Linear relation inside of a triangle
$begingroup$
Let $d(P,l)$ be distance between point $P$ and line $l$. Inside triangle $ABC$ there are points $D$ and $E$, for which
$$
d(D,AC)+d(D,BC)=d(D,AB),
$$
$$
d(E,AC)+d(E,BC)=d(E,AB).
$$
Prove that $d(X,AC)+d(X,BC)=d(X,AB)$ for any point $X$ on segment $DE$.
Looks like there is a laborious analytical proof relying on the coordinate method. But could there be an easier solution, that is also purely geometric?
geometry triangle
asked Dec 25 '18 at 21:23
user75619user75619
344113
$endgroup$
add a comment |
$begingroup$
Let $d(P,l)$ be distance between point $P$ and line $l$. Inside triangle $ABC$ there are points $D$ and $E$, for which
$$
d(D,AC)+d(D,BC)=d(D,AB),
$$
$$
d(E,AC)+d(E,BC)=d(E,AB).
$$
Prove that $d(X,AC)+d(X,BC)=d(X,AB)$ for any point $X$ on segment $DE$.
Looks like there is a laborious analytical proof relying on the coordinate method. But could there be an easier solution, that is also purely geometric?
geometry triangle
asked Dec 25 '18 at 21:23
user75619user75619
344113
$endgroup$
1
$begingroup$
Have you drawn it out? Where did you find this problem statement? What geometric work have you done to crack this?
$endgroup$
– Namaste
Dec 25 '18 at 21:27
$begingroup$
@amWhy Official solution for one of the IMO problems. They use this without giving a proof, simply saying that they rely on linearity of the relation in variable $X$. I have drawn it out, but have no idea how to approach this geometrically...
$endgroup$
– user75619
Dec 25 '18 at 21:35
2
$begingroup$
I can't answer the question of whether there's a purely geometric proof, but do you know what "linearity" means here? That argument is worth understanding and is fast once one is comfortable with the terms.
$endgroup$
– Milo Brandt
Dec 25 '18 at 21:41
$begingroup$
@MiloBrandt I don't. I was actually hoping I would understand it from a geometric proof...
$endgroup$
– user75619
Dec 25 '18 at 21:47
2
$begingroup$
Try to start with this (forget E, D for the moment): exactly one point on AC and exactly one on BC verify the condition, denote the points P, Q. Then explore the segment PQ.
$endgroup$
– user376343
Dec 25 '18 at 23:47
add a comment |
$begingroup$
Let $d(P,l)$ be distance between point $P$ and line $l$. Inside triangle $ABC$ there are points $D$ and $E$, for which
$$
d(D,AC)+d(D,BC)=d(D,AB),
$$
$$
d(E,AC)+d(E,BC)=d(E,AB).
$$
Prove that $d(X,AC)+d(X,BC)=d(X,AB)$ for any point $X$ on segment $DE$.
Looks like there is a laborious analytical proof relying on the coordinate method. But could there be an easier solution, that is also purely geometric?
geometry triangle
asked Dec 25 '18 at 21:23
user75619user75619
344113
$endgroup$
Let $d(P,l)$ be distance between point $P$ and line $l$. Inside triangle $ABC$ there are points $D$ and $E$, for which
$$
d(D,AC)+d(D,BC)=d(D,AB),
$$
$$
d(E,AC)+d(E,BC)=d(E,AB).
$$
Prove that $d(X,AC)+d(X,BC)=d(X,AB)$ for any point $X$ on segment $DE$.
Looks like there is a laborious analytical proof relying on the coordinate method. But could there be an easier solution, that is also purely geometric?
geometry triangle
geometry triangle
asked Dec 25 '18 at 21:23
user75619user75619
344113
asked Dec 25 '18 at 21:23
user75619user75619
344113
asked Dec 25 '18 at 21:23
user75619user75619
344113
asked Dec 25 '18 at 21:23
user75619user75619
344113
asked Dec 25 '18 at 21:23
user75619user75619
344113
344113
1
$begingroup$
Have you drawn it out? Where did you find this problem statement? What geometric work have you done to crack this?
$endgroup$
– Namaste
Dec 25 '18 at 21:27
$begingroup$
@amWhy Official solution for one of the IMO problems. They use this without giving a proof, simply saying that they rely on linearity of the relation in variable $X$. I have drawn it out, but have no idea how to approach this geometrically...
$endgroup$
– user75619
Dec 25 '18 at 21:35
2
$begingroup$
I can't answer the question of whether there's a purely geometric proof, but do you know what "linearity" means here? That argument is worth understanding and is fast once one is comfortable with the terms.
$endgroup$
– Milo Brandt
Dec 25 '18 at 21:41
$begingroup$
@MiloBrandt I don't. I was actually hoping I would understand it from a geometric proof...
