2. Cevians
a. In the triangle shown to the right with one cevian, all segment
lengths between two points of intersection are integers. What is
the smallest possible sum of all segment lengths?
m
a
t
We assumed that the repeated integers would be symmetric, and that
this would require right triangles, which led us to two 3-4-5 triangles
sharing the cevian of length 4, for a total of 20. However, this was a
terrible assumption; it turns out that there is a solution to Stewart’s
Theorem based on a 4-4-2 triangle divided into a 2-2-1 triangle and a
4-3-2 triangle.
c
n
b
b. In the triangle shown to the right with one cevian, all segment lengths between two
points of intersection are distinct integers. What is the smallest possible sum of all segment
lengths?
Stewart’s Theorem says that 𝑎2 𝑛 + 𝑏 2 𝑚 = 𝑡 2 𝑐 + 𝑚𝑐𝑛. We substituted 𝑐 = 𝑚 + 𝑛 and solved
𝑎2 𝑛+𝑏 2 𝑚−𝑚2 𝑛−𝑛2 𝑚
for t, getting 𝑡 = √
𝑚+𝑛
. a, b, m, and n could be as small as 1, 2, 3, and 4, but t
must be smaller than at least one of a or b, so really a, b, m, and n must include an integer at last
as large as 5. At this point, we plugged it into a spreadsheet, and got a solution that was a 4-8-10
triangle divided into a 2-4-5 and a 5-6-10, for a total of 27.
c. In the triangle with three concurrent cevians shown to the right,
all segment lengths around the perimeter are integers. What is the
smallest possible value of the perimeter?
Symmetry in an equilateral triangle with sides of length 2 would make
all perimeter segments 1, for an answer of 6.
b
c
a
d
f
e
d. In the triangle with three concurrent cevians shown to the right, all segment lengths
around the perimeter are distinct integers. What is the smallest possible value of the
perimeter?
𝑎
𝑐
𝑒
Ceva’s Theorem says that 𝑏 × 𝑑 × 𝑓 = 1, although there are many ways to express this.
𝑎𝑐𝑒
Examining 𝑏𝑑𝑓 = 1 and considering that the smallest answer would use 1, 2, 3, 4, 5, and 6, you
soon realize that wherever you put the 5, there is no other multiple of 5 that can cancel it to get
us to 1. This will be true for other primes, as well… you need to have an even number of
multiples of each prime factor in your set. 1, 2, 3, 4, 6, 8 gets us pretty close, but we have an odd
number (seven) of 2’s in the prime factorizations. 1, 2, 3, 4, 6, 9 should work, as it has four 2’s
and four 3’s. We can do
2×3×6
1×4×9
= 1, but can we make a real triangle with these numbers? Yes:
9 + 2 = 11, 6 + 4 = 10, and 3 + 1 = 4 satisfies the triangle inequality, so the answer is 1 +
2 + 3 + 4 + 6 + 9 = 25.
e. In the triangle with three concurrent cevians shown above and to the right, the areas of
all regions are integers. What is the smallest possible value of the total area?
Again, symmetry in an equilateral triangle with an area of 6 would make all regions have an area
of 1, for an answer of 6.
f. In the triangle with three concurrent cevians shown above and to the
right, the areas of all regions are distinct integers. What is the smallest
possible value of the total area?
b
c
a
f
d
e
Ceva’s Theorem talks about ratios of the segments around the perimeter, but those ratios
𝑎
𝑎+𝑓+𝑒
between segments are equal to ratios between areas. E.g. 𝑏 in the first diagram is equal to 𝑏+𝑐+𝑑
in this diagram. So, we need to arrange areas within the triangle so that the products of these
ratios of the sums of areas cancel to 1. Instead of worrying about the individual values of a, b, c,
etc., we just focused on the sums. The smallest possible answer would be 1 + 2 + 3 + 4 + 5 +
6
7
8
9
10
6 = 21, which would allow ratios of sums of 15, 14, 13, 12, 11, and their reciprocals. Note that we
6
need to balance our prime factors again. If we use 15, we need to use
15
6
10
or 11 to cancel the factor
of 5, but we’d need a 1 for the former and something else with 11 for the latter, so we cannot use
6
8
10
. We cannot use 13 or 11 for similar reasons.
15
7
1
9
3
= 2 and 12 = 4, but these cannot work with
14
their reciprocals to satisfy Ceva’s Theorem, so 21 cannot be the answer.
We thought we had a solution of 22 with a-f being 4, 1, 7, 3, 2, 5, which produced sum ratios of
10 12
11
, , and 11. However, at the last minute we realized that in addition to the sum ratios, the
12 10
smaller areas must also be in ratios that satisfy Ceva’s Theorem. For example, in our solution
𝑎+𝑓+𝑒
𝑏+𝑐+𝑑
11
𝑎
= 11 = 1, so 𝑏 should also be 1, but instead it’s 4! We sought disputes on this part, but
none were received in the one-week dispute window, so we dropped this part for the purposes of
scoring.
After the dispute window had closed, Westlake school kept at it and finally found solutions to
this problem, although they may not be the smallest. Starting with solutions to part d,
multiplying each area by an integer, and subjecting these areas to the sum-of-areas ratio
requirements, they found a triangle with sub-areas of 14, 28, 63, 189, 108, and 18 in clockwise
order, for an answer of 420!