Church Window Design

Last week while I was waiting to pick up my kids from their school carpool line, I noticed that all the windows in the church building had a certain design.  Despite being different widths and heights, each window was symmetrical, where both left and right sides curved up to a point, the outermost curves continued on to meet at another point, and there was a circle in the top middle that was just the right size to perfectly touch all the other curves in one spot each.

I began to wonder about that middle circle.  How did the architect find the correct size of that circle so that it was tangential to the other curves?  Was it by trial and error?  Or was there a formula for it?  Could a general formula be found for it?

To simplify things, let’s assume that all the curves were made by circles (and not by some other conic section like a parabola, ellipse, or hyperbola).  The four different curves can then be constructed by using four congruent circles as guidelines. 

Each window can then be defined by two variables: the radius r of the four congruent circles, and the distance w between the two nonsymmetrical arcs (or half the width of the whole window).  Since each window design is symmetrical, the center of the middle circle with the unknown radius x will lie on the vertical line of symmetry.  We can now construct two right triangles with the following sides as pictured below:

The left-side triangle has a hypotenuse of r – x and legs r – w and y, which makes the Pythagorean equation y² + (r – w)² = (r – x)² or y² = (r – x)² – (r – w)².  The right-side triangle has a hypotenuse of r + x and legs r and y, which makes the Pythagorean equation y² + r² = (r + x)² or y² = (r + x)² – r².  Therefore,

y2 = (r – x)2 – (r – w)2 = (r + x)2 – r2	(Pythagorean’s Theorem)
(r2–2rx+x2) – (r2–2rw+w2) = (r2+2rx+x2) – r2	(expand)
-2rx + x2 + 2rw – w2 = 2rx + x2	(simplify)
-2rx + 2rw – w2 = 2rx	(subtract x2)
2rw – w2 = 4rx	(add 2rx)
x = (2rw – w^2)/4r	(divide 4r)

The radius x of the middle circle is therefore

where r is the radius of one of the larger four congruent circles and w is half the width of the whole window.  So if the radius of one of the larger circles is 3 feet, and half the width of the window is 2 feet, then the radius of the middle circle would be x = ^{(2rw – w^2)}/_4r = ^{(2(3)2 – 2^2)}/₄₍₃₎ = ⁸/₁₂ feet or 8 inches.  Or if the radius of one of the larger circles is 8 feet, and half the width of the window is 4 feet, then the radius of the middle circle would be x = ^{(2rw – w^2)}/_4r = ^{(2(8)4 – 4^2)}/₄₍₈₎ = ⁴⁸/₃₂ = ³/₂ feet or 18 inches.

As a side note, calculus lovers should recognize that if r remains constant, the maximum radius x of the middle circle occurs when w = r.  This is because ^dx/_dw = ¹/_4r(2r – 2w), and setting this to zero gives the solution w = r.  Geometrically, w = r occurs when the top of the window does not come to a point but is a smooth semicircle.

There are probably other church window designs that have similar properties.  For this particular design, the size of the middle circle can be calculated exactly.

Kissing Circles (Three Circles and a Line)

Circles that are mutually tangent to each other are called “kissing circles” because they barely touch each other (or “kiss”) at one point.  There are several different diagrams that have kissing circles, but this article will focus on the specific case of three circles and a line that are all mutually tangent to each other, as in the diagram below:

When three circles and a line are mutually tangent to each other, the relationship between the three radii of each circle is

The proof can be found by solving the three right triangles formed by using the segments from one center of a circle to the next as hypotenuses, as pictured below in gray:

The bottom left gray triangle makes the Pythagorean equation (r₁ – r₃)² + x₁² = (r₁ + r₃)², which simplifies to

(r1 – r3)2 + x12 = (r1 + r3)2	(Pythagorean’s Theorem)
r12 – 2r1r3 + r32 + x12 = r12 + 2r1r3 + r32	(expand)
-2r1r3 + x12 = 2r1r3	(subtract r12 and r32)
x12 = 4r1r3	(add 2r1r3)
x1 = 2√(r1r3)	(square root)

