Prove this is a right triangle in a sphere

[Sho Kano]

Homework Statement

Let P be a point on the sphere with center O, the origin, diameter AB, and radius r. Prove the triangle APB is a right triangle

Homework Equations

|AB|^2 = |AP|^2 + |PB|^2
|AB}^2 = 4r^2

The Attempt at a Solution

Not sure if showing the above equations are true is the correct way to do this problem

 

[LCKurtz]

Homework Statement

Let P be a point on the sphere with center O, the origin, diameter AB, and radius r. Prove the triangle APB is a right triangle

Homework Equations

|AB|^2 = |AP|^2 + |PB|^2
|AB}^2 = 4r^2

The Attempt at a Solution

Not sure if showing the above equations are true is the correct way to do this problem

HInt: Try drawing a picture of triangle ABP with O at the center of AB. What do you know about OA, OB, and OP?

 

[Ray Vickson]

Homework Statement

Let P be a point on the sphere with center O, the origin, diameter AB, and radius r. Prove the triangle APB is a right triangle

Homework Equations

|AB|^2 = |AP|^2 + |PB|^2
|AB}^2 = 4r^2

The Attempt at a Solution

Not sure if showing the above equations are true is the correct way to do this problem

The triangle ABP lies in a plane through the center of the sphere, so turns the three-dimensional problem into a 2-dimensional (planar) problem involving a circle of radius $r$.

 

[Sho Kano]

Here's a very bad drawing (I forgot to insert the origin in, but let's just say it's there) (and let's say that the angle between BP and BA is x)

Is this sufficient?
|AB|^2 = |AP|^2 + |PB|^2
2r = d
|AP| = dsin(x)
|PB| = dcos(x)
|AB|^2 = d^2

d^2 = d^2 (cos(x)^2 + sin(x)^2)
d^2 = d^2

 

[LCKurtz]

View attachment 105538
Here's a very bad drawing (I forgot to insert the origin in, but let's just say it's there) (and let's say that the angle between BP and BA is x)

Is this sufficient?
|AB|^2 = |AP|^2 + |PB|^2

You can't start with the Pythagorean theorem because you don't know that is a right triangle. That is what you are supposed to prove.

2r = d
|AP| = dsin(x)
|PB| = dcos(x)
|AB|^2 = d^2

d^2 = d^2 (cos(x)^2 + sin(x)^2)
d^2 = d^2

You aren't trying to prove $d^2 = d^2$.

 

[Sho Kano]

You can't start with the Pythagorean theorem because you don't know that is a right triangle. That is what you are supposed to prove.
You aren't trying to prove $d^2 = d^2$.

But if it satisfies the pythagorean theorem then that must mean it's a right triangle right?

 

[LCKurtz]

At the risk of repeating myself: You DON'T KNOW it satisfies the Pythagorean theorem because you don't know $x$ is a right angle.

 

[RUber]

What methods and/or tools have you been learning recently that would be applicable to this problem?

Are you allowed to use the inscribed angle theorem?
https://en.wikipedia.org/wiki/Inscribed_angle

Or would using isosceles triangle rules be more appropriate?

 

[Sho Kano]

What methods and/or tools have you been learning recently that would be applicable to this problem?

Are you allowed to use the inscribed angle theorem?
https://en.wikipedia.org/wiki/Inscribed_angle

Or would using isosceles triangle rules be more appropriate?

I'm taking vector calculus and our teacher assigned this question for homework. I guess the methods we have been learning is the dot product and cross product and various vector properties.

 

[Sho Kano]

At the risk of repeating myself: You DON'T KNOW it satisfies the Pythagorean theorem because you don't know $x$ is a right angle.

I don't know where to start. Could you give me a hint? I know that the magnitude of OP where P is any Point in the circle is r.

 

[LCKurtz]

I don't know where to start. Could you give me a hint? I know that the magnitude of OP where P is any Point in the circle is r.

This problem doesn't require any calculus at all. You have a triangle with isosceles sub-triangles. Label the angles. Almost anything you write down about the various angles should lead you to the solution.

