vec{d'} | &=& 0 \\ |(\lambda \cdot \vec{a} - \frac{ \vec{a} }{2} + \frac{ \vec{c} }{2}) \times (\mu \cdot \vec{c} - \frac{ \vec{c} }{2} + \frac{ \vec{a} }{2} )| &=& 0 \\ \lambda \mu | \vec{a}\times \vec{c} | - \frac{\lambda}{2 } |\vec{a}\times \vec{c} | + \underbrace{\frac{\lambda}{2 } |\vec{a}\times {\vec{a} }| }_{\text{area}=0} \\ - \frac{\mu}{2 } |\vec{a}\times \vec{c} | + \frac{1}{4 } |\vec{a}\times \vec{c} | - \underbrace{\frac{1}{4 } |\vec{a}\times \vec{a}| }_{\text{area}=0}\\ + \underbrace{\frac{\mu}{2 } |\vec{c}\times \vec{c} | }_{\text{area}=0} - \underbrace{\frac{1}{4 } |\vec{c}\times \vec{c} | }_{\text{area}=0} + \frac{1}{4 } |\vec{c}\times \vec{a} | &=& 0 \\ \lambda \mu | \vec{a}\times \vec{c} | - \frac{\lambda}{2 } |\vec{a}\times \vec{c} | - \frac{\mu}{2 } |\vec{a}\times \vec{c} | \\ + \frac{1}{4 } |\vec{a}\times \vec{c} | + \frac{1}{4 } |\vec{c}\times \vec{a} | &=& 0 \\ \lambda \mu | \vec{a}\times \vec{c} | - \frac{\lambda}{2 } |\vec{a}\times \vec{c} | - \frac{\mu}{2 } |\vec{a}\times \vec{c} | \\ + \frac{1}{4 } |\vec{a}\times \vec{c} | - \frac{1}{4 } |\vec{a}\times \vec{c} | &=& 0 \\ \lambda \mu | \vec{a}\times \vec{c} | - \frac{\lambda}{2 } |\vec{a}\times \vec{c} | - \frac{\mu}{2 } |\vec{a}\times \vec{c} | &=& 0 \\ \hline \underbrace{| \vec{a}\times \vec{c} |}_{\ne 0} \underbrace{(\lambda \mu- \frac{\lambda}{2 }- \frac{\mu}{2 } )}_{=0} &=& 0 \\ \mathbf{ \lambda \mu- \frac{\lambda}{2}- \frac{\mu}{2} }&\mathbf{=}& \mathbf{0}\\ \end{array} \)

 

\(\begin{array}{rcll} \boxed{~ \begin{array}{lrcl} & \lambda \mu- \frac{\lambda}{2}- \frac{\mu}{2} & = & 0 \\ & 2\lambda \mu &=& \lambda - \mu \\ \hline & 2\lambda -1 &=& \frac{\lambda}{\mu} \\ & \color{red} \mu & \color{red}=& \color{red}\frac{\lambda}{2\lambda -1} \\ \text{or } & 2\mu -1 &=& \frac{\mu}{\lambda} \\ & \color{red} \lambda & \color{red}=& \color{red}\frac{\mu}{2\mu -1} \\ \text{or } & (2\lambda -1)(2\mu -1) &=& 1 \\ \end{array} ~} \end{array}\)

 

IV. Solution

\(\begin{array}{rcll} \vec{a} &=& \vec{B} - \vec{C}\\ \vec{c} &=& \vec{B} - \vec{A}\\ \vec{GE} = \vec{G} - \vec{E} &=& \lambda \cdot \vec{a} \\ \vec{HE} = \vec{H} - \vec{E} &=& \mu \cdot \vec{c} \\ \hline \boxed{~ \begin{array}{rcll} \vec{GE} = \vec{G} - \vec{E} &=& \lambda \cdot \vec{a} \\ \vec{HE} = \vec{H} - \vec{E} &=& \frac{\lambda}{2\lambda -1} \cdot \vec{c} \\ \vec{GE} = \vec{G} - \vec{E} &=& \lambda \cdot ( \vec{B} - \vec{C} ) \\ \vec{HE} = \vec{H} - \vec{E} &=& \frac{\lambda}{2\lambda -1} \cdot ( \vec{B} - \vec{A} ) \\ d \parallel d' \text{ or } \vec{(GA)} \parallel \vec{(HC)} \end{array} ~} \end{array} \)

 

laugh

heureka  Nov 23, 2015
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4+0 Answers

 #1
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ABC

Guest Nov 20, 2015
 #2
avatar+18715 
+20
Best Answer

ABC is a triangle. D, E and F are the respective middles of segments [AB], [AC] and [BC].

