# Tag Archives: pythagorean

## Circle and Square

Here’s another great geometry + algebra problem, posed by Megan Schmidt and pitched by Justin Aion to some students in his Geometry class.

Following is the problem as Justin posed it yesterday.

Justin described the efforts of three of his students’ on his his ‘blog.  Following is my more generalized approach.  Don’t read further if you want to solve this problem for yourself!

My first instinct in any case like this is build it in a dynamic geometry package and play.  Using my TI-Nspire, without loss of generality, I graphed a circle centered at the origin, constructed a tangent segment at the bottom of the circle centered on the y-axis, and then used that segment to construct a square.  I recognized that the locus of the upper right corners of all such squares would form a line.

That made it clear to me that for any circle, there was a unique square that intersected the circle three times as Megan had posed.

Seeing this and revealing its algebraic bias, my mind conceived an algebraic solution.  Assuming the radius of the circle is R, the equation of my circle is $x^2+y^2=R^2$ making the lower y-intercept of the circle $(0,-R)$.  That made $y=2x-R$ the locus line containing the upper right corner of the square.

To find generic coordinates of the upper right corner of the square in terms of R, I just needed to solve the system of equations containing the circle and the line.  That’s easy enough to compute by hand if you can handle quadratic algebra.  That manipulation is not relevant right now, so my Nspire CAS’s version is:

The output confirms the two intersections are $(0,-R)$ and the unknown at $\displaystyle \left( \frac{4R}{5} , \frac{3R}{5} \right)$.

Because of the horizontal symmetry of the square with respect to the y-axis, the system solution shows that the generic length of the side of the square is $\displaystyle 2\left( \frac{4R}{5} \right) = \frac{8R}{5}$.  The circle’s y-intercept at $(0,-R)$ means the generic diameter of the circle is $2R$.

Therefore, the generic ratio of the circle’s diameter to the square’s side length is

$\displaystyle \frac{diameter}{side} = \frac{2R}{(8R)/5} = \frac{5}{4}$.

And this is independent of the circle’s radius!  The diameter of the circle is always $\frac{5}{4}$ of the square’s side.

CONCLUSION:

For Megan’s particular case with a side length of 20, that gives a circle diameter of 25, confirming Justin’s students’ solution.

Does anyone have a different approach?  I’m happy to compile and share all I get.

AN ASIDE:

While not necessary for the generalized solution, it was fun to see a 3-4-5 right triangle randomly appear in Quadrant 1.

## Transformations II and a Pythagorean Surprise

In my last post, I showed how to determine an unknown matrix for most transformations in the xy-plane and suggested that they held even more information.

Given a pre-image set of points which can be connected to enclose one or more areas with either clockwise or counterclockwise orientation.  If a transformation T represented by matrix $[T]= \left[ \begin{array}{cc} A & C \\ B & D \end{array}\right]$ is applied to the pre-image points, then the determinant of $[T]$, $det[T]=AD-BC$, tells you two things about the image points.

1. The area enclosed by similarly connecting the image points is $\left| det[T] \right|$ times the area enclosed by the pre-image points, and
2. The orientation of the image points is identical to that of the pre-image if $det[T]>0$, but is reversed if $det[T]<0$.  If $det[T]=0$, then the image area is 0 by the first property, and any question about orientation is moot.

In other words, $det[T]$ is the area scaling factor from the pre-image to the image (addressing the second half of CCSSM Standard NV-M 12 on page 61 here), and the sign of $det[T]$ indicates whether the pre-image and image have the same or opposite orientation, a property beyond the stated scope of the CCSSM.

Example 1: Interpret $det[T]$ for the matrix representing a reflection over the x-axis, $[T]=\left[ r_{x-axis} \right] =\left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right]$.

From here, $det[T]=-1$.  The magnitude of this is 1, indicating that the area of an image of an object reflected over the line $y=x$ is 1 times the area of the pre-image—an obviously true fact because reflections preserve area.

Also, $det \left[ r_{x-axis} \right]<0$ indicating that the orientation of the reflection image is reversed from that of its pre-image.  This, too, must be true because reflections reverse orientation.

Example 2: Interpret $det[T]$ for the matrix representing a scale change that doubles x-coordinates and triples y-coordinates, $[T]=\left[ S_{2,3} \right] =\left[ \begin{array}{cc} 2 & 0 \\ 0 & 3 \end{array} \right]$.

For this matrix, $det[T]=+6$, indicating that the image’s area is 6 times that of its pre-image area, while both the image and pre-image have the same orientation.  Both of these facts seem reasonable if you imagine a rectangle as a pre-image.  Doubling the base and tripling the height create a new rectangle whose area is six times larger.  As no flipping is done, orientation should remain the same.

Example 3 & a Pythagorean Surprise:  What should be true about  $det[T]$ for the transformation matrix representing a generic rotation of $\theta$ units around the origin,  $[T]=\left[ R_\theta \right] = \left[ \begin{array}{cc} cos( \theta ) & -sin( \theta ) \\ sin( \theta ) & cos( \theta ) \end{array} \right]$ ?

