Circles of Apollonius of a Triangle

Let’s investigate all triangles in which one side length is twice another side length and the third side length is a constant length of 9. Some triangles that fit these requirements would include ones with integer side lengths (4, 8, 9), (5, 10, 9), (6, 12, 9), (7, 14, 9), and (8, 16, 9):

Placing each of these triangles one on top of the other, it would appear that the third vertex follows a circular path:

and can be animated as follows:

Is this path always a circle, or is this just a coincidence with the values that we picked? 

This is no coincidence.  The Greek mathematician Apollonius showed that any circle can be defined as the set of points with a constant ratio of distances to two set points.  Conversely, we can prove that a set of points which maintains a constant non-one ratio of distances to two fixed points will always form a circle. 

If the constant ratio is k = ^b/_a and the distance between the two fixed points is c so that a < b < c, then we can make △ABC with side lengths a, b, and c and with opposite angles ∠A, ∠B, and ∠C, so that ∠B is at the origin and side c is along the x-axis so that the coordinate of A is (c, 0).

Then the locus of points C that are ak away from B and bk away from A can be expressed by the distance equation as x² + y² = a²k² and (x – c)² + y² = b²k².  Rearranging and combining gives a²b²k² = b²x² + b²y² = a²(x – c)² + a²y², and further rearranging gives (x + ^{a^2 c}/_{b^2 – a^2})² + y² = (^abc/_{b^2 – a^2})², a circle equation with a radius of r = ^abc/_{b^2 – a^2}. Therefore, vertex C will always lie on the path of a circle.

In fact, three circles of Apollonius can be drawn for any scalene triangle, one for each vertex:

Using the same argument as above, the radii of the three circles of Apollonius will be r₁ = ^abc/_{c^2 – a^2}, r₂ = ^abc/_{c^2 – b^2}, and r₃ = ^abc/_{b^2 – a^2}.  

Since ¹/_r1 = ¹/_abc(c² – a²), ¹/_r2 = ¹/_abc(c² – b²), and ¹/_r3 = ¹/_abc(b² – a²), and since ¹/_abc(c² – a²) = ¹/_abc(b² – a²) + ¹/_abc(c² – b²), we can derive the elegantly compact relation:

where r₁ < r₂ and r₁ < r₃.

Another way to write this equation is r₂^-1 + r₃^-1 = r₁^-1, which places it in an increasing list of geometric equations that are in the form of xⁿ + yⁿ = zⁿ, the most famous being a² + b² = c², the Pythagorean Theorem.  Other equations in that form include the relationship between the major axis, minor axis, and focal length of ellipses and hyperbolas (b² + c² = a² and a² + b² = c²), the radii of three mutually tangent circles and a line  (r₁^-½ + r₂^-½ = r₃^-½), and, of course, the addition of segments (a¹ + b¹ = c¹).

It is amazing that over a thousand years ago Apollonius showed that any circle can be defined as the set of points with a constant ratio of distances to two set points.  It is also amazing that the simple relationship of ¹/_r1 = ¹/_r2 + ¹/_r3 is true for the radii of the three circles of Apollonius for any scalene triangle, despite the complicated steps in between.

The Imaginary Part of a Parabola

Parabolas are commonly taught in most high school algebra classes. All parabolas are curves defined by a quadratic equation and are symmetrical, open in one direction, and have a vertex.  Generally, a parabola defined by a quadratic equation in standard form y = ax² + bx + c has a line of symmetry at x = ^-b/_2a, opens upward if a is positive (but downward if a is negative), and has a vertex at (^-b/_2a, ^{-b^2 + 4ac}/_4a).  For example, a parabola defined by the quadratic equation y = x² – 6x + 13 have variables a = 1, b = -6, and c = 13, which means it has a line of symmetry at x = ^-b/_2a = ^--6/_2·1 = 3, opens upward because a is positive, and has a vertex at (^-b/_2a, ^{-b^2 + 4ac}/_4a) = (^--6/_2·1, ^{-(-6)^2 + 4·1·13}/_4·1) = (3, 4). 

Some other points that are on the parabolic curve can be found by using the given equation.  For example, in the equation y = x² – 6x + 13, when x = 0, y = 0² – 6·0 + 13 = 13, so (0, 13) is on the curve, and when x = 1, y = 1² – 6·1 + 13 = 8, so (1, 8) is on the curve.  Similar calculations can be made to show that (2, 5), (3, 4), (4, 5), (5, 8), (6, 13), (and so on) are also on this parabolic curve.

