Euler does an Almost Pythagorean Theorem

A while back I posted several times about relations that reminded me of the Pythagorean Theorem. After several requests I rushed out a short paper of the ones I could remember best.

Today I found another while reading one of the wonderful articles by Ed Sandifer that were regularly featured in the MAA, "How Euler Did It."
This particular one was is titled "Beyond Isosceles Triangles".

This is about Euler's paper E324 -- Proprietates triangulorum, quorum anguli certam inter se tenent rationem (Properties of triangles for which certain angles have a ratio between themselves) which is not yet translated at The Euler Archive.

"We know lots about triangles for which Angle A = Angle B. Such triangles are isosceles, and we have known at least since Euclid that Angle A = Angle B exactly when a = b. In 1765, Euler studied a generalization of this situation. What happens if Angle B is some multiple of Angle A ?"

The one that caught my eye says "if the sides of a triangle satisfy the relation ac = bb- aa , then Angle B = 2 Angle A ." ..
I prefer it better as a² + ac = b².

Students should know a couple of candidates in which one angle is twice another.  The 30, 60. 90 for example, and an isosceles right triangle should both be candidates to confirm. 



Euler goes on to prove that when Angle B is three times Angle A, then (b² -a²)(b - a) =ac² . Beautiful, but not quite so "Pythagorean"..

The geometry is clever, and probably clear enough for a really good high school geometry student to handle. Too many diagrams to copy here, so give it a read.
Sandifer says Euler continues through a ratio of 5 to 1, spots a pattern and extends the results to 13.

Sounds like a neat class project to me.  

Sums of Squares Problems

Now even my ex-students are trying to hang me up with "Pythagorean-Like" Problems.  Today I got these two in an email. Since the problems date back to Lewis Carroll (whom  my student's assume I must  have gone to school with since I am so very old).

 I knew the problems and the solutions so I will leave them for now as problems for the reader, and post a solution in a day or two; hopefully with a litlte of the history. 
They are suitable for HS level so everyone is invited. Feel free to submit your solutions.

For any two positive integers X and Y, twice the sum of their squares can be expressed as the sum of two squares.  For example 2(4² + 7²) =    2(16+49) = 2(65) = 130..but 130 = (11²+3²)   

And the followup??? Prove that three times the sum of two squares can be expressed as the sum of four squares. 

Enjoy

A Problem with a Plethora of Pleasing Results

Two squares, and it is almost Pythagorean.....

I had a student ask me this one. He thought I might want to add it to my almost Pythagorean list... I do, but could I prove it's true?   He asked:
If one square is inside another with corresponding sides parallel and the vertices of the inner square are connected in order to the sides of the outer square, then the sum of the squares of the distances of the connections from opposite vertices will be equal.  For the image below,  GD² + EB² = FC² + HA². 

We played around at the board for awhile and finally proved it with nothing fancier than the Pythagorean Theorem and simple algebra.  My method was to draw a horizontal segment (labled R for right shift) to mark how far the inner square is moved from the left edge of the outer square and then a vertical one, D, for down-shift. Then with lables of L for the large squares side and  S for the small squares side (not creative, but easy to remember) I could write the length of all four segments.  (DG)² for instance would be D² + (R+S². We expanded out the squares of all the binomials and trinomials (and he corrected me when I wrote D for R or got a negative as the product of two negatives) and then crossed off mathching parts until everything was gone..  the two expansions were the same alphabet soup of algebraic variabless.  QED. 

One student witnessing our work commented that we had covered the board and never written a number.  I seized the moment to explain what power that was, and began to expound on the implications for possible generalizations... since L, S, R, and D could be any positive (actually two could have been negative) value, and had no relative order or size, S did not have to be smaller than L... and R did not have to be smaller than L... conclusion??? It must also be true that the squared distances are equal even if the one square is not inside the other.  It could be partly inside, or totally outside and either could be larger. 
AND.... pausing for breath... S could be very, very small..... approaching zero, so the degenerate case would  have the smaller square as a point....giving another corollary :
For a square and a point on the same plane, the square of the distances from the point to one pair of opposite vertices will have the same total as the square of the distances to the other pair. (Why have I never seen this theorem?... Is it in textbooks?)    And I propose that the condition that the point is in the same plane is unnecessary.  In fact that seems almost trival since it only adds twice the square of the perpendicular distance from the point to the plane of the square... if my mind is working correctly this morning.  If that is true, that also means that for any square based pyramid, the squares of the sums of the lengths of two opposite edges from the base to the vertex will also be equal.


