I follow the Five Triangles ‘blog for cool math problems. A recent one proved particularly nice.
At first I wasn’t sure this situation was invariant. I didn’t see how fixing three triangle areas guaranteed a fixed quadrilateral area. Not seeing an immediate general solution approach, I reasoned that if there was a solution, it worked for multiple overall configurations. If it worked in general, then it must also work for any particular case I chose, so I made the cevians perpendicular. That made each of the given area triangles right. I modeled that by constructing the overall triangle with the cevian intersection at the origin and the legs of the given area triangles along the coordinate axes.
There are many ways to do this, but I reasoned that if there was a single answer, then any one of them would work. A right triangle with legs of length 8 and 5 would have area 20. Constructing that triangle in GeoGebra fixed the lengths of the legs of the other two triangles and the hypotenuses of the area 8 & 15 triangles intersected at a Quadrant II point. Here’s my construction.
I overlayed a polygon to create the quadrilateral and measured its area directly. For fun, I also wrote algebraic equations for lines CB and DA, found the coordinates of point F by solving the 2×2 linear system, used that to derive the area of , and determined the area of the quadrilateral from that.
While I realized that this approach was just a single case of the given problem, it absolutely convinced me that the solution was unique. Once the area 20 triangle was defined (whether or not the triangle was right), a side and the area of each of the other two given triangles is known. That meant the heights of the triangles would be determined and thereby the location of the quadrilateral’s fourth vertex. So, I knew without a doubt that the unknown area was , but I didn’t know a general solution.
Chronology of the General Solution
While I worked more on the problem, I also pitched it to my Twitter network and asked a colleague at my school, Tatiana Yudovina, if she was interested in the problem. Next is Tatiana’s initial solution, followed by my generic Geogebra construction, and a much shorter solution Tatiana created. My conclusion takes the problem to a more generic state and raises some potential extensions.
Tatiana’s First Solution:
Leveraging the fact that triangles with the same base have equivalent height and area ratios, she created a system of equations that solved to eventually determine the quadrilateral’s area.
My Generic GeoGebra Solution:
While Tatiana was working on her algebraic answer, I was creating a dynamic version on GeoGebra. I built the area 20 triangle by first drawing a segment AB and measuring its length, a. That meant the height of this triangle, h, was given by . Then I constructed a perpendicular line to AB and used the “Segment with Fixed Length” tool and defined the length using the generic length of h as defined above to create segment AC. This worked because GeoGebra defined the length of AB as a variable as shown below.
I used the “Compass” tool to create a circle with radius AC through the perpendicular line created earlier. Point D is the intersection of the circle and the normal line. I then constructed a perpendicular to AD through D and placed a random point E on this new line. Point E was the requisite height above AB to guarantee that always had area 20 which I confirmed by drawing the triangle and computing its area.
I hid AC, the circle, and both normals. Segment AB was a completely independent object, and point E was free to move along the second “height” normal. I measured AE and repeated the previous construction to create the area 15 triangle. Because BE was part of a cevian, I drew line BE to determine point J on the normal defining the final vertex of the area 15 triangle.
Again, I hid all of my constructions and repeated the process to create the final vertex, K, of the area 8 triangle off side BE of the area 20 triangle. Extending segments AJ and BK defined point L, the final vertex of the quadrilateral. Laying a quadrilateral in the figure let me compute its area. Moving points A, B, and E around the screen and seeing the areas remain fixed is pretty compelling evidence that the quadrilateral’s area is always 27, and Tatiana’s proof showed why. You can play with my final construction on GeoGebra Tube here.
Then Tatiana emailed me a much shorter proof.
Tatiana’s Short Solution:
Reversing the logic of her first solution, Tatiana reasoned that equivalent-altitude triangles had equal base and area ratios.
And the sum of X and Y gave the quadrilateral’s area.
This problem was entertaining both in the solution and the multiple ways we found it. Creating the dynamic construction gave insights into the critical features of the problem.
Here are some potential extensions I developed for this problem. I haven’t fully explored any of them yet, hoping some of my geometry students this year might take up the exploration challenge. I’d love to hear if any of my readers have any further suggestions.
- It might be interesting to create an even more dynamic construction with the areas of the three given triangles defined by sliders.
- Can the quadrilateral’s area be expressed as a closed-form function of the areas of the three given triangles.
- What happens on the boundaries of this problem? That is, what happens if one of the side triangles was a degenerate with area 0? What would happen to the quadrilateral? Would would be the corresponding affect on the area formula from extension 2?
- Extending 3 even further, if both given side triangles were degenerates with area 0, it seems that the area formula from extension 2 should collapse to the area of the final given non-zero triangle, but does it?
Thanks again, Five Triangles, for another great problem!