# Tag Archives: law of sines

## Lovely or Tricky Triangle Question?

In addition to not being drawn to scale and asking for congruence anyway, I like this problem because it potentially forces some great class discussions.

One responder suggested using the Law of Sines (LoS) to establish an isosceles triangle.  My first thought was that was way more sophisticated than necessary and completely missed the fact that the given triangle information was SSA.

My initial gut reaction was this SSA setup was a “trick” ambiguous case scenario and no congruence was possible, but I couldn’t find a flaw in the LoS logic. After all, LoS fails when attempting to find obtuse angles, but the geometry at play here clearly makes angles B and C both acute.  That meant LoS should work, and this was actually a determinate SSA case, not ambiguous.  I was stuck in a potential contradiction.  I was also thinking with trigonometry–a far more potent tool than I suspected was necessary for this problem.

“Stuck” moments like this are GOLDEN for me in the classroom.  I could imagine two primary student situations here.  They either  1) got a quick “proof” without recognizing the potential ambiguity, or 2) didn’t have a clue how to proceed.  There are many reasons why a student might get stuck here, all of which are worth naming and addressing in a public forum.  How can we NAME and MOVE PAST situations that confuse us?  Perhaps more importantly, how often do we actually recognize when we’re in the middle of something that is potentially slipperier than it appears to be on the surface?

PROBLEM RESOLUTION:

I read later that some invoked the angle bisector theorem, but I took a different path.  I’m fond of a property I asked my geometry classes to prove last year .

If any two of a triangle’s 1) angle bisector, 2) altitude, and 3) median coincide, prove that the remaining segment does, too, and that whenever this happens, the triangle will be isosceles with its vertex at the bisected angle.

Once I recognized that the angle bisector of angle BAC was also the median to side BC, I knew the triangle was isosceles.  The problem was solved without invoking any trigonometry or any similarity ratios.

Very nice problem with VERY RICH discussion potential.  Thanks for the tweet, Mr. Noble.

For more conversation on this, check out this Facebook conversation.

## Squares and Octagons

Following is a really fun problem Tom Reardon showed my department last May as he led us through some TI-Nspire CAS training.  Following the introduction of the problem, I offer a mea culpa, a proof, and an extension.

THE PROBLEM:

Take any square and construct midpoints on all four sides.
Connect the four midpoints and four vertices to create a continuous 8-pointed star as shown below.  The interior of the star is an octagon.  Construct this yourself using your choice of dynamic geometry software and vary the size of the square.

Compare the areas of the external square and the internal octagon.

You should find that the area of the original square is always 6 times the area of the octagon.

I thought that was pretty cool.  Then I started to play.

MINOR OBSERVATIONS:

Using my Nspire, I measured the sides of the octagon and found it to be equilateral.

As an extension of Tom’s original problem statement, I wondered if the constant square:octagon ratio occurred in any other quadrilaterals.  I found the external quadrilateral was also six times the area of the internal octagon for parallelograms, but not for any more general quadrilaterals.  Tapping my understanding of the quadrilateral hierarchy, that means the property also holds for rectangles and rhombi.

MEA CULPA:

Math teachers always warn students to never, ever assume what they haven’t proven.  Unfortunately, my initial exploration of this problem was significantly hampered by just such an assumption.  I obviously know better (and was reminded afterwards that Tom actually had told us that the octagon was not equiangular–but like many students, I hadn’t listened).   After creating the original octagon, measuring its sides and finding them all equivalent, I errantly assumed the octagon was regular.  That isn’t true.

That false assumption created flaws in my proof and generalizations.  I discovered my error when none of my proof attempts worked out, and I eventually threw everything out and started over.  I knew better than to assume.  But I persevered, discovered my error through back-tracking, and eventually overcame.  That’s what I really hope my students learn.

THE REAL PROOF:

Goal:  Prove that the area of the original square is always 6 times the area of the internal octagon.

Assume the side length of a given square is $2x$, making its area $4x^2$.

The octagon’s area obviously is more complicated.  While it is not regular, the square’s symmetry guarantees that it can be decomposed into four congruent kites in two different ways.  Kite AFGH below is one such kite.