Now that I was on the roll, I kept playing after the kids wandered away...

What about if the sides of the squares are not parallel.  Playing around with Geogebra, I have determined that that is not a necessary condition.  The original theorem is true for any two squares in the plane however oriented... that is, they do not have to have their sides parallel.  (proof needed, submit yours and save me some time, I'll amend this part)

what about rectangles.??..... If we inscribe a square in a rectangle, then the degenerate case (the inner square is a point) is still true, but with a square and a rectangle the vertex to vertex sums are not equal... in fact, I have discovered that they will differ by a constant value (no matter where the square is or how it is rotated ) that is equal to the product of twice the side of the square times the diffence in the sides of the rectangle.  This property, however, is not preserved when the square is rotated inside the rectangle.  I think that there would be a different constant difference for any orientation of the corresponding sides of the square and rectangle (untested wild conjecture).

While trying to search for the history of the problem, I found a similar relation about parallelograms In the MAA reprint of Jacques Hadamard's Geometry, a similar problem appears on page

I worked out this "constant distance" and found that the difference  is equal to twice the product of the length times the width times the cosine of the acute angle between them, 2*L*W Cos(theta) ("Holy Law of Cosines, Batman!").  I think this proof would be suitable for HS students.  Again, there seems there would be no need for this point to be in the plane of the parallelogram. 

If we replace the square by a rhombus, and sum the squares of the distances from a point to a pair of opposite vertices, I believe the two sums will differer by s^2.  I have not  had time to prove this  yet (offers accepted). 

While I'm in the mood to make conjectures, If the two squares lie in parallel planes then the sum of squares will almost cerainly still be equal. I imagine a physical construction of the two squares in the plane, with elastic chords connecting the vertices.  When I lift one square out of the plane and translate it along a line perpendicular to the original plane I am adding a distance of 2* h^2 to each pair of vertices sum.  Since the two squares could have been in any orientation originally, it seems that no special precaution about the orientation of the squares would be needed.  This would seem to have application to skew prisms, frustums of pyramids and skew frustums (I'm not sure we have such a definiton in geometry, but I mean a frustum in which one base is not similarly aligned with the other when projected on the same plane. 

 I suspect that if the two squares are not even in parallel planes, the sum of the squares of segments containing alternate vertices will be equal also.  No time to attack that yet.

A first postscript:  It seems that for any even sided regular polygon, the sums of squares of distances from a point to pairs of opposite vertices would be equal also since any two pairs of these opposite verties would form a rectangle.   Likewise, if two even sided regular polygons have opposite vertices connected in sequential order, the same property is preserved as for squares. 

Complex and Other Conjugates

We had covered the Fundamental Theorem of Algebra in pre-calc and after a few days of trying to make sure they understood how to decide how many real and non-real solutions there were to a real valued polynomials and that they could be factored over linear and quadratic factors , I was trying to get them to see how easy it was to work back and forth from the quadratic polynomial and the complex zeros. Finally I told them, "It's almost like the Pythagorian Theorem. a²+ b² = c " [They will later learn that this is really just the product of two equal pythagorean results for each of the conjugates, but not yet]

I showed them that the complex solutions always came in conjugate pairs, a + bi and a - bi... and that if they knew the zeros, they could create a quadratic which had those solutions. It seems that students are no longer exposed to what I call the sum and product property of quadratic zeros... that if the zeros of a quadratic are r and s, a quadratic with those zeros is given by x² - (r+s)x + (rs)... (If you knew the zeros were at x=3 and x= -1, then a quadratic with those zeros would be x² - (3-1)x + (3*-1) or x² - (2)x -3).

They had to get it somewhere if they were ever taught to factor, but it never seems to find purchase in their mathematical memory. I pointed out that for complex conjugates, r+s was real since the complex +bi and - bi eliminated each other, and so r+s was just twice the real coefficient of either conjugate. I connected this to the graph by showing that this "a" in the a+bi was the axis of symmetry of the graph. For example a quadratic with complex zeros at 3+2i and 3-2i will have a linear term of -(3+3)x [assuming we use the easiest case with A=1]. This also means that the of the quadratic will have its vertex at a point with an x value of 3.

It is also easy to see the constant term of the quadratic. If you multiply (a+bi)(a-bi) the distribution gives a² +abi - abi - b²i²... As with the sum, the two imaginary terms will cancel out, and since i²=-1, we see that the product of the two complex conjugates will always be a² + b². So for the example 3+2i and 3-2i, the product of the conjugates is just 9+4=13. Putting this in as the constant term, we see that the quadratic x² - 6x + 13 has zeros at 3+2i and 3-2i.

