Show that if two triangles built on parallel lines, with equal bases have the same perimeter only if they are...
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Show that if two triangles built on parallel lines, with equal bases have the same perimeter only if they are congruent.
Prove that lines intersecting parallel similar triangles are concurrentDoes proving that two lines are parallel require a postulate?proving that $BC' parallel B'C$Without using angle measure how do I prove two lines are parallel to the same line are parallel to each other?Two congruent segments does have the same length?Two triangles cirumcribed a conic problemShow that two parallel lines have the same direction vector from a different definition of parallel lines.Proof: Two triangles have the same ratio of length for each corresponding side then they are similarIf the heights of two triangles are proportional then prove that they are similiarIf ratio of sides of two triangles is constant then the triangles have the same angles
$begingroup$
Show that if two triangles built on parallel lines, as shown above, with |AB|=|A'B'| have the same perimeter only if they are congruent.
I've tried proving by contradiction:
Suppose they are not congruent but have the same perimeter, then either
|AC|$neq$|A'C| or |BC|$neq$ |B'C'|.
Let's say |AC|$neq$|A'C'|, and suppose that |AC| $lt$ |A'C'|.
If |BC|=|B'C'| then the triangles would be congruent which is false from my assumption.
If |BC| $gt$ |B'C'| then |A'C'| + |B'C'| $gt$ |AC| + |BC| which is false because their perimeters should be equal.
On the last possible case, |BC|$gt$|B'C'| I got stuck. I can't find a way to show that it is false.
How can I show that the last case is false?
geometry euclidean-geometry
asked yesterday
BanBan
703
$endgroup$
add a comment |
$begingroup$
Show that if two triangles built on parallel lines, as shown above, with |AB|=|A'B'| have the same perimeter only if they are congruent.
I've tried proving by contradiction:
Suppose they are not congruent but have the same perimeter, then either
|AC|$neq$|A'C| or |BC|$neq$ |B'C'|.
Let's say |AC|$neq$|A'C'|, and suppose that |AC| $lt$ |A'C'|.
If |BC|=|B'C'| then the triangles would be congruent which is false from my assumption.
If |BC| $gt$ |B'C'| then |A'C'| + |B'C'| $gt$ |AC| + |BC| which is false because their perimeters should be equal.
On the last possible case, |BC|$gt$|B'C'| I got stuck. I can't find a way to show that it is false.
How can I show that the last case is false?
geometry euclidean-geometry
asked yesterday
BanBan
703
$endgroup$
$begingroup$
But you have gone to the case AC<A'C', so from BC=B'C' you don't get congruent.
$endgroup$
– coffeemath
23 hours ago
$begingroup$
Yeah, sorry. Let's just say that |A'C'| + |B'C'| > |AC| + |BC| for that one too, which is false.
$endgroup$
– Ban
22 hours ago
$begingroup$
If $BC=B'C'$ and $AB=A'B'$ then the perimeters are not equal. I hope someone posts an elementary solution not involving conics or functions.
$endgroup$
– BPP
48 mins ago
add a comment |
$begingroup$
Show that if two triangles built on parallel lines, as shown above, with |AB|=|A'B'| have the same perimeter only if they are congruent.
I've tried proving by contradiction:
Suppose they are not congruent but have the same perimeter, then either
|AC|$neq$|A'C| or |BC|$neq$ |B'C'|.
Let's say |AC|$neq$|A'C'|, and suppose that |AC| $lt$ |A'C'|.
If |BC|=|B'C'| then the triangles would be congruent which is false from my assumption.
If |BC| $gt$ |B'C'| then |A'C'| + |B'C'| $gt$ |AC| + |BC| which is false because their perimeters should be equal.
On the last possible case, |BC|$gt$|B'C'| I got stuck. I can't find a way to show that it is false.
