Tag Archives: system of equations

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.

square2

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.

square3

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?

Chemistry, CAS, and Balancing Equations

Here’ s a cool application of linear equations I first encountered about 20 years ago working with chemistry colleague Penney Sconzo at my former school in Atlanta, GA.  Many students struggle early in their first chemistry classes with balancing equations.  Thinking about these as generalized systems of linear equations gives a universal approach to balancing chemical equations, including ionic equations.

This idea makes a brilliant connection if you teach algebra 2 students concurrently enrolled in chemistry, or vice versa.

FROM CHEMISTRY TO ALGEBRA

Consider burning ethanol.  The chemical combination of ethanol and oxygen, creating carbon dioxide and water:

C_2H_6O+3O_2 \longrightarrow 2CO_2+3H_2O     (1)

But what if you didn’t know that 1 molecule of ethanol combined with 3 molecules of oxygen gas to create 2 molecules of carbon dioxide and 3 molecules of water?  This specific set coefficients (or multiples of the set) exist for this reaction because of the Law of Conservation of Matter.  While elements may rearrange in a chemical reaction, they do not become something else.  So how do you determine the unknown coefficients of a generic chemical reaction?

Using the ethanol example, assume you started with

wC_2H_6O+xO_2 \longrightarrow yCO_2+zH_2O     (2)

for some unknown values of w, x, y, and z.  Conservation of Matter guarantees that the amount of carbon, hydrogen, and oxygen are the same before and after the reaction.  Tallying the amount of each element on each side of the equation gives three linear equations:

Carbon:  2w=y
Hydrogen:  6w=2z
Oxygen:  w+2x=2y+z

where the coefficients come from the subscripts within the compound notations.  As one example, the carbon subscript in ethanol ( C_2H_6O ) is 2, indicating two carbon atoms in each ethanol molecule.  There must have been 2w carbon atoms in the w ethanol molecules.

This system of 3 equations in 4 variables won’t have a unique solution, but let’s see what my Nspire CAS says.  (NOTE:  On the TI-Nspire, you can solve for any one of the four variables.  Because the presence of more variables than equations makes the solution non-unique, some results may appear cleaner than others.  For me, w was more complicated than z, so I chose to use the z solution.)

chem1

All three equations have y in the numerator and denominators of 2.  The presence of the y indicates the expected non-unique solution.  But it also gives me the freedom to select any convenient value of y I want to use.  I’ll pick y=2 to simplify the fractions.  Plugging in gives me values for the other coefficients.

chem2

Substituting these into (2) above gives the original equation (1).

VARIABILITY EXISTS

Traditionally, chemists write these equations with the lowest possible natural number coefficients, but thinking of them as systems of linear equations makes another reality obvious.  If 1 molecule of ethanol combines with 3 molecules of hydrogen gas to make 2 molecules of carbon dioxide and 3 molecules of water, surely 10 molecule of ethanol combines with 30 molecules of hydrogen gas to make 20 molecules of carbon dioxide and 30 molecules of water (the result of substituting y=20 instead of the y=2 used above).

You could even let y=1 to get z=\frac{3}{2}, w=\frac{1}{2}, and x=\frac{3}{2}.  Shifting units, this could mean a half-mole of ethanol and 1.5 moles of hydrogen make a mole of carbon dioxide and 1.5 moles of water.  The point is, the ratios are constant.  A good lesson.

ANOTHER QUICK EXAMPLE:

Now let’s try a harder one to balance:  Reacting carbon monoxide and hydrogen gas to create octane and water.

wCO + xH_2 \longrightarrow y C_8 H_{18} + z H_2 O

Setting up equations for each element gives

Carbon:  w=8y
Oxygen:  w=z
Hydrogen:  2x=18y+2z

I could simplify the hydrogen equation, but that’s not required.  Solving this system of equations gives

chem3

Nice.  No fractions this time.  Using y=1 gives w=8, x=17, and z=8, or

8CO + 17H_2 \longrightarrow C_8 H_{18} + 8H_2 O

Simple.

EXTENSIONS TO IONIC EQUATIONS:

Now let’s balance an ionic equation with unknown coefficients a, b, c, d, e, and f:

a Ba^{2+} + b OH^- + c H^- + d PO_4^{3-} \longrightarrow eH_2O + fBa_3(PO_4)_2

In addition to writing equations for barium, oxygen, hydrogen, and phosphorus, Conservation of Charge allows me to write one more equation to reflect the balancing of charge in the reaction.

