# Tag Archives: CAS

## 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!

## Quadratics + Tangent = ???

Here’s a very pretty problem I encountered on Twitter from Mike Lawler 1.5 months ago.

I’m late to the game replying to Mike’s post, but this problem is the most lovely combination of features of quadratic and trigonometric functions I’ve ever encountered in a single question, so I couldn’t resist.  This one is well worth the time for you to explore on your own before reading further.

My full thoughts and explorations follow.  I have landed on some nice insights and what I believe is an elegant solution (in Insight #5 below).  Leading up to that, I share the chronology of my investigations and thought processes.  As always, all feedback is welcome.

WARNING:  HINTS AND SOLUTIONS FOLLOW

Investigation  #1:

My first thoughts were influenced by spoilers posted as quick replies to Mike’s post.  The coefficients of the underlying quadratic, $A^2-9A+1=0$, say that the solutions to the quadratic sum to 9 and multiply to 1.  The product of 1 turned out to be critical, but I didn’t see just how central it was until I had explored further.  I didn’t immediately recognize the 9 as a red herring.

Basic trig experience (and a response spoiler) suggested the angle values for the tangent embedded in the quadratic weren’t common angles, so I jumped to Desmos first.  I knew the graph of the overall given equation would be ugly, so I initially solved the equation by graphing the quadratic, computing arctangents, and adding.

Insight #1:  A Curious Sum

The sum of the arctangent solutions was about 1.57…, a decimal form suspiciously suggesting a sum of $\pi/2$.  I wasn’t yet worried about all solutions in the required $[0,2\pi ]$ interval, but for whatever strange angles were determined by this equation, their sum was strangely pretty and succinct.  If this worked for a seemingly random sum of 9 for the tangent solutions, perhaps it would work for others.

Unfortunately, Desmos is not a CAS, so I turned to GeoGebra for more power.

Investigation #2:

In GeoGebra, I created a sketch to vary the linear coefficient of the quadratic and to dynamically calculate angle sums.  My procedure is noted at the end of this post.  You can play with my GeoGebra sketch here.

The x-coordinate of point G is the sum of the angles of the first two solutions of the tangent solutions.

Likewise, the x-coordinate of point H is the sum of the angles of all four angles of the tangent solutions required by the problem.

Insight #2:  The Angles are Irrelevant

By dragging the slider for the linear coefficient, the parabola’s intercepts changed, but as predicted in Insights #1, the angle sums (x-coordinates of points G & H) remained invariant under all Real values of points A & B.  The angle sum of points C & D seemed to be $\pi/2$ (point G), confirming Insight #1, while the angle sum of all four solutions in $[0,2\pi]$ remained $3\pi$ (point H), answering Mike’s question.

The invariance of the angle sums even while varying the underlying individual angles seemed compelling evidence that that this problem was richer than the posed version.

Insight #3:  But the Angles are bounded

The parabola didn’t always have Real solutions.  In fact, Real x-intercepts (and thereby Real angle solutions) happened iff the discriminant was non-negative:  $B^2-4AC=b^2-4*1*1 \ge 0$.  In other words, the sum of the first two positive angles solutions for $y=(tan(x))^2-b*tan(x)+1=0$ is $\pi/2$ iff $\left| b \right| \ge 2$, and the sum of the first four solutions is $3\pi$ under the same condition.  These results extend to the equalities at the endpoints iff the double solutions there are counted twice in the sums.  I am not convinced these facts extend to the complex angles resulting when $-2.

I knew the answer to the now extended problem, but I didn’t know why.  Even so, these solutions and the problem’s request for a SUM of angles provided the insights needed to understand WHY this worked; it was time to fully consider the product of the angles.

Insight #4:  Finally a proof

It was now clear that for $\left| b \right| \ge 2$ there were two Quadrant I angles whose tangents were equal to the x-intercepts of the quadratic.  If $x_1$ and $x_2$ are the quadratic zeros, then I needed to find the sum A+B where $tan(A)=x_1$ and $tan(B)=x_2$.

