# Tag Archives: geometry

## 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?

## Painting and Probability

Here’s a cool probability problem the start of which is accessible to middle and high school students with careful reasoning.  It was posted by Alexander Bogomolny on Cut the Knot a week ago. I offer what I think is a cool extension of the problem following the initial credits.

The next day Mike Lawler tweeted another phenomenal video solution worked out with his boys during Family Math.

Mike’s videos show his boys thinking through simpler discrete cases of the problem and their eventual case-by-case solution to the nxnxn question.  The last video on the page gives a lovely alternative solution completely bypassing the case-by-case analyses.  This is also “Solution 1” on Alexander’s page.

EXTENDING THE PROBLEM:

When I first encountered the problem, I wanted to think about it before reading any solutions.  As with Mike’s boys, I was surprised early on when the probability for a 2x2x2 cube was $\displaystyle \frac{1}{2}$ and the probability for a 3x3x3 cube was $\displaystyle \frac{1}{3}$.  That was far too pretty to be a coincidence.  My solution exactly mirrored the nxnxn case-by-case analysis in Mike’s videos:  the probability of rolling a painted red face up from a randomly selected smaller cube is $\displaystyle \frac{1}{n}$.

Surely something this simple and clean could be generalized.  Since a cube can be considered a “3-dimensional square”, I re-phrased Alexander’s question into two dimensions.  The trickier part was thinking what it would mean to “roll” a 2-dimensional shape.

The outside of an nxn square is painted red and is chopped into $n^2$ unit squares.  The latter are thoroughly mixed up and put into a bag.  One small square is withdrawn at random from the bag and spun on a flat surface.  What is the probability that the spinner stops with a red side facing you?

Shown below is a 4×4 square, but in all sizes of 2-dimensional squares, there are three possible types in the bag:  those with 2, 1, or 0 sides painted.

I solved my first variation case by case.  In any nxn square,

• There are 4 corner squares with 2 sides painted.  The probability of picking one of those squares and then spinning a red side is $\displaystyle \frac{4}{n^2} \cdot \frac{2}{4} = \frac{2}{n^2}$.
• There are $4(n-2)$ edge squares not in a corner with 1 side painted.  The probability of picking one of those squares and then spinning a red side is $\displaystyle \frac{4(n-2)}{n^2} \cdot \frac{1}{4} = \frac{n-2}{n^2}$.
• All other squares have 0 sides painted, so the probability of picking one of those squares and then spinning a red side is 0.
• Adding the probabilities for the separate cases gives the total probability:  $\displaystyle \frac{2}{n^2}+\frac{n-2}{n^2}=\frac{1}{n}$

After reading Mike’s and Alexander’s posts, I saw a much easier approach.

• Paint all 4 edges of an nxn square, and divide each painted edge into n painted unit segments.  This creates $4 \cdot n$ total painted small segments.
• Decompose the nxn original square into $n^2$ unit squares.  Each unit square has 4 edges giving $4 \cdot n^2$ total edges.
• Because every edge of every unit square is equally likely to be spun, the total probability of randomly selecting a smaller square and spinning a red side is $\displaystyle \frac{4n}{4n^2}=\frac{1}{n}$.

The dimensions of the “square” don’t seem to matter!

WARNING:

Oliver Wendell Holmes noted, “A mind that is stretched by a new experience can never go back to its old dimensions.”  The math after this point has the potential to stretch…

EXTENDING THE PROBLEM MORE:

I now wondered whether this was a general property.

In the 2-dimensional square, 1-dimensional edges were painted and the square was spun to find the probability of a red edge facing.  With the originally posed cube, 2-dimensional faces were painted and the cube was tossed to find the probability of an upward-facing red face.  These cases suggest that when a cube of some dimension with edge length n is painted, is decomposed into unit cubes of the original dimension, and is spun/tossed to show a cube of one smaller dimension, then the probability of getting a painted smaller-dimensional cube of is always $\displaystyle \frac{1}{n}$, independent of the dimensions occupied by the cube.

Going beyond the experiences of typical middle or high school students, I calculated this probability for a 4-dimensional hypercube (a tesseract).

• The exterior of a tesseract is 8 cubes.  Ignore the philosophical difficulty of what it means to “paint” (perhaps fill?) an entire cube.  After all, we’re already beyond the experience of our 3-dimensions.
• Paint/fill all 8 cubes on the surface of the tesseract, and divide each painted cube into $n^3$ painted unit cubes.  This creates $8 \cdot n^3$ total painted unit cubes.
• Decompose the original tesseract into $n^4$ unit tesseracts.  Each unit tesseract has 8 cubes giving $8 \cdot n^4$ total unit cubes.
• Because every unit cube on every unit tesseract is equally likely to be “rolled”, the total probability of randomly selecting a smaller tesseract and rolling a red cube is $\displaystyle \frac{8n^3}{8n^4}=\frac{1}{n}$.

The probability is independent of dimension!

