Featured 4 days ago

Smallest possible x-co-ordinate of

solution 2 :

Given a circle

let

as

slope of

Let co-ordinates of

Hence, smallest possible x-co-ordinate of

Featured 1 week ago

#"using the "color(blue)"Intersecting chords theorem"#

#"When 2 chords intersect inside a circle, the product of"#

#"their segments are equal"#

#rArr18(n+1)=9(n+7)#

#"divide both sides by 9"#

#rArr2(n+1)=n+7#

#rArr2n+2=n+7#

#"subtract "n" from both sides"#

#2n-n+2=7#

#rArrn+2=7#

#"subtract 2 from both sides"#

#rArrn=7-2=5#

Featured 1 week ago

See process below

Equation for

Equation for

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

Featured 1 week ago

The general equation of a circle is:

Where:

Circle A

If B is translated by

Centre:

Circle B:

By using the distance between the centres and the radii we can deduce the following:

Let:

If:

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:

PLOT:

Featured 5 days ago

option (1)

**Q92**

Given that G is the centroid of

AP median is produced such that

As

Again

So

Now

and

So

So

So

Area of

Featured 4 days ago

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

We have

That's the general solution; let's apply it to

Uh, the one with the least x coordinate is

Check:

We check the squared distance to each point

Featured 3 days ago

See below.

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.

Featured 2 days ago

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 **bisector** is through the midpoint of AC:

That meets

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

We're assuming

What about when

How about

For

Unless

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.

Featured 2 days ago

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 **Law of Sines**:

So we must find one of the sides and the sine of its opposite angle, say

The line

So we have found **Law of cosine**, which claims that, in any triangle:

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,

Featured yesterday

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.

Hence

As A and B are different points then

So

Now it is given that

So

Inserting

Now

Inserting

So length of the Latus rectum will be

**A shortcut**

As

So