How can I show that the last case is false?
geometry euclidean-geometry
asked yesterday
BanBan
703
$endgroup$
Show that if two triangles built on parallel lines, as shown above, with |AB|=|A'B'| have the same perimeter only if they are congruent.
I've tried proving by contradiction:
Suppose they are not congruent but have the same perimeter, then either
|AC|$neq$|A'C| or |BC|$neq$ |B'C'|.
Let's say |AC|$neq$|A'C'|, and suppose that |AC| $lt$ |A'C'|.
If |BC|=|B'C'| then the triangles would be congruent which is false from my assumption.
If |BC| $gt$ |B'C'| then |A'C'| + |B'C'| $gt$ |AC| + |BC| which is false because their perimeters should be equal.
On the last possible case, |BC|$gt$|B'C'| I got stuck. I can't find a way to show that it is false.
How can I show that the last case is false?
geometry euclidean-geometry
geometry euclidean-geometry
asked yesterday
BanBan
703
asked yesterday
BanBan
703
asked yesterday
BanBan
703
asked yesterday
BanBan
703
asked yesterday
BanBan
703
703
$begingroup$
But you have gone to the case AC<A'C', so from BC=B'C' you don't get congruent.
$endgroup$
– coffeemath
23 hours ago
$begingroup$
Yeah, sorry. Let's just say that |A'C'| + |B'C'| > |AC| + |BC| for that one too, which is false.
$endgroup$
– Ban
22 hours ago
$begingroup$
If $BC=B'C'$ and $AB=A'B'$ then the perimeters are not equal. I hope someone posts an elementary solution not involving conics or functions.
$endgroup$
– BPP
48 mins ago
add a comment |
$begingroup$
But you have gone to the case AC<A'C', so from BC=B'C' you don't get congruent.
$endgroup$
– coffeemath
23 hours ago
$begingroup$
Yeah, sorry. Let's just say that |A'C'| + |B'C'| > |AC| + |BC| for that one too, which is false.
$endgroup$
– Ban
22 hours ago
$begingroup$
If $BC=B'C'$ and $AB=A'B'$ then the perimeters are not equal. I hope someone posts an elementary solution not involving conics or functions.
$endgroup$
– BPP
48 mins ago
$begingroup$
But you have gone to the case AC<A'C', so from BC=B'C' you don't get congruent.
$endgroup$
– coffeemath
23 hours ago
$begingroup$
But you have gone to the case AC<A'C', so from BC=B'C' you don't get congruent.
$endgroup$
– coffeemath
23 hours ago
$begingroup$
Yeah, sorry. Let's just say that |A'C'| + |B'C'| > |AC| + |BC| for that one too, which is false.
$endgroup$
– Ban
22 hours ago
$begingroup$
Yeah, sorry. Let's just say that |A'C'| + |B'C'| > |AC| + |BC| for that one too, which is false.
$endgroup$
– Ban
22 hours ago
$begingroup$
If $BC=B'C'$ and $AB=A'B'$ then the perimeters are not equal. I hope someone posts an elementary solution not involving conics or functions.
$endgroup$
– BPP
48 mins ago
$begingroup$
If $BC=B'C'$ and $AB=A'B'$ then the perimeters are not equal. I hope someone posts an elementary solution not involving conics or functions.
$endgroup$
– BPP
48 mins ago
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Imagine that $AB$ is fixed on the bottom line and $C$ varies from way off left to way off right. The perimeter of the triangle is a decreasing function until triangle $ABC$ is isosceles, then increasing. It's clear from the symmetry that it takes on every value greater than its minimum value at just two points symmetrical with respect to the perpendicular bisector of $AB$.
answered 23 hours ago
Ethan BolkerEthan Bolker
45.8k553120
$endgroup$
1
$begingroup$
This argument is clear except between the two right-angled cases, when $angle ABC$ or $angle CAB$ is a right angle. Then it's not clear whether the perimeter is increasing or decreasing, as one of the sides is getting longer while the other is getting shorter. Calculus or algebra (reduction to quadratic) could be used here. As such, this is only a partial solution
$endgroup$
– man on laptop
4 hours ago
$begingroup$
@manonlaptop I thought about that case. It's intuitively clear that at any point the longer leg is changing faster than the shorter leg. (As you commented, you only need that observation when the triangle is acute.) You could provide a rigorous algebraic proof. I'm sure you could do it synthetically with Euclid.