More formally,

The exterior of a d-dimensional hypercube with edge length n is painted red and is chopped into $n^d$ unit d-dimensional hypercubes.  The latter are put into a bag of sufficient dimension to hold them and thoroughly mixed up.  A unit d-dimensional hypercube is withdrawn at random from the bag and tossed.  The probability that the unit d-dimensional hypercube lands with a red (d-1)-dimensional hypercube showing is $\displaystyle \frac{1}{n}$ .

PROOF:

• The exterior of a d-dimensional hypercube is comprised of 2d (d-1)-dimensional hypercubes of dimension (d-1).  Paint/fill all 2d surface hypercubes and divide each painted (d-1)-dimensional hypercube into $n^{d-1}$ painted unit hypercubes.  This creates $2d \cdot n^{d-1}$ total painted unit hypercubes.
• Decompose the original tesseract into $n^d$ unit d-dimensional hypercubes.  Each unit d-dimensional hypercube has 2d surface (d-1)-dimensional hypercubes giving $2d \cdot n^d$ total surface unit d-dimensional hypercubes.
• Because every unit (d-1)-dimensional hypercube on the surface of every unit d-dimensional hypercube is equally likely to be “rolled”, the total probability of randomly selecting a unit d-dimensional hypercube and rolling a (d-1)-dimensional red-painted hypercube is $\displaystyle \frac{2d \cdot n^{d-1}}{2d \cdot n^d}=\frac{1}{n}$.

## 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!

## From a Square to Ratios to a System of Equations

Here’s another ratio problem from @Five_Triangles, this time involving triangle areas bounded by a square.

Don’t read further until you’ve tried this for yourself. It’s a fun problem that, at least from my experience, doesn’t end up where or how I thought it would.

INITIAL THOUGHTS

I see two big challenges here.

First, the missing location of point P is especially interesting, but is also likely to be quite vexing for many students. This led me to the first twist I found in the problem: the introduction of multiple variables and a coordinate system. Without some problem-solving experience, I don’t see that as an intuitive step for most middle school students. Please don’t interpret this as a knock on this problem, I’m simply agreeing with @Five_Triangle’s assessment that this problem is likely to be challenging for middle school students.

The second challenge I found emerged from the introduction the coordinate system: an underlying 2×2 system of equations. There are multiple ways to tackle a solution to a linear system, but this strikes me as yet another high hurdle for younger students.

Finally, I’m a bit surprised by my current brain block on multiple approaches for this problem. I suspect I’m blinded here by my algebraic bias in problem solving; surely there are approaches that don’t require this. I’d love to hear any other possibilities.

POINT P VARIES

Because I was given properties of point P and not its location, the easiest approach I could see was to position the square on the xy-plane with point B at the origin, $\overline{AB}$ along the y-axis, and $\overline{BC}$ along the x-axis. That gave my point P coordinates (x,y) for some unknown values of x & y.

The helpful part of this orientation is that the x & y coordinates of P are automatically the altitudes of $\Delta ABP$ and $\Delta BCP$, respectively. The altitudes of the other two triangles are determined through subtraction.

AREA RATIOS BECOME A LINEAR SYSTEM

From here, I used the given ratios to establish one equation in terms of x & y.

$\displaystyle \frac{\Delta ABP}{\Delta DAP} = \frac{\frac{1}{2}*12*x}{\frac{1}{2}*12*(12-y)} = \frac{3}{4}$

Of course, since all four triangles have the same base lengths, the given area ratios are arithmetically equivalent to corresponding height ratios.  I used that to write a second equation.

$\displaystyle \frac{\Delta BCP}{\Delta CDP} = \frac{y}{12-x} = \frac{1}{3}$

Simplifying terms and clearing denominators leads to $4x=36-3y$ and $3y=12-x$, respectively.

A VERY INTERESTING insight at this point is that there is an infinite number of locations within the square at which each ratio is true.  Specifically, the $\Delta ABP : \Delta DAP = 3:4$ ratio is true everywhere along the line 4x=36-3y.  This problem constrains us to only the points within the square with vertices (0,0), (12,0), (12,12), and (0,12), but setting that aside, anywhere along the line 4x=36-3y would satisfy the first constraint.  The same is true for the second line and constraint.

I think it would be very interesting for students to construct this on dynamic geometry software (e.g., GeoGebra or the TI-Nspire) and see the ratio remain constant everywhere along either line even though the triangle areas vary throughout.

Together, these lines form a 2×2 system of linear equations with the solution to both ratios being the intersection point of the two lines.  There are lots of ways to do this; I wonder how a typical 6th grader would tackle them. Assuming they have the algebraic expertise, I’d have work them by hand and confirm with a CAS.

The question asks for the area of $\Delta ABP = \frac{1}{2}*12*x = 6*8 = 48$.

PROBLEM VARIATIONS

Just two extensions this time.  Other suggestions are welcome.