$endgroup$
– user75619
Dec 25 '18 at 21:47
2
$begingroup$
Try to start with this (forget E, D for the moment): exactly one point on AC and exactly one on BC verify the condition, denote the points P, Q. Then explore the segment PQ.
$endgroup$
– user376343
Dec 25 '18 at 23:47
add a comment |
1
$begingroup$
Have you drawn it out? Where did you find this problem statement? What geometric work have you done to crack this?
$endgroup$
– Namaste
Dec 25 '18 at 21:27
$begingroup$
@amWhy Official solution for one of the IMO problems. They use this without giving a proof, simply saying that they rely on linearity of the relation in variable $X$. I have drawn it out, but have no idea how to approach this geometrically...
$endgroup$
– user75619
Dec 25 '18 at 21:35
2
$begingroup$
I can't answer the question of whether there's a purely geometric proof, but do you know what "linearity" means here? That argument is worth understanding and is fast once one is comfortable with the terms.
$endgroup$
– Milo Brandt
Dec 25 '18 at 21:41
$begingroup$
@MiloBrandt I don't. I was actually hoping I would understand it from a geometric proof...
$endgroup$
– user75619
Dec 25 '18 at 21:47
2
$begingroup$
Try to start with this (forget E, D for the moment): exactly one point on AC and exactly one on BC verify the condition, denote the points P, Q. Then explore the segment PQ.
$endgroup$
– user376343
Dec 25 '18 at 23:47
1
1
$begingroup$
Have you drawn it out? Where did you find this problem statement? What geometric work have you done to crack this?
$endgroup$
– Namaste
Dec 25 '18 at 21:27
$begingroup$
Have you drawn it out? Where did you find this problem statement? What geometric work have you done to crack this?
$endgroup$
– Namaste
Dec 25 '18 at 21:27
$begingroup$
@amWhy Official solution for one of the IMO problems. They use this without giving a proof, simply saying that they rely on linearity of the relation in variable $X$. I have drawn it out, but have no idea how to approach this geometrically...
$endgroup$
– user75619
Dec 25 '18 at 21:35
$begingroup$
@amWhy Official solution for one of the IMO problems. They use this without giving a proof, simply saying that they rely on linearity of the relation in variable $X$. I have drawn it out, but have no idea how to approach this geometrically...
$endgroup$
– user75619
Dec 25 '18 at 21:35
2
2
$begingroup$
I can't answer the question of whether there's a purely geometric proof, but do you know what "linearity" means here? That argument is worth understanding and is fast once one is comfortable with the terms.
$endgroup$
– Milo Brandt
Dec 25 '18 at 21:41
$begingroup$
I can't answer the question of whether there's a purely geometric proof, but do you know what "linearity" means here? That argument is worth understanding and is fast once one is comfortable with the terms.
$endgroup$
– Milo Brandt
Dec 25 '18 at 21:41
$begingroup$
@MiloBrandt I don't. I was actually hoping I would understand it from a geometric proof...
$endgroup$
– user75619
Dec 25 '18 at 21:47
$begingroup$
@MiloBrandt I don't. I was actually hoping I would understand it from a geometric proof...
$endgroup$
– user75619
Dec 25 '18 at 21:47
2
2
$begingroup$
Try to start with this (forget E, D for the moment): exactly one point on AC and exactly one on BC verify the condition, denote the points P, Q. Then explore the segment PQ.
$endgroup$
– user376343
Dec 25 '18 at 23:47
$begingroup$
Try to start with this (forget E, D for the moment): exactly one point on AC and exactly one on BC verify the condition, denote the points P, Q. Then explore the segment PQ.
$endgroup$
– user376343
Dec 25 '18 at 23:47
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The proof is as simple as it gets:
It's given that:
$$d_1+d_2=d_3$$
$$e_1+e_2=e_3$$
Introduce: $DX=u$ for some point $Xin DE$.
$$x_1=d_1+frac{u}{DE}(e_1-d_1)$$
$$x_2=d_2+frac{u}{DE}(e_2-d_2)$$
$$x_1+x_2=d_1+frac{u}{DE}(e_1-d_1)+d_2+frac{u}{DE}(e_2-d_2)=$$
$$x_1+x_2=(d_1+d_2)+frac{u}{DE}((e_1+e2)-(d_1+d2))=$$
$$x_1+x_2=d_3+frac{u}{DE}(e_3-d_3)=$$
$$x_1+x_2=x_3$$
Yes, it's all about "linearity".
answered Dec 26 '18 at 14:40
OldboyOldboy
8,66211036
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The proof is as simple as it gets:
It's given that:
$$d_1+d_2=d_3$$
$$e_1+e_2=e_3$$
Introduce: $DX=u$ for some point $Xin DE$.