The bottom right gray triangle makes the Pythagorean equation (r₂ – r₃)² + x₂² = (r₂ + r₃)², which similarly simplifies to x₂ = 2√(r₂r₃), and the top gray triangle makes (r₁ – r₂)² + x₃² = (r₁ + r₂)², which simplifies to x₃ = 2√(r₁r₂).  Since x₁ + x₂ = x₃,

x1 + x2 = x3	(angle addition)
2√(r1r3) + 2√(r2r3) = 2√(r1r2)	(substitution)
√(r1r3) + √(r2r3) = √(r1r2)	(divide by 2)
√r3(√r1 + √r2) = √(r1r2)	(factor √r3)
√r3 = √(r1r2)/(√r1 + √r2)	(divide by (√r1 + √r2))
1/√r3 = (√r1 + √r2) /√(r1r2)	(rearrange)
1/√r3 = √r1/√(r1r2) + √r2/√(r1r2)	(distribute denominator)
1/√r3 = 1/√r2 + 1/√r1	(simplify)
1/√r1 + 1/√r2 = 1/√r3	(rearrange)

The mathematician Descartes studied the case of four kissing circles and derived the equation

where k₁ = ¹/_r1, k₂ = ¹/_r2, k₃ = ¹/_r3, and k₄ = ¹/_r4.

Four Kissing Circles

The derivation for this formula involves solving a quadratic system of three equations of three variables and is beyond the scope of this article.  However, we can use Descartes’ Theorem to as another proof for ¹/_√r1 + ¹/_√r2 = ¹/_√r3, because the case of three kissing circles and a line is actually a specific case in which the line has a curvature of k₃ = 0.  Therefore,

k4 = k1 + k2 + k3 ± 2√(k1k2 + k2k3 + k1k3)	(Descartes’ Theorem)
k4 = k1 + k2 + k3 + 2√(k1k2 + k2k3 + k1k3)	(equation for inner circle)
k4 = k1 + k2 + (0) + 2√(k1k2 + k2(0) + k1(0))	(k3 = 0)
k4 = k1 + k2 + 2√(k1k2)	(simplify)
k4 = k1 + 2√(k1k2) + k2	(rearrange)
k4 = (√k1 + √k2)2	(factor)
√k4 = √k1 + √k2	(square root)
1/√r4 = 1/√r2 + 1/√r1	(substitute)
1/√r1 + 1/√r2 = 1/√r4	(rearrange)

The relationship between the three radii can also be written as r₁^-½ + r₂^-½ = r₃^-½, or a^-½ + b^-½ = c^-½, which is fascinatingly similar to relationship of the three sides in a right angle, which according to Pythagorean’s Theorem is a² + b² = c².  The only difference between the two equations is the exponents, which are opposite reciprocals.

Now that we have established the relationship of the radii for three kissing circles and a line, we can investigate some specialized cases that produce some unexpected sequences.  It should be noted that if two radii are the reciprocal of a square integer, then the third radius will be the reciprocal of the sum of those two integers squared.  In other words, if r₁ = ¹/_m2 and r₂ = ¹/_n2, then r₃ = ¹/_{(m + n)}2.  This can be shown algebraically:

1/√r1 + 1/√r2 = 1/√r3	(radius relationship)
r1-½ + r2-½ = r3-½	(rewrite exponents)
(1/m2)-½ + (1/n2)-½ = r3-½	(substitute  r1 and r2)
m + n = r3-½	(simplify exponents)
(m + n)-2 = r3	(negative square both sides)
r3 = 1/(m + n)2	(rewrite exponent)

So if two kissing circles have a radii of 1, then r₁ = 1 = ¹/₁2 and m = 1 and r₂ = 1 = ¹/₁2 and n = 1, and the third kissing circle that is in between them will have a radius r₃ = ¹/_{(m + n)}2 = ¹/_{(1 + 1)}2 = ¹/₂2.  Continuing on in one direction, the circle between this new circle with radius ¹/₂2 and the original circle ¹/₁2 will have a radius of ¹/_{(2 + 1)}2 = ¹/₃2, and the circle between this new circle with radius ¹/₃2 and the original circle ¹/₁2 will have a radius of ¹/_{(3 + 1)}2 = ¹/₄2, and so on, so that sequence of new radii for each successive circle is r = ¹/₁2, ¹/₂2, ¹/₃2, ¹/₄2, … which are the square reciprocals of the counting numbers.