$endgroup$
– Ethan Bolker
3 hours ago
add a comment |
$begingroup$
Fix $ A' $ and $ B' $. As $ AB = A'B' $ is fixed, the points $ C' $ for which $ ABC $ has the same perimeter as $ A'B'C' $ are the points for which $ AC + BC = A'C' + B'C' $. You recognize here the definition of an ellipse of focus $ A' $ and $ B' $. Hence the locus of $ C' $ is an ellipse.
Finally, $ C' $ is in the meantime on an ellipse and on a line. These two have two intersections which give the directly and indirectly congruents triangles.
The easiest way to uncover your last case is using the ellipse argument.
answered 23 hours ago
AstaulpheAstaulphe
765
$endgroup$
add a comment |
$begingroup$
As an alternative proof, because the triangles are built on parallel lines, they have the same area. Using Heron's Formula
$$
A = frac{1}{4}sqrt{(AB + AC + BC)(-AB + AC + BC)(AB - AC + BC)(AB + AC - BC)}
$$
and a bit of algebra, you can show that either $AC = A'C'$ and $BC = B'C'$ or $AC = B'C'$ and $BC = A'C'$. In both cases $ABC cong A'B'C'$.
answered 23 hours ago
eyeballfrogeyeballfrog
7,202633
$endgroup$
add a comment |
$begingroup$
Let's say the distance between the two lines is $1$. Put an $x$ axis on the lower line, and a $y$ axis through the first point in the triangle. This ensures that the bottom left point of the triangle has coordinate $(0,0)$. Place the $(1,0)$ coordinate on the bottom right corner of the triangle.
If the third point of the triangle is on $(x,1)$, then the diameter is $$f(x) = sqrt{1 + x^2} + sqrt{1 + (1-x)^2} + 1$$.
Observe that $f(x)$ has a line of symmetry at $x=0.5$. In other words, if you do the substitution $u = 1-x$ you get the same function.
Next observe by plotting or by differentiation that the function is monotonically decreasing when $x < 0.5$ and increasing when $x > 0.5$.
By the previous paragraph, if a triangle exists with a certain diameter, at most only one other triangle can exist with that diameter. Moreover, the paragraph previous to that says that this other triangle can be reflected at the line $x=0.5$ to yield the first.
Q.E.D.
answered 4 hours ago
man on laptopman on laptop
5,83111538
$endgroup$
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
$begingroup$
Imagine that $AB$ is fixed on the bottom line and $C$ varies from way off left to way off right. The perimeter of the triangle is a decreasing function until triangle $ABC$ is isosceles, then increasing. It's clear from the symmetry that it takes on every value greater than its minimum value at just two points symmetrical with respect to the perpendicular bisector of $AB$.
answered 23 hours ago
Ethan BolkerEthan Bolker
45.8k553120
$endgroup$
1
$begingroup$
This argument is clear except between the two right-angled cases, when $angle ABC$ or $angle CAB$ is a right angle. Then it's not clear whether the perimeter is increasing or decreasing, as one of the sides is getting longer while the other is getting shorter. Calculus or algebra (reduction to quadratic) could be used here. As such, this is only a partial solution
$endgroup$
– man on laptop
4 hours ago
$begingroup$
@manonlaptop I thought about that case. It's intuitively clear that at any point the longer leg is changing faster than the shorter leg. (As you commented, you only need that observation when the triangle is acute.) You could provide a rigorous algebraic proof. I'm sure you could do it synthetically with Euclid.