A line d passing through A intersects with (DE) in G, a line d' passing through C intersects with (EF) in H.

Under what condition are lines (AG) and (CH) parallel?

You can get to an hypothesis, then prove it, using the Cartesian coordinate system of origin E. Good luck!

Nota Bene: The problem was originally in French, I had to translate it by myself. I apologize if the translation is wrong or if there are mistakes in notations or something. If you cannot do the exercise because of this, say it in the comments, I'll do my best to fix it. Thanks.

 

I. Ths sides of the triangle ABC are: a = (BC), b = (AC) , c = (AB)

 

II.  (DE) \(\parallel\) (BC) and (EF) \(\parallel\) (AB).

Proof:

The angles of the triangle are: A = BAC,  B = CBA, C = ACB

1. Cos-Rule:

Let x = (DE) then: \(\begin{array}{rcll} (1): & \quad x^2 &=& (\frac{b}{2})^2 + (\frac{c}{2})^2 - 2\frac{b}{2}\frac{c}{2}\cdot\cos{(A)} \qquad | \qquad \cdot 4\\ & 4x^2 &=& b^2 +c^2 - 2bc\cdot\cos{(A)} \\ (2): & a^2 &=& b^2 +c^2 - 2bc\cdot\cos{(A)} \\ \hline \text{compare:} & 4x^2 &=& a^2 \\ \text{we find:} & x &=& \frac{a}{2}\\ \text{or} & (DE) &=& \frac{(BC)}{2} \end{array}\)

 

2. Sin-Rule:

Let \(\epsilon\) = angle ADE then: \(\begin{array}{rcll} (1): & \quad \frac{\sin{(\epsilon)} } {\frac{b}{2} } &=& \frac{\sin{(A)} } { \frac{a}{2} } \\ & \quad \frac{\sin{(\epsilon)} } { b } &=& \frac{\sin{(A)} } { a } \\ (2): & \quad \frac{\sin{(B)} } { b } &=& \frac{\sin{(A)} } { a } \\ \hline \text{compare:} & \sin{(\epsilon)} &=& \sin{(B)}\\ \text{we find:} & \epsilon &=& B\\ \end{array}\)

 

so we have (DE) \(\parallel\) (BC)

\(\begin{array}{rcll} \text{permute and we have: } \quad (EF) &=& \frac{(AB)}{2} = \frac{(c)}{2} \\ \text{and angle } CFE &=& B\\ \text{so we have also } \quad \mathbf{(FE) \parallel (AB)} \end{array}\)

 

III.