From the coefficients of the given quadratic, I knew $x_1+x_2=tan(A)+tan(B)=9$ and $x_1*x_2=tan(A)*tan(B)=1$.

Employing the tangent sum identity gave

$\displaystyle tan(A+B) = \frac{tan(A)+tan(B)}{1-tan(A)tan(B)} = \frac{9}{1-1}$

and this fraction is undefined, independent of the value of $x_1+x_2=tan(A)+tan(B)$ as suggested by Insight #2.  Because tan(A+B) is first undefined at $\pi/2$, the first solutions are $\displaystyle A+B=\frac{\pi}{2}$.

Insight #5:  Cofunctions reveal essence

The tangent identity was a cute touch, but I wanted something deeper, not just an interpretation of an algebraic result.  (I know this is uncharacteristic for my typically algebraic tendencies.)  The final key was in the implications of $tan(A)*tan(B)=1$.

This product meant the tangent solutions were reciprocals, and the reciprocal of tangent is cotangent, giving

$\displaystyle tan(A) = \frac{1}{tan(B)} = cot(B)$.

But cotangent is also the co-function–or complement function–of tangent which gave me

$tan(A) = cot(B) = tan \left( \frac{\pi}{2} - B \right)$.

Because tangent is monotonic over every cycle, the equivalence of the tangents implied the equivalence of their angles, so $A = \frac{\pi}{2} - B$, or $A+B = \frac{\pi}{2}$.  Using the Insights above, this means the sum of the solutions to the generalization of Mike’s given equation,

$(tan(x))^2+b*tan(x)+1=0$ for x in $[0,2\pi ]$ and any $\left| b \right| \ge 2$,

is always $3\pi$ with the fundamental reason for this in the definition of trigonometric functions and their co-functions.  QED

Insight #6:  Generalizing the Domain

The posed problem can be generalized further by recognizing the period of tangent: $\pi$.  That means the distance between successive corresponding solutions to the internal tangents of this problem is always $\pi$ each, as shown in the GeoGebra construction above.

Insights 4 & 5 proved the sum of the angles at points C & D was $\pi/2$.  Employing the periodicity of tangent,  the x-coordinate of $E = C+\pi$ and $F = D+\pi$, so the sum of the angles at points E & F is $\frac{\pi}{2} + 2 \pi$.

Extending the problem domain to $[0,3\pi ]$ would add $\frac{\pi}{2} + 4\pi$ more to the solution, and a domain of $[0,4\pi ]$ would add an additional $\frac{\pi}{2} + 6\pi$.  Pushing the domain to $[0,k\pi ]$ would give total sum

$\displaystyle \left( \frac{\pi}{2} \right) + \left( \frac{\pi}{2} +2\pi \right) + \left( \frac{\pi}{2} +4\pi \right) + \left( \frac{\pi}{2} +6\pi \right) + ... + \left( \frac{\pi}{2} +2(k-1)\pi \right)$

Combining terms gives a general formula for the sum of solutions for a problem domain of $[0,k\pi ]$

$\displaystyle k * \frac{\pi}{2} + \left( 2+4+6+...+2(k-1) \right) * \pi =$

$\displaystyle = k * \frac{\pi}{2} + (k)(k-1) \pi =$

$\displaystyle = \frac{\pi}{2} * k * (2k-1)$

For the first solutions in Quadrant I, $[0,\pi]$ means k=1, and the sum is $\displaystyle \frac{\pi}{2}*1*(2*1-1) = \frac{\pi}{2}$.

For the solutions in the problem Mike originally posed, $[0,2\pi]$ means k=2, and the sum is $\displaystyle \frac{\pi}{2}*2*(2*2-1) = 3\pi$.

I think that’s enough for one problem.

APPENDIX

My GeoGebra procedure for Investigation #2:

• Graph the quadratic with a slider for the linear coefficient, $y=x^2-b*x+1$.
• Label the x-intercepts A & B.
• The x-values of A & B are the outputs for tangent, so I reflected these over y=x to the y-axis to construct A’ and B’.
• Graph y=tan(x) and construct perpendiculars at A’ and B’ to determine the points of intersection with tangent–Points C, D, E, and F in the image below
• The x-intercepts of C, D, E, and F are the angles required by the problem.
• Since these can be points or vectors in Geogebra, I created point G by G=C+D.  The x-intercept of G is the angle sum of C & D.
• Likewise, the x-intercept of point H=C+D+E+F is the required angle sum.