$endgroup$
– Ethan Bolker
3 hours ago
add a comment |
$begingroup$
Imagine that $AB$ is fixed on the bottom line and $C$ varies from way off left to way off right. The perimeter of the triangle is a decreasing function until triangle $ABC$ is isosceles, then increasing. It's clear from the symmetry that it takes on every value greater than its minimum value at just two points symmetrical with respect to the perpendicular bisector of $AB$.
answered 23 hours ago
Ethan BolkerEthan Bolker
45.8k553120
$endgroup$
1
$begingroup$
This argument is clear except between the two right-angled cases, when $angle ABC$ or $angle CAB$ is a right angle. Then it's not clear whether the perimeter is increasing or decreasing, as one of the sides is getting longer while the other is getting shorter. Calculus or algebra (reduction to quadratic) could be used here. As such, this is only a partial solution
$endgroup$
– man on laptop
4 hours ago
$begingroup$
@manonlaptop I thought about that case. It's intuitively clear that at any point the longer leg is changing faster than the shorter leg. (As you commented, you only need that observation when the triangle is acute.) You could provide a rigorous algebraic proof. I'm sure you could do it synthetically with Euclid.
$endgroup$
– Ethan Bolker
3 hours ago
add a comment |
$begingroup$
Imagine that $AB$ is fixed on the bottom line and $C$ varies from way off left to way off right. The perimeter of the triangle is a decreasing function until triangle $ABC$ is isosceles, then increasing. It's clear from the symmetry that it takes on every value greater than its minimum value at just two points symmetrical with respect to the perpendicular bisector of $AB$.
answered 23 hours ago
Ethan BolkerEthan Bolker
45.8k553120
$endgroup$
Imagine that $AB$ is fixed on the bottom line and $C$ varies from way off left to way off right. The perimeter of the triangle is a decreasing function until triangle $ABC$ is isosceles, then increasing. It's clear from the symmetry that it takes on every value greater than its minimum value at just two points symmetrical with respect to the perpendicular bisector of $AB$.
answered 23 hours ago
Ethan BolkerEthan Bolker
45.8k553120
answered 23 hours ago
Ethan BolkerEthan Bolker
45.8k553120
answered 23 hours ago
Ethan BolkerEthan Bolker
45.8k553120
answered 23 hours ago
Ethan BolkerEthan Bolker
45.8k553120
45.8k553120
1
$begingroup$
This argument is clear except between the two right-angled cases, when $angle ABC$ or $angle CAB$ is a right angle. Then it's not clear whether the perimeter is increasing or decreasing, as one of the sides is getting longer while the other is getting shorter. Calculus or algebra (reduction to quadratic) could be used here. As such, this is only a partial solution
$endgroup$
– man on laptop
4 hours ago
$begingroup$
@manonlaptop I thought about that case. It's intuitively clear that at any point the longer leg is changing faster than the shorter leg. (As you commented, you only need that observation when the triangle is acute.) You could provide a rigorous algebraic proof. I'm sure you could do it synthetically with Euclid.
$endgroup$
– Ethan Bolker
3 hours ago
add a comment |
1
$begingroup$
This argument is clear except between the two right-angled cases, when $angle ABC$ or $angle CAB$ is a right angle. Then it's not clear whether the perimeter is increasing or decreasing, as one of the sides is getting longer while the other is getting shorter. Calculus or algebra (reduction to quadratic) could be used here. As such, this is only a partial solution
$endgroup$
– man on laptop
4 hours ago
$begingroup$
@manonlaptop I thought about that case. It's intuitively clear that at any point the longer leg is changing faster than the shorter leg. (As you commented, you only need that observation when the triangle is acute.) You could provide a rigorous algebraic proof. I'm sure you could do it synthetically with Euclid.
