An example of a Math Olympiad Problem
Question
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.
(d-1) respectively. We are told that ” 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.” This gives us,
(a – 1)^2 = (b – 1)^2 + (c – 1)^2 + (d – 1)^2 …(1)
For the statement, “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.
” means,
(a + 1)^2 + (d + 1)^2 = (b + 1)^2 + (c + 1)^2 ….(2)
We use the “identity” below to “simplify” things :
( x + 1 ) ^2 = ( x – 1 )^2 + 4x.
Just set x = a,b,c,d respectively to get :
( a + 1 ) ^2 = ( a – 1 )^2 + 4a.
( b + 1 ) ^2 = ( b – 1 )^2 + 4b.
( c + 1 ) ^2 = ( c – 1 )^2 + 4c.
( d + 1 ) ^2 = ( d – 1 )^2 + 4d.
Put these into (2) to obtain :
(a + 1)^2 + (d + 1)^2 = (b + 1)^2 + (c + 1)^2 becomes :
(a – 1)^2 + 4a + (d -1)^2 + 4d = (b -1)^2 + 4b + (c – 1)^2 + 4c
Call this last expression as ( 4 ).
We get from (4), (a – 1)^2 equals the expression below.:
(b -1)^2 + 4b + (c – 1)^2 + 4c – [ 4a + (d -1)^2 + 4d ]
We get from (1), (a – 1)^2 equals the expression below :
(b – 1)^2 + (c – 1)^2 + (d – 1)^2
“equate” (a -1)^2 in (4) & in (1) above to get,
(b -1)^2 + 4b + (c – 1)^2 + 4c – [ 4a + (d -1)^2 + 4d ]
= (b – 1)^2 + (c – 1)^2 + (d – 1)^2
“Note that (b -1)^2 and (c -1)^2 appears on both sides of the equation above and can be cancelled”… simplifying, the above will reduce to :
4b + 4c – [ 4a + (d -1)^2 + 4d ]
= (d – 1)^2
or,
4 ( b + c ) = (d – 1)^2 + [ 4a + (d -1)^2 + 4d ]
4 ( b + c ) = 2*(d – 1)^2 + 4 ( a + d )
2 ( b + c – a – d ) = ( d – 1 )^2 …(5)
Consider the “last” equation (5) above. d -1 even means that
d must be odd. d is in [2,16] Thus d >= 3 , that is “d is bigger or equal to 3”. Since the “ages are distinct”, Thus c >= 4,
b >= 5 & a >= 6. Similarly, since a <= 16, b <= 15, c <= 14 and d <= 13.
In (5) & the information above awe have,
b + c – a – d <= 15 + 14 – 6 – 3 = 20.
So, ( d – 1 )^2 = 2 ( b + c – a – d ) <= 2*40 = 80
That is (d – 1)^2 <= 80 or (d – 1) <= 8.95 or
d <= 9.95. Since d is odd, thus the “possibilities” for d are
d = 3 , 5, 7, 9.
We consider each of the 4 cases based on the value of d separately.
Case 1 ( d = 3 ).
(5) becomes,
2 ( b + c – a – 3 ) = ( 3 – 1 )^2 = 4
b + c – a = 5 or b + c = a + 5.
Substitute a = ( b + c – 5 ) into (1) and simplify to get,
15 + bc = 5* ( b + c ) means 5 divides bc. Consider for “simplicity” 15 + xy = 5*(x + y) instead. With b > c implies,
max ( x,y) = b and min(x,y) = c. With this “assumption” it helps us because now we can “assume without loss of generality that” let x be a multiple of 5. Since in [2,16] the multiples of 5 are : 5,10 & 15. We set x = 5, 10 , and 15 to get the corresponding values of y. For x = 5, equation becomes
3 + y = 5 + y giving the “contradiction” that 3 = 5!. For x = 10, we get 3 + 2y = 10 + y or y = 7. For x = 15, we get,
3 + 3y = 15 + y or y = 6.
Hence solutions are (x,y) = (10, 7) & ( 15 , 6) Since b > c, this implies, (b,c) = (10,7) or (15,6).
For (b,c) = (10,7) , b + c = a + 5 gives a = 12.
Solution here is (a,b,c,d) = (12,10,7,3).
For (b,c) = (15,6), b + c = a + 5 gives a = 16.
Solution here is (a,b,c,d) = (16,15,6,3).
Case 2 ( d = 5)
(5) becomes,
2 ( b + c – a – 5 ) = ( 5 – 1 )^2
b + c = a + 13
We put a = b + c – 13 into (1) and simplify to get,
89 + bc = 13 ( b + c ). Use above ( b + c ) = 13 + a to get,
89 + bc = 13 ( a + 13 ) = 13a + 169.
bc = 13a + 80.
Note that since a >= 6,
13a + 80 >= 13*6 + 80 = 158. and by the last equality above,
implies bc >= 158. Call this (6).
If b <= 13 implies c <= 12 and bc <= 12*13 = 156 < 158 contradicting (6). So b >= 14 and earlier we showed b <= 15. So either b = 14 or 15. We proceed below :
If b = 14,
89 + bc = 13 ( b + c ). becomes 89 + 14c = 13*(14 + c ) giving
89 + 14c = 182 + 13c , that is , c = 93 [ not possible for us because in our case c <=16 ].
If b = 15,
89 + bc = 13 ( b + c ). becomes 89 + 15c = 13*(15 + c )giving 89 + 15c = 195 + 13c and that is, c = 53 [ not possible for us because in our case c <=16 ].
Thus there are NO feasible solutions that satisfy our problem in this case.
Case 3 ( d = 7 ).
(5) becomes,
2 ( b + c – a – 7 ) = ( 7 – 1 )^2
b + c = a + 25
Now, from a >= 6, its a + 25 >= 6 + 25 = 31. So, b + c >= 31. But since b <= 15 & c <= 14, we get b + c <= 29 which contradicts “b+c>=31” above. Hence no solutions in this case either.
Case 4 ( d = 9 )
(5) becomes,
2 ( b + c – a – 9 ) = ( 9 – 1 )^2
b + c = a + 41 implies b + c > 41 but we saw from above that
b + c <= 29; Hence a contradiction and no solutions for this case as well.
To conclude the only solutions are as follows:
(a,b,c,d) = (12,10,7,3) or (16,15,6,3).
To “check”, use a calculator and consider for both cases of solution above :
For (a,b,c,d) = (12,10,7,3),
11^2 = 9^2 + 6^2 + 2^2 and
13^2 + 4^2 = 11^2 + 8^2
For (a,b,c,d) = (16,15,6,3),
15^2 = 14^2 + 5^2 + 2^2 and
17^2 + 4^2 = 16^2 + 7^2
