Find the sides of the right triangle so that their sum is minimized

[MatinSAR]

Homework Statement

Find the sides of the right triangle so that their sum is minimized.

Relevant Equations

$a^2+b^2=c^2$

Hello. We know that $a^2+b^2=c^2$ and we want to minimize $a+b$.

$$b= \sqrt {c^2-a^2}$$ $$ \dfrac {d}{da} (a+\sqrt {c^2-a^2})=0$$ $$ 1-\dfrac {a}{ \sqrt {c^2-a^2}}=0 $$ This gives $$a=\dfrac {c}{\sqrt 2}$$

But it doesn't work for c=5. I know a=3 and b=4 minimize a+b.

 

[Hill]

Homework Statement: Find the sides of the right triangle so that their sum is minimized.

What is a constraint? Without a constraint the answer is $a=b=0$...

I know a=3 and b=4 minimize a+b.

How do you know this? $3+4=7$, but if you take $a=5, b=0$ you get $5$.

 

[MatinSAR]

What is a constraint? Without a constraint the answer is $a=b=0$...

Suppose $c=5$.
Then both $a$ and $b$ shoud be between $0$ and $5$.

How do you know this? $3+4=7$, but if you take $a=5, b=0$ you get $5$.

But length of a triangle cannot be $0$.

 

[Hill]

Both a and b shoud be between 0 and 5.

$a=b=0$ satisfies this.

But length of a triangle cannot be 0.

I am not sure, what "length of a triangle" means, but you can take e.g. $a=4.99, b=0.316$ to get $a+b=5.3<7$.

 

[MatinSAR]

$a=b=0$ satisfies this.

By between, I meant not equal. ($0<a<5$ and $0<b<5$)

I am not sure, what "length of a triangle" means, but you can take e.g. $a=4.99, b=0.316$ to get $a+b=5.3<7$.

Sorry I meant length of the leg of the triangle. But those a and b doesn't satisfy $a^2+b^2=25$. Yes?

 

[Hill]

You should take time to think. Look:

By between, I meant not equal.

Then take $a=b=0.0001$.

But those a and b doesn't satisfy a2+b2=25. Yes?

They do. If you need to be precise, take $a=4.99, b=\sqrt {25-4.99^2}$.

 

[MatinSAR]

Then take $a=b=0.0001$.

It doesn't satisfy $a^2+b^2=25$.

They do. If you need to be precise, take $a=4.99, b=\sqrt {25-4.99^2}$.

That's it! Thanks for your help @Hill ...

So there is no answer. Because we can choose $a=4.99....9$ and get $b=\sqrt {25-4.99....9^2}$.
($a \to 5$ then $b \to 0$)

 

[Hill]

It doesn't satisfy $a^2+b^2=25$.

That's it! Thanks for your help @Hill ...

So there is no answer. Because we can choose $a=4.99....9$ and get $b=\sqrt {25-4.99....9^2}$.
($a \to 5$ then $b \to 0$)

This would be the answer if the statement were "Find the sides of the right triangle so that their sum is minimized for a given hypothenuse $c=5$." The bold part is a constraint.

Are you sure this is the right statement?

 

[MatinSAR]

This would be the answer if the statement were "Find the sides of the right triangle so that their sum is minimized for a given hypothenuse $c=5$." The bold part is a constraint.

Are you sure this is the right statement?

It is a part of a physics problem. The question should be wrong.

Imagine a person who wants to reach a point 5 kilometers away from their current location in the southwest direction. They can only move along streets that run either north-south or east-west. What is the shortest distance they must travel to reach their destination?

I was trying to solve this when that question came to my mind.

If by sothwest they meant 45 degrees with west then we acn solve ...

 

[SammyS]

Homework Statement: Find the sides of the right triangle so that their sum is minimized.
Relevant Equations: $a^2+b^2=c^2$

Hello. We know that $a^2+b^2=c^2$ and we want to minimize $a+b$.

$$b= \sqrt {c^2-a^2}$$ $$ \dfrac {d}{da} (a+\sqrt {c^2-a^2})=0$$ $$ 1-\dfrac {a}{ \sqrt {c^2-a^2}}=0 $$ This gives $$a=\dfrac {c}{\sqrt 2}$$

But it doesn't work for c=5. I know a=3 and b=4 minimize a+b.