Square Reciprocals of Counting Numbers

Similarly, if we start with two kissing circles that have a radii of 1 but alternate directions left and right for every other circle, the sequence becomes r = ¹/₁2, ¹/₁2, ¹/_{(1 + 1)}2 =¹/₂2, ¹/_{(1 + 2)}2 =¹/₃2, ¹/_{(2 + 3)}2 =¹/₅2, ¹/_{(3 + 5)}2 =¹/₈2, … or r = ¹/₁2, ¹/₁2, ¹/₂2, ¹/₃2, ¹/₅2, ¹/₈2, … which are the square reciprocals of the Fibonacci numbers.

                                                                                                                                                  

Square Reciprocals of Fibonacci Numbers

One last specialized case of kissing circles and a line that should be mentioned are Ford circles, named after the twentieth century mathematician Lester R. Ford.  Starting with two circles that have radii of r = ½, place the center of one circle directly above the 0 on the number line and place the center of the other circle directly above the 1.  The new kissing circle formed in between these two circles will have a center directly above x = ¹/₂.  Furthermore, the circle between the circles x = 0 and x = ¹/₂ is x = ¹/₃, and the circle between the circles x = ¹/₂ and x = 1 is x = ²/₃.  This process can be continued on indefinitely, where it is possible to place a new circle at the x-value of any irreducible fraction ^p/_q.

                        

Ford Circles

Additionally, it can be proved that the kissing circle between any x = ^p1/_q1 and x = ^p2/_q2 is x = ^{(p1 + p2)}/ _{(q1 + q2)}.  For example, the kissing circle between the circles x = ¹/₂ and x = ²/₃ is x = ^{(1 + 2)}/_{(2 + 3)} = ³/₅. 

                                      

To prove this, we must first recognize that the first two outer Ford circles both have x-values in the form of x = ^p/_q and radii in the form of r = ¹/_2q2.  (For the left circle, p₁ = 0 and q₁ = 1, so that x = ^p1/_q1 = 0 and r = ¹/_2q12 = ¹/₂.  For the right circle, p₂ = 1 and q₂ = 1, so that x = ^p2/_q2 = 1 and r = ¹/_2q22 = ¹/₂.)  It can then be proved algebraically that any new kissing circle produced by two circles with both x-values in the form of x = ^p/_q and radii in the form of r = ¹/_2q2 also have these properties, specifically x = ^{(p1 + p2)}/ _{(q1 + q2)} and r = ¹/_{2(q1 +q2)}2.  First, to prove that r = ¹/_{2(q1 +q2)}2:

1/√r1 + 1/√r2 = 1/√r3	(radius relationship)
r1-½ + r2-½ = r3-½	(rewrite exponents)
(1/2q12)-½ + (1/2q22)-½ = r3-½	(subst r1 and r2)
2½q1 + 2½q2 = r3-½	(simplify exponents)
2½(q1 + q2) = r3-½	(factor)
2-1(q1 + q2)-2 = r3	(negative square both sides)
r3 = 1/2(q1 +q2)2	(rewrite exponent)

To prove that x = ^{(p1 + p2)}/ _{(q1 + q2)}, recall from above that x₃ is the distance between the centers of the two outside kissing circles, which in this case is x₃ = ^p2/_q2 – ^p1/_q1, and that x₁ is the distance between the centers of an outside and inside kissing circle, which in this case we are trying to prove that x₁ = ^{(p1 + p2)}/ _{(q1 + q2)} – ^p1/_q1.  Also recall from above that x₁ = 2√(r₁r₃), x₃ = 2√(r₁r₂), and √r₃ = ^√(r1r2)/_{(√r1 + √r2)}.  Therefore:

x1 = 2√(r1r3)	(given)
x1 = 2√r1√r3	(distribute)
x1 = 2√r1√(r1r2)/(√r1 + √r2)	(subst √r3 = √(r1r2)/(√r1 + √r2))
x1 = √r1x3/(√r1 + √r2)	(subst x3 = 2√(r1r2))
x1 = √(1/2q12)(p2/q2 – p1/q1) / (√(1/2q12) + √(1/2q22))	(subst r1 = 1/2q12, r2 = 1/2q22, and x3 = p2/q2 – p1/q1)
x1 = 1/q1(p2/q2 – p1/q1) / (1/q1 + 1/q2)	(simplify)
x1 = (q1p2 – q2p1)/q1(q2 + q1)	(multiply by q1q1q2/q1q1q2)
x1 = (q1p2 + q1p1 – q1p1 – q2p1)/q1(q2 + q1)	(add and subtract q1p1)
x1 = q1(p2 + p1) – p1(q1 – q2)/q1(q2 + q1)	(factor)
x1 = q1(p2 + p1)/q1(q2 + q1) – p1(q1 – q2)/q1(q2 + q1)	(distribute denominator)
x1 = (p2 + p1)/(q2 + q1) – p1/q1	(simplify)