 

[Sho Kano]

This problem doesn't require any calculus at all. You have a triangle with isosceles sub-triangles. Label the angles. Almost anything you write down about the various angles should lead you to the solution.

By sub-triangles, you mean the triangle inscribed within the bigger one? How do I know that those triangles are isosceles?

 

[LCKurtz]

By sub-triangles, you mean the triangle inscribed within the bigger one? How do I know that those triangles are isosceles?

Here is what you wrote in post #10:

"I don't know where to start. Could you give me a hint? I know that the magnitude of OP where P is any Point in the circle is r."

 

[Sho Kano]

Here is what you wrote in post #10:

"I don't know where to start. Could you give me a hint? I know that the magnitude of OP where P is any Point in the circle is r."

Don't see how that automatically makes a triangle inscribed within triangle APB an isosceles. I don't see it.

 

[LCKurtz]

Draw the picture I suggested in post #2, including the point O. The smaller triangles all have the point O as a vertex. Use the ideas you have been given. I can't do any more for you without working the problem for you.

 

[RUber]

Look at it as two triangles, AOP and POB. Let one angle at O (say angle AOP) have measure z, then what is the measure of angle POB?
Notice that AO is a radius, and OP is a radius. This should give you a clue about what LCKurtz was talking about.
Remember that isosceles triangles have equal base angles and the sum of interior angles sum to 180.
This should be enough info for you to do the proof geometrically.

Using vectors, if you use the unit circle with diameter points at (-1,0) and (1,0) for simplicity, then for any point where $\theta \neq 0, \pi$, your point P is at $\cos \theta \hat x + \sin \theta \hat y$. You can use the dot product of the vectors describing the legs to show that the vectors are orthogonal.
You could generalize to a non-unit circle by using diameter points at (-r,0) and (r,0) and point P at $r\cos \theta \hat x + r\sin \theta \hat y$.

 

[Sho Kano]

Think I got it, from the figure on post #4
PA\cdot PB=0\\ PA=OA-OP\\ PB=OB-OP\\ (OA-OP)\cdot (OB-OP)=0\\ OA\cdot OB-OA\cdot OP-OP\cdot OB+OP\cdot OP=0\\ OA=-OB,\quad OB\cdot OB={ r }^{ 2 },\quad OP\cdot OP={ r }^{ 2 }\\ -{ r }^{ 2 }-OA\cdot OP-OP\cdot OB+{ r }^{ 2 }=0\\ OP\cdot (-OA-OB)=0\\ OP\cdot (-OA-(-OA))=0\\ OP\cdot (O)=0

 

[RUber]

Think I got it, from the figure on post #4
PA\cdot PB=0\\ PA=OA-OP\\ PB=OB-OP\\ (OA-OP)\cdot (OB-OP)=0\\ OA\cdot OB-OA\cdot OP-OP\cdot OB+OP\cdot OP=0\\ OA=-OB,\quad OB\cdot OB={ r }^{ 2 },\quad OP\cdot OP={ r }^{ 2 }\\ -{ r }^{ 2 }-OA\cdot OP-OP\cdot OB+{ r }^{ 2 }=0\\ OP\cdot (-OA-OB)=0\\ OP\cdot (-OA-(-OA))=0\\ OP\cdot (O)=0

It looks like you are starting with what you want to prove. Other than that, I think your expansion and logic look good.

 

[ehild]

Think I got it, from the figure on post #4

If you do what LCKurtz (Post #15) suggested and draw the line OP, it cuts the triangle into two isosceles ones. The angle APB is β+γ. What is its value?

 

[Sho Kano]

If you do what LCKurtz (Post #15) suggested and draw the line OP, it cuts the triangle into two isosceles ones. The angle APB is β+γ. What is its value?
View attachment 105648

I see why they are isosceles now, then
2γ = α
2γ + 2β = 180
γ + β = 90
This way is much faster haha

 

[RUber]

The fastest way I have found is to use the inscribed angle theorem. Not much to show there, though.
"The inscribed angle theorem states that an angle θ inscribed in a circle is half of the central angle 2θ that subtends the same arc on the circle."
A diameter is an example of a 180 degree central angle, so the inscribed angle at point P must be 1/2 of that.