Euler Line

Most Geometry classes go over the different properties of triangles, including the orthocenter, centroid, and circumcenter.  The orthocenter is the point of concurrency of the three altitudes (a line segment through a vertex and perpendicular to a line containing the opposite side).

Orthocenter O of ∆ABC

The centroid is the point of concurrency of the three medians (a line segment through a vertex and through the midpoint of the opposite side).  One property of a centroid is that it cuts each median in a 2:1 ratio (see here for the proof).

Centroid P of ∆ABC

The circumcenter is the point of concurrency of the three perpendicular bisectors (a line segment through a vertex and through the perpendicular bisector of the opposite side). 

Circumcenter Q of ∆ABC

A little known fact about these three points of concurrency is that the orthocenter, centroid, and circumcenter always lie on the same line (with the exception of an equilateral triangle, in which the orthocenter, centroid, and circumcenter are actually the same point).  This was proved in 1765 by the Swiss mathematician Leonhard Euler and is consequently named after him.  Here are some examples of the Euler Line in different triangles:

Orthocenter O, Centroid P, and

Circumcenter Q of Acute ∆ABC

Orthocenter O, Centroid P, and

Circumcenter Q of Right ∆ABC

Orthocenter O, Centroid P, and

Circumcenter Q of Obtuse ∆ABC

A clever proof for the Euler Line can be found here and involves proving the similarity of two triangles containing the orthocenter O, centroid P, and circumcenter Q, or more specifically, the similarity of ∆OAP and ∆QDP in the following diagram:

First, we must recognize that in the next diagram, Q is both the circumcenter of ∆ABC and the orthocenter of ∆DEF.  Also, because D, E, and F are midpoints of the sides of ∆ABC , ∆ABC is similar to ∆DEF at a 2:1 ratio.

Therefore, since ∆ABC and ∆DEF are at 2:1 ratio, the vertex to orthocenter of ∆ABC (segment AO) and the vertex to orthocenter of ∆DEF (segment DQ) are also at a 2:1 ratio (side: AO ~ DQ).  Second, as mentioned previously, the centroid cuts the median at a 2:1 ratio, which means segments AP and DP are also at a 2:1 ratio (side: AP ~ DP).  Finally, ÐOAP is congruent to ÐQDP because they are alternate interior angles to the parallel lines formed by AO and DQ (angle: ÐOAP @ ÐQDP).  This is sufficient to prove that ∆OAP is similar to ∆QDP by SAS similarity.  Since the triangles are similar, ÐAOP is congruent to ÐDQP, which makes OP and QP segments of the same transversal OQ.  In other words, O, P, and Q (the orthocenter, centroid, and circumcenter) lie on the same line.

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.

Solving a Quadratic with a Straightedge and Compass

In most high school algebra classes, students learn to solve a quadratic ax² + bx + c = 0 using a variety of different methods.  The most common methods taught are factoring, graphing, completing the square, or using the quadratic equation.  However, a method for solving a quadratic using a straightedge and compass is described near the end of the book Number: The Language of Science by Tobias Dantzig (first published in 1930).

To solve the quadratic equation ax² + bx + c = 0 using a straightedge and compass, let p = -^b/_a and q = ^c/_a.  Use a compass and straightedge to plot out the coordinates U(0, 1), P(p, 0), Q(0, q), R(p, q), and S(p, 1) and then construct rectangle UQRS and its diagonals.  Then, using where the diagonals of the rectangle intersect at point C, construct circle C with radius CU.  Circle C will then intersect the x-axis at the two solutions X₁ and X₂.

For example, let’s say you wanted to solve the quadratic x² – 6x + 5 = 0.  Then a = 1, b = -6, and c = 5, and p = -^b/_a = 6 and q = ^c/_a = 5.  Use a compass and straightedge to plot out the coordinates U(0, 1), P(6, 0), Q(0, 5), R(6, 5), and S(6, 1) and then construct rectangle UQRS and its diagonals.  Then, using where the diagonals of the rectangle intersect at point C(3, 3), construct circle C with radius CU.  Circle C intersects the x-axis at x = 1 and x = 5, which are the two solutions to x² – 6x + 5 = 0. 