Judging by your method of solution, you have the constraint that the hypotenuse has a fixed length.

For $c=5$, your solution gives $a=\dfrac {5}{\sqrt 2}\approx 3.5355\,.\$ So that $b=\dfrac {5}{\sqrt 2}\$ as well. This gives $a+b=5\,\sqrt{2\,}\approx 7.071\,.\$

This is actually the MAXimum for $a+b$.

There is no minimum for $a+b$. Mathematically, the minimum for $a+b$ is $5$, but that requires either $a$ or $b$ to be zero, which is nonsense as a length.

 

[MatinSAR]

Judging by your method of solution, you have the constraint that the hypotenuse has a fixed length.

Yes.

For $c=5$, your solution gives $a=\dfrac {5}{\sqrt 2}\approx 3.5355\,.\$ So that $b=\dfrac {5}{\sqrt 2}\$ as well. This gives $a+b=5\,\sqrt{2\,}\approx 7.071\,.\$

This is actually the MAXimum for $a+b$.

There is no minimum for $a+b$. Mathematically, the minimum for $a+b$ is $5$, but that requires either $a$ or $b$ to be zero, which is nonsense as a length.

It's clear now. Thank you for your time.

 

[FactChecker]

There is no minimum for $a+b$. Mathematically, the minimum for $a+b$ is $5$, but that requires either $a$ or $b$ to be zero, which is nonsense as a length.

I am not sure what you mean by "nonsense as a length". Do you mean that it is not a triangle if one side has zero length? I guess that is not a triangle, but then you can conclude that the smaller one side is, the better.
After all, the shortest distance between two points (the two ends of c) is a straight line. That means that one side would equal c and the other side would have zero length.

 

[pasmith]

It is a part of a physics problem. The question should be wrong.

Imagine a person who wants to reach a point 5 kilometers away from their current location in the southwest direction. They can only move along streets that run either north-south or east-west. What is the shortest distance they must travel to reach their destination?

I was trying to solve this when that question came to my mind.

If by sothwest they meant 45 degrees with west then we acn solve ...

That is what southwest means.

 

[FactChecker]

$$b= \sqrt {c^2-a^2}$$ $$ \dfrac {d}{da} (a+\sqrt {c^2-a^2})=0$$ $$ 1-\dfrac {a}{ \sqrt {c^2-a^2}}=0 $$ This gives $$a=\dfrac {c}{\sqrt 2}$$

This is a good way to find the interior extrema, but don't forget about the endpoints of a=0 and a=c. In fact, you have found the interior maximum at $b=a=\dfrac {c}{\sqrt 2}$.

But it doesn't work for c=5. I know a=3 and b=4 minimize a+b.

As others have said, this is wrong.

 

[MatinSAR]

That is what southwest means.

I'm not sure if the person who asked it considered this.

As others have said, this is wrong.

Yes. The problem was with the question. Or at least with my understanding of the question.

Imagine a person who wants to reach a point 5 kilometers away from their current location in the southwest direction. They can only move along streets that run either north-south or east-west. What is the shortest distance they must travel to reach their destination?

I was trying to solve this when that question came to my mind.

If by sothwest they meant 45 degrees with west then we acn solve ...

This.

@FactChecker and @pasmith Thank you for your time.

 

[pasmith]

Also to add: for an arbitrary triangle with vertices A, B and C, the sides must satisfy |AB| + |BC| \geq |AC| together with the two cyclic permutations ot that, with equality if and only if the vertices all fall on the same straight line (which isn't really a triangle). Thus if A and C are fixed then it is not possible to choose B so as to minimize |AB| + |BC| subhect to |AB| &gt; 0 and |BC| &gt; 0.

To answer the actual question: To get to a point 5km to the southwest, one must move 5/\sqrt{2}\,\mathrm{km} south and 5/\sqrt{2}\,\mathrm{km} west. It follows that if one is constrained to move only north-south or east-west then the minimum distance is 5\sqrt{2}\,\mathrm{km}. There is no unique path which attains this; any path which satisfies the requirement of only moving southwards or westwards will attain it. See taxicab metric.

 

[MatinSAR]

To answer the actual question: To get to a point 5km to the southwest, one must move 5/\sqrt{2}\,\mathrm{km} south and 5/\sqrt{2}\,\mathrm{km} west.