Barium:  a = 3f
Oxygen:  b +4d = e+8f
Hydrogen:  b+c=2e
Phosphorus:  d=2f
CHARGE (+/-):  2a-b-c-3d=0

Solving the system gives

chem4

Now that’s a curious result.  I’ll deal with the zeros in a moment.  Letting d=2 gives f=1 and a=3, indicating that 3 molecules of ionic barium combine with 2 molecules of ionic phosphate to create a single uncharged molecule of barium phosphate precipitate.

The zeros here indicate the presence of “spectator ions”.  Basically, the hydroxide and hydrogen ions on the left are in equal measure to the liquid water molecule on the right.  Since they are in equal measure, one solution is

3Ba^{2+}+6OH^- +6H^-+2PO_4^{3-} \longrightarrow 6H_2O + Ba_3(PO_4)_2

CONCLUSION:

You still need to understand chemistry and algebra to interpret the results, but combining algebra (and especially a CAS) makes it much easier to balance chemical equations and ionic chemical equations, particularly those with non-trivial solutions not easily found by inspection.

The minor connection between science (chemistry) and math (algebra) is nice.

As many others have noted, CAS enables you to keep your mind on the problem while avoiding getting lost in the algebra.

Student Quadratic Creativity

I’m teaching Algebra 2 this summer for my school.  In a recent test on quadratic functions, I gave a question I thought would be a little different, but still reachable for those willing to make connections or exert a little creativity.

Write a system of quadratic functions that has exactly one solution:  (1,1).

Their handheld graphing calculators were allowed.  Some students definitely had difficulty with the challenge, some gave a version of the answer I expected, and one adopted a form I knew was possible, but doubted anyone would actually find during a test situation.

I show my students’ solutions below.  But before you read on, can you give your own solution?

SOLUTION ALERT!  Don’t read further if you want to find your own solution.


WHAT I EXPECTED

We’ve had many discussions in class about the power of the Rule of 4–that math ideas can be expressed numerically, graphically, algebraically, and verbally.  When you get stumped in one representation, being able to shift to a different form is often helpful.  That could mean a different algebraic representation, or a different Rule of 4 representation altogether.

The question is phrased verbally asking for an algebraic answer.  But it asks about a solution to a system of equations.  I hoped my students would recall that the graphical version of a system solution is equivalent to the point(s) where the graphs of the equations intersected.  In my mind, the easiest way to do this is to write quadratic functions with coincident vertices.  And this is most easily done in vertex form.  The cleanest answer I ever got to this question was

quad1

A graphical representation verifies the solution.

quad3

Another student recognized that if two parabolas shared a vertex, but had different “slopes”, their only possible point of intersection was exactly the one the question required.  Here’s a graphical version of her answer.

quad2

From these two, you can see that there is actually an infinite number of correct solutions.  And I was asking them for just one of these!  🙂

WHAT I KNEW, BUT DIDN’T EXPECT

Another way to solve this question makes use of the geometry of quadratic graphs.  If two quadratics have the same leading coefficients, they are the same graph, intersect exactly once, or never intersect.  This is a very non-trivial idea for most students.  While I’m not convinced the author of the following solution had this in mind when he answered the question, his solution works because of that fact.  Here’s what J wrote on last week’s test and its graph.

quad4

quad5

J used more equations than he needed, but had he restricted himself to just two equations, I’m not sure the lovely pattern would have been so obvious.

This is a very different (and super cool) answer than what I expected my students to produce.  Lesson re-learned:  Challenge your students, give them room to express creativity and individuality, and be prepared to be amazed by them.

NEXT STEPS

J’s answer actually opens the door to other avenues of exploration.

  1. Can you generalize the form of all of J’s equations, essentially defining a family of quadratics?  Can you prove that all members of your generalization satisfy the question posed and that no other answers are possible?
  2. Can you find forms of other generalized families of quadratic functions whose only solution is (1,1)?
  3. Notice that there were two types of solutions above:  A) those with coincident vertices and different lead coefficients and B) those with identical lead coefficients and different vertices.  Are these the only types of quadratics that can answer this question?  That is, is there a system of quadratics with (1,1) as the only solution that have identical vertices and lead coefficients?  Could both be different and (1,1) be the only solution?
  4. If I relax the requirement that the quadratics be functions, what other types of quadratics are possible?  [This could be a very nice calculus question!]

For my part, I’m returning to some of these questions this week to stretch and explore my student’s creativity and problem-solving.

I’d love to hear what you or your students discover.

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.

circle1

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!

SOLUTION ALERT!!!

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.

circle2

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.

circle6

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:

circle4

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.

circle5

Systems of lines

Here’s an interesting variation of a typical (MS) problem I found by following the Five Triangles ‘blog: http://fivetriangles.blogspot.com/2013/09/97-no-triangle.html .

