## Wednesday, December 9, 2015

### Unizor - Geometry2D - Apollonius Problems - Lines and Circles

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

The problems below are easily solvable using the transformation of inversion (symmetry relative to a circle). We recommend to review a previous lecture dedicated to this topic.

The main technique we will use is transformation of a circle passing through a center of an inversion circle into a straight line.

Here is a quick reminder of how inversion works.

An inversion circle with radius R and center O is given. Then any point P (other than center O) is transformed into point P' such that P' lies on a ray from center O to given point P and OP·OP'=R²

We have proven that any circle passing through a center of an inversion circle is transformed by an inversion into a straight line and a circle not passing through a center of inversion is transformed into a circle.

Apollonius Problems -

Lines and Circles

Problem LLC

Construct a circle tangential to two given lines, a and b, and tangential to a given circle c, presuming that both lines are located outside of circle c.

Solution

Analysis:

Assume, our circle is constructed. It is tangential to lines a and b and to circle c.

We can easily reduce this problem to the one we solved before - construction of a circle tangential to two lines and passing through a point.

To accomplish this, increase the radius of a presumably constructed circle by the radius of a given circle and shift both given lines, a and b, away from a given circle in between them by the same value to new positions - a' and b'.

Now a new circle, concentric with the one we have to construct and having a radius greater than that by a radius of a given circle c, will be tangential to lines a' and b' and passing through a center of a given circle c.

Construction:

From a center of circle c drop a perpendicular onto line a and extend it by the radius of circle c beyond its base on line a to point A. Draw line a' parallel to line a through point A.

From a center of circle c drop a perpendicular onto line b and extend it by the radius of circle c beyond its base on line b to point B. Draw line b' parallel to line b through point B.

Construct a circle tangential to lines a' and b' and passing through a center of c.

Decrease the radius of this new circle by a radius of circle c leaving a center in place. This is a required circle.

Problem LCC

Construct a circle tangential to a given line a and tangential to two given circles, c and d, presuming that none of the given elements lies inside the other.

Solution

Analysis:

Assume, our circle is constructed. It is tangential to line a and to circles c and d.

Increase the radius of a presumably constructed circle by the smaller radius among two given circles (let's say, it's circle c), shift line a away from circle c parallel to itself by this same radius to a new positions a', and concentrically shrink circle d by the same radius to a new circle d'.

Now a new circle, concentric with the one we have to construct and having a radius greater than that by a radius of a given circle c, will be tangential to line a', to circle d' and passing through a center of a given circle c.

We can construct it by introducing an inversion with a center at the center of circle c that transforms line a' into some circle aa, circle d' - also into some circle dd, and the circle concentric to the one we need to construct - into a line tangential to these two circles. aa and dd.

Construction:

From a center of circle c drop a perpendicular onto line a and extend it by the radius of circle c beyond its base on line a to point A. Draw line a' parallel to line a through point A.

Draw a circle d' concentric to circle d with a radius smaller than original by a radius of circle c.

Draw an inversion circle with a center at the center of circle c and invert relative to it line a' into circle aa and circle d' into circle dd.

Construct a line tangential to circles aa and dd.

Invert this line into a circle.

Decrease the radius of this new circle by a radius of circle c leaving a center in place. This is a required circle.

## Tuesday, December 8, 2015

### Unizor - Geometry2D - Apollonius Problems - Points and Circles

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

The problems below are easily solvable using the transformation of inversion (symmetry relative to a circle). We recommend to review a previous lecture dedicated to this topic.

The main technique we will use is transformation of a circle passing through a center of an inversion circle into a straight line.

Here is a quick reminder of how inversion works.

An inversion circle with radius R and center O is given. Then any point P (other than center O) is transformed into point P' such that P' lies on a ray from center O to given point P and OP·OP'=R²

We have proven that any circle passing through a center of an inversion circle is transformed by an inversion into a straight line and a circle not passing through a center of inversion is transformed into a circle.

Apollonius Problems -

Points and Circles

Problem PPC

Construct a circle passing through two given points, A and B, and tangential to a given circle c, presuming that both points are located outside of circle c.

Solution

Analysis:

Assume, our circle is constructed. It passes through points A and B and is tangential to circle c.

Using point A as a center of inversion and any radius draw a circle that will serve as an inversion circle.

Transform all elements relative to this inversion circle.

Point B will transform into point B', circle c will be transformed into circle c' and a circle we assumed as constructed (passing through a center of inversion A) will be transformed into a line that will be a tangent from point B' to circle c'.

Construction:

Draw an inversion circle q with center at point A and some radius R.

Construct images of point B and circle c relative to inversion circle q.

Draw a tangent from point B' to circle c'.

Transform a tangent through inversion relative to our inversion circle q. The resulting circle would be tangential to circle c and pass through points A and B.