I’m returning to my ‘blog after a prolonged absence.  My next several posts will explore ideas I shared and learned at the USACAS-10 conference hosted at Hawken School last weekend.

Finding equations for quadratic functions has long been a staple of secondary mathematics.  Historically, students are given information about some points on the graph of the quadratic, and efficient students typically figure out which form of the equation to use.  This post from my Curious Quadratics a la CAS presentation explores a significant mindset change that evolves once computer algebra enters the learning environment.

HISTORIC BACKGROUND:

Students spend lots of time (too much?) learning how to manipulate algebraic expressions between polynomial forms.  Whether distributing, factoring, or completing the square, generations of students have spent weeks changing quadratic expressions between three common algebraic forms

Standard:  $y=a*x^2+b*x+c$

Factored:  $y=a*(x-x_1)(x-x_2)$

Vertex:  $y=a*(x-h)^2+k$

many times without ever really knowing why.  I finally grasped deeply the reason for this about 15 years ago in a presentation by Bernhard Kutzler of Austria.  Poorly paraphrasing Bernhard’s point, he said in more elegant phrasing,

We change algebraic forms of functions because different forms reveal different properties of the function and because no single form reveals everything about a function.

While any of what follows could be eventually derived from any of the three quadratic forms, in general the Standard Form explicitly gives the y-intercept, Factored Form states x-intercepts, and Vertex Form “reveals” the vertex (duh).  When working without electronic technology, students can gain efficiency by choosing to work with a quadratic form that blends well with given information.  To demonstrate this, here’s an example of the differences between non-tech and CAS approaches.

COMPARING APPROACHES:

For an example, determine all intercepts and the vertex of the parabola that passes through $(10, 210)$$(5, 40)$, and $(-2, -30)$.

NON-TECH:  Not knowing anything about the points, use Standard form, plug in all three points, and solve the resulting system.

$y=a*x^2+b*x+c$
$210 = 100a+10b+c$
$40 = 25a+5b+c$
$-30 = 4a-2b+c$

Use any approach you want to solve this 3×3 system to get $a=2$, $b=4$, and $c=-30$.

That immediately gives the y-intercept at -30.  Factoring to $y=2(x+5)(x-3)$ or using the Quadratic Formula reveals the x-intercepts at -5 and 3.  Completing the square or leveraging symmetry between the known x-intercepts gives the vertex at $(-1,-32)$.  Some less-confident students find all of the hinted-at manipulations in this paragraph burdensome or even daunting.

CAS APPROACH:  By declaring the form you want/need, you can directly get any information you require.  In the next three lines on my Nspire CAS, notice that the only difference in my commands is the form of the equation I want in the first part of the command.  Also notice my use of lists to simplify substitution of the given points.

The last line’s output gave two solutions only because I didn’t specify which of x1 and x2 was the larger x-intercept, so my Nspire gave me both.

The -30 y-intercept appears in the first output, the vertex in the second, and the x-intercepts in the third.  Any information is equally simple to obtain.

CONCLUSION:

In the end, it’s all about knowing what you want to find and how to ask questions of the tools you have available.  Understanding the algebra behind the solutions is important, but endless repetition of these tasks is not helpful, even though it may be easy to test.

Instead, focus on using what you know, explore for patterns, and ask good questions.  …And teach with a CAS!

## Binomial Expansion Variation

Several years ago, I posed on this ‘blog a problem I learned from Natalie Jackucyn:

For some integers A, B, and n, one term of the expansion of $(Ax+By)^n$ is $27869184x^5y^3$.  What are the values of A, B, and n?

In this post, I reflect for a moment on what I’ve learned from the problem and outline a solution approach before sharing a clever alternative solution one of my students this year leveraged through her CAS-enabled investigation.

