# Tag Archives: triangle

## Inscribed Triangle Challenge

@MathCeyhun posed an interesting geometry problem yesterday.

Even more interesting is that, as @MathCeyhun noted in a subsequent tweet, none of the posted solutions is correct.  There have been a few posted answers, but no solutions, so I thought I’d give it a try.

OBSERVATIONS

• The perpendicular bisector of each chord was given, and perpendicular bisectors of chords always lie on radii of the circle.
• If r is the radius of the circle, then the lengths of the extensions of the perpendicular bisectors are r-1, r-2, and r-3.
• Nothing given guarantees anything special about the triangle, so I assumed it was scalene.  I called the side lengths 2x, 2y, and 2z to simplify the bisection labels.
• Adding the bisector extensions, the radii to the vertices, and segment names and labels gave me this.

[Irrelevant to this problem, but I just realized by looking at this image that every triangle can be dissected into three isosceles triangles with congruent sides and a shared vertex point at the triangle’s circumcenter.  Pretty.]

SETTING UP MY SOLUTION

Each isosceles triangle is bisected by the perpendicular bisector of its base from which I extracted three relationships from the Pythagorean theorem.

$x^2+(r-1)^2=r^2 \longrightarrow x^2=(2r-1) \cdot 1$

$y^2+(r-2)^2=r^2 \longrightarrow y^2=(2r-2) \cdot 2$

$z^2+(r-3)^2=r^2 \longrightarrow z^2=(2r-3) \cdot 3$

[The relationship between the half-sides, the extension of the radius beyond the triangle, and the radius is another pretty pattern.]

That gives 3 equations in 4 variables.  I needed one more to solve….

The area of $\Delta ABC$ can be expressed two ways:  as the sum of the areas of the isosceles triangles, and using Heron’s formula.  From the areas of the isosceles triangles,

$Area( \Delta ABC) = \frac{1}{2}(2x)(r-1) + \frac{1}{2}(2y)(r-2) + \frac{1}{2}(2z)(r-3)$
$Area( \Delta ABC) = x \cdot (r-1) + y \cdot (r-2) + z \cdot (r-3)$

The sides of $\Delta ABC$ are 2x, 2y, and 2z, so its semiperimeter is x+y+z and Heron’s formula gives its area as

$Area( \Delta ABC) = \sqrt{(x+y+z)(-x+y+z)(x-y+z)(x+y-z)}$.

The area of a given triangle is unique, so the two different area expressions are equivalent, giving a fourth equation.

SOLVING A SYSTEM & ANSWERING THE QUESTION

With four equations in four variables, I had a system of equations.  The algebra was messy, so I invoked my CAS to crunch it for me.

The question asked for the area of the triangle, so I just substituted my values back into the area formulas.

And 17.186… is clearly not one of the choices in the original problem.

A PLEA…

Recognizing the perpendicular bisectors, seeing all the right triangles, and connecting the multiple ways to describe the area of a triangle made this solution reasonably easy to find with the help of my computer algebra system (CAS), but I know the background algebra is, at best, cumbersome.  I hope there’s a more elegant solution, but I don’t see it.  Can anyone offer a suggestion?

Either way … this is definitely becoming a challenge problem for my Quantitative Reasoning class this coming week!

## Pythagorean Investigation

Here’s a challenge @jamestanton tweeted yesterday:

Visually, Tanton is asking if there is an integer right triangle (like the standard version shown on the left below) for which the integer triangle on the right exists.

The algebraic equivalent to this question is, for some $a^2+b^2=c^2$, does there exist a Natural number d so that $b^2+c^2=d^2$?

I invoked Euclid’s formula in my investigation to show that there is no value of d to make this possible.  I’d love to hear of any other proof variations.

INVOKING EUCLID’S FORMULA

For any coprime natural numbers m & n where $m>n$ and $m-n$ is odd, then every primitive Pythagorean triple can be generated by $\left\{ m^2 - n^2, 2mn, m^2 + n^2 \right\}$.

For any Natural number kevery Pythagorean triple can be generated by $\left\{ k \cdot \left( m^2 - n^2 \right), k \cdot \left( 2mn \right), k \cdot \left( m^2 + n^2 \right) \right\}$.