Rotations preserve area without reversing orientation, so $det\left[ R_\theta \right]$ should be +1.  Using this fact and computing the determinant gives

$det \left[ R_\theta \right] = cos^2(\theta ) + sin^2(\theta )=+1$ .

In a generic right triangle with hypotenuse C, leg A adjacent to acute angle $\theta$, and another leg B, this equation is equivalent to $\left( \frac{A}{C} \right) ^2 + \left( \frac{B}{C} \right) ^2 = 1$, or $A^2+B^2=C^2$, the Pythagorean Theorem.  There are literally hundreds of proofs of this theorem, and I suspect this proof has been given sometime before, but I think this is a lovely derivation of that mathematical hallmark.

Conclusion:  While it seems that these two properties about the determinants of transformation matrices are indeed true for the examples shown, mathematicians hold out for a higher standard.   I’ll offer a proof of both properties in my next post.

## Generalized Pythagoras through Vectors

Here’s a proof of the Pythagorean Theorem by way of vectors.  Of course, if your students already know vectors, they’re already way past the Pythagorean Theorem, but I thought Richard Pennington‘s statement of this on LinkedIn gave a pretty and stunningly brief (after all the definitions) proof of one of mathematics’ greatest equations.

Let O be the origin, and let $\overrightarrow A$ and $\overrightarrow B$ be two position vectors starting at O. The vector from $\overrightarrow A$ to $\overrightarrow B$ is simply $\overrightarrow {B-A}$, which I will call $\overrightarrow C$. Using properties of dot products,

$\overrightarrow C\cdot\overrightarrow C = \overrightarrow {B-A}\cdot\overrightarrow {B-A} = \overrightarrow B\cdot\overrightarrow B-2\overrightarrow A\cdot\overrightarrow B+\overrightarrow A\cdot\overrightarrow A$

The dot product of a vector with itself is the square of its magnitude, so

$|\overrightarrow C|^2=|\overrightarrow B|^2-2|\overrightarrow A||\overrightarrow B|cos\theta+|\overrightarrow A|^2$

where $\theta$ is the angle between $\overrightarrow A$ and $\overrightarrow B$.

This is the Law of Cosines–in my classes, I call it the generalized Pythagorean Theorem for all triangles.  If $\theta=\frac{\pi}{2}$, then $\overrightarrow A$ and $\overrightarrow B$ are the legs of a right triangle with hypotenuse $\overrightarrow C$ which makes $cos(\theta)=cos(\frac{\pi}{2})=0$ and

$|\overrightarrow C|^2=|\overrightarrow A|^2+|\overrightarrow B|^2$

Pretty.

## Three Little Geometry Problems

Here are three variations of geometry problems I got from @jamestanton on Twitter.

1. The numerical measure of a rectangle’s area and perimeter are equal (P=A) (obviously the units are different).  If the rectangle’s sides have integer lengths, what are the dimensions of the rectangle?
2. The numerical measure of a box’s surface area and volume are equal (V=SA).  If the box’s sides have integer lengths, what are its dimensions?
3. The numerical measure of a right triangle’s area and perimeter are equal (P=A).  If the triangle’s sides have integer lengths, what are its dimensions?

To give a complete solution to a math problem, remember that you must

1. show that your proffered solution(s) is (are) correct, and
2. shows that no other solutions exist.

Find convincing arguments that you have found all of the solutions for each.  While my solutions are shown below, I eagerly welcome suggestions for any other approaches.

STOP!!!
DO NOT READ ANY FURTHER IF YOU WANT TO SOLVE THESE PROBLEMS ON YOUR OWN.

Problem 1:  $P=A$ for a rectangle
Let the a=length and b=width.  Without loss of generality, assume $a\le b$.  Then, $a\cdot b=2(a+b) \Longrightarrow \frac{1}{2}=\frac{1}{a}+\frac{1}{b}$ which implies $2.

If $a=3, b=6$.  If $a=4, b=4$.  Thus, the only rectangles for which $P=A$ are a 3×6 and a square with side 4.

Problem 2:  $P=SA$ for a box
Let the a=length, b=width, and c=height.  Without loss of generality, assume $a\le b\le c$.  Then, $a\cdot b\cdot c=2(a\cdot b+a\cdot c+b\cdot c) \Longrightarrow \frac{1}{2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ which implies $2.

It wasn’t worth it to find all these by hand, so I wrote a quick spreadsheet to find all c values for given a and b values under the condition $a\le b\le c$.

Therefore, there are 8 such boxes with integer dimensions: 3x7x42, 3x8x24, 3x9x18, 3x10x15, 4x5x20, 4x6x12, 5x5x10, and 6x6x6.