WHAT I LEARNED BEFORE THIS YEAR

Mostly, I’ve loved this problem for its “reversal” of traditional binomial expansion problems that typically give A, B, and n values and ask for either complete expansions or specific terms of the polynomial.  Both of these traditional tasks are easily managed via today’s technology.  In Natalie’s variation, neither the answer nor how you would proceed are immediately obvious.

The first great part of the problem is that it doesn’t seem to give enough information.  Second, it requires solvers to understand deeply the process of polynomial expansion.  Third, unlike traditional formulations, Natalie’s version doesn’t allow students to avoid deep thinking by using technology.

In the comments to my original post, Christopher Olah and a former student, Bryan Spellman, solved the problem via factoring and an Excel document, respectively.  Given my algebraic tendencies, I hadn’t considered Bryan’s Excel “search” approach, but one could relatively easily program Excel to provide an exhaustive search.  I now think of Bryan’s approach as a coding approach to a reasonably efficient search of the sample space of possible solutions.  Most of my students’ solutions over the years essentially approach the problem the same way, but less efficiently, by using one-case-at-a-time expansions via CAS commands until they stumble upon good values for A, B, and n.  Understandably, students taking this approach typically become the most frustrated.

Christopher’s approach paralleled my own.  The x and y exponents from the expanded term show that n=5+3=8.  Expanding a generic $(Ax+By)^8$ then gives a bit more information.  From my TI-Nspire CAS,

so there are 56 ways an $x^5y^3$ term appears in this expansion before combining like terms (explained here, if needed).  Dividing the original coefficient by 56 gives $a^5b^3=497,664$, the coefficient of $x^5y^3$.

The values of a and b are integers, so factoring 497,664 shows these coefficients are both co-multiples of 2 and 3, but which ones?  In essence, this defines a system of equations.  The 3 has an exponent of 5, so it can easily be attributed to a, but the 11 is not a multiple of either 5 or 3, so it must be a combination.  Quick experimentation with the exponents leads to $11=5*1+3*2$, so $2^1$ goes to a and $2^2$ goes to b.  This results in $a=3*2=6$ and $b=2^2=4$.

WHAT A STUDENT TAUGHT ME THIS YEAR

After my student, NB, arrived at $a^5b^3=497,664$ , she focused on roots–not factors–for her solution.  The exponents of a and b suggested using either a cubed or a fifth root.

The fifth root would extract only the value of a if b had only singleton factors–essentially isolating the a and b values–while the cubed root would extract a combination of a and b factors, leaving only excess a factors inside the radical.  Her investigation was simplified by the exact answers from her Nspire CAS software.

From the fifth root output, the irrational term had exponent 1/5, not the expected 3/5, so b must have had at least one prime factor with non-singular multiplicity.  But the cubed root played out perfectly.   The exponent–2/3–matched expectation, giving a=6, and the coefficient, 24, was the product of a and b, making b=4.  Clever.

EXTENSIONS & CONCLUSION

Admittedly, NB’s solution would have been complicated if the parameter was composed of something other than singleton prime factors, but it did present a fresh, alternative approach to what was becoming a comfortable problem for me.  I’m curious about exploring other arrangements of the parameters of $(Ax+By)^n$ to see how NB’s root-based reasoning could be extended and how it would compare to the factor solutions I used before.  I wonder which would be “easier” … whatever “easier” means.

As a ‘blog topic for another day, I’ve learned much by sharing this particular problem with several teachers over the years.  In particular, the initial “not enough information” feel of the problem statement actually indicates the presence of some variations that lead to multiple solutions.  If you think about it, NB’s root variation of the solution suggests some direct paths to such possible formulations.  As intriguing as the possibilities here are, I’ve never assigned such a variation of the problem to my students.

As I finish this post, I’m questioning why I haven’t yet taken advantage of these possibilities.  That will change. Until then, perhaps you can find some interesting or alternative approaches to the underlying systems of equations in this problem.  Can you create a variation that has multiple solutions?  Under what conditions would such a variation exist?  How many distinct solutions could a problem like this have?