$$x_1=d_1+frac{u}{DE}(e_1-d_1)$$
$$x_2=d_2+frac{u}{DE}(e_2-d_2)$$
$$x_1+x_2=d_1+frac{u}{DE}(e_1-d_1)+d_2+frac{u}{DE}(e_2-d_2)=$$
$$x_1+x_2=(d_1+d_2)+frac{u}{DE}((e_1+e2)-(d_1+d2))=$$
$$x_1+x_2=d_3+frac{u}{DE}(e_3-d_3)=$$
$$x_1+x_2=x_3$$
Yes, it's all about "linearity".
answered Dec 26 '18 at 14:40
OldboyOldboy
8,66211036
$endgroup$
add a comment |
$begingroup$
The proof is as simple as it gets:
It's given that:
$$d_1+d_2=d_3$$
$$e_1+e_2=e_3$$
Introduce: $DX=u$ for some point $Xin DE$.
$$x_1=d_1+frac{u}{DE}(e_1-d_1)$$
$$x_2=d_2+frac{u}{DE}(e_2-d_2)$$
$$x_1+x_2=d_1+frac{u}{DE}(e_1-d_1)+d_2+frac{u}{DE}(e_2-d_2)=$$
$$x_1+x_2=(d_1+d_2)+frac{u}{DE}((e_1+e2)-(d_1+d2))=$$
$$x_1+x_2=d_3+frac{u}{DE}(e_3-d_3)=$$
$$x_1+x_2=x_3$$
Yes, it's all about "linearity".
answered Dec 26 '18 at 14:40
OldboyOldboy
8,66211036
$endgroup$
add a comment |
$begingroup$
The proof is as simple as it gets:
It's given that:
$$d_1+d_2=d_3$$
$$e_1+e_2=e_3$$
Introduce: $DX=u$ for some point $Xin DE$.
$$x_1=d_1+frac{u}{DE}(e_1-d_1)$$
$$x_2=d_2+frac{u}{DE}(e_2-d_2)$$
$$x_1+x_2=d_1+frac{u}{DE}(e_1-d_1)+d_2+frac{u}{DE}(e_2-d_2)=$$
$$x_1+x_2=(d_1+d_2)+frac{u}{DE}((e_1+e2)-(d_1+d2))=$$
$$x_1+x_2=d_3+frac{u}{DE}(e_3-d_3)=$$
$$x_1+x_2=x_3$$
Yes, it's all about "linearity".
answered Dec 26 '18 at 14:40
OldboyOldboy
8,66211036
$endgroup$
The proof is as simple as it gets:
It's given that:
$$d_1+d_2=d_3$$
$$e_1+e_2=e_3$$
Introduce: $DX=u$ for some point $Xin DE$.
$$x_1=d_1+frac{u}{DE}(e_1-d_1)$$
$$x_2=d_2+frac{u}{DE}(e_2-d_2)$$
$$x_1+x_2=d_1+frac{u}{DE}(e_1-d_1)+d_2+frac{u}{DE}(e_2-d_2)=$$
$$x_1+x_2=(d_1+d_2)+frac{u}{DE}((e_1+e2)-(d_1+d2))=$$
$$x_1+x_2=d_3+frac{u}{DE}(e_3-d_3)=$$
$$x_1+x_2=x_3$$
Yes, it's all about "linearity".
answered Dec 26 '18 at 14:40
OldboyOldboy
8,66211036
answered Dec 26 '18 at 14:40
OldboyOldboy
8,66211036
answered Dec 26 '18 at 14:40
OldboyOldboy
8,66211036
answered Dec 26 '18 at 14:40
OldboyOldboy
8,66211036
8,66211036
add a comment |
add a comment |
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1
$begingroup$
Have you drawn it out? Where did you find this problem statement? What geometric work have you done to crack this?
$endgroup$
– Namaste
Dec 25 '18 at 21:27
$begingroup$
@amWhy Official solution for one of the IMO problems. They use this without giving a proof, simply saying that they rely on linearity of the relation in variable $X$. I have drawn it out, but have no idea how to approach this geometrically...
$endgroup$
– user75619
Dec 25 '18 at 21:35
2
$begingroup$
I can't answer the question of whether there's a purely geometric proof, but do you know what "linearity" means here? That argument is worth understanding and is fast once one is comfortable with the terms.
$endgroup$
– Milo Brandt
Dec 25 '18 at 21:41
$begingroup$
@MiloBrandt I don't. I was actually hoping I would understand it from a geometric proof...
$endgroup$
– user75619
Dec 25 '18 at 21:47
2
$begingroup$
Try to start with this (forget E, D for the moment): exactly one point on AC and exactly one on BC verify the condition, denote the points P, Q. Then explore the segment PQ.
$endgroup$
– user376343
Dec 25 '18 at 23:47