$endgroup$
– Ethan Bolker
3 hours ago
1
1
$begingroup$
This argument is clear except between the two right-angled cases, when $angle ABC$ or $angle CAB$ is a right angle. Then it's not clear whether the perimeter is increasing or decreasing, as one of the sides is getting longer while the other is getting shorter. Calculus or algebra (reduction to quadratic) could be used here. As such, this is only a partial solution
$endgroup$
– man on laptop
4 hours ago
$begingroup$
This argument is clear except between the two right-angled cases, when $angle ABC$ or $angle CAB$ is a right angle. Then it's not clear whether the perimeter is increasing or decreasing, as one of the sides is getting longer while the other is getting shorter. Calculus or algebra (reduction to quadratic) could be used here. As such, this is only a partial solution
$endgroup$
– man on laptop
4 hours ago
$begingroup$
@manonlaptop I thought about that case. It's intuitively clear that at any point the longer leg is changing faster than the shorter leg. (As you commented, you only need that observation when the triangle is acute.) You could provide a rigorous algebraic proof. I'm sure you could do it synthetically with Euclid.
$endgroup$
– Ethan Bolker
3 hours ago
$begingroup$
@manonlaptop I thought about that case. It's intuitively clear that at any point the longer leg is changing faster than the shorter leg. (As you commented, you only need that observation when the triangle is acute.) You could provide a rigorous algebraic proof. I'm sure you could do it synthetically with Euclid.
$endgroup$
– Ethan Bolker
3 hours ago
add a comment |
$begingroup$
Fix $ A' $ and $ B' $. As $ AB = A'B' $ is fixed, the points $ C' $ for which $ ABC $ has the same perimeter as $ A'B'C' $ are the points for which $ AC + BC = A'C' + B'C' $. You recognize here the definition of an ellipse of focus $ A' $ and $ B' $. Hence the locus of $ C' $ is an ellipse.
Finally, $ C' $ is in the meantime on an ellipse and on a line. These two have two intersections which give the directly and indirectly congruents triangles.
The easiest way to uncover your last case is using the ellipse argument.
answered 23 hours ago
AstaulpheAstaulphe
765
$endgroup$
add a comment |
$begingroup$
Fix $ A' $ and $ B' $. As $ AB = A'B' $ is fixed, the points $ C' $ for which $ ABC $ has the same perimeter as $ A'B'C' $ are the points for which $ AC + BC = A'C' + B'C' $. You recognize here the definition of an ellipse of focus $ A' $ and $ B' $. Hence the locus of $ C' $ is an ellipse.
Finally, $ C' $ is in the meantime on an ellipse and on a line. These two have two intersections which give the directly and indirectly congruents triangles.
The easiest way to uncover your last case is using the ellipse argument.
answered 23 hours ago
AstaulpheAstaulphe
765
$endgroup$
add a comment |
$begingroup$
Fix $ A' $ and $ B' $. As $ AB = A'B' $ is fixed, the points $ C' $ for which $ ABC $ has the same perimeter as $ A'B'C' $ are the points for which $ AC + BC = A'C' + B'C' $. You recognize here the definition of an ellipse of focus $ A' $ and $ B' $. Hence the locus of $ C' $ is an ellipse.
Finally, $ C' $ is in the meantime on an ellipse and on a line. These two have two intersections which give the directly and indirectly congruents triangles.
The easiest way to uncover your last case is using the ellipse argument.
answered 23 hours ago
AstaulpheAstaulphe
765
$endgroup$
Fix $ A' $ and $ B' $. As $ AB = A'B' $ is fixed, the points $ C' $ for which $ ABC $ has the same perimeter as $ A'B'C' $ are the points for which $ AC + BC = A'C' + B'C' $. You recognize here the definition of an ellipse of focus $ A' $ and $ B' $. Hence the locus of $ C' $ is an ellipse.
Finally, $ C' $ is in the meantime on an ellipse and on a line. These two have two intersections which give the directly and indirectly congruents triangles.
