# Monthly Archives: March 2016

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

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

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

INITIAL THOUGHTS

I see two big challenges here.

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

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

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

POINT P VARIES

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

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

AREA RATIOS BECOME A LINEAR SYSTEM

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

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

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

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

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

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

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

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

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

PROBLEM VARIATIONS

Just two extensions this time.  Other suggestions are welcome.

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

FURTHER EXTENSIONS?

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

## 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.