Smallest possible x-co-ordinate of
solution 2 :
Given a circle
Let co-ordinates of
Hence, smallest possible x-co-ordinate of
#"using the "color(blue)"Intersecting chords theorem"#
#"When 2 chords intersect inside a circle, the product of"#
#"their segments are equal"#
#"divide both sides by 9"#
#"subtract "n" from both sides"#
#"subtract 2 from both sides"#
See process below
A traslation is a isometric transformation, then the image for B in this movement is other circle of radius 6 and center in
Obviously doesn`t overlap both circles and the minimum distance will lay in the line joining both centers
Lets calculate line equation
This line intercept to both circles in 4 points (see graph)
Distance is calculated by the standard formula (euclidean distance)
Lets see these points:
The minimum distance will be
The general equation of a circle is:
If B is translated by
By using the distance between the centres and the radii we can deduce the following:
Using the distance formula:
Sum of radii:
So the circles intersect at two points or one is contained in the other. This can be tested by noticing that if the diameter of the smaller circle is less than the radius of the larger then the smaller circle is contained in the larger one.
Diameter of smaller circle is
Radius of larger circle is
So smaller circle is contained in the larger.
To find the shortest distance:
Given that G is the centroid of
AP median is produced such that
We're oddly given a circle in this problem and awkward wording asking for the smallest possible x coordinate.
I'm not sure what the circle has to do with anything. We could list equations endlessly for circles containing these two points. Forget about the circle.
Given one side, there will be two equilateral triangles with that as a side. Let's rewrite the problem:
Find all possible third vertices
There are a few different ways to do this. I'd lean toward complex numbers, but I probably should try to keep this simpler than that.
Let's review. The altitude
We just need to go
The midpoint D of AB is
The direction vector from A to B is
For the perpendicular direction vector we swap and negate one:
Now schematically what we're doing is
That's the general solution; let's apply it to
Uh, the one with the least x coordinate is
We check the squared distance to each point
This question is a little ambiguous. I'm guessing you mean a volume of revolution. If this is the case then a rectangle rotated about a line could be a solid cylinder or a hollow cylinder ( Tube) if the line of rotation is parallel or perpendicular to the sides of the rectangle. If the line of rotation is neither parallel nor perpendicular then a vast number of different solids could result. A big problem here is on the definition. Strictly a prism has a base that is a polygon and faces that are flat, but you could say a cylinder has a polygonal base with an in finite number of sides and an infinite number of faces.
Here are two rotations of a rectangle around different lines.
Hey, this got featured before it was really done. It's still not done.
This is another in my series where I generalize one of these old questions.
"The triangle's perpendicular bisectors" is not a term we hear that often. Each side has a perpendicular bisector, the perpendicular through the midpoint. It will occasionally intersect the opposite vertex (in which case it's the altitude of an isosceles triangle) but usually it will intersect exactly one of the other sides.
I've rewritten the question to put one vertex at the origin and to just ask for the perpendicular bisector of the opposite side.
The sides of the triangle are given parametrically as
The perpendicular family to
Two equations in two unknowns.
We're more interested in
The roles aren't quite symmetrical so we repeat the process for the meet of the bisector and
Yes I know this is getting long Socratic. It's an involved problem.
The numerators are the same in
We want to solve
Let's first assume
What about when
Still not there.
We need to show the denominators have opposite signs.
The triangle inequality says
The dot product and the sum of squares have a close relationship. We expand
We still haven't shown we pass through at most one side or the vertex. It remains to show
Still more to do.
The area of the circle is aproximately
First things first, let us visualise the points we have:
These points may look collinear, but they are not.
Now, let's draw the triangle:
And finally, the circle:
We can finally start applying formulae and such.
The area of a circle, denoted
In our case, the radius we search is the radius of the circumscribed triangle,
But how do we find
So we must find one of the sides and the sine of its opposite angle, say
So we have found
and its analogues. We are doing this to find the value of
We got to find
We can see that
From the Fundamental property of Trigonometry, we have:
If you're wondering why
Finally, we can find
So the area of the circle,
Let the equation of the parabola in standard form be
Let the coordinates of one end point A of the given focal chord AB be
AS and SB line segments are on same straight line. So their slopes must be same.
As A and B are different points then
Now it is given that
So length of the Latus rectum will be