1. What’s the ratio of the area of $\Delta BCP : \Delta DAP$ at the point P that satisfies both ratios??
It’s not 1:4 as an errant student might think from an errant application of the transitive property to the given ratios.  Can you show that it’s actually 1:8?
2. If a random point is chosen within the square, is that point more likely to satisfy the area ratio of $\Delta ABP : \Delta DAP$ or the ratio of $\Delta BCP : \Delta CDP$?
The first ratio is satisfied by the line 4x=36-3y which intersects the square on the segment between (9,0) and (0,12).  At the latter point, both triangles are degenerate with area 0.  The second ratio’s line intersects the square between (12,0) and (0,4).  As the first segment is longer (how would a middle schooler prove that?), it is more likely that a randomly chosen point would satisfy the $\Delta ABP : \Delta DAP$  ratio.  This would be a challenging probability problem, methinks.

FURTHER EXTENSIONS?

What other possibilities do you see either for a solution to the original problem or an extension?

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.

## 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.

## Squares and Octagons, A compilation

My last post detailed my much-too-long trigonometric proof of why the octagon formed by connecting the midpoints and vertices of the edges of a square into an 8-pointed star is always 1/6 of the area of the original square.

My proof used trigonometry, and responses to the post on Twitter  and on my ‘blog showed many cool variations.  Dave Radcliffe thought it would be cool to have a compilation of all of the different approaches.  I offer that here in the order they were shared with me.

Method 1:  My use of trigonometry in a square.  See my original post.

Method 2:  Using medians in a rectangle from Tatiana Yudovina, a colleague at Hawken School.

Below, the Area(axb rectangle) = ab = 16 blue triangles, and
Area(octagon) = 4 blue triangles – 2 red deltas..

Now look at the two green, similar triangles.  They are similar with ratio 1/2, making

Area(red delta) = $\displaystyle \frac{b}{4} \cdot \frac{a}{6} = \frac{ab}{24}$, and

Area(blue triangle) = $\displaystyle \frac{1}{16} ab$

So, Area(octagon) = $\displaystyle 2 \frac{ab}{24}-4\frac {ab}{16}=\frac{1}{6}ab$.

QED

Method 3:  Using differences in triangle areas in a square (but easily extended to rectangles)from @Five_Triangles (‘blog here).

Full solution here.

Method 4:  Very clever shorter solution using triangle area similarity in a square also from @Five_Triangles (‘blog here).

Full second solution here.

Method 5:  Great option Using dilated kitesfrom Dave Radcliffe posting as @daveinstpaul.

Full pdf and proof here.

Method 6:  Use fact that triangle medians trisect each other from Mike Lawler posting as @mikeandallie.

Tweet of solution here.

Method 7:  Use a coordinate proof on a specific square from Steve Ingrassia, a colleague at Hawken School.  Not a quick proof like some of the geometric solutions, but it’s definitely different than the others.

If students know the formula for finding the area of any polygon using its coordinates, then they can prove this result very simply with nothing more than simple algebra 1 techniques.   No trig is required.

The area of polygon with vertices (in either clockwise or counterclockwise order, starting at any vertex) of $(x_1, y_1)$, $(x_2, y_2)$, …, $(x_n, y_n)$ is

$\displaystyle Area = \left| \frac{(x_1y_2-x_2y_1)+(x_2y_3-x_3y_2)+...+(x_{n-1}y_n-x_ny_{n-1})}{2} \right|$

Use a 2×2 square situated with vertices at (0,0), (0,2), (2,2), and (2,0).  Construct segments connecting each vertex with the midpoints of the sides of the square, and find the equations of the associated lines.

• L1 (connecting (0,0) and (2,1):    y = x/2
• L2 (connecting (0,0) and (1,2):   y=2x
• L3 (connecting (0,1) and (2,0):  y= -x/2 + 1
• L4 (connecting (0,1) and (2,2):  y= x/2 + 1
• L5 (connecting (0,2) and (1,0):  y = -2x + 2
• L6 (connecting (0,2) and (2,1):  y= -x/2 + 2
• L7 (connecting (1,2) and (2,0):  y = -2x + 4
• L8 (connecting (2,2) and (1,0):  y = 2x – 2

The 8 vertices of the octagon come at pairwise intersections of some of these lines, which can be found with simple substitution:

• Vertex 1 is at the intersection of L1 and L3:   (1, 1/2)
• Vertex 2 is at the intersection of L3 and L5:  (2/3, 2/3)
• Vertex 3 is at the intersection of L2 and L5:  (1/2, 1)
• Vertex 4 is at the intersection of L2 and L4:  (2/3, 4/3)
• Vertex 5 is at the intersection of L4 and L6:  (1, 3/2)
• Vertex 6 is at the intersection of L6 and L7:  (4/3, 4/3)
• Vertex 7 is at the intersection of L7 and L8:  (3/2, 1)
• Vertex 8 is at the intersection of L1 and L8:  (4/3, 2/3)

Using the coordinates of these 8 vertices in the formula for the area of the octagon, gives

$\displaystyle \frac{ \left| 1/3 +1/3+0+(-1/3)+(-2/3)+(-1/3)+0 \right|}{2} = \frac{2}{3}$

Since the area of the original square was 4, the area of the octagon is exactly 1/6th of the area of the square.