## 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.

Mike Lawler posted this on Twitter a couple weeks ago.

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

$(x-4)+24+(x-4)+(3x+2)=88$
$x=14$

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

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

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

First a diversion:

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

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

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

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

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

Trying to Salvaging the Problem:

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

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

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

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

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

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

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

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

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

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

Checking types:

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

CONCLUSION

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

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

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

## Problems in Time

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

Seeing that problem, I wondered

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

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

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

Anyone?

## 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.

## 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.

## Probability, Polynomials, and Sicherman Dice

Three years ago, I encountered a question on the TI-Nspire Google group asking if there was a way to use CAS to solve probability problems.  The ideas I pitched in my initial response and follow-up a year later (after first using it with students in a statistics class) have been thoroughly re-confirmed in my first year teaching AP Statistics.  I’ll quickly re-share them below before extending the concept with ideas I picked up a couple weeks ago from Steve Phelps’ session on Probability, Polynomials, and CAS at the 64th annual OCTM conference earlier this month in Cleveland, OH.

BINOMIALS:  FROM POLYNOMIALS TO SAMPLE SPACES

Once you understand them, binomial probability distributions aren’t that difficult, but the initial conjoining of combinatorics and probability makes this a perennially difficult topic for many students.  The standard formula for the probability of determining the chances of K successes in N attempts of a binomial situation where p is the probability of a single success in a single attempt is no less daunting:

$\displaystyle \left( \begin{matrix} N \\ K \end{matrix} \right) p^K (1-p)^{N-K} = \frac{N!}{K! (N-K)!} p^K (1-p)^{N-K}$

But that is almost exactly the same result one gets by raising binomials to whole number powers, so why not use a CAS to expand a polynomial and at least compute the $\displaystyle \left( \begin{matrix} N \\ K \end{matrix} \right)$ portion of the probability?  One added advantage of using a CAS is that you could use full event names instead of abbreviations, making it even easier to identify the meaning of each event.

The TI-Nspire output above shows the entire sample space resulting from flipping a coin 6 times.  Each term is an event.  Within each term, the exponent of each variable notes the number of times that variable occurs and the coefficient is the number of times that combination occurs.  The overall exponent in the expand command is the number of trials.  For example, the middle term– $20\cdot heads^3 \cdot tails^3$ –says that there are 20 ways you could get 3 heads and 3 tails when tossing a coin 6 times. The last term is just $tails^6$, and its implied coefficient is 1, meaning there is just one way to flip 6 tails in 6 tosses.

The expand command makes more sense than memorized algorithms and provides context to students until they gain a deeper understanding of what’s actually going on.

FROM POLYNOMIALS TO PROBABILITY

Still using the expand command, if each variable is preceded by its probability, the CAS result combines the entire sample space AND the corresponding probability distribution function.  For example, when rolling a fair die four times, the distribution for 1s vs. not 1s (2, 3, 4, 5, or 6) is given by

The highlighted term says there is a 38.58% chance that there will be exactly one 1 and any three other numbers (2, 3, 4, 5, or 6) in four rolls of a fair 6-sided die.  The probabilities of the other four events in the sample space are also shown.  Within the TI-Nspire (CAS or non-CAS), one could use a command to give all of these probabilities simultaneously (below), but then one has to remember whether the non-contextualized probabilities are for increasing or decreasing values of which binomial outcome.

Particularly early on in their explorations of binomial probabilities, students I’ve taught have shown a very clear preference for the polynomial approach, even when allowed to choose any approach that makes sense to them.

TAKING POLYNOMIALS FROM ONE DIE TO MANY

Given these earlier thoughts, I was naturally drawn to Steve Phelps “Probability, Polynomials, and CAS” session at the November 2014 OCTM annual meeting in Cleveland, OH.  Among the ideas he shared was using polynomials to create the distribution function for the sum of two fair 6-sided dice.  My immediate thought was to apply my earlier ideas.  As noted in my initial post, the expansion approach above is not limited to binomial situations.  My first reflexive CAS command in Steve’s session before he share anything was this.

By writing the outcomes in words, the CAS interprets them as variables.  I got the entire sample space, but didn’t learn gain anything beyond a long polynomial.  The first output– $five^2$ –with its implied coefficient says there is 1 way to get 2 fives.  The second term– $2\cdot five \cdot four$ –says there are 2 ways to get 1 five and 1 four.  Nice that the technology gives me all the terms so quickly, but it doesn’t help me get a distribution function of the sum.  I got the distributions of the specific outcomes, but the way I defined the variables didn’t permit sum of their actual numerical values.  Time to listen to the speaker.

