The area of a triangle within a triangle

I’ve been confronted by what seems to be a very difficult geometry problem.

The questions asks how much smaller triangle DEF is from triangle ABC. Lines are constructed on each side at the 1/3 point (of the side) and go to a corresponding vertex. The lines essentially split each side of the triangle into 2 corresponding line segments, 1/3 and 2/3 the length of the entire side. The lines meet up to form the triangle DEF. ABC is not an equilateral triangle.

I attempted to use Heron’s formula to find the area of ABC, but xyz for the sides didn’t get me too far. The link is an illustration of the problem.

Thanks in advance for any help.

You should probably try to convince us that it’s not for homework or else I doubt you’ll get any help…

Well, it’s not homework, but my teacher did propose it to us. I don’t get anything for answering it, but it has me interested. The entire math department at my school is trying to solve it but to no avail.

I believe that this will work:

Call D the point on the side opposite A, E opposite B, and F opposite C. Express the coordinates in barycentric coordinates D = 1/3 B + 2/3 C, E = 1/3 C + 2/3 A, F = 1/3 A + 2/3 C. Using the equations found here calculate the area of the triangles ABE, BCF, and ACD. Then, from ABE calculate the area of AEF, from BCF calculate the area of BDF, and from ACD calculate the area of CDE. The area of the triangle DEF will then be ABC - AEF - BDF - CDE.

Punoquillads is on the right track, though I think he needs to relabel the points opposite the vertices, since DEF is taken by the triangle in the center. Call the point G where the line from vertex A meets the opposite side BC; similarly, point H is on the side opposite vertex B, and I is opposite C.

For an explanation that avoids barycentric coordinates, consider all the “one third” sized triangles (ABG, BCH, and CAI). Each of these have area 1/3 that of ABC, so taken separately and added they will have an area equal to ABC. Of course they don’t match ABC because they overlap; the three smaller triangles attached to each vertex (ADI, BEG, and CFH) are all counted twice in the summation. Consequently, the sum of area of these three small triangle must equal the area of the interior triangle DEF.

If you assume ABC is equilateral (against the rules, I know), it should be obvious in this case that all the little corner triangles are congruent. By carefully looking at the angles formed by the lines, it’s fairly easy to prove each little corner triangle is similar (though flipped) to the larger “one third size” triangle it shares its most acute vertex with (i.e. ADI ~ ABG, BEG ~ BCH, CFH ~ CHI). Also, these littler triangles have sides which are 1/3 that of their larger similars (i.e. AI = 1/3(AB), BG = 1/3(BC), CH=1/3(CA)), so they must be 1/9 the area of the larger similar, which themselves are 1/3 the size of ABC. Consequentially, the littler triangles have area 1/27(ABC). Adding the three of them up and noting that together their area equals the area of the central triangle DEF, this means the area of DEF must be 1/9(ABC).

Now you could argue that your teachers wouldn’t assign the problem if it didn’t have a single solution for all such triangles, so this is a rather sneaky “proof” that DEF = 1/9(ABC). I’d love to find something more general though…

Solve the problem for an equilateral triangle A’B’C’ (with interior triangle D’E’F’). The ratio of the area of triangle D’E’F’ to the area of triangle A’B’C’ will be the same as the ratio of the area of triangle DEF to triangle ABC. This is true because there exists an affine map between any two triangles in the plane.

Thx ultrafilter; this fact justifies the generalization of the equilateral case solved above.

1/9 sounds low. Consider the triangle with points A = (0,3) B = (3,0) C = (-3,0). It’s pretty simple math to show that its area is 9. The corresponding triangle DEF would have D = (-1,0) E = (-1,2) F = (2,1). Its area is 3.

Ah, I think I see. One set isn’t 1/3 of the larger triangle, they’re 2/3 of the larger triangle, meaning the parts that you cut off are each 2/9 of ABC, so all 3 parts are 6/9 = 2/3 of ABC, leaving the area of DEF = 1/3 of ABC.

Perhaps you’re misunderstanding the problem. Take a look at the diagram again; they’re looking for the area of the triangle DEF formed by the intersection of the trisectors themselves, not the triangle formed by the points where they intersect with the sides.

For the triangle you give, the intersection points are (-12/14, 3/7), (-3/5, 9/5), and (9/7, 6/7).

The bolded part isn’t quite right. The sides AI:AB are in the ratio 1:3, but AI and AB are not corresponding sides of the triangles ADI ~ ABG, so you cannot conclude that the areas are as 1:9. The correct answer (for the ratio of areas of DEF and ABC) is strictly between 1/9 and 1/3. I have a cute proof-by-picture that involves no barycentric coordinates. Hint: place A, B, C cleverly on a triangular lattice.

Crap, you;'re right…back to the drawing board…

OK, thinking about this on the drive home helped. Again, referring to the diagram and defining point G where the line from vertex A meets the opposite side BC; similarly, point H is on the side opposite vertex B, and I is opposite C.

Let’s assume ABC is equilateral. Triangle ABG is then similar to ADI (although flipped, as Omphaloskeptic pointed out).

Now in triangle ABG, is the length of side AB is X, the length of side BG is (1/3)X, and between them is the known angle of 60 degrees. We can therefore compute the length of side AG; it is (7/9)X. Comparing ABG to the similar ADI, AG corresponds to AI, which by definition is (1/3)X. Therefore, the two similar triangles have sides in the ratio of 7:3.

Using this information, one can either compute the length of the sides in the embedded equilateral triangle DEF (the answer is (3/7)X) or use the fact that the total area of the little corner triangles sum to the area of the embedded equilateral. Either way, the area of DEF will be (3/7)^2 the area of ABC, or 9/49 as big.

Hope that does it…

Almost; you’re missing a square root in your application of the law of cosines.

Not that it’s probably what you really want, but this seems to be a special case of Routh’s Theorem .


How did we arrive at (7/9)X? The post after you mentions the law of cosines. My trig is rusty, but I thought we needed a right angle for that.

When dividing a 60 degree angle as they have so that the proportions are 1/3x and 2/3x for the side the line segment intersects, it would leave us with a 20 degree sliver, yes? That means with the known 60 degree angle, we have a 100 degree angle left over (not 90).

Or is my trig even rustier than I thought?

You don’t need a right angle to use the cosine law. The cosine law reduces to Pythagoras’ Theorem when you have a right angle.

I think this helped the most, thanks.

And thanks to everyone for their input.

Is this cute enough? It didn’t help me. :slight_smile:

Haha, yeah. You can just calculate the answer, by brute force, for a specific triangle with specific side lengths and then apply the thing about affine transforms. You can prove Routh’s theorem the same way. (But please don’t just use it directly! That’s so cheap!)

I love it! Totally out-of-the-box math thinking.