The generator term $k \cdot \left( m^2 + n^2 \right)$ must be the original hypotenuse (side c), but either $k \cdot \left( m^2 - n^2 \right)$ or $k \cdot \left( 2mn \right)$ can be side b.  So, if Tanton’s scenario is true, I needed to check two possible cases.  Does there exist a Natural number d such that

$d^2 = \left( k \cdot \left( m^2 - n^2 \right) \right)^2 + \left( k \cdot \left( m^2 + n^2 \right) \right)^2 = 2k^2 \left( m^4 + n^4 \right)$

or

$d^2 = \left( k \cdot \left( 2mn \right) \right)^2 + \left( k \cdot \left( m^2 + n^2 \right) \right)^2 = k^2 \left( m^4 +6m^2n^2 + n^4 \right)$

is true?

EVALUATING THE POSSIBILITIES

For the first equation, there is a single factor of 2 on the right, and there is no way to extract an odd number of factors of 2 from $\left( m^4 + n^4 \right)$ or $k^2$, so $2k^2 \left( m^4 + n^4 \right)$ can’t represent a perfect square.

For the second equation, there is no way to factor $\left( m^4 +6m^2n^2 + n^4 \right)$ over Integers, so $k^2 \left( m^4 +6m^2n^2 + n^4 \right)$ can’t be a perfect square either.

Since neither equation can create a perfect square, there is no Natural value of d that makes {b, c, d} a Pythagorean triple.  Tanton’s challenge is impossible.

Does anyone have a different approach?

## Inscribed Right Angle Proof Without Words

Earlier this past week, I assigned the following problem to my 8th grade Geometry class for homework.  They had not explored the relationships between circles and inscribed angles, so I added dashed auxiliary segment AD as a hint.

What follows first is the algebraic solution I expected most to find and then an elegant transformational explanation one of my students produced.

PROOF 1:

Given circle A with diameter BC and point D on the circle.  Prove triangle BCD is a right triangle.

After some initial explorations on GeoGebra sliding point D around to discover that its angle measure was always $90^{\circ}$ independent of the location of D, most successful solutions recognized congruent radii AB, AC, and AD, creating isosceles triangles CAD and BAD.  That gave congruent base angles x in triangle CAD, and y in BAD.

The interior angle sum of a triangle gave $(x)+(x+y)+(y)=180^{\circ}$, or $m \angle CDB = x+y = 90^{\circ}$, confirming that BCD was a right triangle.

PROOF 2:

Then, one student surprised us.  She marked the isosceles base angles as above before rotating $\Delta BCD$ $180^{\circ}$ about point A.

Because the diameter rotated onto itself, the image and pre-image combined to form an quadrilateral with all angles congruent.  Because every equiangular quadrilateral is a rectangle, M had confirmed BCD was a right triangle.

CONCLUSION:

I don’t recall seeing M’s proof before, but I found it a delightfully elegant application of quadrilateral properties.  In my opinion, her rotation is a beautiful proof without words solution.

Encourage freedom, flexibility of thought, and creativity, and be prepared to be surprised by your students’ discoveries!

Most of my thinking about teaching lately has been about the priceless, timeless value of process in problem solving over the ephemeral worth of answers.  While an answer to a problem puts a period at the end of a sentence, the beauty and worth of the sentence was the construction, word choice, and elegance employed in sharing the idea at the heart of the sentence.

Just as there are many ways to craft a sentence–from cumbersome plodding to poetic imagery–there are equally many ways to solve problems in mathematics.  Just as great writing reaches, explores, and surprises, great problem solving often starts with the solver not really knowing where the story will lead, taking different paths depending on the experience of the solver, and ending with even more questions.

I experienced that yesterday reading through tweets from one of my favorite middle and upper school problem sources, Five Triangles.  The valuable part of what follows is, in my opinion, the multiple paths I tried before settling on something productive.  My hope is that students learn the value in exploration, even when initially unproductive.

At the end of this post, I offer a few variations on the problem.

The Problem

Try this for yourself before reading further.  I’d LOVE to hear others’ approaches.