He suggested using a common variable, X, for all faces with the value of each face expressed as an exponent.  That is, a standard 6-sided die would be represented by $X^1+X^2+ X^3+X^4+X^5+X^6$ where the six different exponents represent the numbers on the six faces of a typical 6-sided die.  Rolling two such dice simultaneously is handled as I did earlier with the binomial cases.

NOTE:  Exponents are handled in TWO different ways here.  1) Within a single polynomial, an exponent is an event value, and 2) Outside a polynomial, an exponent indicates the number of times that polynomial is applied within the specific event.  Coefficients have the same meaning as before.

Because the variables are now the same, when specific terms are multiplied, their exponents (face values) will be added–exactly what I wanted to happen.  That means the sum of the faces when you roll two dice is determined by the following.

Notice that the output is a single polynomial.  Therefore, the exponents are the values of individual cases.  For a couple examples, there are 3 ways to get a sum of 10 $\left( 3 \cdot x^{10} \right)$, 2 ways to get a sum of 3 $\left( 2 \cdot x^3 \right)$, etc.  The most commonly occurring outcome is the term with the largest coefficient.  For rolling two standard fair 6-sided dice, a sum of 7 is the most common outcome, occurring 6 times $\left( 6 \cdot x^7 \right)$.  That certainly simplifies the typical 6×6 tables used to compute the sums and probabilities resulting from rolling two dice.

While not the point of Steve’s talk, I immediately saw that technology had just opened the door to problems that had been computationally inaccessible in the past.  For example, what is the most common sum when rolling 5 dice and what is the probability of that sum?  On my CAS, I entered this.

In the middle of the expanded polynomial are two terms with the largest coefficients, $780 \cdot x^{18}$ and $780 \cdot x^{19}$, meaning a sums of 17 and 18 are the most common, equally likely outcomes when rolling 5 dice.  As there are $6^5=7776$ possible outcomes when rolling a die 5 times, the probability of each of these is $\frac{780}{7776} \approx 0.1003$, or about 10.03% chance each for a sum of 17 or 18.  This can be verified by inserting the probabilities as coefficients before each term before CAS expanding.

With thought, this shouldn’t be surprising as the expected mean value of rolling a 6-sided die many times is 3.5, and $5 \cdot 3.5 = 17.5$, so the integers on either side of 17.5 (17 & 18) should be the most common.  Technology confirms intuition.

ROLLING DIFFERENT DICE SIMULTANEOUSLY

What is the distribution of sums when rolling a 4-sided and a 6-sided die together?  No problem.  Just multiply two different polynomials, one representative of each die.

The output shows that sums of 5, 6, and 7 would be the most common, each occurring four times with probability $\frac{1}{6}$ and together accounting for half of all outcomes of rolling these two dice together.

A BEAUTIFUL EXTENSION–SICHERMAN DICE

My most unexpected gain from Steve’s talk happened when he asked if we could get the same distribution of sums as “normal” 6-sided dice, but from two different 6-sided dice.  The only restriction he gave was that all of the faces of the new dice had to have positive values.  This can be approached by realizing that the distribution of sums of the two normal dice can be found by multiplying two representative polynomials to get

$x^{12}+2x^{11}+3x^{10}+4x^9+5x^8+6x^7+5x^6+4x^5+3x^4+2x^3+x^2$.

Restating the question in the terms of this post, are there two other polynomials that could be multiplied to give the same product?  That is, does this polynomial factor into other polynomials that could multiply to the same product?  A CAS factor command gives

Any rearrangement of these eight (four distinct) sub-polynomials would create the same distribution as the sum of two dice, but what would the the separate sub-products mean in terms of the dice?  As a first example, what if the first two expressions were used for one die (line 1 below) and the two squared trinomials comprised a second die (line 2)?

Line 1 actually describes a 4-sided die with one face of 4, two faces with 3s, and one face of 2.  Line 2 describes a 9-sided die (whatever that is) with one face of 8, two faces of 6, three faces of 4, two faces of 2, and one face with a 0 ( $1=1 \cdot x^0$).  This means rolling a 4-sided and a 9-sided die as described would give exactly the same sum distribution.  Cool, but not what I wanted.  Now what?

