Monthly Archives: December 2014

Saving a Quadrilateral Problem

Mike Lawler posted this on Twitter a couple weeks ago.

The sad part of the problem is that it has no realistic solution as posed.  From the way the diagram is drawn, the unlabeled side was likely intended to be congruent to its opposite side, making its value=x-4.  Adding the four sides and equating to the given perimeter was probably the intended solution.  This approach gives


That’s nice enough, and a careless solver would walk away (see later in this post).  But you should always check your solution.  If x=14, then the longer base is 3(14)+2=44 and the two congruent sides are each 10.  That appears to be OK until you add the two congruent sides to the shorter base:  10+24+10=44.  Unfortunately, that’s the same length as the longer base, so this particular quadrilateral would have height=0.

Alas, this problem, as initially defined, creates a degenerate quadrilateral, but you would know this only if you checked down a couple layers–something I suspect most students (and obviously some problem writers) would miss.  Unless a class has explicitly addressed degenerate forms, I don’t think this is a fair question as written.

Even so, I wondered if the problem could be saved.  It wasn’t an isosceles quadrilateral in the formulation suggested by its unknown writer, but I wondered if there was a way to save it.  My following attempts all keep the given side labels, but assume the figure is not drawn to scale.

First a diversion:

Some don’t realize that the definition of a trapezoid is not a 100% settled issue in mathematics.  I posted on this almost three years ago (here) and got a few surprisingly fierce responses.

The traditional camp holds to Euclid’s definition that a trapezoid is a quadrilateral with exactly one pair of parallel sides.  I always found it interesting that this is the only quadrilateral Euclid restrictively defined in the Elements.

The other camp defines a trapezoid as a quadrilateral with at least one pair of parallel sides.  I’ve always liked this camp for two reasons.  First, this inclusive definition is more consistent with all of the other inclusive quadrilateral definitions.  Second, it allows greater connections between types.

Most students eventually learn that “a square is a rectangle, but a rectangle is not necessarily a square.”  This is a logical result of the inclusive definition of a rectangle.  Following that reasoning, two equivalent statements possible only from an inclusive definition of a trapezoid are “a parallelogram is a trapezoid, but a trapezoid is not necessarily a parallelogram,” and “a rectangle is an isosceles trapezoid, but an isosceles trapezoid is not necessarily a rectangle.”

Much more importantly, the inclusive definitions create what I call a “cascade of properties.”  That is, any properties held by any particular quadrilateral are automatically inherited by every quadrilateral in the direct line below it in the Quadrilateral Hierarchy.

Trying to Salvaging the Problem:

Recovery attempt 1.  My first attempt to save the problem–which I first thought would be a satisfactory moment of potentially controversial insight using the inclusive trapezoid definition–ended in some public embarrassment for me.  (I hope I recover that with this post!)

Under the inclusive definition, squares and rectangles are also isosceles quadrilaterals.  I wondered if the quadrilateral could be a rectangle.  If so, opposite sides would be equal, giving 24=3x+2, or x=\frac{22}{3}.  That creates a rectangle with sides \frac{10}{3} and 24.  I was pleased to have “saved” the problem and quickly tweeted out this solution.  Unfortunately, I forgot to double check the perimeter requirement of 88–a point someone had to point out on Twitter.  I know better than to make unchecked public pronouncements.  Alas.  The given problem can’t be saved by treating it as a rectangle.

Recovery attempt 2.  Could this be a square?  If so, all sides are equal and x-4=24 \longrightarrow x=20, but this doesn’t match the x-value found from the rectangle in the first recovery attempt.  This is important because squares are rectangles.

The given information doesn’t permit a square.  That means the given information doesn’t permit any form of non-degenerate isosceles trapezoid.  Disappointing.

Attempt 3–Finally Recovered.  What if this really was an isosceles trapezoid, but not as drawn?  Nothing explicit stated in the problem prevents me from considering the x-4 and the unlabeled sides parallel, and the other two congruent as shown.


So, 24=3x+2 \longrightarrow x=\frac{22}{3} as before with the rectangle, making \left( \frac{22}{3} \right) -4 = \frac{10}{3} the last labeled side.  The sum of these three sides is \frac{154}{3}, so the last side must be \frac{110}{3} for the overall perimeter to be 88.  The sum of the smallest of three of these is greater than the 4th, so the degenerate problem that scuttled Attempt #1 did not happen here.

So, there is a solution, x=\frac{22}{3}, that satisfies the problem as stated, but how many students would notice the first degenerate case, and then read the given figure as not to scale before finding this answer?  This was a poorly written problem.

In the end, the solution for x that I had posted to Twitter turned out to be correct … but not for the reasons I had initially claimed.

Attempt 4–Generalizing.  What if the problem was rephrased to make it an exploration of quadrilateral properties?  Here’s a suggestion I think might make a dandy exploration project for students.