First Thoughts and Inherent Variability

My teaching career has been steeped in transformations, and I’ve been playing with origami lately, so my brain immediately translated the setup:

Fold vertex A of equilateral triangle ABC onto side BC.  Let segment DE be the resulting crease with endpoints on sides AB and AC with measurements as given above.

So DF is the folding image of AD and EF is the folding image of AE.  That is, ADFE is a kite and segment DE is a perpendicular bisector of (undrawn) segment AF.  That gave $\Delta ADE \cong \Delta FDE$ .

I also knew that there were lots of possible locations for point F, even though this set-up chose the specific orientation defined by BF=3.

Lovely, but what could I do with all of that?

Trigonometry Solution Eventually Leads to Simpler Insights

Because FD=7, I knew AD=7.  Combining this with the given DB=8 gave AB=15, so now I knew the side of the original equilateral triangle and could quickly compute its perimeter or area if needed.  Because BF=3, I got FC=12.

At this point, I had thoughts of employing Heron’s Formula to connect the side lengths of a triangle with its area.  I let AE=x, making EF=x and $EC=15-x$.  With all of the sides of $\Delta EFC$ defined, its perimeter was 27, and I could use Heron’s Formula to define its area:

$Area(\Delta EFC) = \sqrt{13.5(1.5)(13.5-x)(x-1.5)}$

But I didn’t know the exact area, so that was a dead end.

Since $\Delta ABC$ is equilateral, $m \angle C=60^{\circ}$ , I then thought about expressing the area using trigonometry.  With trig, the area of a triangle is half the product of any two sides multiplied by the sine of the contained angle.  That meant $Area(\Delta EFC) = \frac{1}{2} \cdot 12 \cdot (15-x) \cdot sin(60^{\circ}) = 3(15-x) \sqrt3$.

Now I had two expressions for the same area, so I could solve for x.

$3\sqrt{3}(15-x) = \sqrt{13.5(1.5)(13.5-x)(x-1.5)}$

Squaring both sides revealed a quadratic in x.  I could do this algebra, if necessary, but this was clearly a CAS moment.

I had two solutions, but this felt WAY too complicated.  Also, Five Triangles problems are generally accessible to middle school students.  The trigonometric form of a triangle’s area is not standard middle school fare.  There had to be an easier way.

A Quicker Ending

Thinking trig opened me up to angle measures.  If I let $m \angle CEF = \theta$, then $m \angle EFC = 120^{\circ}-\theta$, making $m \angle DFB = \theta$, and I suddenly had my simple breakthrough!  Because their angles were congruent, I knew $\Delta CEF \sim \Delta BFD$.

Because the triangles were similar, I could employ similarity ratios.

$\frac{7}{8}=\frac{x}{12}$
$x=10.5$

And that is one of the CAS solutions by a MUCH SIMPLER approach.

Extensions and Variations

Following are five variations on the original Five Triangles problem.  What other possible variations can you find?

1)  Why did the CAS give two solutions?  Because $\Delta BDF$ had all three sides explicitly given, by SSS there should be only one solution.  So is the 13.0714 solution real or extraneous?  Can you prove your claim?  If that solution is extraneous, identify the moment when the solution became “real”.

2)  Eliminating the initial condition that BF=3 gives another possibility.  Using only the remaining information, how long is $\overline{BF}$ ?

$\Delta BDF$ now has SSA information, making it an ambiguous case situation.  Let BF=x and invoke the Law of Cosines.

$7^2=x^2+8^2-2 \cdot x \cdot 8 cos(60^{\circ})$
$49=x^2-8x+64$
$0=(x-3)(x-5)$

Giving the original BF=3 solution and a second possible answer:  BF=5.

3)  You could also stay with the original problem asking for AE.

From above, the solution for BF=3 is AE=10.5.  But if BF=5 from the ambiguous case, then FC=10 and the similarity ratio above becomes

$\frac{7}{8}=\frac{x}{10}$
$x=\frac{35}{4}=8.75$

4)  Under what conditions is $\overline{DE} \parallel \overline{BC}$ ?

5)  Consider all possible locations of folding point A onto $\overline{BC}$.  What are all possible lengths of $\overline{DE}$?