(Note:  If you sign up on this or other ‘blogs, you can get lots of problems emailed to you every time they are added.)

INITIAL SOLUTION

I know this question can absolutely be solved without using technology, but when a colleague asked if it was appropriate to use technology here (my school is one-to-one with tablet laptops), I thought it would be cool to share with her the ease and power of Desmos.  You can enter the equations from the problem exactly as given (no need to solve for y), or you can set up a graph in advance for your students and email them a direct link to an already-started problem.

If you follow this link, you can see how I used a slider (a crazy-simple addition on Desmos) to help students discover the missing value of a.

DesmosSolve

FOLLOW-UP

I suggest in this case that playing with this problem graphically would grant insight for many students into the critical role (for this problem) of the intersection point of the two explicitly defined lines.  With or without technology support, you could then lead your students to determine the coordinates of that intersection point and thereby the value of a.

Keeping with my CAS theme, you could determine those coordinates using GeoGebra’s brand new CAS View:

GeogebraCAS

Substituting the now known values of x and y into the last equation in the problem gives the desired value of a.

NOTE:  I could have done the sliders in GeoGebra, too, but I wanted to show off the ease of my two favorite (and free!) online math tools.

CONCLUSION

Thoughts?  What other ideas or problems could be enhanced by a properly balanced use of technology?

As an extension to this particular problem, I’m now wondering about the area of triangle formed for any value of a.  I haven’t played with it yet, but it looks potentially interesting.  I see both tech and non-tech ways to approach it.

Non-Calculus approach to Invariable Calculus Project

I shared my posts (here and here) on the Invariable Calculus Project in the AP Calculus Community.  Gary Litvin posted a response within the Community offering there a great non-calculus alternative solution to the original problem of the area of the triangle formed by the x- and y-axes and any Quadrant I tangent line to \displaystyle \frac{1}{x}.  Here’s a paraphrase of Gary’s approach.

Let \displaystyle \frac{x}{a} + \frac{y}{b}=1 be any Quadrant I tangent line to \displaystyle y=\frac{1}{x}.  (In case you don’t recognize it, this tangent equation uses the intercept form of a line–a is the x-intercept and b is the y-intercept.)  Because the line intersects the parabola in a single point, we can find that point by solving the system of equations defined by the two equations.  Substituting for y gives

\displaystyle \frac{x}{a} + \frac{\frac{1}{x}}{b}=1.

This is equivalent to x^2 - a \cdot x+\frac{a}{b}=0, a quadratic.  We could determine the value of x using the quadratic formula.  Because there is only one solution to this equation (there is only one point of intersection, the point of tangency), the discriminant must be zero.  That means

\displaystyle (-a)^2 - 4\cdot \left( \frac{a}{b} \right)=0

which can be rearranged to give ab=4 (a=0 is extraneous).  Therefore, the area of the triangle formed by the tangent line to \displaystyle y=\frac{1}{x} and the coordinate axes is \displaystyle Area=\frac{1}{2} ab=2 no matter what the point of tangency.

Shiny.

Exploring Sequences and Lines

Here’s another favorite problem that could be used for any middle or high school students who’ve been exposed to both arithmetic sequences and linear equations.

There is a family of lines, Ax+By+C=0, for which \left\{ A, B, C \right\} can be any arithmetic sequence.  What do all members of this family have in common?

As with most great math problems, the problem is simply stated and can be approached from many different perspectives.  In the end, this one isn’t that difficult to crack, but the solution (at least to me) is not intuitively obvious from the problem statement. If you’ve not seen this before, please try it before reading further.

WARNING:  Problem Solution Follows

MOST COMMON STUDENT APPROACH:  Given the problem’s phrasing, most begin by writing out a few possible equations using different arithmetic sequences, typically with positive common differences.  After a few tries, most get a little frustrated as no obvious pattern emerges from the equations.

It is this point that is typically the most difficult for me as a teacher.  I want to help anyone who struggles, but “help” often means allowing others the room to struggle and to find ways of solving problems on their own.  Only when a student or group gets really frustrated do I sometimes ask, “Are there any other ways you can look at this problem or at your work?”

Eventually, most decide to graph their equations to see if anything pops out.  The following graph was submitted this past semester by one of my precalculus students using the free online Desmos calculator.