Therefore, the area of the octagon is 4 times the area of AFGH.  One way to express the area of any kite is $\frac{1}{2}D_1\cdot D_2$, where $D_1$ and $D_2$ are the kite’s diagonals. If I can determine the lengths of $\overline{AG}$ and $\overline {FH}$, then I will know the area of AFGH and thereby the ratio of the area of the square to the area of the octagon.

The diagonals of every kite are perpendicular, and the diagonal between a kite’s vertices connecting its non-congruent sides is bisected by the kite’s other diagonal.  In terms of AFGH, that means $\overline{AG}$ is the perpendicular bisector of $\overline{FH}$.

The square and octagon are concentric at point A, and point E is the midpoint of $\overline{BC}$, so $\Delta BAC$ is isosceles with vertex A, and $\overline{AE}$ is the perpendicular bisector of $\overline{BC}$.

That makes right triangles $\Delta BEF \sim \Delta BCD$.  Because $\displaystyle BE=\frac{1}{2} BC$, similarity gives $\displaystyle AF=FE=\frac{1}{2} DC=\frac{x}{2}$.  I know one side of the kite.

Let point I be the intersection of the diagonals of AFGH.  $\Delta BEA$ is right isosceles, so $\Delta AIF$ is, too, with $m\angle{IAF}=45$ degrees.  With $\displaystyle AF=\frac{x}{2}$, the Pythagorean Theorem gives $\displaystyle IF=\frac{x}{2\sqrt{2}}$.  Point I is the midpoint of $\overline{FH}$, so $\displaystyle FH=\frac{x}{\sqrt{2}}$.  One kite diagonal is accomplished.

Construct $\overline{JF} \parallel \overline{BC}$.  Assuming degree angle measures, if $m\angle{FBC}=m\angle{FCB}=\theta$, then $m\angle{GFJ}=\theta$ and $m\angle{AFG}=90-\theta$.  Knowing two angles of $\Delta AGF$ gives the third:  $m\angle{AGF}=45+\theta$.

I need the length of the kite’s other diagonal, $\overline{AG}$, and the Law of Sines gives

$\displaystyle \frac{AG}{sin(90-\theta )}=\frac{\frac{x}{2}}{sin(45+\theta )}$, or

$\displaystyle AG=\frac{x \cdot sin(90-\theta )}{2sin(45+\theta )}$.

Expanding using cofunction and angle sum identities gives

$\displaystyle AG=\frac{x \cdot sin(90-\theta )}{2sin(45+\theta )}=\frac{x \cdot cos(\theta )}{2 \cdot \left( sin(45)cos(\theta ) +cos(45)sin( \theta) \right)}=\frac{x \cdot cos(\theta )}{\sqrt{2} \cdot \left( cos(\theta ) +sin( \theta) \right)}$

From right $\Delta BCD$, I also know $\displaystyle sin(\theta )=\frac{1}{\sqrt{5}}$ and $\displaystyle cos(\theta)=\frac{2}{\sqrt{5}}$.  Therefore, $\displaystyle AG=\frac{x\sqrt{2}}{3}$, and the kite’s second diagonal is now known.

So, the octagon’s area is four times the kite’s area, or

$\displaystyle 4\left( \frac{1}{2} D_1 \cdot D_2 \right) = 2FH \cdot AG = 2 \cdot \frac{x}{\sqrt{2}} \cdot \frac{x\sqrt{2}}{3} = \frac{2}{3}x^2$

Therefore, the ratio of the area of the square to the area of its octagon is

$\displaystyle \frac{area_{square}}{area_{octagon}} = \frac{4x^2}{\frac{2}{3}x^2}=6$.

QED

EXTENSIONS:

This was so nice, I reasoned that it couldn’t be an isolated result.

I have extended and proved that the result is true for other modulo-3 stars like the 8-pointed star in the square for any n-gon.  I’ll share that very soon in another post.

I proved the result above, but I wonder if it can be done without resorting to trigonometric identities.  Everything else is simple geometry.   I also wonder if there are other more elegant approaches.

Finally, I assume there are other constant ratios for other modulo stars inside larger n-gons, but I haven’t explored that idea.  Anyone?