## Next Steps from a Triangle

Watching the news a couple mornings ago, an impossible triangle appeared on the screen.  Hopefully some readers might be able to turn some first ideas a colleague and I had into a great applied geometry lesson.  What follows are some teacher thoughts.  My colleagues and I hope to cultivate classes where students become curious enough to raise some of these questions themselves.

WHAT’S WRONG?

At first glance, the labeling seems off.  In Euclidean geometry, the Triangle Inequality says the sum of the lengths of any two sides of a triangle must exceed the length of the third side.  Unfortunately, the shorter two sides sum to 34 miles, so the longest side of 40 miles seems physically impossible.  Someone must have made a typo.  Right?

But to dismiss this as a simple typo would be to miss out on some spectacular mathematical conversations.  I’m also a big fan of taking problems or situations with prima facie flaws and trying to recover either the problem or some aspects of it (see two of previous posts here and here).

WHAT DOES APPROXIMATELY MEAN?

Without confirming any actual map distances, I first was drawn to the vagueness of the approximated side lengths.  Was it possible that this triangle was actually correct under some level of round-off adjustment?  Hopefully, students would try to determine the degree of rounding the graphic creator used.  Two sides are rounded to a multiple of 10, but the left side appears rounded to a nearest integer with two significant digits.  Assuming the image creator was consistent (is that reasonable?), that last side suggests the sides were rounded to the nearest integer.  That means the largest the left side could be would be 14.5 miles and the bottom side 20.5 miles.  Unfortunately, that means the third side can be no longer than 14.5+20.5=35 miles.  Still not enough to justify the 40 miles, but this does open one possible save.

But what if all three sides were measured to the nearest 10 instead of my assumed ones place?  In this case the sides would be approximately 10, 20, and 40.  Again, this looks bad at first, but a 10 could have been rounded from a 14.9, a 20 from a 24.9, making the third side a possible 14.9+24.9=39.8, completely justifying a third side of 40.    This wasn’t the given labeling, but it would have potentially saved the graphic’s legitimacy.

GEOMETRY ALTERNATIVE

Is there another way the triangle might be correct?  Rarely do pre-collegiate geometry classes explore anything beyond Euclidean geometry.  One of my colleagues, Steve, proposed spherical geometry:

Does the fact that the earth is round play a part in these seemingly wrong values (it turns out “not really”… Although it’s not immediately clear, the only way to violate the triangle inequality in spherical geometry is to connect point the long way around the earth. And based on my admittedly poor geographical knowledge, I’m pretty sure that’s not the case here!)

SHORTEST DISTANCE

Perhaps students eventually realize that the distances involved are especially small relative to the Earth’s surface, so they might conclude that the Euclidean geometry approximation in the graphic is likely fine.

Then again, why is the image drawn “as the crow flies”?  The difficult mountainous terrain in upstate New York make surface distances much longer than air distances between the same points.  Steve asked,

in the context of this problem (known location of escaped prisoners), why is the shortest distance between these points being shown? Wouldn’t the walking/driving distance by paths be more relevant?  (Unless the prisoners had access to a gyrocopter…)

The value of a Euclidean triangle drawn over mountainous terrain has become questionable, at best.

FROM PERIMETER TO AREA

I suspect the triangle awkwardly tried to show the distances the escapees might have traveled.  Potentially interesting, but when searching for a missing person in the mountains–the police and news focus at the time of the graphic–you don’t walk the perimeter of the suspected zone, you have to explore the area inside.

A day later, I saw the search area around Malone, NY shown as a perfect circle.  (I wish I had grabbed that image, too.).  Around the same time, the news reported that the search area was 22 square miles.

• Was the authorities’ 22 measure an approximation of a circle’s area, a polygon based on surface roads, or some other shape?
• Going back to the idea of a spherical triangle, Steve hoped students would ask if they could “compute that from just knowing the side lengths? Is there a spherical Herons Formula?”
• If the search area was a more complicated shape, could you determine its area through some sort of decomposition into simpler shapes?  Would spherical geometry change how you approach that question?  Steve wondered if any students would ask, “Could we compute that from just knowing the side lengths? Is there a spherical Herons Formula?
• At one point near the end of the search, I hear there were about 1400 police officers in the immediate vicinity searching for the escapee.  If you were directing the search for a prison escapee or a lost hiker, how would you deploy those officers?  How long would it take them to explore the entire search zone?  How would the shape of the potential search zone affect your deployment plan?
• If you spread out the searchers in an area, what is the probability that an escapee or missing person could avoid detection?  How would you compute such a probability?
• Ultimately, I propose that Euclidean or spherical approximations seriously underestimated the actual surface area?  The dense mountainous terrain significantly complicated this search.  Could students extrapolate a given search area shape to different terrains?  How would the number of necessary searchers change with different terrains?
• I think there are some lovely openings to fractal measures of surface roughness in the questions in the last bullet point.