ALines1

Two lines intersecting in a point is common.  Three or more in a single point almost always indicates something interesting.  Seven lines through a single point is screaming for attention!  From this graph, all lines in this family apparently contain the point (1,-2).  That seems a bit random until you investigate further, but pattern recognition is just half of the battle for a mathematician.  Now that something interesting has been discovered, a mathematician wants to know if this is a fluke or something inherent to all members of the family.

NOTE:  All graphs my students have produced over the years have always contained specific equations.  I don’t know that it’s any more enlightening, but I did create on Desmos a slider version of the graphs of this family with one slider for the initial term (A) and a second for its common difference (d).

UNIQUE SOLUTION METHODS FROM STUDENTS:

All successful solutions eventually rewrite the sequence \left\{ A,B,C \right\} to \left\{ A,A+d,A+2d \right\} where d is the common difference for a generic arithmetic sequence with initial term A.

Method I: After discovering the common point of intersection, most plug it into the left side of the equation and simplify to get

Ax+By+C=A\cdot 1+\left( A+d\right)\cdot -2+\left( A+2d \right)=0.

Because the left side reduces to zero for all generic arithmetic sequences, \left\{ A,A+d,A+2d \right\}, (1,-2) must be common to all members of this family.

A few students aren’t comfortable dealing with 0=0, so these tend to plug in x=1 and solve for y to get y=-2, proving that the y-coordinate for x=1 for all members of this family is always -2.

Method II:  A few students every year work algebraically from properties of arithmetic sequences.  For any arithmetic sequence, \left\{ A,B,C \right\}, \frac{A+C}{2}=B.  This rewrites to 1\cdot A-2\cdot B+C=0, so whenever \left( x,y \right)=\left(1,-2 \right), then Ax+By+C=0 is a fundamental property of all arithmetic sequences.

Personally, I think this method gets closest to explaining why the point (1,-2) is the common characteristic of this family.

Method III:  This year, I had a student take an approach I’d never seen before.  She defined one arithmetic sequence as \left\{ a,a+d,a+2d \right\} and another as \left\{ m,m+n,m+2n \right\} for any real values of a,d,m, and n.  This leads to a system of equations: a\cdot x+(a+d)\cdot y+(a+2d)=0 and m\cdot x+(m+n)\cdot y+(m+2n)=0 .  If you have some younger students or if all the variables make you nervous, the solution is available via Wolfram Alpha.

Still, this system is pretty easy to manipulate.  If you multiply the first equation by m and the second by a, the x-terms will eliminate with subtraction, giving

m\cdot((a+d)\cdot y+(a+2d))-a\cdot((m+n)\cdot y+(m+2n))=0.

Solving for y shows that all of the coefficients simplify surprisingly easily.

((ma+md)-(am+an))\cdot y=-(ma+2md)+(am+2an)
(md-an)\cdot y = -2\cdot (md-an)\Longrightarrow y=-2

From here, determining x=1 is easy, proving the relationship.

SOLUTIONS and APPROACHES NOT YET OFFERED BY STUDENTS:

Approach A:  High school students don’t often think about ways to simplify problem situations, especially at the beginning of problems.  One way I did that for this problem in later class discussions was to recognize that it one of the terms in the arithmetic sequence was 0, you didn’t need to deal with nearly as many terms. For example, if your sequence was {1,0,-1}, the linear equation would be x-1=0.  Similarly, the sequence \left\{ 01,2 \right\} leads to y+2=0.  Obviously, the only thing these two lines have in common is the point (1,-2).  A proof of the property must still be established, but this is one of the fastest ways I’ve seen to identify the central property.

Approach B:  A purely algebraic approach to this problem could redefine the arithmetic sequence as \left\{ a,a+d,a+2d\right\}as before, giving:

a\cdot x+(a+d)\cdot y+(a+2d)=0

Collecting like terms gives

(x+y+1)\cdot a+(y+2)\cdot d=0.

The values of a and d must remain as parameters to include all possible arithmetic sequences.  Because the equation always equals 0, the coefficients of a and d are both 0, making y=-2 (for the coefficient of d) and therefore x=1.

EXTENSION:

We once had a test question at the end of the unit containing this exercise.  Basically, it reminded students that they had discovered that all lines Ax+By+C=0, for which \left\{ A, B, C \right\} can be any arithmetic sequence contained the point (1,-2).  It then asked for an equation of a family of linear functions using the same arithmetic \left\{ A, B, C \right\} that all contained the point (1,2).

The two most common responses we’ve seen involve a reflection or a vertical translation.  (1,-2) can become (1,2) by reflecting over the x-axis, so making the y-values negative would do the trick:  Ax-By+C=0.  Similarly, (1,-2) can become (1,2) by translating up 4 units, giving Ax+B(y-4)+C=0.