\(\begin{array}{rcll} \vec{a} &=& \vec{B} - \vec{C}\\ \vec{c} &=& \vec{B} - \vec{A}\\ \vec{GE} = \vec{G} - \vec{E} &=& \lambda \cdot \vec{a} \\ \vec{HE} = \vec{H} - \vec{E} &=& \mu \cdot \vec{c} \\ \vec{AE} = \vec{A} - \vec{E} &=& \frac{ \vec{a} }{2} - \frac{ \vec{c} }{2} \\ \vec{CE} = \vec{C} - \vec{E} &=& \frac{ \vec{c} }{2} - \frac{ \vec{a} }{2} \\ \hline \vec{d} &=& \vec{GE} - \vec{AE}\\ \vec{d'} &=& \vec{HE} - \vec{CE}\\ \hline \vec{d} = \lambda \cdot \vec{a} - \frac{ \vec{a} }{2} + \frac{ \vec{c} }{2} \\ \vec{d'} = \mu \cdot \vec{c} - \frac{ \vec{c} }{2} + \frac{ \vec{a} }{2} \\ \mathbf{If } \quad |\vec{d} \times \vec{d'} | = 0 \quad \text{ then } \quad \vec{d} \parallel \vec{d'}\\ \hline | \vec{d} \times \vec{d'} | &=& 0 \\ |(\lambda \cdot \vec{a} - \frac{ \vec{a} }{2} + \frac{ \vec{c} }{2}) \times (\mu \cdot \vec{c} - \frac{ \vec{c} }{2} + \frac{ \vec{a} }{2} )| &=& 0 \\ \lambda \mu | \vec{a}\times \vec{c} | - \frac{\lambda}{2 } |\vec{a}\times \vec{c} | + \underbrace{\frac{\lambda}{2 } |\vec{a}\times {\vec{a} }| }_{\text{area}=0} \\ - \frac{\mu}{2 } |\vec{a}\times \vec{c} | + \frac{1}{4 } |\vec{a}\times \vec{c} | - \underbrace{\frac{1}{4 } |\vec{a}\times \vec{a}| }_{\text{area}=0}\\ + \underbrace{\frac{\mu}{2 } |\vec{c}\times \vec{c} | }_{\text{area}=0} - \underbrace{\frac{1}{4 } |\vec{c}\times \vec{c} | }_{\text{area}=0} + \frac{1}{4 } |\vec{c}\times \vec{a} | &=& 0 \\ \lambda \mu | \vec{a}\times \vec{c} | - \frac{\lambda}{2 } |\vec{a}\times \vec{c} | - \frac{\mu}{2 } |\vec{a}\times \vec{c} | \\ + \frac{1}{4 } |\vec{a}\times \vec{c} | + \frac{1}{4 } |\vec{c}\times \vec{a} | &=& 0 \\ \lambda \mu | \vec{a}\times \vec{c} | - \frac{\lambda}{2 } |\vec{a}\times \vec{c} | - \frac{\mu}{2 } |\vec{a}\times \vec{c} | \\ + \frac{1}{4 } |\vec{a}\times \vec{c} | - \frac{1}{4 } |\vec{a}\times \vec{c} | &=& 0 \\ \lambda \mu | \vec{a}\times \vec{c} | - \frac{\lambda}{2 } |\vec{a}\times \vec{c} | - \frac{\mu}{2 } |\vec{a}\times \vec{c} | &=& 0 \\ \hline \underbrace{| \vec{a}\times \vec{c} |}_{\ne 0} \underbrace{(\lambda \mu- \frac{\lambda}{2 }- \frac{\mu}{2 } )}_{=0} &=& 0 \\ \mathbf{ \lambda \mu- \frac{\lambda}{2}- \frac{\mu}{2} }&\mathbf{=}& \mathbf{0}\\ \end{array} \)

 

\(\begin{array}{rcll} \boxed{~ \begin{array}{lrcl} & \lambda \mu- \frac{\lambda}{2}- \frac{\mu}{2} & = & 0 \\ & 2\lambda \mu &=& \lambda - \mu \\ \hline & 2\lambda -1 &=& \frac{\lambda}{\mu} \\ & \color{red} \mu & \color{red}=& \color{red}\frac{\lambda}{2\lambda -1} \\ \text{or } & 2\mu -1 &=& \frac{\mu}{\lambda} \\ & \color{red} \lambda & \color{red}=& \color{red}\frac{\mu}{2\mu -1} \\ \text{or } & (2\lambda -1)(2\mu -1) &=& 1 \\ \end{array} ~} \end{array}\)

 

IV. Solution

\(\begin{array}{rcll} \vec{a} &=& \vec{B} - \vec{C}\\ \vec{c} &=& \vec{B} - \vec{A}\\ \vec{GE} = \vec{G} - \vec{E} &=& \lambda \cdot \vec{a} \\ \vec{HE} = \vec{H} - \vec{E} &=& \mu \cdot \vec{c} \\ \hline \boxed{~ \begin{array}{rcll} \vec{GE} = \vec{G} - \vec{E} &=& \lambda \cdot \vec{a} \\ \vec{HE} = \vec{H} - \vec{E} &=& \frac{\lambda}{2\lambda -1} \cdot \vec{c} \\ \vec{GE} = \vec{G} - \vec{E} &=& \lambda \cdot ( \vec{B} - \vec{C} ) \\ \vec{HE} = \vec{H} - \vec{E} &=& \frac{\lambda}{2\lambda -1} \cdot ( \vec{B} - \vec{A} ) \\ d \parallel d' \text{ or } \vec{(GA)} \parallel \vec{(HC)} \end{array} ~} \end{array} \)

 

laugh

heureka  Nov 23, 2015
 #3
avatar+91051 
+10

Brilliant Heureka :))

You have got my 5 points but you should add your own as well :

In fact, you should for all your answers    laugh 

Melody  Nov 23, 2015
 #4
avatar+886 
+10

Thanks heureka; your answer seems (and is) exact.

You got 20/20 plus a brownie:

EinsteinJr  Nov 23, 2015

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