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, along the y-axis, and 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 and , 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.

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.

Simplifying terms and clearing denominators leads to and , 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 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 .

**PROBLEM VARIATIONS **

Just two extensions this time. Other suggestions are welcome.

**What’s the ratio of the area of 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?

**If a random point is chosen within the square, is that point more likely to satisfy the area ratio of or the ratio of ?**

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 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?

You write:

>”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.”

This Y6 problem was not intended to be solved with variables (Y7-9), a coordinate system (Y7), linear systems (Y8?), or algebra (Y8-9). It also doesn’t require knowing the location of point P.

It’s a ratio problem (Y6), and by solving it otherwise, the gist of this problem—recognising a simple but important relationship between all four triangles (also Y6)—is missed. It can be solved with an arithmetic approach similar to the party ratios problem, plus the addition of the geometric relationship alluded to above. The problem’s intent is to pull together two disparate Y6 skills to create “cognitive stress”.

A previous essay on “cognitive stress”:

https://docs.google.com/document/d/1BnmvpBJ1lQdRwpGKNwS4I-d0cyFPfan8n4ngDjDuB5g/edit

That the problem can be solved algebraically is happenstance. Another way of putting it is: algebra works because this problem is fairly basic. We have more complex Y6 ratio problems that will bring algebra to its knees. Indeed, an over-reliance on algebra can lead to an impenetrable morass of variables and equations, while comparing ratios remains fast and simple.

>”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.”

We would, too, but to further illustrate our point, we will follow up eventually with an example of an algebra-crushing Y6-skills ratio problem.