ERROR ANALYSIS

Ultimately, we hope students would ask

• What caused the graphic’s errors?  Based on analyses above and some Google mapping, we think “a liberal interpretation of the “approximately” label on each leg might actually be the culprit.”  What do the triangle inequality violations suggest about round-off errors or the use of significant digits?
• The map appeared to be another iteration of a map used a few days earlier.  Is it possible that compounded rounding errors were partially to blame?
• Surely the image’s designer new the triangle was an oversimplification of the reality.  Assuming so, why was this graphic used anyway?  Does it have any news value?  Could you design a more meaningful infographic?

APPRECIATION

Many thanks to Steve Earth for his multiple comments and thoughts that helped fill out this post.

## 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.

## 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.

PROOF:

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.

SUPPORTING PROOF 1:

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

SUPPORTING PROOF 2:
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.

## Star angles

Given a 5-pointed star like the one shown below, what is the sum of the angles in the outer “arms” of the star?

A variation on this problem showed up on a contest our students took last week.  I think there are multiple solution approaches to this, but the exploration and discovery are well worth it.  I offer a generalization at the end.

While not permitted on the contest, my first instinct was to construct the star using dynamic geometry software.  The image below is my result using the TI-Nspire; other softwares would work equally well.

After dragging around the star tips, watching the angles change while the sum didn’t, it was pretty clear that the sum was fixed at 180 degrees.  Nice, but why?

Here’s my no-technology solution from the day of the contest.

I saw the figure as an interior pentagon with exterior triangles off each edge.  Using this, I knew that ALL of the angles in the figure must add to $3\cdot 180$ (for the pentagon) plus $5\cdot 180\deg$ (for the 5 triangles), for a total $8\cdot 180$ degrees.  Then I noticed that there were 10 linear pairs of angles where the triangles met the pentagon, two at each pentagonal vertex.  Subtracting these from the initial sum would leave just the desired star point angles, but unfortunately would subtract the interior pentagonal angles twice, so the pentagon would need to be added back to re-balance.  That left the measure of the start point angles for any orientation to be:

$8\cdot 180$ (0riginal) $-10\cdot 180$ (10 linear pairs) $+ 3\cdot 180$ (rebalance pentagon)

For a total of $1\cdot 180 = 180$ degrees.

I’d love to hear other approaches.

EXTENDING:

There’s some lovely geometry and arithmetic overlaps in the construction of stars.  That’s perhaps another post for another time.  For now, I’ll extend this to stars with 5 or more points as stars with fewer points aren’t possible with non-degenerate internal polygons like above.

Using the same logic as above, imagine an n-pointed star comprised of an interior n-gon with its edges extended to form n exterior triangles.  My goal now is to find the sum of the n exterior star point angles.

As before, the n triangle angles sum to $n\cdot 180$ degrees and the interior n-gon’s angles sum to $(n-2)\cdot 180$ degrees for a total interior angle measure of $(2n-2)\cdot 180$ degrees.

There are $2n$ linear pairs at the junctures between the n-gon and its exterior triangles.  Subtracting their sum, $(2n\cdot 180)$ degrees, from total angle sum subtracts the n-gon’s angles twice, so adding the n-gon’s angles, $(n-2)\cdot 180$ degrees, back in once gives the desired sum.

$(2n-2)\cdot 180 - 2n\cdot 180 + (n-2)\cdot 180 = (n-4)\cdot 180$ degrees

So, for any n-pointed star with $n\ge 5$ as defined above, the sum of the angles in the star’s exterior arms is $(n-4)\cdot 180$ degrees.  When $n=5$, we get the star point angle sum of 180 degrees from earlier.