This gives another way to solve a quadratic equation with complex roots. If we have a function x² - B x + C that we know has complex solutions, we know that the real coefficient, a, of the conjugate pair will be -B/2 (shades of the quadratic formula)... and we know that a² + b² = C, so we can figure out b pretty simply. If we had a quadratic such as x² - 8 x + 25 we would quickly see that the real coefficient is -(-8)/2 = 4... and know that 4² + b² = 25, so b must be 3, and the two solutions are at 4 +/- 3i. I finally admit to the students that this is just a variant of the quadratic formula, and sooner or later someone will ask if it is not possible to do the same thing with a quadratic with real zeros... I let them suggest a quadratic, and only press for the use of an even B in the linear coefficient for ease of explanation. For example with x² - 6 x + 5 (ok, I picked an easy one).. we would see that the "a" part was three again... but now we have 3² + b^2 = 5.

If we remember that b was the imaginary coefficient, it makes sense that the b^2 would be negative if the zeros were real... Kids translate that into, "just subtract instead of add, and take the square root of that for the "b" value. So 9 - 4 = 5, and the square root of 4 is two. Our real solutions are 3 +/- 2; or 5 and 1.
At this point I show them that with real rational coefficients, not only do non-real zeros come in conjugate pairs, but so do the real irrational zeros. If we had x² - 8 x + 10; the solutions of 4 +/- sqrt(16-10) would be the conjugate irrational solutions.

If we graph y=x² - 6x + 13, we notice that not only is the vertex at x=3, but its y coordinate is at 4... the square of 2, which was the b part of our complex conjugates... It seems that functions with complex solutions a +/- bi, the vertex is at (a, b²). This also works for real valued zeros if we with the a +/- b approach above. The vertex of x² -6x + 5 would be at (3, - 2²).

I'm not sure I would try to drag a really weak class throuh all this, but it often helps bright kids see how interconnected the zeros, factors, and graphs of all quadratics are, not just the complex numbers.

An Interesting Triangle Property

I'm considering this one for my "Almost Pythagorean" file. I came across this recently and found it interesting. The Pythagorean Theorem is actually a property about triangles with 90 degree angles; but this one is a property of all triangles that contain a 60 degree angle at vertex A.

The "Then" implied by the above is that

The derivation is not too difficult and might well be presented to a good pre-calc student as a challenge. If you don't want me to spoil it, stop reading now until you have tried it.... It makes me wonder what we could come up with if we started with a thirty-degree angle.

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and yet Another "Almost Pythagorean" Relation

I have written frequently about relationships that, for one reason or another, reminded me of the Pythagorean Theorem. (see here, and here, and here for example).

After trying to write up all the ones I could think of, out of the blue I got a nice note from professor Robin Whitty who maintains the excellent "Theorem of the Day" web site. (If you are a high school teacher or student, it's a must see.) He had prepared a page on Des Cartes' Circle Theorem, and had included a link to my page on the subject (I told you it was a good site). It was one more "almost Pythagorean" theorem. It's easiest to right if we consider the bend of a circle as the reciprocal of the radius, b= 1/r. With that notation adjustment, then for four circles with bends b₁, b₂,b₃,b₄, Des Cartes' Circle theorem says that one-half the square of the sum of the bends is equal to the sum of the squares of the bends.

Another "Almost Pythagorean" Relationship

A while back Arjen Dijksman at "Physics Intuitions" posted a blog reflecting on my post about Almost Pythagorean Triangles . Then he went me one better with a new relation that is pretty incredible. In essence, he shows that for ANY triangle, ABC with opposite sides a, b, and c, there is a relation a²+b²=c²+2 t², where t length of the tangent from point C to the circle with diameter AB.

He has some nice clear graphics to show the development.. check it out..and with a change of sign the equation generalizes to triangles where c lies inside the circle with diameter AB.

This is, of course, too close to the law of cosines to be ignored.. and it is easy to see that t² = a*b*cos(C)... in fact as soon as we set t²=AC*DC, and recognize that angle BDC is a right angle, we have Cos(C)= DC/BC... thus AC*DC = AC*BC*(DC/BC)= ab Cos(C).