## Roots of Complex Numbers without DeMoivre

Finding roots of complex numbers can be … complex.

This post describes a way to compute roots of any number–real or complex–via systems of equations without any conversions to polar form or use of DeMoivre’s Theorem.  Following a “traditional approach,” one non-technology example is followed by a CAS simplification of the process.

Most sources describe the following procedure to compute the roots of complex numbers (obviously including the real number subset).

• Write the complex number whose root is sought in generic polar form.  If necessary, convert from Cartesian form.
• Invoke DeMoivre’s Theorem to get the polar form of all of the roots.
• If necessary, convert the numbers from polar form back to Cartesian.

As a very quick example,

Compute all square roots of -16.

Rephrased, this asks for all complex numbers, z, that satisfy  $z^2=-16$.  The Fundamental Theorem of Algebra guarantees two solutions to this quadratic equation.

The complex Cartesian number, $-16+0i$, converts to polar form, $16cis( \pi )$, where $cis(\theta ) = cos( \theta ) +i*sin( \theta )$.  Unlike Cartesian form, polar representations of numbers are not unique, so any full rotation from the initial representation would be coincident, and therefore equivalent if converted to Cartesian.  For any integer n, this means

$-16 = 16cis( \pi ) = 16 cis \left( \pi + 2 \pi n \right)$

Invoking DeMoivre’s Theorem,

$\sqrt{-16} = (-16)^{1/2} = \left( 16 cis \left( \pi + 2 \pi n \right) \right) ^{1/2}$
$= 16^{1/2} * cis \left( \frac{1}{2} \left( \pi + 2 \pi n \right) \right)$
$= 4 * cis \left( \frac{ \pi }{2} + \pi * n \right)$

For $n= \{ 0, 1 \}$, this gives polar solutions, $4cis \left( \frac{ \pi }{2} \right)$ and $4cis \left( \frac{ 3 \pi }{2} \right)$ .  Each can be converted back to Cartesian form, giving the two square roots of -16:  $4i$ and $-4i$.  Squaring either gives -16, confirming the result.

I’ve always found the rotational symmetry of the complex roots of any number beautiful, particularly for higher order roots.  This symmetry is perfectly captured by DeMoivre’s Theorem, but there is arguably a simpler way to compute them.

NEW(?) NON-TECH APPROACH:

Because the solution to every complex number computation can be written in $a+bi$ form, new possibilities open.  The original example can be rephrased:

Determine the simultaneous real values of x and y for which $-16=(x+yi)^2$.

Start by expanding and simplifying the right side back into $a+bi$ form.  (I wrote about a potentially easier approach to simplifying powers of i in my last post.)

$-16+0i = \left( x+yi \right)^2 = x^2 +2xyi+y^2 i^2=(x^2-y^2)+(2xy)i$

Notice that the two ends of the previous line are two different expressions for the same complex number(s).  Therefore, equating the real and imaginary coefficients gives a system of equations:

Solving the system gives the square roots of -16.

From the latter equation, either $x=0$ or $y=0$.  Substituting $y=0$ into the first equation gives $-16=x^2$, an impossible equation because x & y are both real numbers, as stated above.

Substituting $x=0$ into the first equation gives $-16=-y^2$, leading to $y= \pm 4$.  So, $x=0$ and $y=-4$ -OR- $x=0$ and $y=4$ are the only solutions–$x+yi=0-4i$ and $x+yi=0+4i$–the same solutions found earlier, but this time without using polar form or DeMoivre!  Notice, too, that the presence of TWO solutions emerged naturally.

Higher order roots could lead to much more complicated systems of equations, but a CAS can solve that problem.

CAS APPROACH:

Determine all fourth roots of $1+2i$.

