IMO SOLUTIONS 2020

Posted by: admin Comments: 0 0 Post Date: February 7, 2020

SOLUTIONS

  1. In how many ways can you arrange four 2-digit numbers in an increasing order and in each arrangement all digits from 1 to 8 must appear only once?

(For example: 12, 34, 56, 78 , it just one arrangement)

 

Solution.

If we order 2-digit numbers in increasing order and we use only digits from 1 to 8 once. All digits must appear in any order so we should look at first digits of each number because second digit cannot affect.  For example, 19 is less than 20. First digit of first number can be 1,2,3,4 and 5.

If the first digit of first number is 1 and first digit of second number is 2, then there are 4!(5+4+3+2+1) =24×15

If the first digit of first number is 1 and first digit of second number is 3, then there are 4!(4+3+2+1) =24×10

If the first digit of first number is 1 and first digit of second number is 4, then there are 4!(3+2+1) =24×6

If the first digit of first number is 1 and first digit of second number is 5, then there are 4!(2+1) =24×3

If the first digit of first number is 1 and first digit of second number is 6, then there are 4!(1) =24×1

24×(15 + 10 + 6 + 3 +1 )

If the first digit of first number is 2 and first digit of second number is 3, then there are 4!(4+3+2+1) =24×10

If the first digit of first number is 2 and first digit of second number is 4, then there are 4!(3+2+1) =24×6

If the first digit of first number is 2 and first digit of second number is 5, then there are 4!(2+1) =24×3

If the first digit of first number is 2 and first digit of second number is 6, then there are 4! =24×1

24×(10 + 6 + 3 +1 )

If the first digit of first number is 3 and first digit of second number is 4, then there are 4!(3+2+1) =24×6

If the first digit of first number is 3 and first digit of second number is 5, then there are 4!(2+1) =24×3

If the first digit of first number is 3 and first digit of second number is 6, then there are 4! =24×1

24×(6 + 3 +1 )

If the first digit of first number is 4 and first digit of second number is 5, then there are 4!(2+1) =24×3

If the first digit of first number is 4 and first digit of second number is 6, then there are 4! =24×1

24×(3 +1 )

If the first digit of first number is 5 and first digit of second number is 6, then there are 4! =24×1

The total number of arrangement is 24×(1×15 + 2×10 + 3×6 + 4×3 +5×1) = 24×70 = 1680

 

  1. 2. Let ABC be acute-angled triangle and let D, E, and F be the feet of altitudes from A,B, and C to sides BC, CA, and AB, respectively. Denote by wB and wC the incircles of triangles BDF and CDE, and let these circles be tangent to segments DF and DE at M and N, respectively. Let line MN meet circles wB and wC again at P ¹ M and Q ¹ N, respectively.

Prove that MP = NQ

  1. 3. Solution: Let N have the digits abc and let N/11 have the digits xy. Then if the sum of the squares of its digits is equal to N/11 + 1, we have two cases:

 

Case 1: a = x , b = x + y , c = y. We obtain the equation

x 2 + (x + y)2 + y2  = 10x + y + 1 Þ  2x 2 + 2y2 + 2xy = 10x + y + 1

The LHS is even, therefore on the RHS we must have y Îí1, 3, 5, 7, 9ý.

Considering each possible value of y we get the following equations which we solve for integer solutions:

 

If y = 1 we obtain the equation 2x2 – 8x = 0 Þ  x(x – 4) = 0 Þ x = 0 or x = 4.

This yields N = 451.

 

If y = 3 we obtain the equation 2x2 – 4x + 14 = 0 Þ   x2 – 2x + 7 = 0.

The discriminant is –24, so the equation has no real-valued solutions.

 

If y = 5 we obtain the equation 2x2 + 44 = 0 Þ x2 = –22. Clearly the equation has no real-valued solutions.

 

If y = 7 we obtain the equation  2x2 + 4x + 90 = 0 Þ   x2 + 2x + 45 = 0.

