The Dramatic Conclusion of POTW

Cutting the form at the vertices into three trapezoids of height 2, the areas are 9*2, 10*2, and 11*2, since the middle line is the average of the inner and outer line. So the total area should be 60. I think this applies to any shape of a constant width: Area = Width * Length of central line

Cthulhu, I got a different answer, and sadly my work was a lot more complex. Here it is:

For this exercise, side A = 11, B = 10, and C = 9. Vertical height of the middle triangle = H. We will also use S = the semi-perimeter of the middle triangle = (A+B+C)/2. I will use D and E to represent the sides of the smaller and larger triangles corresponding to A. Finally, I will use X to represent the angle between sides A and B.
1. Use known area of middle triangle to determine its height.
-Area = 0.5 base*H but also = SQRT[S*(S-A)*(S-B)*(S-C)]. We can calculate the latter with the information given, Area = SQRT(1800). For reference, this is about 42.43 though we don’t need the actual number.
-Taking A to be the base, H = 2*SQRT(1800)/11. For reference, this is about 7.713 though again we don’t need the actual number.
2. Derive some useful geometrical relationships.
-From the description of the problem, we know the smaller triangle has a height of H-1 and the larger has a height of H+1.
-We also know from geometry that sine(X) = H/A = (H-1)/D = (H+1)/E. From this, we can derive that D = A*(H-1)/H and E = A*(H+1)/H. Since we used side A as the base of our middle triangle, D and G will also be the bases of our smaller and larger triangles.
-We also know from geometry that the areas of the smaller and larger triangle are 0.5*(H-1)*D and 0.5*(H+1)*E, respectively.
3. Solve for area of smaller triangle.
-Area = 0.5*(H-1)*[A*(H-1)/H] = 0.5*(H-1)^2*A/H. This equals 11/2*[(H^2-2H+1)/H] or 11/2*(H-2+1/H).
4. Solve for area of larger triangle.
-Area = 0.5*(H+1)*(A(H+1)/H) or eventually = 11/2*(H+2+1/H).
5. Solve for shaded area
-This is the larger triangle’s area minus the small triangle’s area. The first and third factors of each polynomial cancel out and we end up with 11/2*[2-(-2)] or just 22.

Final observation: this seems like a lot of work just to get an answer of A*2. Which makes me think there is probably an easier way to do it.

Oops, I just realized that my #2, first subpoint is blatantly wrong. Time to rework.

Cthulhu,

I don't have a valid solution (yet) but I can see an error in yours. The three triangles can easily be proven to be equiangular. Therefore, they are similar. Thus, in your trapezoids the known side is not the arithmetic mean of the two unknown sides. For any set of corresponding sides, the ratio of the smaller unknown to the known side is equal to the ratio of the known side to the larger unknown side. Arbitrarily using the known value of 11, and letting a and b be the smaller and larger unknowns respectively, a/11 = 11/b which gives ab=121. Thus, the known side is the geometric mean of the unknowns, not the arithmetic mean. (The only way that the geometric mean and arithmetic mean of two numbers can be equal is if the two numbers are equal; that is clearly not the case in this problem).

@Sean T: For any trapezoid, its width at half its height is always the arithmetic mean of the two parallel sides, and the given sides are all by construction at the half of the height of each trapezoid. Your a/11=11/b argument on the other hand is faulty, as you are trying to use similarity arguments on sides from three different similar triangles, but those arguments only work on pairs of triangles. That it is wrong is easy to see if you make the outer triangle larger. Applying your argument, b would now at the same time be longer than before, but retain its ratio a/11=11/b, which obviously can't be true.

Cthulhu,

You are correct, and your original solution is as well. I erred in assuming that the ratio of the corresponding sides of the smaller and middle triangles must be equal to the ratio of the sides of the middle and large triangles. That is not true, so my comment was incorrect.