The easiest way to uncover your last case is using the ellipse argument.
answered 23 hours ago
AstaulpheAstaulphe
765
answered 23 hours ago
AstaulpheAstaulphe
765
answered 23 hours ago
AstaulpheAstaulphe
765
answered 23 hours ago
AstaulpheAstaulphe
765
765
add a comment |
add a comment |
$begingroup$
As an alternative proof, because the triangles are built on parallel lines, they have the same area. Using Heron's Formula
$$
A = frac{1}{4}sqrt{(AB + AC + BC)(-AB + AC + BC)(AB - AC + BC)(AB + AC - BC)}
$$
and a bit of algebra, you can show that either $AC = A'C'$ and $BC = B'C'$ or $AC = B'C'$ and $BC = A'C'$. In both cases $ABC cong A'B'C'$.
answered 23 hours ago
eyeballfrogeyeballfrog
7,202633
$endgroup$
add a comment |
$begingroup$
As an alternative proof, because the triangles are built on parallel lines, they have the same area. Using Heron's Formula
$$
A = frac{1}{4}sqrt{(AB + AC + BC)(-AB + AC + BC)(AB - AC + BC)(AB + AC - BC)}
$$
and a bit of algebra, you can show that either $AC = A'C'$ and $BC = B'C'$ or $AC = B'C'$ and $BC = A'C'$. In both cases $ABC cong A'B'C'$.
answered 23 hours ago
eyeballfrogeyeballfrog
7,202633
$endgroup$
add a comment |
$begingroup$
As an alternative proof, because the triangles are built on parallel lines, they have the same area. Using Heron's Formula
$$
A = frac{1}{4}sqrt{(AB + AC + BC)(-AB + AC + BC)(AB - AC + BC)(AB + AC - BC)}
$$
and a bit of algebra, you can show that either $AC = A'C'$ and $BC = B'C'$ or $AC = B'C'$ and $BC = A'C'$. In both cases $ABC cong A'B'C'$.
answered 23 hours ago
eyeballfrogeyeballfrog
7,202633
$endgroup$
As an alternative proof, because the triangles are built on parallel lines, they have the same area. Using Heron's Formula
$$
A = frac{1}{4}sqrt{(AB + AC + BC)(-AB + AC + BC)(AB - AC + BC)(AB + AC - BC)}
$$
and a bit of algebra, you can show that either $AC = A'C'$ and $BC = B'C'$ or $AC = B'C'$ and $BC = A'C'$. In both cases $ABC cong A'B'C'$.
answered 23 hours ago
eyeballfrogeyeballfrog
7,202633
answered 23 hours ago
eyeballfrogeyeballfrog
7,202633
answered 23 hours ago
eyeballfrogeyeballfrog
7,202633
answered 23 hours ago
eyeballfrogeyeballfrog
7,202633
7,202633
add a comment |
add a comment |
$begingroup$
Let's say the distance between the two lines is $1$. Put an $x$ axis on the lower line, and a $y$ axis through the first point in the triangle. This ensures that the bottom left point of the triangle has coordinate $(0,0)$. Place the $(1,0)$ coordinate on the bottom right corner of the triangle.
If the third point of the triangle is on $(x,1)$, then the diameter is $$f(x) = sqrt{1 + x^2} + sqrt{1 + (1-x)^2} + 1$$.
Observe that $f(x)$ has a line of symmetry at $x=0.5$. In other words, if you do the substitution $u = 1-x$ you get the same function.
Next observe by plotting or by differentiation that the function is monotonically decreasing when $x < 0.5$ and increasing when $x > 0.5$.
By the previous paragraph, if a triangle exists with a certain diameter, at most only one other triangle can exist with that diameter. Moreover, the paragraph previous to that says that this other triangle can be reflected at the line $x=0.5$ to yield the first.