In conclusion, the relationship between the three kissing circles and a line is ¹/_√r1 + ¹/_√r2 = ¹/_√r3.  This can be rewritten as r₁^-½ + r₂^-½ = r₃^-½, which is fascinatingly similar to Pythagorean’s Theorem.  Furthermore, it can be shown that the radii of a series of kissing circles under two kissing unit circles and a line in one direction is a sequence of square reciprocals of the counting numbers, and in alternating directions is a sequence of square reciprocals of the Fibonacci numbers.  Finally, Ford circles are kissing circles in which it is possible to place a new circle at the x-value of any irreducible fraction ^p/_q, and any successive kissing circle between x = ^p1/_q1 and x = ^p2/_q2 will have a new x-value of x = ^{(p1 + p2)}/ _{(q1 + q2)}. 

Area of a Triangle

Every school-age student knows that the area of a triangle is one half the base times its height, or

Area = ½bh

A rather clever proof for this is to show that two identical triangles can be cut and rotated into a single rectangle of the same base and height, and since two identical triangles would have an area of A = bh, one triangle would have an area of A = ½bh. 

However, what if the height or base of a triangle is not given?  Can the area still be found?  You may recall from geometrical congruency proofs that a triangle can be defined by a combination of its angles and sides, namely side-angle-side (SAS), side-side-side (SSS), angle-side-angle (ASA) or angle-angle-side (AAS).  Therefore, there must be an area formula for each of these congruencies.

SAS Triangle Area Formula

If two sides and an included angle of a triangle are given (SAS), the area is

Area = ½ a b sin C

where a and b are the two sides and ∠C is the measure of the included angle.  This formula is often taught in high school trigonometry, and is handy for answering some SAT and ACT questions.

The proof is fairly straightforward.  In the diagram above, sin C = ^h/_a, or h = a sin C.  Substituting this into A = ½bh, we get A = ½b(a sin C), or A = ½ a b sin C.

SSS Triangle Area Formula (Heron’s Formula)

If all three sides of a triangle are given (SSS), the area is

where a, b, and c are the three sides, and s = ½(a + b + c).  This formula is also known as Heron’s Formula and is somewhat obscure in high school math curricula, likely because of the complexity of the derivation.

Heron’s formula can be derived by manipulating Pythagorean’s Theorem from the two right angle triangles in the above diagram.  For the left-hand right triangle,

b12 + h2 = a2

(Pythagorean’s Theorem)

and for the right-hand right triangle,

b22 + h2 = c2	(Pythagorean’s Theorem)
(b – b1)2 + h2 = c2	(since b = b1 + b2)
b2 – 2b1b + b12 + h2 = c2	(expand)
b12 + h2 = c2 – b2 + 2b1b	(rearrange)

This means that b₁² + h² is equal to both a² and c² – b² + 2b₁b, so

a2 = c2 – b2 + 2b1b	(transitive)
2b1b = a2 + b2 – c2	(rearrange)
b1 = 1/2b(a2 + b2 – c2)	(divide)

Going back to the left-hand triangle formula, substituting b₁, and solving and simplifying h:

b12 + h2 = a2	(Pythagorean’s Theorem)
(1/2b(a2 + b2 – c2))2 + h2 = a2	(substitution)
h2 = a2 – (1/2b(a2 + b2 – c2))2	(subtraction)
h = √(a2 – (1/2b(a2 + b2 – c2))2)	(square root)
h = √(a2 – 1/4b^2(a2 + b2 – c2)2)	(square)
h = √(1/4b^24a2b2 – 1/4b^2(a2 + b2 – c2)2)	(common denominator)
h = √(1/4b^2(4a2b2 – (a2 + b2 – c2)2))	(factor out 1/4b^2)
h = 1/2b√(4a2b2 – (a2 + b2 – c2)2)	(square root 1/4b^2)
h = 1/2b√((2ab – (a2 + b2 – c2)) (2ab + (a2 + b2 – c2)))	(difference of squares)
h = 1/2b√((2ab – a2 – b2 + c2) (2ab + a2 + b2 – c2))	(distribute)
h = 1/2b√((c2 – (a2 – 2ab + b2)) ((a2 + 2ab + b2) – c2))	(rearrange, factor)
h = 1/2b√((c2 – (a – b)2)((a + b)2 – c2))	(factor)
h = 1/2b√((c – (a – b))(c + (a – b)) ((a + b) – c)((a + b) + c))	(difference of squares)
h = 1/2b√((c – a + b)(c + a – b) (a + b – c)(a + b + c))	(distribute negative)
h = 1/2b√((-a + b + c)(a – b + c) (a + b – c)(a + b + c))	(rearrange)
h = 1/2b√((a+b+c – 2a)(a+b+c – 2b) (a+b+c – 2c)(a+b+c))	(rearrange)
h = 1/2b√((16)½(a+b+c – 2a)½(a+b+c – 2b) ½(a+b+c – 2c)½(a+b+c))	(rearrange)
h = 1/2b√(16(½(a+b+c) – a)(½(a+b+c) – b) (½(a+b+c) – c)(½(a+b +c))	(distribute)
h = 1/2b√(16(s – a)(s – b)(s – c)s)	(substitute s = ½(a + b + c))
h = 1/2b4√(s(s – a)(s – b)(s – c))	(square root, rearrange)
h = 2/b√(s(s – a)(s – b)(s – c))	(simplify)

Lastly, substituting h into the area formula:

A = ½bh	(area formula)
A = ½b(2/b√(s(s – a)(s – b)(s – c)))	(substitute)
A = √(s(s – a)(s – b)(s – c))	(simplify)

which is Heron’s Formula.

ASA Triangle Area Formula

If two angles and an included side of a triangle are given (ASA), the area is

which is also equivalent to

The proof involves the SAS Triangle Area Formula (A = ½ a b sin C), the Law of Sines (^{sin A}/_a = ^{sin B}/_b = ^{sin C}/_c), and the trigonometric identity sin(180° – x ) = sin x.  From the Law of Sines, a = ^{b sin A}/_{sin B}.  Substituting this into the SAS Triangle Area Formula we get:

A = ½ a b sin C	(SAS area formula)
A = ½ (b sin A/sin B) b sin C	(substitute)
A = b^2 sin A sin C/2 sin B	(simplify)
A = b^2 sin A sin C/2 sin (180° – (A + C))	(angle sum of triangle)
A = b^2 sin A sin C/2 sin (A + C)	(trig identity)

The secondary formula can be derived by expanding sin (A + C) and dividing the numerator and denominator by cos A cos C:

A = b^2 sin A sin C/2 (sin A cos C + cos A sin C)	(sine addition formula)
A = b^2 tan A tan C /2 (tan A + tan C)	(divide numerator and denominator by cos A cos C)

AAS Triangle Area Formula

If two angles and a non-included side of a triangle are given (AAS), the area is

The proof is similar to the ASA area proof, and also involves the SAS Triangle Area Formula (A = ½ a b sin C), the Law of Sines (^{sin A}/_a = ^{sin B}/_b = ^{sin C}/_c), and the trigonometric identity sin(180° – x ) = sin x.  From the Law of Sines, a = ^{b sin A}/_{sin B}.  Substituting this into the SAS Triangle Area Formula we get:

A = ½ a b sin C	(SAS area formula)
A = ½ (b sin A/sin B) b sin C	(substitute)
A = b^2 sin A sin C/2 sin B	(simplify)
A = b^2 sin A sin (180° – (A + B))/2 sin B	(angle sum of triangle)
A = b^2 sin A sin (A + B)/2 sin B	(trig identity)

Conclusion

The area of a triangle can be expressed in many different ways other than the traditional formula of Area = ½bh.  If two sides and an included angle are given (SAS), the area can be expressed as Area = ½ a b sin C.  If all three sides are given (SSS), the area can be expressed as Area = √(s(s – a)(s – b)(s – c)), which is also known as Heron’s Formula.  If two angles and an included side of a triangle are given (ASA), the area can be expressed as either Area = ^{b^2 sin A sin C}/_{2 sin (A + C)} or Area = ^{b^2 tan A tan C} /_{2 (tan A + tan C)}.  Finally, if two angles and a non-included side of a triangle are given (AAS), the area can be expressed as Area = ^{b^2 sin A sin (A + B)}/_{2 sin B}.