This formula is itself not so interesting, but that one can know this general result from manipulating nothing more than the angle properties of triangles, polygons, and linear pairs is nice and completely accessible to geometry students of many levels and ages.

Again, I’d love to hear (and will post/share) any other solution approaches.

## Midpoints, midpoints, everywhere!

I didn’t encounter the Quadrilateral Midpoint Theorem (QMT) until I had been teaching a few years.  Following is a minor variation on my approach to the QMT this year plus a fun way I leveraged the result to introduce similarity.

In case you haven’t heard of it, the surprisingly lovely QMT says that if you connect, in order, the midpoints of the four sides of a quadrilateral–any quadrilateral–even if the quadrilateral is concave or if its sides cross–the resulting figure will always be a parallelogram.

This is a cool and easy property to explore on any dynamic geometry software package (GeoGebra, TI-Nspire, Cabri, …).

SKETCH OF THE TRADITIONAL PROOF:  The proof is often established through triangle similarity:  Whenever you connect the midpoints of two sides of a triangle, the resulting segment will be parallel to and half the length of the triangle’s third side.  Draw either diagonal in the quadrilateral to create two triangles.  Connecting the midpoints of the other two sides of each triangle creates two congruent parallel sides, so the quadrilateral connecting all four midpoints must be a parallelogram.

NEW APPROACH THIS YEAR:  I hadn’t yet led my class into similarity, but having just introduced coordinate proofs, I tried an approach I’d never used before.  I assigned a coordinate proof of the QMT.  I knew the traditional approach existed, but I wanted them to practice their new technique.  From a lab in December, they already knew the result of the QMT, but they hadn’t proved it.

PART I:  Let quadrilateral ABCD be defined by the points , A=(a,b), B=(c,d), C=(e,f),  and D=(g,h).  There are several ways to prove that the midpoints of ABCD are the vertices of a parallelogram.  Provide one such coordinate proof.

All groups quickly established the midpoints of the four sides:  $AB_{mid}=\left( \frac{a+c}{2},\frac{b+d}{2} \right)$$BC_{mid}=\left( \frac{c+e}{2},\frac{d+f}{2} \right)$$CD_{mid}=\left( \frac{e+g}{2},\frac{f+h}{2} \right)$, and $DA_{mid}=\left( \frac{g+a}{2},\frac{h+b}{2} \right)$.  From there, my students took three approaches to the final proof, each relying on a different sufficiency condition for parallelograms.

The most common was to show that opposite sides were parallel.  $\displaystyle slope \left( AB_{mid} \text{ to } BC_{mid} \right) = \frac{\frac{a-e}{2}}{\frac{b-f}{2}}=\frac{a-e}{b-f}$ and $\displaystyle slope \left( CD_{mid} \text{ to } DA_{mid} \right) =\frac{a-e}{b-f}$, making those two midpoint segments parallel.  Likewise, $\displaystyle slope \left( BC_{mid} \text{ to } CD_{mid} \right) =$ $\displaystyle slope \left( DA_{mid} \text{ to } AB_{mid} \right) = \frac{c-g}{d-h}$, proving the other opposite side pair also was parallel.  With both pairs of opposite sides parallel, the midpoint quadrilateral was necessarily a parallelogram.

I had two groups leverage the fact that the diagonals of parallelograms were mutually bisecting.    $\displaystyle midpoint \left( AB_{mid} \text{ to } CD_{mid} \right) = \left( \frac{a+c+e+g}{4},\frac{b+d+f+h}{4}\right) =$ $= midpoint \left( BC_{mid} \text{ to } DA_{mid} \right)$.  QED.

One student even proved that opposite sides were congruent.

While it was not readily available for my students this year, I can imagine allowing CAS for these manipulations if I use this activity in the future.

EXTENDING THE QMT TO SIMILARITY:  For the next stage, I asked my students to explains what happens when the QMT is applied to degenerate quadrilaterals.

PART II:  You could think of triangles as being degenerate quadrilaterals when two quadrilateral vertices coincide to make one side of the quadrilateral have side length 0.  Apply this to generic quadrilateral ABCD from above where points A and D coincide to create triangle BCD.  Use this to explain how the segment connecting the midpoints of any two sides of a triangle is related to the third side of the triangle.

