# 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!

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## 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!

## Value Process over Answers

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.