Problem PCC

Construct a circle passing through a given points, A and tangential to two given circles, c and d, presuming that none of the given elements lies inside the other.

Solution

Analysis:

Assume, our circle is constructed. It passes through point A and is tangential to circles c and d.

Using point A as a center of inversion and any radius draw a circle that will serve as an inversion circle.

Transform all elements relative to this inversion circle.

Circle c will be transformed into circle c', circle d will be transformed into circle d', and a circle we assumed as constructed (passing through a center of inversion A) will be transformed into a line that will be a tangent to both circles c' and d'.

Construction:

Draw an inversion circle q with center at point A and some radius R.

Construct images of circles c and d relative to inversion circle q.

Draw a tangent to circles c' and d'.

Transform a tangent through inversion relative to our inversion circle q. The resulting circle would be tangential to circles c and d.

## Monday, December 7, 2015

### Unizor - Geometry2D - Inversion

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

Inversion

Before addressing construction problems of Apollonius, where a circle is one of the given elements, to which a constructed circle must be tangential, we will define a special transformation on a plane that allows to convert these types of problems into problems that involve only lines and points.

In other words, we will try to simplify our task using this transformation and to reduce its solution to a known solution with only points and lines given.

This is a rather clever method which is not easy to come up with by yourself. Yet, it is relatively simple and allows easily to solve Apollonius problems that involve given circles.

The transformation we would like to address is called inversion or symmetry relatively to a circle.

Any transformation of the points on a plane prescribes how each point is transformed. For instance, symmetry relatively to a straight line transforms all points lying on one side of a plane (let's call it here "left of the symmetry line") to points on another side ("right of the symmetry line") and all points from the "right" side to corresponding points on the "left". This transformation has a rule that the source and the image points lie on a perpendicular to a symmetry line and on the same distance from it. Points on the symmetry line are transformed into themselves.

The definition of inversion is somewhat similar, but slightly more complex. We divide the plane by a circle of certain radius R with a center O (inversion circle) into two parts - inside and outside of this circle.

All points of the inside area (except a center of an inversion circle) are transformed into points outside and all outside points are transformed to inside. The rule of transformation is as follows.

For any point P inside of an inversion circle we connect a center of an inversion circle with this point by a ray and find on this ray on the outside of a circle such a point P' that

OP·OP' = R²

Similarly, for any point P' outside of an inversion circle we find point P inside that satisfies the same condition.

Obviously, if point P inside an inversion circle is transformed into point P', then this point P' is transformed by an inversion to point P. Also obvious is that all points on the inversion circle are transformed into themselves. These properties qualify the transformation of inversion to be called "symmetry relative to a circle".

The closer point P lies to a center - the farther from a center will be P' to preserve the main rule of inversion.

The center of an inversion circle does not participate in the transformation, though sometimes one might say that a center is transformed into "infinity", which only means that, as we move point P infinitely closer to a center, its image P' moves infinitely far from a center.

Inversion of a Straight Line

Theorem

Any straight line lying outside of an inversion circle is transformed into a circle inside the inversion circle that passes through a center of inversion.

Inversion of a Circle

Lets consider a circle of radius r concentric with an inversion circle and lying inside it. For any point P on it the segment OP will have a length r. Therefore, the distance from a center of an inversion circle to an image of this point P' must be R²/r for the main rule of inversion to hold. As point P goes around a circle of radius r, its image P' should always be on a distance R²/r from point O - that is, all image points must lie on a circle of radius R²/r concentric with an inversion circle.

We see now that a circle, concentric with an inversion circle is transformed into another concentric circle.

The interesting theorem we are going to prove is that the fact of transformation of a circle into a circle is much broader and expands to circles that are not concentric with an inversion circle. This is the main subject of this lecture.

Theorem

Any circle lying inside an inversion circle and not passing through its center is transformed into a circle outside the inversion circle.

## Tuesday, December 1, 2015

### Unizor - Geometry3D - Spheres - Problems 3

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

Spheres - Problems 3

Just as a reminder, here are a few formulas related to spheres. They are all derived in the previous lectures and are helpful in solving these problems.

Volume of a sphere: 4πR³/3

Surface area of a sphere: 4πR²

Volume of a cap: πH²(3R−H)/3

Surface area of a dome: 2πRH

Volume of a sector: 2πR²H/3

Problem A

Diameter of Jupiter is 11 times greater than diameter of Earth.

Diameter of Mars is half of diameter of Earth.

What is the ratio of surface and volume of Jupiter to those of Mars, assuming all the planets are spherical?

Answer:

The ratio of the surface areas of Jupiter to that of Mars is 484:

SJupiter = 484·SMars

The ratio of their volumes is 10648:

VJupiter = 10648·VMars

Problem B

A cylindrical glass of radius R is filled with water to the middle.

A ball (heavier than water) of a radius r (smaller than R) is dropped into this glass.