That’s equivalent to finding all simultaneous x and y values that satisfy $1+2i=(x+yi)^4$.  Expanding the right side is quickly accomplished on a CAS.  From my TI-Nspire CAS:

Notice that the output is simplified to $a+bi$ form that, in the context of this particular example, gives the system of equations,

Using my CAS to solve the system,

First, note there are four solutions, as expected.  Rewriting the approximated numerical output gives the four complex fourth roots of $1+2i$$-1.176-0.334i$$-0.334+1.176i$$0.334-1.176i$, and $1.176+0.334i$.  Each can be quickly confirmed on the CAS:

CONCLUSION:

Given proper technology, finding the multiple roots of a complex number need not invoke polar representations or DeMoivre’s Theorem.  It really is as “simple” as expanding $(x+yi)^n$ where n is the given root, simplifying the expansion into $a+bi$ form, and solving the resulting 2×2 system of equations.

At the point when such problems would be introduced to students, their algebraic awareness should be such that using a CAS to do all the algebraic heavy lifting is entirely appropriate.

As one final glimpse at the beauty of complex roots, I entered the two equations from the last system into Desmos to take advantage of its very good implicit graphing capabilities.  You can see the four intersections corresponding to the four solutions of the system.  Solutions to systems of implicit equations are notoriously difficult to compute, so I wasn’t surprised when Desmos didn’t compute the coordinates of the points of intersection, even though the graph was pretty and surprisingly quick to generate.

## Define Your Own Math Rule

My friend, Knox S., introduced me to this problem.  According to a post on The Telegraph’s Education page, this was originally  posted on Facebook by Randall Jones.

The first line is fine by the standard rules of arithmetic, but as soon as you read the 2nd and 3rd lines, you know something is amiss.  What could be the output of line 4?

The Telegraph post above claims there are two answers.  Sadly, that post suggests there are only two solutions.  The reality is that there is an infinite number of correct answers.

I first share the two most commonly proffered solutions suggested by the Telegraph as the only answers.  I follow this with Knox’s clever use of an incremental number base.  Finally, I offer a more generalized approach to support my claim of many more solutions.

STANDARD SOLUTIONS

Consistent with the first three lines, the same rule to line 4 “proves” the answer is 40:

While nothing requires it, this approach is recursive.  I’ve not seen anyone say this, but the 40 approach requires the equations to appear in the given order.  If you give the equations in a different order, the rule is no longer consistent.  In particular, if you wanted a 5th line, what would it be?  There’s nothing clear about how to extend this solution.

• THE ANSWER IS 96:  Alternatively, you can multiply the two numbers on the left and add that product to the first number.  This procedure is consistent with the first three lines, so the solution to line 4 must be 96:

The nice thing about this approach is that the solution is explicit, not recursive.  What’s obviously counter-intuitive is why you would first multiply the given numbers, and then why you would add the result to the first number, not the second.  This approach is consistent with the given information, so it is valid.

Unlike the first solution, this multiplicative approach is not commutative.  By this rule, 1+4 yields 5, as shown, but 4+1 would be $4+(4*1)=8$.  Nothing in the problem statement required commutativity, so no worries.

Another good aspect of this algorithm is that the order of the equations is now irrelevant.  It applies no matter what numbers are “added” on the left side of the equation.  This is definitely more satisfying.

CHANGE THE NUMBER BASE

• THE ANSWER IS 201:  Knox noticed that if you changed the number base, you could find another legit pattern.  The first line is standard arithmetic, but how could the next lines be consistent, too?  You know 2+5 doesn’t give 12 in standard base-10 arithmetic, but if you use base-5, $2+5=7=1*5^1+2*5^0=12_5$.

Unfortunately, in base-5, line 1 would be $(1+4)_5=10_5$ and line 3 would be $(3+6)_5=14_5$, both inconsistent.  Knox’s cleverest move was to vary the number base.  The 3rd line is true in base-4; since the 1st line is true in any base larger than five, he found a consistent pattern by applying base-6 to line 1:

Following this pattern, the next line would be base-3, giving 201 as the answer:

The best part of Knox’s solution is that he maintains the addition integrity of the left side.  The down-side is that this approach works for only one more line.  Any 5th line would give a base-2 (binary) answer, and since base-1 does not exist, the problem would end there.

Knox’s approach also allows you to use any numbers you want for the left-hand sums.  But notice that answers depend on where you write the sum.  For example, if (2+5) was in any other line, you would not get 12.  In line 1, $(2+5)_6=11_6$, in line 3, you’d get $(2+5)_4=13_4$.