Problem 3:  $P=A$ for a right triangle
Let the legs be a and b and the hypotenuse be c.  Perhaps there is a way to employ the technique I used on the first two problems, but my first successful solution invoked a variation on Euclid’s formula:  For some integer values of k, m, and n with $m\le n$, $a=k\cdot (m^2-n^2), b=2k\cdot m\cdot n,$ and $c=k\cdot (m^2+n^2)$ will form all Pythagorean triples (although not uniquely).

Because $\frac{1}{2} a\cdot b=a+b+c$, Euclid’s formula gives

$\begin{tabular}{ r c l } $$2$$ & $$=$$ & $$\frac{\displaystyle a\cdot b}{\displaystyle a+b+c}$$ \\ & $$=$$ & $$\frac{\displaystyle [k\cdot (m^2-n^2)]\cdot [2k\cdot m\cdot n]}{\displaystyle [k\cdot (m^2-n^2)]+[2k\cdot m\cdot n]+[k\cdot (m^2+n^2)]}$$ \\ & $$=$$ & $$\frac{\displaystyle 2k^2mn(m^2-n^2)}{\displaystyle 2km(m+n)}$$ \\ & $$=$$ & $$kn(m-n)$$ \end{tabular}$

What started out feeling like a difficult search for unknown side lengths has been dramatically simplified.  $2=kn(m-n)$ can only happen if

1. $k=2, n=1, m-n=1 \rightarrow m=2$ which Euclid’s formula converts to $a=6, b=8, c=10$ .
2. $k=1, n=2, m-n=1 \rightarrow m=3$ which Euclid’s formula converts to $a=5, b=12, c=13$ .
3. $k=1, n=1, m-n=2 \rightarrow m=3$, but this leads to a repeat of the first solution.

Therefore, there are only two right triangles with the property $P=A$:  the 6-8-10 and the 5-12-13 right triangles.

Again, any other solution approaches are encouraged and will be posted.

## Area 10 Squares

Kudos to Dave Gale and chris maths for their great posts about introductory lessons that inspired the questions I pose below. At this point, I don’t have an answer to the query, but I welcome any insights and particularly any other spin-off ideas you may have.

If you have a standard sheet of square grid paper whose dots are exactly 1 unit apart and I ask you to draw a square of area 1, a square of area 4, and a square of area 9, you would probably quickly respond with the following.

Then I ask you to draw a square of area … [deliberate pause] … many immediately begin to think of continuing the pattern to area 16, but instead I ask for a square of area 10. Whether you know the Pythagorean Theorem or just how to compute the areas of squares and triangles, some experimentation hopefully will lead you to some form of the following figure which shows a square with area 10. The real twist for students here is that they need to adjust their point of view from what I’ll call horizontal squares (above) to tilted squares (below).

So, here are my questions:
1) What square areas can be created using square grid paper?
2) What areas of squares can be created more than one way?
3) Is there a largest square area that can NOT be created using square grid paper?

STOP! STOP! STOP!
Do not read any further if you want to work on these questions yourself.

What follows are my musings on these questions and some definite spoilers are included.

Question 1:
Perfect squares (1, 4, 9, 16, 25, 36, …) obviously can be found using increasingly larger horizontal squares. Dave’s ‘blog post gives a great start at the non-perfect, tilted squares.

Image source here.
The information he gives in the image above leads to areas of 2, 5, 8, 10, 13, 17, 18, 20, 25, 26, 29, 32, 34, 37, …

Merging the “tilted” list with the “horizontal list gives 1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 25, 26, 29, 32, 34, 36, 37, …

The missing square areas are 3, 6, 7, 11, 12, 14, 15, 19, 21, 22, 23, 24, …
Whether this list ends depends on the solution to posed question #3.

Question 2:

Comparing the horizontal and tilted lists, the first area that can be found both ways is 25. I briefly thought that was an amazing find until I remembered that the smallest integral Pythagorean Triple is 3-4-5. So a tilted square whose vector position (using Dave’s language) can be expressed as [3,4] can also be created using a horizontal side of 5–the Pythagorean Theorem arises!

There may be other ways to get equivalent square areas (I’d love to hear any if you know some!), but any integral Pythagorean Triple represents a square area that can be represented at least two ways on square grid paper. There are an infinite number of such equivalences.

Question 3:
I don’t know the answer to this, but I think I’m close. I’ll post the problem before I finish it for the fun of letting others into the enjoyment of solving what I think is a cool pre-collegiate level math problem.

I did notice that the missing areas at the end of my discussion of question 1 seems to include a large number of multiples of 3, excluding of course, the horizontal squares with sides that have lengths that are multiples of 3. So is it possible to prove that any area that is
– a multiple of 3, but
– not a perfect square
can not be drawn on square grids?
If so, then there is no maximum area of a square that cannot be drawn using square grid paper.
If not, then the solution to this question may lie in a direction I have not conceived.

Again, I don’t know the answers to questions 1 or 3. Discussion is welcome and encouraged.