I encourage you to construct this using a dynamic geometry package, but here’s the result.

Heres a brief video showing the quadrilateral going degenerate.

Notice the parallelogram still exists and forms two midpoint segments on the triangle (degenerate quadrilateral).  By parallelogram properties, each of these segments is parallel and congruent to the opposite side of the parallelogram, making them parallel to and half the length of the opposite side of the triangle.

CONCLUSION:  I think it critical to teach in a way that draws connections between ideas and units. This exercise made a lovely transition from quadrilaterals through coordinate proofs to the triangle midpoint theorem.

## Two Squares, Two Triangles, and some Circles

Here’s another fun twist on another fun problem from the Five Triangles ‘blog.  A month ago, this was posted.

What I find cool about so many of the Five Triangles problems is that most permit multiple solutions.  I also like that several Five Triangles problems initially appear to not have enough information.  This one is no different until you consider the implications of the squares.

I’ve identified three unique ways to approach this problem.  I’d love to hear if any of you see any others.  Here are my solutions in the order I saw them.  The third is the shortest, but all offer unique insights.

Method 1: Law of Cosines

This solution goes far beyond the intended middle school focus of the problem, but it is what I saw first.  Sometimes, knowing more gives you additional insights.

Because DEF is a line and EF is a diagonal of a square, I know $m\angle CEF=45^{\circ}$, and therefore $m\angle CED=135^{\circ}$.  $\Delta CEF$ is a 45-45-90 triangle with hypotenuse 6, so its leg, CE has measure $\frac{6}{\sqrt{2}}=3\sqrt{2}$.  Knowing two sides and an angle in $\Delta DEC$ means I could apply the Law of Cosines.

$DC^2 = 4^2 + (3\sqrt{2})^2 - 2\cdot (3\sqrt{2}) \cdot \cos(135^{\circ})=58$

Because I’m looking for the area of ABCD,  and that is equivalent to $DC^2$, I don’t need to solve for the length of DC to know the area I seek is 58.

Method 2: Use Technology

I doubt many would want to solve using this approach, but if you don’t see (or know) trigonometry, you could build a solution from scratch if you are fluent with dynamic geometry software (GeoGebra, TI-Nspire, GSP).  My comfort with this made finding the solution via construction pretty straight-forward.

1. Construct segment EF with fixed length 6.
2. Build square CEGF with diagonal EF.  (This can be done several ways.  I was in a transformations mood, so I rotated EF $90^{\circ}$ to get the other endpoints.)
3. Draw line EF  and then circle with radius 4 through point E.
4. Mark point D as the intersection of circle and line EF outside CEGF .
5. Draw a segment through points and C.  (The square of the length of CD is the answer, but I decided to go one more step.)
6. Construct square ABCD with sides congruent to CD.  (Again, there are several ways to do this.  I left my construction marks visible in my construction below.)
7. Compute the area of ABCD.

Here is my final GeoGebra construction.

Method 3: The Pythagorean Theorem

Sometimes, changing a problem can make it much easier to solve.

As soon as I saw the problem, I forwarded it to some colleagues at my school.  Tatiana wrote back with a quick solution.  In the original image, draw diagonal, CG, of square CEGF. Because the diagonals of a square perpendicularly bisect each other, that creates right $\Delta DHC$ with legs 3 and 7.  That means the square of the hypotenuse of $\Delta DHC$ (and therefore the area of the square) can be found via the Pythagorean Theorem.

$DC^2 = 7^2+3^2 = 58$

Method 4: Coordinate Geometry

OK, I said three solutions, and perhaps this approach is completely redundant given the Pythagorean Theorem in the last approach, but you could also find a solution using coordinate geometry.

Because the diagonals of a square are perpendicular, you could construct ECFG with its center at the origin.  I placed point C at (0,3) and point E at (3,0).  That means point D is at (7,0), making the solution to the problem the square of the length of the segment from (0,3) to (7,0).  Obviously, that can be done with the Pythagorean Theorem, but in the image below, I computed number i in the upper left corner of this GeoGebra window as the square of the length of that segment.

Fun.