By now, you should see that any any rule could work so long as you are consistent.  Because standard arithmetic does not apply, solvers should feel free to invoke any functions or algorithms desired.  One way to do this is to think of each line as the inputs (left side) and output (right side) of a three-variable function.

• THE ANSWER IS 96:  One possible function is $z=f(x,y)=a*x^2+b*y^2+c$ for some values of a, b, and c that passes through (1,4,5), (2,5,12), and (3,6,21).  I used my TI-Nspire CAS to solve the resulting system:

That means if x and y are the given left-side numbers and z is the right-side answer, the equation $\frac{1}{3}*x+\frac{2}{3}*y-6=z$ satisfies the first three lines and the answer to line 4 is 96

• THE ANSWER IS $\displaystyle \frac{2574}{29}$:  If you can square the inputs, why not cube them?  That means another possible function is $z=f(x,y)=a*x^3+b*y^3+c$.  My CAS solution of the resulting system leads to the fractional answer:

The first three given equations essentially define three ordered triples–(1,4,5), (2,5,12), and (3,6,21)–so almost any equation you conceive with three unknown coefficients can be used to create a 3×3 system of equations.  The fractional solution for line 4 may not be as satisfying as any of the earlier approaches using only integers, but these last two examples make it clear that there should be an infinite number of solutions.

These last two solutions are especially nice because they are explicit and don’t depend on the order of the given information.  You can choose any two numbers to “add”, and the algorithms will work.

Notice also that all of these functions, except for Knox’s, are non-commutative.  No worries, the problem already broke free of standard rules in line 2.

ONE THAT DIDN’T WORK

The last two examples prove the existence of quadratic and cubic solutions, so why not a linear solution?  In other words, is there a 3D plane in the form $z=a*x+b*y+c$ containing the given points?

Unfortunately, the resulting 3×3 system didn’t solve. The determinant of the coefficient matrix is zero, suggesting an inconsistent or dependent system.  Upon further inspection, subtracting line 1 from line 2 in the planar system gives $a+b=7$.  Similarly, subtracting line 2 from line 3 gives $a+b=9$.  Since both can’t be simultaneously true, the system is inconsistent and has no solution.  It was worth the effort.

CONCLUSION

Since standard arithmetic didn’t apply after the first line and no other restrictions were in play, that opened the door to lots of creativity.  The many different solutions to this problem all hinge on finding some function–any function–that satisfied the first three lines.  Find one of these, and the last line is simple.  That some attempts won’t work is no hinderance.  Even when standard algorithms seem to apply, there is almost always the possibility of some creative twist when working with numerical sequences.

So, whenever you’re faced with a non-standard system, have fun, be creative, and develop something unexpected.

## Party Ratios

I find LOTS of great middle school problems from @Five_Triangles on Twitter.  Their post two days ago was no exception.

The problem requires a little stamina, but can be approached many ways–two excellent criteria for worthy student explorations.  That it has some solid extensions makes it even better.  Following are a few different solution approaches some colleagues and I created.

INITIAL THOUGHTS, VISUAL ORGANIZATION, & A SOLUTION

The most challenging part of this problem is data organization.  My first thoughts were for a 2-circle Venn Diagram–one for gender and one for age.  And these types of Venn Diagrams are often more easily understood, in my experience, in 2×2 Table form with extra spaces for totals.  Here’s what I set up initially.

The ratio of Women:Girls was 11:4, so the 24 girls meant each “unit” in this ratio accounted for 24/4=6 people.  That gave 11*6=66 women and 66+24=90 females.

At this point, my experience working with algebraic problems tempted me to overthink the situation.  I was tempted to let B represent the unknown number of boys and set up some equations to solve.  Knowing that most 6th graders would not think about variables, I held back that instinct in an attempt to discover what a less-experienced mind might try. I present my initial algebra solution below.

The 5:3 Male:Female ratio told me that each “gender unit” represented 90/3=30 people.  That meant there were 5*30=150 men and 240 total people at the party.