More on Vectors in the HS Curriculum

Several colleagues took me to task privately, and unjustly I think (he whined), for suggesting that more stuff (vector topics) should be added to the (already overstuffed) curriculum. My idea was neither to replace the traditional y=f(x) approach commonly used, nor to introduce multiple chapters in vectors throughout the curriculum strand. Instead, I think the integration of a few “15 minute vector asides” at a number of places each year would have the potential to greatly enrich a student’s ability to do those big ideas in math, generalize, synthesize and specialize.

Distance is an abstract concept that is introduced as early as fifth or sixth grade for the number line (maybe earlier) and by grades seven or eight for the coordinate plane. Ok, I’m ready to let the fifth graders stay with Pt A –Pt B for distance on the number line. For the coordinate plane, the usual approach is to resort to a memorized “distance formula” that few of them ever realize is the Pythagorean theorem. Even fewer naturally extend that to three-space, although I admit that would seem pretty trivial. If at about this moment we gave a very brief introduction to vectors The “vector” from Point A to Point B is (B-A). Kids could use vectors to translate points on a two-or three-d grid and immediately realize the relationship between the vectors [3,5] and [-3,-5] . I would think within a single days lesson, most students could write the vectors between points in two-space and three space (how often do grade 6-7-8 students see a three dimensional point, I wonder.). Then the square root of the dot product is an easy way to define distance, and it is immediately defined for ALL dimensions. By day two, students can be finding distances between points in the room located with coordinates in feet or meters and an origin established at one corner of the room. The same students would, I suspect, immediately see slope as a “vector” relationship (although there will always be kids who get hung up on the order, x over y / y over x, as they do now).

One of the places I notice a need for at least a “vector translation” understanding of a line segment shows up in the frequency with which Alg I and II students are asked to find the midpoint of a segment (there is probably even a “midpoint formula” in the book) but are almost never asked to find the point 2/3 of the way from A to B. One leads to a view of math as using rules, and another as using ideas. My goal is to move away from the former and towards the latter.

addendum Addendum to “proportion of a line segment”.
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In a similar way, many students are taught to find the centroid of a triangle, usually by a memorized relationship. The student who can understand and apply the use of a vector approach to lines will be able to see that any point on the plane determined by triangle ABC can be represented as a linear combination P= ra + sb + tc with r+s+t=1 . Again, many students are more at ease with representing this linear combination as P=a + s(b-a) + t( c-a) since it can be seen visually as a translation from the origin to point A, then motions parallel to the two sides meeting at A to locate the point.
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Somewhere in the Alg I/Geometry sequence, they come to that point where they are given three consecutive points on a parallelogram and asked to find the fourth point. I cannot imagine many kids being successful at this who do not take a more-or-less vector approach to the motion along the segments of the parallelogram. But once they see a vector approach to “moving” a point; and if they were familiar with 3-d coordinates, how long would it take to give them four non-coplanar points representing adjacent vertices of a parallelepiped and have them find the other four points.

If we look at two sides of a triangle as vectors, call them a and b, and think of the third side c as the vector a-b, then the square of the length of c is c . c. But replacing c with a-b we get (a-b) . (a-b) and distributing we get a . a + b . b – 2 a . b and keeping in mind that a . b = |a| |b| cos(theta) we have the law of cosines . The distributive nature of the dot product gives the law of cosines as a simple application of the distance formula. And if a and b are perpendicular, their product is zero, and so the Pythagorean theorem also pops out.

A geometry student with an understanding of the dot product can show easily that the sum of the squares of the sides of a parallelogram is equal to the sum of the squares of its diagonals. If we represent the two sides at Point A as vectors a and b, then the diagonal AC is a+b, and the diagonal BD is b-a. The dot product of w=a+b with itself gives |a+b|²= |a|² + |b|² + 2(a.b).
The same approach with v=a-b gives |a-b|²= |a|² + |b|² - 2(a.b), and summing the two we get the result that |v|²+|w|² = |a|² + |b|²+|a|² + |b|² and substituting in a1 and b1 in one pair gives the final result. Understanding the distributive law removes the necessity to use the memorized law of cosines. (I would love to have a student respond that they didn’t remember the law of cosines, but given a moment they could reproduce it.

As a sidebar, I would add that like my recent post about Almost Pythagorean relations this one qualifies as both also Pythagorean, and also as one of those special cases where two wrongs make a right...ie. for sides a, b, c, and d of a parallelogram with diagonals d₁ and d₂ it is NOT true that a²+b²=d₁² and likewise for c²+d²=d₂², but when both equations are added together, the result is true.

A geometry course rich with vector topics would seem to make three-dimensions a much deeper part of the course. It would also, as I hope to show in subsequent posts, allow us to easily extend some of the things we do in two-space.