The discriminant is –176, so the equation has no real-valued solutions.

 

If y = 9 we obtain the equation  2x2 + 8x + 152 = 0 Þ   x2 + 4x +76 = 0.

The discriminant is –288, so the equation has no real-valued solutions.

 

Case 2: a = x + 1, b = x + y – 10, c = y. We obtain the equation

(x + 1)2 + (x + y – 10)2 + y2 = 10x + y + 1 Þ  2x 2 + 2y2 – 2xy – 18x – 20y + 101= 10x + y + 1

 

The LHS is odd, therefore on the RHS we must have y Îí0; 2; 4; 6; 8ý.

Considering each possible value of y we get the following equations which we solve for integer solutions:

 

If y = 0 we obtain the equation 2x2 – 24x + 66 = 0 Þ  x2 – 12x + 33 = 0.

The discriminant is –4, so the equation has no real-valued solutions.

 

If y = 2 we obtain the equation 2x2 – 24x + 48 = 0 Þ  x2 – 10x + 24 = 0.

The discriminant is 12, so the equation has no integer solutions.

 

If y = 4 we obtain the equation 2x2 – 24x + 48 = 0 Þ  x2 – 10x + 24 = 0 Þ x = 6 or

x = 4. The solution x = 6 yields N = 704, while x = 4 does not lead to a sensible integer.

 

If y = 6 we obtain the equation 2x2 – 16x + 46 = 0 Þ  x2 – 8x + 23 = 0.

The discriminant is –28, so the equation has no real-valued solutions.

 

If y = 8 we obtain the equation 2x2 – 12x + 60 = 0 Þ  x2 – 6x + 30 = 0.

The discriminant is –84, so the equation has no real-valued solutions.

 

Thus the integers N are 451 and 704.

 

 

  1. 4. Find all functions f: R ® R such that f(x2) – yf(y) = f(x + y) (f(x) – y) for all

real numbers x and y.

Solution:

Taking y = 0 in the functional equation gives us that f(x2) = f(x)2 for all real x.

In particular, f(0) = f(0)2, and so f(0) Î í0, 1ý.

If f(0) = 1, then taking x = 0 in the functional equation gives us that

1 –  yf(y) = f(y)(1 – y) = f(y) – yf(y) for all real y, and so f(y) = 1 for all real numbers y. We can check that this does indeed satisfy the functional equation.

Now suppose that f(0) = 0.

Since f(–x)2 = f(x2) = f(x)2 for all x, we know that for each x, either f(–x) = –f(x) or f(–x) = f(x). Now taking y = –x in the functional equation gives us that

f(x2) + xf(–x) = 0, and so we derive that f(x)2 = xf(x) or f(x)2 = –xf(x) for each x. This shows that f(x) Îí0, –x, xý for all real numbers x.

 

Since f(x2) = f(x)2, we in fact have that f(x) Îí0, xý for all positive real numbers x. Now suppose that there is a positive real number x such that f(x) = 0.

Then f(x2) = 0, and we find that for all positive y, we have –yf(y) = f(x + y)( –y), and so f(y) = f(x + y).

Suppose that f(y) = y. Then f(x + y) ¹ 0, and so f(x + y) = x + y. But then

y = x + y, a contradiction. Thus we must have that f(y) = 0 for all positive y, and since f(–y) = ±f(y) for all y, we have that f(y) = 0 for all negative values of y as well. Thus f is identically 0, but this does not satisfy the functional equation.

 

We thus have that there is no positive real number x such that f(x) = 0, and so

f(x) = x for all x ³ 0. We now recall that f(x2) + xf(–x) for all x. Since f(x2) = x2, this gives us that f(–x) = –x for all x. We thus find that f(x) = x for all real numbers x, which does indeed satisfy the functional equation. All solutions to the functional equation are thus given by the constant function f(x) = 1, and the identity function f(x) = x.

 

 

 

 

 

 

 

 

 

 

Share this post

Leave a Reply

Your email address will not be published. Required fields are marked *