Factorization gave four distinct sub-polynomials, each with multitude 2.  One die could contain 0, 1, or 2 of each of these with the remaining factors on the other die.  That means there are $3^4=81$ different possible dice combinations.  I could continue with a trail-and-error approach, but I wanted to be more efficient and elegant.

What follows is the result of thinking about the problem for a while.  Like most math solutions to interesting problems, ultimate solutions are typically much cleaner and more elegant than the thoughts that went into them.  Problem solving is a messy–but very rewarding–business.

SOLUTION

Here are my insights over time:

1) I realized that the $x^2$ term would raise the power (face values) of the desired dice, but would not change the coefficients (number of faces).  Because Steve asked for dice with all positive face values.  That meant each desired die had to have at least one x to prevent non-positive face values.

2) My first attempt didn’t create 6-sided dice.  The sums of the coefficients of the sub-polynomials determined the number of sides.  That sum could also be found by substituting $x=1$ into the sub-polynomial.  I want 6-sided dice, so the final coefficients must add to 6.  The coefficients of the factored polynomials of any die individually must add to 2, 3, or 6 and have a product of 6.  The coefficients of $(x+1)$ add to 2, $\left( x^2+x+1 \right)$ add to 3, and $\left( x^2-x+1 \right)$ add to 1.  The only way to get a polynomial coefficient sum of 6 (and thereby create 6-sided dice) is for each die to have one $(x+1)$ factor and one $\left( x^2+x+1 \right)$ factor.

3) That leaves the two $\left( x^2-x+1 \right)$ factors.  They could split between the two dice or both could be on one die, leaving none on the other.  We’ve already determined that each die already had to have one each of the x, $(x+1)$, and $\left( x^2+x+1 \right)$ factors.  To also split the $\left( x^2-x+1 \right)$ factors would result in the original dice:  Two normal 6-sided dice.  If I want different dice, I have to load both of these factors on one die.

That means there is ONLY ONE POSSIBLE alternative for two 6-sided dice that have the same sum distribution as two normal 6-sided dice.

One die would have single faces of 8, 6, 5, 4, 3, and 1.  The other die would have one 4, two 3s, two 2s, and one 1.  And this is exactly the result of the famous(?) Sicherman Dice.

If a 0 face value was allowed, shift one factor of x from one polynomial to the other.  This can be done two ways.

The first possibility has dice with faces {9, 7, 6, 5, 4, 2} and {3, 2, 2, 1, 1, 0}, and the second has faces {7, 5, 4, 3, 2, 0} and {5, 4, 4, 3, 3, 2}, giving the only other two non-negative solutions to the Sicherman Dice.

Both of these are nothing more than adding one to all faces of one die and subtracting one from from all faces of the other.  While not necessary to use polynomials to compute these, they are equivalent to multiplying the polynomial of one die by x and the other by $\frac{1}{x}$ as many times as desired. That means there are an infinite number of 6-sided dice with the same sum distribution as normal 6-sided dice if you allow the sides to have negative faces.  One of these is

corresponding to a pair of Sicherman Dice with faces {6, 4, 3, 2, 1, -1} and {1,5,5,4,4,3}.

CONCLUSION:

There are other very interesting properties of Sicherman Dice, but this is already a very long post.  In the end, there are tremendous connections between probability and polynomials that are accessible to students at the secondary level and beyond.  And CAS keeps the focus on student learning and away from the manipulations that aren’t even the point in these explorations.

Enjoy.

## CAS Conference

Computer algebra systems (CAS) have the potential to revolutionize mathematics education at the middle and secondary level. Experience how CAS can be integrated into Pre-algebra, Algebra 1 & 2, Precalculus, Calculus, Statistics, and Geometry classes.

Consider attending (or maybe even proposing a session) for the 9th USACAS conference, July 18-19, 2015 at Hawken School in Cleveland, OH.

• Discover how secondary and middle school teachers are using CAS in their own classrooms.
• Get classroom tested ideas developed for CAS-enhanced classrooms.
• Learn what other countries are doing with CAS.
• Interact with prominent US and International CAS pioneers.

Information on registration, speakers, presentations, Saturday outing, and hotel can be found at http://usacas.org .

Consider proposing a session at http://bit.ly/USACAS9_speakers .