This discussion is now closed.
Check out other Related discussions
- BMO Prep
- Is it too late for serious progress at BMO1/BMO2
- how many pairs of solutions
- Advice for Maths Olympiads
- BMO2 Preparation
- Cayley Olympiad
- BMO1/BMO2 Preparation
- How to get better at BMO1
- BMO topics
- is olympiads important?
- Senior Maths Challenge
- BMO1 help
- Senior maths challenge
- Should I do BMO1 as a Year 13 Student
- Year 12 | a realistic GYG
- Should I bother putting Cayley/Hamilton/Maclaurin Olympiad medals in Maths PS
- Geometry in bmo & smc
- SMC Prep
- How exactly are marks awarded in BMO??
- Solving Diophantine equations in math competitions
Bmo1 1999 Q1
BMO1
I have four children. The age in years of each child is a
positive integer between 2 and 16 inclusive and all four ages
are distinct. A year ago the square of the age of the oldest
child was equal to the sum of the squares of the ages of the
other three. In one year’s time the sum of the squares of the
ages of the oldest and the youngest will be equal to the sum
of the squares of the other two children.
Decide whether this information is sufficient to determine their
ages uniquely, and find all possibilities for their ages.
positive integer between 2 and 16 inclusive and all four ages
are distinct. A year ago the square of the age of the oldest
child was equal to the sum of the squares of the ages of the
other three. In one year’s time the sum of the squares of the
ages of the oldest and the youngest will be equal to the sum
of the squares of the other two children.
Decide whether this information is sufficient to determine their
ages uniquely, and find all possibilities for their ages.
Woundering how people dealt with this question.
My strat was to take the first part (1 year ago) and find all the poss answers, then cross-refernce with the second part (1 year time) to see which answers held true.
E4 = E3 + E2 + E1
Where E [belongs to set] [4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225]
E4 > E3 > E2 > E1
A bit on number crunching:
225 = 196 + 25 + 4 ---> 15^2 = 14^2 + 5^2 + 2^2
225 = 121 + 100 + 4 ---> 15^2 = 11^2 + 10^2 + 2^2
196 = 144 + 36 + 16 ---> 14^2 = 12^2 + 6^2 + 4^2
169 = 144 + 16 + 9 ---> 13^2 = 12^2 + 4^2 + 9^2
121 = 81 + 36 + 4 ---> 11^2 = 9^2 + 6^2 +2^2
49 = 36 + 9 + 4 ---> 7^2 = 6^2 + 3^2 + 2^2
D4 + D1 = D2 + D3
Where D [belongs to set] [9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289]
If an equation from part1 holds true, then:
(Root[E4] + 2)^2 + (Root[E1] + 2)^2 = (Root[E2] + 2)^2 + (Root[E3] + 2)^2
That is, (see first equation):
17^2 + 4^2 = 16^2 + 7^2 Which is true, therefore the ages 16, 15, 6, and 3 are a possibilty.
Checking each of the equations, only the 1st one and 5th one holds true, therefore the children are ages [16, 15, 6 and 3] OR [12, 10, 7, and 3].
Any other approaches?
If you call the ages now a<b<c<d then you have two equations:
(d-1)^2 = (a-1)^2 + (b-1)^2 + (c-1)^2 (*)
(d+1)^2 + (a+1)^2 = (b+1)^2 +(c+1)^2
Taking one from the other and rearranging you get
d = b + c - (a^2 +1)/2.
This means a is odd. Also from this, as d>c, then b > (a^2+1)/2.
As b =< 20 then a = 3 or 5.
CASE 1: a = 3 and d = b + c -5
Subbing this into (*) and rearranging eventually gives
(b-5)(c-5) = 10.
This has to be the product 2*5 or 1 *10. As b<c then b,c = 7,10 or 6,15.
CASE 2: a = 5 and d = b + c - 13.
Subbing this into (*) and rearranging eventually gives
(b-13)(c-13) = 80.
If we write 80 as r*s at least one of the factors is 8 which would result in b and c being outside the 2...16 range.
So the only solutions are those you found with the computer.
(d-1)^2 = (a-1)^2 + (b-1)^2 + (c-1)^2 (*)
(d+1)^2 + (a+1)^2 = (b+1)^2 +(c+1)^2
Taking one from the other and rearranging you get
d = b + c - (a^2 +1)/2.
This means a is odd. Also from this, as d>c, then b > (a^2+1)/2.
As b =< 20 then a = 3 or 5.
CASE 1: a = 3 and d = b + c -5
Subbing this into (*) and rearranging eventually gives
(b-5)(c-5) = 10.
This has to be the product 2*5 or 1 *10. As b<c then b,c = 7,10 or 6,15.
CASE 2: a = 5 and d = b + c - 13.
Subbing this into (*) and rearranging eventually gives
(b-13)(c-13) = 80.
If we write 80 as r*s at least one of the factors is 8 which would result in b and c being outside the 2...16 range.
So the only solutions are those you found with the computer.
Reply 2
Hmm good idea, i assumed that the (x-1)^2 and +1 would be a messy way, turns out its nice
Lucky for me, i didn't use a computer to do the question, only 15mins of number crunching :P
Lucky for me, i didn't use a computer to do the question, only 15mins of number crunching :P
Related discussions
- BMO Prep
- Is it too late for serious progress at BMO1/BMO2
- how many pairs of solutions
- Advice for Maths Olympiads
- BMO2 Preparation
- Cayley Olympiad
- BMO1/BMO2 Preparation
- How to get better at BMO1
- BMO topics
- is olympiads important?
- Senior Maths Challenge
- BMO1 help
- Senior maths challenge
- Should I do BMO1 as a Year 13 Student
- Year 12 | a realistic GYG
- Should I bother putting Cayley/Hamilton/Maclaurin Olympiad medals in Maths PS
- Geometry in bmo & smc
- SMC Prep
- How exactly are marks awarded in BMO??
- Solving Diophantine equations in math competitions
Latest
Trending
Last reply 3 days ago
Edexcel AS Level Maths May 15th 2025 Pure Paper 1 + Unofficial Mark SchemeMaths Study Help
42
143
Last reply 1 week ago
Further mechanics 2 options overlapping with physics rather than stats?