A mathematician wanted to teach his children the value of cooperation, so he told them the following
"I chose a secret triangle for which the lengths of its sides are all integers.
To you my dear son Charlie, I am giving the triangle's perimeter. And to you, my beloved daughter Ariella, I am giving its area.
Since you are both such talented mathematicians, I'm sure that together you can find the lengths of the triangle's sides."
Instead of working together, Charlie and Ariella had the following conversation after their father gave each of them the information he promised.
Charlie: "Alas, I cannot deduce the lengths of the sides from my knowledge of the perimeter."
Ariella: "I do not know the perimeter, but I cannot deduce the lengths of the sides from just knowing the area. Maybe our father is right and we should cooperate after all."
Charlie: "Oh no, no need. Now I know the lengths of the sides."
Ariella: "Well, now I know them as well."
Find the lengths of the triangle's sides and explain the dialogue above.
This is a nice variation on the Sum and Product Problem, a puzzle featured by Martin Gardner in his Scientific American column in 1979 (as I well recall: picture a household of grad students competing to solve it.) It could be argued that this is an even nicer problem, because it seems that no limit on the integers is required for the problem to have to a unique solution, unlike the Sum and Product Problem.
Here, if we call the sides a, b, c, the perimeter p, and the area A, we have two guiding rules: To make triangles, unique sets of a, b, c must have each side less than the sum of the other two, and the area A is given by Heron's Formula.
One perimeter may be the sum of different a, b, c sets, which I call a p's "multiple constructs", and each set, in turn, yields a different A. Likewise, A may be "ambiguous", meaning: obtainable from different perimeters. The statements given by the participants in the puzzle one by one lead to constraints on the correct (p, A) pair. The pair must have these qualities:
1) p has multiple possible constructs.
2) A is ambiguous. Further, at least two of the p's associated with A each have multiple possible constructs.
3) Of all the constructs of 1), the associated A's are all ambiguous, except one A, which is unambiguous.
4) Of all the p's that are possibly associated with A, only one has both properties 1) and 3) above.
A partial list of the ranked multiplicity of A's and of p's is shown below with the associated program.
Next step: find a satisfying (p, A) pair!
a b c p A
------------------------------
1 1 1 3 0.43
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1 2 2 5 0.97
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1 3 3 7 1.48
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2 2 2 6 1.73
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1 4 4 9 1.98
2 2 3 7 1.98
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1 5 5 11 2.49
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2 3 3 8 2.83
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2 3 4 9 2.90
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1 6 6 13 2.99
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1 7 7 15 3.49
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2 4 5 11 3.80
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2 4 4 10 3.87
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3 3 3 9 3.90
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1 8 8 17 3.99
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3 3 5 11 4.15
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3 3 4 10 4.47
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1 9 9 19 4.49
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2 5 6 13 4.68
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2 5 5 12 4.90
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1 10 10 21 4.99
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3 4 6 13 5.33
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1 11 11 23 5.49
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3 4 4 11 5.56
2 6 7 15 5.56
------------------------------
.
.
.
a b c A p
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1 1 1 0.43 3
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1 2 2 0.97 5
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2 2 2 1.73 6
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1 3 3 1.48 7
2 2 3 1.98 7
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2 3 3 2.83 8
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1 4 4 1.98 9
2 3 4 2.90 9
3 3 3 3.90 9
------------------------------
2 4 4 3.87 10
3 3 4 4.47 10
------------------------------
2 4 5 3.80 11
3 3 5 4.15 11
1 5 5 2.49 11
3 4 4 5.56 11
------------------------------
2 5 5 4.90 12
3 4 5 6.00 12
4 4 4 6.93 12
------------------------------
2 5 6 4.68 13
1 6 6 2.99 13
3 4 6 5.33 13
3 5 5 7.15 13
4 4 5 7.81 13
------------------------------
3 5 6 7.48 14
2 6 6 5.92 14
4 4 6 7.94 14
4 5 5 9.17 14
.
.
.
program her
implicit none
integer a,b,c,p2,cnt,l(4,1000),dum(4),ii,j,k,oldp
real area,p,xl(1000),xdum,olda
cnt=0
oldp=0
olda=0
do a = 1,20
do b = a,20
do 1 c = b,20
if(a.ge.b+c.or.b.ge.a+c.or.c.ge.a+b)goto 1
cnt=cnt+1
p2=a+b+c
p=p2/2.
area = sqrt(p*(p-a)*(p-b)*(p-c))
l(1,cnt)=a
l(2,cnt)=b
l(3,cnt)=c
l(4,cnt)=p2
xl(cnt)=area
1 enddo
enddo
enddo
c
c area sort
c
do j = 1,cnt-1
do k=j+1,cnt
if(xl(j).gt.xl(k))then
xdum=xl(j)
xl(j)=xl(k)
xl(k)=xdum
do ii=1,4
dum(ii)=l(ii,j)
l(ii,j)=l(ii,k)
l(ii,k)=dum(ii)
enddo
endif
enddo
enddo
do k=1,cnt
if(xl(k).ne.olda)then
print 111
111 format(30('-'))
olda=xl(k)
endif
print 333, (l(ii,k),ii=1,4),xl(k)
333 format(4(i4,2x),f6.2)
enddo
print*
print*
print*
c
c perim sort
c
do j = 1,cnt-1
do k=j+1,cnt
if(l(4,j).gt.l(4,k))then
xdum=xl(j)
xl(j)=xl(k)
xl(k)=xdum
do ii=1,4
dum(ii)=l(ii,j)
l(ii,j)=l(ii,k)
l(ii,k)=dum(ii)
enddo
endif
enddo
enddo
do k=1,cnt
if(l(4,k).ne.oldp)then
print 111
oldp=l(4,k)
endif
print 222, (l(ii,k),ii=1,3),xl(k),l(4,k)
222 format(3(i4,2x),f6.2,2x,i4)
enddo
end
Edited on February 15, 2019, 11:22 am
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