However, we can also find imaginary points on any parabolic curve as well.    In the equation y = x² – 6x + 13, when y = 0, then 0 = x² – 6x + 13, and using the quadratic formula x = ^{–b ± √(b^2 – 4ac)}/_2a with variables a = 1, b = -6, and c = 13, yields x = ^{–-6 ± √((-6)^2 – 4·1·13)}/_2·1 = ^{6 ± √(36 – 52)}/₂ = ^{6 ± √-16}/₂ = ^{6 ± 4i}/₂ = 3 ± 2i, so (3 + 2i, 0) and (3 – 2i, 0) should also be on the curve.  Similarly, when y = 3, then 3 = x² – 6x + 13 or 0 = x² – 6x + 10, and using the quadratic formula x = ^{–b ± √(b^2 – 4ac)}/_2a with variables a = 1, b = -6, and c = 13 – 3 = 10, yields x = ^{–-6 ± √((-6)^2 – 4·1·(13 – 3))}/_2·1 = ^{6 ± √(36 – 40)}/₂ = ^{6 ± √-4}/₂ = ^{6 ± 2i}/₂ = 3 ± i, so (3 + i, 3) and (3 – i, 3) should also be on the curve.  Similar calculations can be made to show that (3 + 3i, -5), (3 – 3i, -5), (3 + 4i, -12), (3 – 4i, -12), (and so on) are also on the parabolic curve.

But how can you graph these imaginary points on a coordinate graph?  In an Argand graph, the imaginary part is graphed on a separate axes.  But since both the x-axis and y-axis are already being used, we will have to add a third dimensional z-axis to represent the imaginary part.  This results in the following graph for y = x² – 6x + 13:

It appears that the imaginary part is the same size as the real part, but reflected horizontally at the vertex and then rotated 90° into the third dimension.

Depicting three-dimensional graphs on a two-dimensional medium is difficult to do, even with the help of technology.  To make this easier to draw, let’s rotate the imaginary part back into two dimensions by letting the x-axis serve a double purpose of defining x-values and imaginary values.  This means that in our y = x² – 6x + 13 example, the coordinate (3 – i, 3) would transform to (3 – 1, 3) = (2, 3); the coordinate (3 + i, 3) would transform to (3 + 1, 3) = (4, 3); the coordinate (3 – 2i, 0) would transform to (3 – 2, 0) = (2, 0); the coordinate (3 + 2i, 0) would transform to (3 + 2, 0) = (5, 0); and so on; resulting in the following graph:

This transformation graphically changed the imaginary part to a real part.  We can denote this algebraically by multiplying square root part of the quadratic formula by i.  In other words, the imaginary part of any quadratic equation y = ax² + bx + c (or 0 = ax² + bx + c – y) can be represented by new x-values such that x = ^{–b ± i√(b^2 – 4a(c – y))}/_2a.  Solving for y results in another quadratic equation:

x = –b ± i√(b^2 – 4a(c – y))/2a	(x-values of transformation)
2ax = –b ± i√(b2 – 4a(c – y))	(multiply by 2a)
2ax + b = ±i√(b2 – 4a(c – y))	(add b)
(2ax + b)2 = -1(b2 – 4a(c – y))	(square both sides)
4a2x2 + 4abx + b2 = -b2 + 4ac – 4ay	(distribute)
4ay + 4a2x2 + 4abx + b2 = -b2 + 4ac	(add 4ay)
4ay + 4a2x2 + 4abx = -b2 + 4ac	(subtract b2)
4ay + 4a2x2 + 4abx = 4ac – b2	(rearrange b2)
4ay + 4a2x2 = -4abx + 4ac – 2b2	(subtract 4abx)
4ay = -4a2x2 – 4abx + 4ac – 2b2	(subtract 4a2x2)
y = -ax2 – bx + c – b^2/2a	(divide by 4a)

This proves that the imaginary part of a parabola is another parabola.  (A similar proof can be used to show that the imaginary part of a hyperbola is an ellipse, and that the imaginary part of an ellipse is a hyperbola.)  It also gives us a fast way to forcibly graph the imaginary part of a parabola on a graphing calculator (or some other technology) that would not normally do so.  For example, to graph the full graph of y = x² – 6x + 13, where a = 1, b = -6, and c = 13, we should also graph the imaginary part at the same time using the imaginary transformation equation y = -ax² – bx + c – ^{b^2}_/2a or y = -1x² – (-6)x + 13 – ^{(-6)^2}_/2·1 or y = -x² + 6x – 5.  Graphing y = x² – 6x + 13 (in bold) and y = -x² + 6x – 5 gives the full graph, both real and imaginary, of the parabola:

A great application for graphing the imaginary part of the parabola is to use it as a visual aid for finding the number and types of solutions to a quadratic equation, a common objective in most high school algebra classes.  To find the number and types of solutions to a quadratic equation, students are traditionally taught to calculate the discriminant, which is the part under the square root of the quadratic formula, namely b² – 4ac.  If the discriminant is negative, the square root will result in an imaginary number, and the quadratic formula will yield two imaginary solutions and no real solutions.  If the discriminant is equal to zero, the square root will also be equal to zero, and the quadratic formula will yield one real solution and no imaginary solutions.  Finally, if the discriminant is positive, the square root will also be positive, and the quadratic formula will yield two real solutions and no imaginary solutions.

Discriminant	Solutions
-	0 real, 2 imaginary
0	1 real, 0 imaginary
+	2 real, 0 imaginary

This approach is rather abstract, but the concept can now be reinforced by graphing both real and imaginary parts of the parabola, and examining which part intersects with the x-axis.  For example, let’s say we were asked to find the number and types of solutions to x² – 6x + 13 = 0.  The discriminant is b² – 4ac = (-6)² – 4·1·13 = -16, which is negative, so it will have 0 real solutions and 2 imaginary solutions.  Graphing y = x² – 6x + 13 (with its imaginary transformation equation y = -ax² – bx + c – ^{b^2}_/2a or y = -1x² – (-6)x + 13 – ^{(-6)^2}_/2·1 or y = -x² + 6x – 5) shows that the real part of the parabola does not cross the x-axis but the imaginary part of the parabola crosses the x-axis twice, visually reinforcing the result that there are 0 real solutions and 2 imaginary solutions. 

Using a different example, let’s say we were asked to find the number and types of solutions to x² – 6x + 9 = 0.  Now the discriminant is b² – 4ac = (-6)² – 4·1·9 = 0, so it will have 1 real solution and 0 imaginary solutions.  Graphing y = x² – 6x + 9 (with its imaginary transformation equation y = -ax² – bx + c – ^{b^2}_/2a or y = -1x² – (-6)x + 9 – ^{(-6)^2}_/2·1 or y = -x² + 6x – 9) shows that the vertex crosses the x-axis exactly once.  Recall that the vertex is real, not imaginary, visually reinforcing the result that there is 1 real solution and 0 imaginary solutions. 

Lastly, let’s say we were asked to find the number and types of solutions to x² – 6x + 5 = 0.  Now the discriminant is b² – 4ac = (-6)² – 4·1·5 = 16, which is positive, so it will have 2 real solutions and 0 imaginary solutions.  Graphing y = x² – 6x + 5 (with its imaginary transformation equation y = -ax² – bx + c – ^{b^2}_/2a or y = -1x² – (-6)x + 5 – ^{(-6)^2}_/2·1 or y = -x² + 6x – 13) shows that the real part of the parabola crosses the x-axis twice but the imaginary part of the parabola does not cross the x-axis, visually reinforcing the result that there are 2 real solutions and 0 imaginary solutions. 

x2 – 6x + 13 = 0 0 real solutions 2 imaginary solutions	x2 – 6x + 9 = 0 1 real solution 0 imaginary solutions	x2 – 6x + 5 = 0 2 real solutions 1 imaginary solution

In summary, a quadratic equation is actually comprised of two curves – one real and one imaginary.  The real curve is the traditional parabolic curve.  The imaginary curve is the same size and shares the same vertex as the real curve, but is reflected horizontally and rotated 90° into the third dimension.  Rotating the imaginary curve back into two dimensions helps make it easier to graph both real and imaginary parts of the quadratic equation, and also allows us to quickly determine the number and type of solutions by examining where it intersects with the x-axis.

Areas of Conics

Most of us know area formulas for basic geometrical objects, such as circles, squares, rectangles, and triangles.  (The area of a circle is A = πr², the area of a square is A = s², the area of a rectangle is A = lw, and the area of a triangle is A = ½bh.)  Others of us may even remember learning area formulas of other geometrical objects in high school geometry, such as parallelograms, trapezoids, and equilateral triangles.  (The area of a parallelogram is A = bh, the area of a trapezoid is ½(b₁ + b₂)h, and the area of an equilateral triangle is ¼√3s².)