We can show that this method will always work by finding a Cartesian equation for circle C, setting y equal to zero (for the x-axis), and simplifying the answer to being the quadratic equation (x = ^{(–b ± √(b^2 – 4ac))}/_2a).  The center of circle C is the midpoint of U(0, 1) and R(p, q), which is (^p/₂, ^{(q + 1)}/₂), and the radius of circle C is half the distance between U(0, 1) and R(p, q), which is r = ½√(p² + (q – 1)²).  Therefore, the equation for circle C is (x – ^p/₂)² + (y – ^{(q + 1)}/₂)² = (½√(p² + (q – 1)²))² or when y = 0, (x – ^p/₂)² + (^{(q + 1)}/₂)² = ¼(p² + (q – 1)²).  Applying some algebra, (x – ^p/₂)² + (^{(q + 1)}/₂)² = ¼(p² + (q – 1)²) è (2x – p)² + (q + 1)² = p² + (q – 1)² è (2x – p)² = p² + (q – 1)² – (q + 1)² è (2x – p)² = p² + (q² – 2q + 1) – (q² + 2q + 1) è (2x – p)² = p² – 4q è 2x – p = ±√(p² – 4q) è 2x = p ± √(p² – 4q) è x = ^{(p ± √(p^2 – 4q))}/₂.  Substituting p = -^b/_a and q = ^c/_a as defined above, x = ^{(–b ± √(b^2 – 4ac))}/_2a, which is the quadratic equation.

If the quadratic equation ax² + bx + c = 0 has two imaginary solutions, circle C will not intersect the x-axis.  However, the solution can still be found with a straightedge and compass by performing a few additional steps.  First, construct a perpendicular bisector of QR through C called CM.  Second, construct a tangent OT of circle C.  Third, construct circle O with radius OT.  Circle O will then intersect line CM at the two solutions Z₁ and Z₂, in which the x-coordinate is the real part and the y-coordinate is the imaginary part. 

For example, let’s say you wanted to solve the quadratic x² – 4x + 5 = 0.  Then a = 1, b = -4, and c = 5, and p = -^b/_a = 4 and q = ^c/_a = 5.  As before, use a compass and straightedge to plot out the coordinates U(0, 1), P(4, 0), Q(0, 5), R(4, 5), and S(4, 1) and then construct rectangle UQRS and its diagonals.  Then, using where the diagonals of the rectangle intersect at point C(2, 3), construct circle C with radius CU.  This time, however, circle C does not intersect the x-axis, so construct a perpendicular bisector of QR through C called CM, construct a tangent OT of circle C, and construct circle O with radius OT.  Circle O intersects the line CM at Z₁(2, 1) and Z₂(2, -1), and so the two solutions are x = 2 ± i.

We can also show that these additional steps will always work by combining the Cartesian equation for circle O and the equation of the vertical line CM and showing that its solution (x, ±y) can be entered into x’ = x ± yi to simplify into the quadratic equation (x’ = ^{(–b ± √(b^2 – 4ac))}/_2a).  As mentioned above, the center of circle C is (^p/₂, ^{(q + 1)}/₂), and so the vertical line CM can be represented by x = ^p/₂.  The length of segment CO is the distance between O(0, 0) and C(^p/₂, ^{(q + 1)}/₂), which is CO = ½√(p² + (q + 1)²).  Segment CT is a radius of circle C, which we also know from above is CT = ½√(p² + (q – 1)²).  Since ΔCTO is a right angle triangle, CT² + OT² = CT², or substituting, (½√(p² + (q – 1)²))² + OT² = (½√(p² + (q + 1)²))².  Applying some algebra, (½√(p² + (q – 1)²))² + OT² = (½√(p² + (q + 1)²))² è OT² = ¼(p² + (q + 1)²) – ¼(p² + (q – 1)²) è OT² = ¼(p² + q² + 2q + 1) – ¼(p² + q² – 2q + 1) è OT² = ¼(4q) è OT² = q è OT = √q, which means the radius of circle O is √q.  The Cartesian equation for circle O is then x² + y² = q.  Combining this with x = ^p/₂, (^p/₂)² + y² = q è y² = q – (^p/₂)² è y² = ¼(4q – p²) è y = ±½√(4q – p²).  Now if x = ^p/₂ and y = ±½√(4q – p²), x’ = x + yi = ^p/₂ ± ½√(4q – p²)i = ^p/₂ ± ½√(4q – p²)√(-1) = ^p/₂ ± ½√(p² – 4q) = ^{(p ± √(p^2 – 4q))}/₂.  Substituting p = -^b/_a and q = ^c/_a as defined above, x’ = ^{(–b ± √(b^2 – 4ac))}/_2a, which is the quadratic equation.

Therefore, there exists a method for solving a quadratic with a straightedge and compass.  The steps themselves are fairly simple, but the proof is not.  Even more difficult than the proof must have been the invention of the method itself, which must have been by some creative genius.