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a

t

b

c

n

m

a

b c

e

d

f 

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?

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 !owever, this was a

terrible assumption" it turns out that there is a solution to #tewart$s

%heorem based on a 4-4-2 triangle divided into a 2-2-& triangle and a

4-3-2 triangle

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?

#tewart$s %heorem says that a2n+b2m=t 2 c+mcn We substituted c=m+n  and solved for 

t , getting t =√a2

n+b2

m−m2

n−n2

m

m+n a, b, m, and n could be as small as &, 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 't this point, we plugged it into a spreadsheet, and got a solution that was a 4-(-&0

triangle divided into a 2-4-5 and a 5-)-&0, for a total of 2*

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?

#ymmetry in an equilateral triangle with sides of length 2 would ma+e

all perimeter segments &, for an answer of )

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?

eva$s %heorem says thata

b× c

d ×

 e

f  =1

, although there are many ways to epress this

.aminingacebdf 

 =1 and considering that the smallest answer would use &, 2, 3, 4, 5, and ),

you soon reali/e that wherever you put the 5, there is no other multiple of 5 that can cancel it to

get us to & %his will be true for other primes, as well you need to have an even number of

multiples of each prime factor in your set &, 2, 3, 4, ), ( gets us pretty close, but we have an odd

number 1seven of 2$s in the prime factori/ations &, 2, 3, 4, ), should wor+, as it has four 2$s

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a

b c

ed

f 

and four 3$s We can do2×3×6

1×4×9=1

, but can we ma+e a real triangle with these numbers

es69+2=11

,6+4=10

, and3+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?

'gain, symmetry in an equilateral triangle with an area of ) would ma+e all regions have an area

of &, for an answer of )

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?

eva$s %heorem tal+s about ratios of the segments around the perimeter, but those ratios between

segments are equal to ratios between areas .ga

b  in the first diagram is equal toa+ f +e

b+c+d

in this diagram #o, we need to arrange areas within the triangle so that the products of these

ratios of the sums of areas cancel to & 7nstead of worrying about the individual values of a, b, c,

etc, we 8ust focused on the sums %he smallest possible answer would be

1+2+3+4+5+6=21 , which would allow ratios of sums of6

15 ,7

14 ,8

13 ,9

12 ,

10

11 , and their reciprocals 9ote that we need to balance our prime factors again 7f we use

6

15 , we need to use15

6  or10

11  to cancel the factor of 5, but we$d need a & for the

former and something else with && for the latter, so we cannot use6

15 We cannot use8

13  

or10

11  for similar reasons7

14=1

2  and9

12=3

4 , but these cannot wor+ with their

reciprocals to satisfy eva$s %heorem, so 2& cannot be the answer

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We thought we had a solution of 22 with a-f being 4, &, *, 3, 2, 5, which produced sum ratios of 

10

12 ,12

10 , and11

11 !owever, at the last minute we reali/ed that in addition to the sum

ratios, the smaller areas must also be in ratios that satisfy eva$s %heorem :or eample, in our

solutiona+ f +e

b+c+d=11

11=1

, soa

b  should also be &, but instead it$s 4; We sought disputes on

this part, but none were received in the one-wee+ dispute window, so we dropped this part for the

 purposes of scoring

'fter the dispute window had closed, Westla+e school +ept at it and finally found solutions to

this problem, although they may not be the smallest #tarting with solutions to part d,

multiplying each area by an integer, and sub8ecting these areas to the sum-of-areas ratio

requirements, they found a triangle with sub-areas of &4, 2(, )3, &(, &0(, and &( in cloc+wise

order, for an answer of 420;