Given the quadrilateral with three sides labeled as above, but not drawn to scale, and perimeter 88, what more specific types of quadrilateral could be represented by the figure?

Checking types:

  • Rectangles and squares are already impossible.
  • There is one convoluted isosceles trapezoid possibility detailed in Attempt 3.
  • All four sides can’t be equal with the given information (Recovery attempt 2), so rhombus is eliminated.
  • Recovery attempt 1 showed that opposite sides could be equal, but since they then do not meet the perimeter requirement, a parallelogram is gone.
  • In a kite, there are two adjacent pairs of congruent sides.  There are two ways this could happen:  the unlabeled side could be 24, or it could be equal to 3x+2.
    • If the unlabeled is 24, then x-4=3x+2 \longrightarrow x=-3, an impossible result when plugged back into the given side expressions.
    • If the unlabeled side is 3x+2, then x-4=24 \longrightarrow x=28, making the unlabeled side 86 and the overall perimeter 220–much too large.  The quadrilateral cannot be kite.
  • All that remains is a trapezoid and a generic quadrilateral, for which there are no specific side lengths.  With one side unlabeled and therefore unrestricted, the quadrilateral could be constructed in many different ways so long as all sides are positive.  That means
    • x-4>0 \longrightarrow x>4 and
    • 3x+2>0 \longrightarrow x>-\frac{2}{3}.
    • In a quadrilateral, the sum of any three sides must be between half and all of the overall length of the perimeter.  In this case, 88>24+(x-4)+(3x+2)> \frac{1}{2} 88 \longrightarrow 5.5<x<16.5.
    • Putting all three of these together, you can have a trapezoid OR a generic quadrilateral for any 5.5<x<16.5.


The given information CAN define an isosceles trapezoid, but the route to and form of the solution is far more convoluted than I suspect the careless question writer intended.  Sans the isosceles trapezoid requirement, this figure can define only a generic quadrilateral or a trapezoid, and only then for values of x where 5.5<x<16.5.

Trying to make this problem work, despite its initial flaws, turned out to be a fun romp through a unit on quadrilateral classifications.  Running through all of the possibilities, the properties of quadrilaterals, and narrowing down the outcomes might make this problem variation a worthwhile student project, albeit very challenging for many, I think.

I just wish for the students the original problem hadn’t been so poorly written.  But if that had happened, I would have missed out on some fun.  Hopefully it will be worthwhile for some of your students.

Problems in Time

Here’s an easy enough challenge problem for students from Math Counts that I found on Twitter via Mathmovesu (@mathmovesu).

Seeing that problem, I wondered

On a 12-hour digital clock, at how many times during a 24-hour day will all of the digits showing the time be a palindrome?

Are all solutions to the original question automatically solutions to the palindrome variation?

What other questions could we ask here?  I’m particularly interested in questions students might develop.  After all, teachers shouldn’t be the only ones thinking, creating, and extending.


Unexpected Proof of the Pythagorean Theorem

Following is a neat discovery of an alternative proof of the Pythagorean Theorem resulting from the multiple solutions to the Circle and Square problem.  I’m sure someone has seen this before, as there are literally 100s of unique proofs of the Pythagorean Theorem, but this one was new to me.

The intersecting chord relationships in a circle can be proven using only similar triangles.  Proofs of these are at the bottom of this post, if needed.  Using only those, you can prove the Pythagorean Theorem.


The image below–a revision of the diagram from my previous post–shows diameter DE in circle C.  Chord AB is a side of the given square from the Circle and Square problem and is bisected by symmetry into two segments, each of length a.  Let  be the radius of circle C.  Let the portion of DE from point C to chord AB have length b.  Because AB is a chord bisected by diameter DE, two right triangles are created, as shown.


AB and DE are intersecting chords, so a \cdot a = (r-b) \cdot (r+b).  Expanding the right side and moving the b^2 term to the other side gives the Pythagorean Theorem.

Short and sweet once the chord relationships are established.


In the image below, AB and CD are any two chords intersecting at point E.  Vertical angles give \angle DEA \cong \angle BEC .  Because \angle ADE and \angle CBE are inscribed angles sharing arc AC, they are also congruent.


That means \Delta ADE \sim \Delta CBE, which gives \displaystyle \frac{x}{w} = \frac{y}{z}, or x \cdot z = w \cdot y.  QED

Show that if a diameter bisects a chord, the diameter and chord are perpendicular.  Start with the vertical diameter of circle C bisecting chord AB.


It should be straightforward to show \Delta ADC \cong \Delta BDC by SSS.  That means  corresponding angles \angle ADC \cong \angle BDC; as they also from a linear pair, those angles are both right, and the proof is established.

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.


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.


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