D'oh, I solved the wrong problem -- when I read "middle triangle" I thought it was referring to the innermost triangle, and I didn't actually read the rest of the problem because the diagram was self explanatory (I thought). Anyway, using the measurements given on the innermost triangle, I think the shaded area works out to exactly 36 but I don't have time to check my math or to post it. Feel free to try that problem and see what you get.

Also, "concentric" is ambiguous when it comes to triangles. Which center(s) do they share? My intuition says they share at least incenter and orthocenter with the given constraints, but again no time to check. Any others?

Scratch that -- arithmetic error. I get 60+25*sqrt(2)

This is a bit of a cheat, so I am not posting tha actual calculations.

After a few false starts, I figured that there was some known relations between the sides, the altitudes, and the area of a triangle. Looking in Wikipedia, I found the article on "Altitude (triangle)" - http://en.wikipedia.org/wiki/Altitude_(triangle) . This article has a formula that relates the altitudes to the sides, so the altitudes can be calculated.

The inner triangle has altitudes one less than the middle triangle, and the outer triange has altitudes one greater than the middle tiangle.

The article also has a formula that relates the altitudes to the area. Since we can calcuate the lengths of the altitudes of the inner and outer triangle, we can get the areas of those two, and subtract.

As the formulas are rather complex, involving recipricals, I just wrote a progran to figure out the area, and got, numerically, about 19.716.

The altitude differences between triangles is greater than 1 -- it's a difference of 1 at the bases, but there's a difference at the vertices to account for as well.

Another Matt - Oops, you're right. Somewhere along the way I confused the altitude with the distance from the orthocenter to the side.

Jason -- I'm curious how much the students need to show to get credit. Would the Cthulhu's trapezoid approach be sufficient as is, or do they have to do simpler things like prove that the figures are indeed trapezoids (e.g by showing that similar vertices of the three triangles all lie on a line)? How much do you grant as obvious with this kind of thing?

As the sides are equidistant parallel lines, it is easy to show that the vertices fall on the respective angle bisectors, so the three triangles are concentric with regards to ther incenter. I skipped that step in my solution for brevity, but I think it can be regarded as quite obvious.

Yep. Now I'm pretty sure I'm wrong that they share orthocenter as well.

Cthulhu's approach in comment one is precisely the one I had in mind, and any student who gave me that would get full marks. Cthulhu is also right that, given the diagram, it seems clear that it is the incenter that was intended, but I certainly could have been more precise in my phrasing.

There's another approach that no one has mentioned yet. You start with the same three trapezoids, but then you cut them up and rearrange them into a long, thin parallelogram. The area of a parallelogram is base times height. In this case the height is two, while a base must have length 9+10+11=30, for a total area of sixty.

Cthulhu’s approach in comment one is precisely the one I had in mind, and any student who gave me that would get full marks. Cthulhu is also right that, given the diagram, it seems clear that it is the incenter that was intended, but I certainly could have been more precise in my phrasing.

There’s another approach that no one has mentioned yet. You start with the same three trapezoids, but then you cut them up and rearrange them into a long, thin parallelogram. The area of a parallelogram is base times height. In this case the height is two, while a base must have length 9+10+11=30, for a total area of sixty.

Here's a diagram of the problem I had been trying to solve, in which the innermost triangle is given edges of lengths 9, 10, and 11.

https://www.dropbox.com/s/3h93razp4bvnkgm/POTW.jpeg?dl=0

The measurements of the next triangle out form the basis for the trapezoid solution. If the base of the smallest triangle is 10, then the altitude cuts it into lengths 3 and 7; the altitude itself is 6*sqrt(2).

One way to approach getting the measurements of the next triangle out is by using similar triangles to extend the vertices of the base of the inner triangle to those of the outer. The next triangle out is in the ratio (1+sqrt(2)/4):1 to the given triangle, so applying the trapezoid procedure yields 60 + 60*sqrt(2)/4 = 60 + 15*sqrt(2)