This is what I've done. But another person suggusted that sides can be $3$ and $4$ which has left me puzzled.

There is no unique path which attains this; any path which satisfies the requirement of only moving southwards or westwards will attain it. See taxicab metric.

Interesting. Thank you for your help @pasmith .

 

[pasmith]

This is what I've done. But another person suggusted that sides can be $3$ and $4$ which has left me puzzled.

I think they have seen that the hypoteneuse is 5km and have therefore assumed, entirely incorrectly, that the triangle is the well-known right-angled triangle with sides 3 km, 4 km and 5km.

 

[Hill]

Imagine a person who wants to reach a point 5 kilometers away from their current location in the southwest direction. They can only move along streets that run either north-south or east-west. What is the shortest distance they must travel to reach their destination?

What would be the answer if that person starts near the North Pole?

 

[MatinSAR]

What would be the answer if that person starts near the North Pole?

I don't understand your question. What's the difference? The person wants to reach a point in southeast direction still. Is it important where he starts?

 

[Hill]

I don't understand your question. What's the difference? The person wants to reach a point in southeast direction still. Is it important where he starts?

Yes. If he started very close to the North Pole and went SW 5 km, he could even circle the Earth several times. To get to the same point by going W and S, he would need to go just a bit W or E and then straight S. The second way would be shorter than the first.

 

[MatinSAR]

Yes. If he started very close to the North Pole and went SW 5 km, he would circle the Earth several times. To get to the same point by going W and S, he would need to go just a bit W and then straight S. The second way would be shorter than the first.

This is because of the fact that The Earth is not flat. Yes? I think I understand what you meant...

 

[Bystander]

This is what I've done. But another person suggusted that sides can be 3 and 4 which has left me puzzled.

Integer/whole number/pythagorean triples: https://en.wikipedia.org/wiki/Pythagorean_triple .

 

[FactChecker]

I have gotten very confused now about what the full description of the problem is and do not feel like reading through the entire thread to find out.
Is it that the solution must be integers?
Is it that the solution must be along fixed coordinate directions?
Is this on the Earth, with spherical coordinates?

 

[MatinSAR]

Is it that the solution must be integers?

Of course not.

Is it that the solution must be along fixed coordinate directions?

This question is from a book related to high school. I think we can suppose that the person walks in a 2D plane and he has 4 directions(North,West,...) to go.

Is this on the Earth, with spherical coordinates?

No.

Integer/whole number/pythagorean triples: https://en.wikipedia.org/wiki/Pythagorean_triple .

Thank you for your time.

 

[FactChecker]

This question is from a book related to high school. I think we can suppose that the person walks in a 2D plane and he has 4 directions(North,West,...) to go.

Do you mean that you must go strictly N, S, E, or W on each leg? Doesn't that just give you two feasible solutions?

 

[MatinSAR]

Do you mean that you must go strictly N, S, E, or W on each leg?

Yes. He only walks in south direction or west direction or ...

Doesn't that just give you two feasible solutions?

If we define south west as 45 degrees with west, then we can solve the question as described in post #16.

 

[Hill]

we can suppose that the person walks in a 2D plane and he has 4 directions(North,West,...) to go

Assuming 2D plane is not enough. Here is how 2D plane near the North Pole looks like:

The rays go South, the circles go East-West.

 

[MatinSAR]

Assuming 2D plane is not enough. Here is how 2D plane near the North Pole looks like:

It's not enough for the question you mentioned, But I guess it's enough to solve the main question which is related to high school. Do you agree? He doesn't walk that much(only 5 km) so the question doesn't want us to cosider that he is walking on a sphere and ...

The rays go South, the circles go East-West.

Sorry, But I cannot understand this part of your post.

 

[Hill]

It's not enough for the question you mentioned, But I guess it's enough to solve the main question which is related to high school. Do you agree?

No, I do not. What would be enough is assuming that North-South directions and East-West directions are mutually orthogonal straight lines in 2D plane.

I cannot understand this part of your post

If one starts at any point on that map and goes South, he follows a ray which goes from the North Pole and passes through the starting point.
If he goes East or West, he follows a circle which has its center at The North Pole and which passes through the starting point.