Solution Notes
The problem was solved by Joseph DeVincentis and Stephen Meskin. The smallest example is the triangle with sides (7, 17, 13). If we leave the integers, we can get real number solutions such as (√3, √4, √5). Any three consecutive integers can be used here. This follows from the Pythagorean comment below, and this √3-√4-√5 example was the first example found (by Lee Sallows). So (1, √2, √3) is another example.
The similarity problem of Macalester POW 1205 reduces to a variation on Pythagorean triangles. Using some computer algebra, and easily-found formulas for the length of a median of a triangle, I found that if (a, b, c) is a Pythagorean triangle (with a < b < c), then (b − a, b + a, c) satisfies the similarities of problem, provided the new triple is indeed a triangle. And vice versa. The new triple (A, B, C) also satisfies A² + B² == 2C². Using the √3-√4-√5 triangle leads to (1, 7, 5), which is not a triangle. Using (5, 12, 13) leads to (7, 17, 13), which is in fact the smallest triangle satisfying the given similarities.
Here is a link to download Meskin's solution:
s1205.pdf
And here is DeVincentis's solution:
Let a, b, c be the lengths indicated by 1, 2, and 3 hash marks in the diagram (which is slightly different than the usual triangle convention, but helps avoid a lot of ugly fractions here). Let d, e, f, g, h, i be the lengths of the segments shared by T1 and T2, by T2 and T3, ..., by T6 and T1, respectively.
For 4 of the similarities given, the smaller triangle shares an angle with the larger triangle. The other angles cannot match if we simply scale up the small triangle while keeping this angle fixed. In all cases, the sides of the two triangles opposite the shared angle are on lines which intersect at one of the vertices. So we have to flip over and scale up the small triangle to make this work. Even in the special cases where a subtriangle is isosceles and there might be other ways of matching them up, they will be equivalent to this one.
This means that:
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[T1] c/i = 2a/d = (f + i)/c
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[T2] (d + g)/a = b/e = 2a/d
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[T5] 2b/h = a/g = (h + e)/b
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[T6] c/i = 2b/h = (f + i)/c
The repetitions of c/i, 2a/d, 2b/h, and (f + i)/c here mean that all 8 distinct ratios given are the same.
c/i = b/e= a/g = 2a/d = 2b/h = (h + e)/b = (d + g)/a = (f + i)/c
Use the first 4 equalities to solve for e, g, d, and h in terms of a, b, c, and i:
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e = bi/c
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g = ai/c
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d = 2ai/c
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h = 2bi/c
Now plug these into the (h + e)/b and (d + g)/a terms. Both cases yield 3i/c. So c² = 3i² or i = c/√3. Then use this to eliminate i from the previous equations, yielding:
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e = b/sqrt(3)
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g = a/sqrt(3)
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d = 2a/sqrt(3)
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h = 2b/sqrt(3)
Finally, (f + i)/c = 3i/c yields f = 2i = 2c/√3.
Now, the centroid always divides each median in a 2:1 ratio, so the 2:1 ratio is expected. So the median f + i, which connects C to the midpoint of 2c, is of length (√3)c. But d + g — which connects B to the midpoint of 2b — is of length (√3)a, and e + h is of length (√3)b.
Apollonius' theorem allows us to relate a, b, and c using these median lengths:
4a² + 4b² = 2(3c² + c²)
4a² + 4c² = 2(3a² + b²)
4b² + 4c² = 2(3b² + a²)
All of these reduce to a² + b² = 2c².
Now in addition to this, we also have two more similarity relationships to resolve.
a, b/√(3), 2c/√3 is similar to 2c, b, (√3)a in some order. b, 2c/sqrt(3), a/sqrt(3) is similar to a, 2c, sqrt(3) b in some order. Both of these are similar without any further manipulations; just multiply the side lengths of the first triangle by √3.
So look for solutions to a² + b² = 2c², and this will simultaneously provide all the needed median lengths to provide all the similarities. This is easy to search with simple brute force.
The smallest solution which makes a valid triangle is 7, 17, 13 (13 = c), which appears to be the triangle used in the example. 7, 23, 17 and 17, 31, 25 are also solutions, as are 23, 47, 37 and 31, 49, 41, which makes me think there's probably some generating rule for these solutions which uses the legs from one solution to generate other solutions.
I made a demo that essentially solves any jug-pouring problem (NB: you need to have Mathematica or download the free Wolfram CDF Player to view this dynamically):
Solving Decanting Problems by Graph Theory
This demo is only for three jugs, but the code could be modified to handle the general case. It shows the graph that arises in barycentric coordinates. One can move the start and finish states with the mouse, and watch the shortest path change.
[Back to Problem 1205]
© Copyright 2015 Stan Wagon. Reproduced with permission.