Then, the 4:1 Adult:Children ratio showed how to age-divide every group of 5 partygoers.  With 240/5=48 such groups, there were 48 children and 4*48=192 adults.  Subtracting the already known 66 women gave the requested answer:  192-66=126 men.

While this Venn Diagram/Table approach made sense to me, I was concerned that it was a moderately sophisticated and not quite intuitive problem-solving technique for younger middle school students.

WHAT WOULD A MIDDLE SCHOOLER THINK?

A middle school teaching colleague, Becky, offered a different solution I could see students creating.

Completely independently, she solved the problem in exactly the same order I did using ratio tables to manage the scaling at each step instead of my “unit ratios”.  I liked her visual representation of the 4:1 Adults:Children ratio to find the number of adults, which gave the requested number of men.  I suspect many more students would implicitly or explicitly use some chunking strategies like the visual representation to work the ratios.

WHY HAVE JUST ONE SOLUTION?

Math problems involving ratios can usually be opened up to allow multiple, or even an infinite number of solutions.  This leads to some interesting problem extensions if you eliminate the “24 girls” restriction.  Here are a few examples and sample solutions.

What is the least number of partygoers?

For this problem, notice from the table above that all of the values have a common factor of 6.  Dividing the total partygoers by this reveals that 240/6=40 is the least number.  Any multiple of this number is also a legitimate solution.

Interestingly, the 11:4 Women:Girls ratio becomes explicitly obvious when you scale the table down to its least common value.

My former student and now colleague, Teddy, arrived at this value another way.  Paraphrasing, he noted that the 5:3 Male:Female ratio meant any valid total had to be a multiple of 5+3=8.  Likewise, the 4:1 Adult:Child ratio requires totals to be multiples of 4+1=5.  And the LCM of 8 & 5 is 40, the same value found in the preceding paragraph.

What do all total partygoer numbers have in common?

As explained above, any multiple of 40 is a legitimate number of partygoers.

If the venue could support no more than 500 attendees, what is the maximum number of women attending?

12*40=480 is the greatest multiple of 40 below 500.  Because 480 is double the initial problem’s total, 66*2=132 is the maximum number of women.

Note that this can be rephrased to accommodate any other gender/age/total target.

Under the given conditions, will the number of boys and girls at the party ever be identical?

As with all ratio problems, larger values are always multiples of the least common solution.  That means the number of boys and girls will always be identical or always be different.  From above, you can deduce that the numbers of boys and girls at the party under the given conditions will both be multiples of 4.

What variations can you and/or your students create?

RESOLVING THE INITIAL ALGEBRA

Now to the solution variation I was initially inclined to produce.  After initially determining 66 women from the given 24 girls, let B be the unknown number of boys.  That gives B+24 children.  It was given that adults are 4 times as numerous as children making the number of adults 4(B+24)=4B+96.  Subtracting the known 66 women leaves 4B+30 men.  Compiling all of this gives

The 5:3 Male:Female ratio means $\displaystyle \frac{5}{3} = \frac{5B+30}{90} \longrightarrow B=24$, the same result as earlier.

ALGEBRA OVERKILL

Winding through all of that algebra ultimately isn’t that computationally difficult, but it certainly is more than typical 6th graders could handle.

But the problem could be generalized even further, as Teddy shared with me.  If the entire table were written in variables with W=number of women, M=men, G=girls, and B=boys, the given ratios in the problem would lead to a reasonably straightforward 4×4 system of equations.  If you understand enough to write all of those equations, I’m certain you could solve them, so I’d feel confident allowing a CAS to do that for me.  My TI-Nspire gives this.

And that certainly isn’t work you’d expect from any 6th grader.

CONCLUSION

Given that the 11:4 Women:Girls ratio was the only “internal” ratio, it was apparent in retrospect that all solutions except the 4×4 system approach had to find the female values first.  There are still several ways to resolve the problem, but I found it interesting that while there was no “direct route”, every reasonable solution started with the same steps.

Thanks to colleagues Teddy S & Becky M for sharing their solution proposals.