Other than the circle, the area of conics are neglected in most geometry textbooks, even though the Cartesian equation for each conic are included in most algebra textbooks.  The area of a parabola is A = ²/₃bh (where b and h are the base and height of the rectangle that contains it).

The area of an ellipse is A = πab (where a and b are the major and minor axes of the ellipse). 

The area of a hyperbola is A = ab(^h+a/_a√(( ^h+a/_a)² – 1) – ln|^h+a/_a + √(( ^h+a/_a)² – 1)|) (where a and b are the axes of the hyperbola and h is the height).
 

The area formulas for the parabola and ellipse are relatively simple, the equation for the hyperbola is not.

To prove the area of a parabola, we can use integration.  The Cartesian equation for a parabola with its vertex at the origin is y = ax², which means another point on the parabola would be (b, ab²). 

The area of a rectangle with opposite vertices on the origin and (b, ab²) is A = b∙ab² = ab³.  The area of the parabola inside this rectangle is A = ∫₀^b ab² – ax² dx = ab²x – ¹/₃ax³] ₀^b = ab²b – ¹/₃ab³ = ab³ – ¹/₃ab³ = ²/₃ab³.  (The left half of the parabola would be the same by symmetry.)  Therefore, the area of the parabola is ²/₃ of the area of the rectangle that contains it, or A = ²/₃bh.

To prove the area of an ellipse, we can compare its Cartesian equation to the Cartesian equation of a circle.  The Cartesian equation for an ellipse centered at the origin where a and b are the major and minor axes is ^{x^2}/_a^2 + ^{y^2}/_b^2 = 1, which rearranged is y = ^b/_a√(a² – x²), and the Cartesian equation for a circle centered at the origin where a is a radius is x² + y² = a², which rearranged is y = √(a² – x²), so the difference in heights is a factor of ^b/_a. 

    

So if the area of a circle is A = πr², or in this case A = πa², the area of an ellipse must be A = ^b/_aπa² = πab.  (Technically speaking, the area of an ellipse is 4∫₀^{a b}/_a√(a² – x²)dx = ^b/_a 4∫₀^a √(a² – x²)dx = ^b/_aπa² = πab.)

To prove the area of the hyperbola, we can use integration once again.  The Cartesian equation for a horizontal hyperbola centered at the origin where a and b are the major and minor axes is ^{x^2}/_a^2 – ^{y^2}/_b^2 = 1, which rearranged is y = ^b/_a√(x² – a²). 

The area is then A = 2∫_a^{h+a b}/_a√(x² – a²)dx.  Using a right triangle with legs a and √(x² – a²) and hypotenuse x, tan q = ^{√(x^2 – a^2)}/_a or √(x² – a²) = a tan q, and sec q = ^x/_a or x = a sec q, which means dx = a sec q tan q dq. 

Using substitution, A = 2∫_sec-1(a)^{sec-1(h+a) b}/_a a tan q a sec q tan q dq = 2ab∫₀^{sec-1((h+a)/a)} tan² q sec q dq = 2ab∫₀^{sec-1((h+a)/a)} (sec² q – 1)sec q dq = 2ab∫₀^{sec-1((h+a)/a)} sec³ q – sec q dq.  Integrating results in 2ab[½ sec q tan q + ½ ln|sec q + tan q| – ln|sec q + tan q|] ₀^{sec-1((h+a)/a)}  = ab[sec q tan q – ln|sec q + tan q|] ₀^{sec-1((h+a)/a)}.  Since sec(sec^-1(x)) = x and tan(sec^-1(x)) = √(x² – 1), A = ab(^h+a/_a√(( ^h+a/_a)² – 1) – ln|^h+a/_a + √(( ^h+a/_a)² – 1)|).

Although the area of a hyperbola is not an easy formula, both area formulas for a parabola and an ellipse are relatively simple.  The area of an ellipse can also be used to show the special case of a circle where a = b = r (A = πab = π∙r∙r = πr²), and the area formulas of conics nicely correspond with the Cartesian equations of conics learned in algebra.  It makes one wonder why the area formulas for the parabola and ellipse are typically not included in a high school geometry curriculum or textbooks.