By how much the height of the water in a glass will rise?

Answer:

The water will rise by

h = 4r³/(3R²)

Problem C

An air-filled ball made of some heavier than water material is floating in the water such that exactly half of it is above the water.

According to Archimedes' principle of flotation, the weight of a ball equals to the weight of the water it displaced.

The ball's material unit weight is B, the water's unit weight is W.

Assume that the air inside the ball is weightless.

The radius of a ball is R

Construct the equation to solve to determine the width of the ball's wall.

Solution:

Assume that the width of the ball's wall is X. Then the internal radius of the ball is R−X.

The ball's weight equals to its unit weight B multiplied by the volume occupied by its wall, which is a difference between the volumes of two spheres, one has an external diameter of a ball, another is inside it and has its internal diameter:

B·[4πR³/3−4π(R−X)³/3]=

= 4πB[R³−(R−X)³]/3

This weight should be equal to the weight of displaced water, which has a unit weight W and the volume equal to half of the volume of a ball:

W·2πR³/3

The above two weights, that of the ball and that of the displaced water, must be equal, which produces the following equation for X:

4πB·[R³−(R−X)³]/3 =

= 2πW·R³/3

Simplifying this equation, we get

2B·[R³−(R−X)³] = W·R³

Generally speaking, this is an equation of the third degree and can be solved numerically.

Problem D

A sphere is inscribed into a regular tetrahedron, it's tangential to all its four faces.

Prove that the radius of this sphere is 1/4th of the altitude of a tetrahedron.

### Unizor - Geometry2D - Apollonius Problems - Points and Lines

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

Generally speaking, all problems related to construction of a circle tangential to three other geometrical objects, points, straight lines or circles, is referred to as one of the Apollonius problems. Obviously, there is a history behind it. We will not talk about this history and, for those interested, the Web is your best source for details.

Here we will spend some time to address these problems using some old, previously addressed, and some new techniques, wherever they are applicable.

Apollonius Problems 1

Problem PPP

Construct a circle passing through three given points, A, B and C, not lying on the same line.

Solution

This problem had been addressed before. Here is a recap.

A locus of points equidistant from two given points is a line passing through a midpoint of a segment connecting these two points and perpendicular to it (prove it!).

If three points A, B and C are not lying on the same line, the midpoint perpendicular to AB intersects the midpoint perpendicular to BC (prove it!), and their intersection is equidistant from all three points - A, B and C.

Incidentally, the consequence of this is that a perpendicular to a midpoint of AC also passes through the same intersection point, so all three midpoint perpendiculars intersecting at one point - a center of a circle passing through three given points (prove it!).

Problem PPL

Construct a circle passing through two given points, A and B, and tangential to a straight line d, presuming that both points are located on the same side from the line.

Solution

Solution is based on the Tangent-Secant Power theorem that states that, if from a point P outside a circle drawn a tangent with a point of tangency T and a secant intersecting a circle at points A and B, then

PA·PB = PT² (prove it!).

Given PA and PB, we can find PT as an altitude towards a hypotenuse in a right triangle, where the base of this altitude divides a hypotenuse into segments equal to PA and PB.

Knowing that, in case AB is not parallel to line d, we find their intersection P and find PT, thereby determining a point of tangency of line d and a circle.

Once this point T is found, two segment bisectors for segments AB and AT would give a center of a circle.

If AB ∥ d, point T lies on intersection of d and segment AB's bisector.

Problem PLL

Construct a circle passing through a given point A and tangential to two straight lines p and q.

Solution

A solution can be based on a transformation of scaling.

Consider a general case of an angle formed by lines p and q with a vertex M and point A inside this angle.

Assume a circle tangential to lines p and q is constructed and it passes through point A.

Connect vertex M with point A and consider a scaling of this picture relative to center of scaling positioned at vertex M with any factor. Obviously, lines p and q will be transformed into themselves, an old circle tangential to these lines will be transformed into a new circle also tangential to the same lines, and point A on an old circle will be transformed into another point A' lying on line MA an on a new circle.

Based on this, a construction can be started with any circle tangential to lines p and q (its center is, obviously, on a bisector of an angle formed by these lines). Consider a line MA and its intersection with this circle at point A' (two points will lead to two solutions).

Now all we have to do is apply a scaling that transforms point A' into point A. This scaling will transform a circle we started with into the one needed.

A case of parallel lines p and q is trivial since we can easily determine the radius of a circle we want to construct - it is half the distance between these lines.

Problem LLL

Construct a circle tangential to three straight lines p, q and r.

Solution

Lines should not be all parallel. If two are, the case is simple since we know the radius of a circle. So, consider a general case of three lines forming a triangle.

In this general case we deal with a problem of inscribing a circle into a triangle. As we know, it's center lies on intersection of its angles' bisectors.

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