Q.E.D.
answered 4 hours ago
man on laptopman on laptop
5,83111538
$endgroup$
add a comment |
$begingroup$
Let's say the distance between the two lines is $1$. Put an $x$ axis on the lower line, and a $y$ axis through the first point in the triangle. This ensures that the bottom left point of the triangle has coordinate $(0,0)$. Place the $(1,0)$ coordinate on the bottom right corner of the triangle.
If the third point of the triangle is on $(x,1)$, then the diameter is $$f(x) = sqrt{1 + x^2} + sqrt{1 + (1-x)^2} + 1$$.
Observe that $f(x)$ has a line of symmetry at $x=0.5$. In other words, if you do the substitution $u = 1-x$ you get the same function.
Next observe by plotting or by differentiation that the function is monotonically decreasing when $x < 0.5$ and increasing when $x > 0.5$.
By the previous paragraph, if a triangle exists with a certain diameter, at most only one other triangle can exist with that diameter. Moreover, the paragraph previous to that says that this other triangle can be reflected at the line $x=0.5$ to yield the first.
Q.E.D.
answered 4 hours ago
man on laptopman on laptop
5,83111538
$endgroup$
add a comment |
$begingroup$
Let's say the distance between the two lines is $1$. Put an $x$ axis on the lower line, and a $y$ axis through the first point in the triangle. This ensures that the bottom left point of the triangle has coordinate $(0,0)$. Place the $(1,0)$ coordinate on the bottom right corner of the triangle.
If the third point of the triangle is on $(x,1)$, then the diameter is $$f(x) = sqrt{1 + x^2} + sqrt{1 + (1-x)^2} + 1$$.
Observe that $f(x)$ has a line of symmetry at $x=0.5$. In other words, if you do the substitution $u = 1-x$ you get the same function.
Next observe by plotting or by differentiation that the function is monotonically decreasing when $x < 0.5$ and increasing when $x > 0.5$.
By the previous paragraph, if a triangle exists with a certain diameter, at most only one other triangle can exist with that diameter. Moreover, the paragraph previous to that says that this other triangle can be reflected at the line $x=0.5$ to yield the first.
Q.E.D.
answered 4 hours ago
man on laptopman on laptop
5,83111538
$endgroup$
Let's say the distance between the two lines is $1$. Put an $x$ axis on the lower line, and a $y$ axis through the first point in the triangle. This ensures that the bottom left point of the triangle has coordinate $(0,0)$. Place the $(1,0)$ coordinate on the bottom right corner of the triangle.
If the third point of the triangle is on $(x,1)$, then the diameter is $$f(x) = sqrt{1 + x^2} + sqrt{1 + (1-x)^2} + 1$$.
Observe that $f(x)$ has a line of symmetry at $x=0.5$. In other words, if you do the substitution $u = 1-x$ you get the same function.
Next observe by plotting or by differentiation that the function is monotonically decreasing when $x < 0.5$ and increasing when $x > 0.5$.
By the previous paragraph, if a triangle exists with a certain diameter, at most only one other triangle can exist with that diameter. Moreover, the paragraph previous to that says that this other triangle can be reflected at the line $x=0.5$ to yield the first.
Q.E.D.
answered 4 hours ago
man on laptopman on laptop
5,83111538
edited 45 mins ago
answered 4 hours ago
man on laptopman on laptop
5,83111538
answered 4 hours ago
man on laptopman on laptop
5,83111538
answered 4 hours ago
man on laptopman on laptop
5,83111538
5,83111538
add a comment |
add a comment |
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$begingroup$
But you have gone to the case AC<A'C', so from BC=B'C' you don't get congruent.
$endgroup$
– coffeemath
23 hours ago
$begingroup$
Yeah, sorry. Let's just say that |A'C'| + |B'C'| > |AC| + |BC| for that one too, which is false.
$endgroup$
– Ban
22 hours ago
$begingroup$
If $BC=B'C'$ and $AB=A'B'$ then the perimeters are not equal. I hope someone posts an elementary solution not involving conics or functions.
$endgroup$
– BPP
48 mins ago