14th Balkan Mathematical Olympiad Problems 1997

14th Balkan Mathematical Olympiad Problems 1997

A1.  ABCD is a convex quadrilateral. X is a point inside it. XA² + XB² + XC² + XD² is twice the area of the quadrilateral. Show that it is a square and that X is its center.

A2.  A collection of m subsets of X = {1, 2, ... , n} has the property that given any two elements of X we can find a subset in the collection which contains just one of the two. Prove that n ≤ 2^m.

A3.  Two circles C and C' lying outside each other touch at T. They lie inside a third circle and touch it at X and X' respectively. Their common tangent at T intersects the third circle at S. SX meets C again at P and XX' meets C again at Q. SX' meets C' again at U and XX' meets C' again at V. Prove that the lines ST, PQ and UV are concurrent.

A4.  Find all real-valued functions on the reals which satisfy f( xf(x) + f(y) ) = f(x)² + y for all x, y. 

Solutions

14th Balkan 1997 Problem 1

ABCD is a convex quadrilateral. X is a point inside it. XA² + XB² + XC² + XD² is twice the area of the quadrilateral. Show that it is a square and that X is its center.

Solution

XA² + XB² ≥ 2·XA·XB with equality iff XA = XB. But XA·XB ≥ 2 area XAB with equality iff ∠AXB = 90^o. Hence (XA² + XB²) + (XB² + XC²) + (XC² + XD²) + (XD² + XA²) ≥ 4 (area XAB + area XBC + area XCD + area XDA) with equality iff ∠AXB = ∠BXC = ∠CXD = ∠DXA = 90^o and XA = XB = XC = XD. But we are given that equality holds. Hence AXC and BXD are straight line segments with X as their midpoint. Hence ABCD is a square center X.

14th Balkan 1997 Problem 2

A collection of m subsets of X = {1, 2, ... , n} has the property that given any two elements of X we can find a subset in the collection which contains just one of the two. Prove that n ≤ 2^m.

Solution

Let the subsets be Y₁, ... , Y_m. Define f: X → {0, 1, 2, ... , 2^m-1} by f(x) = the binary number a₁a₂ ... a_m where the binary digit a_i is 1 iff x belongs to Y_i. Then we claim that if x ≠ y we must have f(x) ≠ f(y). For if f(x) = f(y) then we could not find A_i such that just one of x, y was in A_i (since their binary expansions would agree in the ith digit). So f is injective. But its range has 2^m elements, so its domain cannot have more than 2^m elements. In other words n ≤ 2^m.

14th Balkan 1997 Problem 3

Two circles C and C' lying outside each other touch at T. They lie inside a third circle and touch it at X and X' respectively. Their common tangent at T intersects the third circle at S. SX meets C again at P and XX' meets C again at Q. SX' meets C' again at U and XX' meets C' again at V. Prove that the lines ST, PQ and UV are concurrent.

Solution

The similarity center X takes P to S and Q to X', so PQ is parallel to SX'. Similarly, UV is parallel to SX. Extend PQ and UV to meet at K. Then SUKP is a parallelogram so KU = SP and KP = SU. Also KQV and SX'X are similar triangles. So KQ/KV = SX'/SX.

Since ST is tangent to C and C', we have SP.SX = ST² = SU. SX'. Hence SX'/SX = SP/SU. Hence KQ/KV = SP/SU = KU/KP, so KP.KQ = KU.KV.

If K does not lie on ST, then ST must cut one of the segments KP and KU. Suppose it is KU. So let ST meet KU at A. Extend PK to meet ST at B. Then KQ.KP < BQ.BP = BT² < AT² = AT.AU < KV.KU. Contradiction. Similarly, we get a contradiction if ST cuts KP. Hence K must lie on ST. 

14th Balkan 1997 Problem 4

Find all real-valued functions on the reals which satisfy f( xf(x) + f(y) ) = f(x)² + y for all x, y.

Solution Answer: (1) f(x) = x for all x; (2) f(x) = -x for all x.

Put x = 0, then f(f(y)) = f(0)² + y. Put y = -f(0)² and k = f(y). Then f(k) = 0. Now put x = y = k. Then f(0) = 0 + k, so k = f(0). Put y = k, x = 0, then f(0) = f(0)² + k, so k = 0. Hence f(0) = 0.

Put x = 0, f(f(y)) = y (*). Put y = 0, f(xf(x)) = f(x)² (**). Put x = f(z) in (**), then using f(z) = x, we have f(zf(z)) = z². Hence z² = f(z)² for all z (***). In particular, f(1) = 1 or -1. Suppose f(1) = 1. Then putting x = 1 in the original relation we get f(1 + f(y) ) = 1 + y. Hence (1 + f(y) )² = (1 + y)². Hence f(y) = y for all y.

Similarly if f(1) = -1, then putting x = 1 in the original relation we get f(-1 + f(y) ) = 1 + y. Hence (-1 + f(y) )² = (1 + y)², so f(y) = -y for all y.

Finally, it is easy to check that f(x) = x does indeed satisfy the original relation, as does f(x) = -x.