Home > Geometry, IMO, Inequalities, Olympiad > IMO 1995 Day 1

IMO 1995 Day 1

I decided to test myself on solving some IMO problems. Here are my solutions for IMO 1995 Day 1. Apart of the last problem which seemd a little more tricky, the first two were pretty straightforward.

Problem 1 Let { A,B,C,D} be four distinct points on a line, in that order. The circles with diameters { AC} and { BD} intersect at { X} and { Y}. The line { XY} meets { BC} at { Z}. Let { P} be a point on the line { XY} other than { Z}. The line { CP} intersects the circle with diameter { AC} at { C} and { M}, and the line { BP} intersects the circle with diameter { BD} at { B} and { N}. Prove that the lines { AM,DN,XY} are concurrent.

Solution: Denote by {C_{AC},C_{BD},C_{AP},C_{DP}} the circles with respective diameters. It is immediate that {XY} is the radical axis of {C_{AC}} and {C_{BD}} and of {C_{AP},C_{PD}}. Moreover, {AM} is the radical axis of the circles {C_{AC}} and {C_{AP}} and {DN} is the radical axis of the circles {C_{BD}} and {C_{PD}}. The radical axis of two circles is characterized by the fact it is the locus of the points which have the same power with respect to the two circles.

Denote by {K} the intersection of {AM} and {XY}. By the above facts we have that {K} has the same power with respect to the circles {C_{AC},C_{BD},C_{AP},C_{DP}}, and as a consequence, it is on the radical axis of the circles {C_{BD}} and {C_{PD}} which is {DN}. This proves that the three lines are concurrent.

Problem 2. Let { a}, { b}, { c} be positive real numbers such that { abc = 1}. Prove that

\displaystyle \frac {1}{a^{3}\left(b + c\right)} + \frac {1}{b^{3}\left(c + a\right)} + \frac {1}{c^{3}\left(a + b\right)}\geq \frac {3}{2}.

Solution: Since {abc=1} the inequality is equivalent to

\displaystyle \sum_{cyc} \frac{(bc)^2}{ab+ac} \geq \frac{3}{2}.

Applying Cauchy-Schwarz and AM-GM we get

\displaystyle \sum_{cyc} \frac{(bc)^2}{ab+ac} \geq \frac{(\sum_{cyc} bc)^2}{2\sum_{cyc}bc}=\frac{\sum_{cyc} bc}{2} \geq \frac{3}{2}.

Problem 3. Determine all integers { n > 3} for which there exist { n} points { A_{1},\cdots ,A_{n}} in the plane, no three collinear, and real numbers { r_{1},\cdots ,r_{n}} such that for { 1\leq i < j < k\leq n}, the area of { \Delta A_{i}A_{j}A_{k}} is { r_{i} + r_{j} + r_{k}}.

Solution: We note right away that if {A_i,A_j,A_k,A_l} (in this order) correspond to vertices of a convex quadrilateral then

\displaystyle r_i+r_k=r_l+r_j \ \ (\star)

For {n=4} the answer is of course yes. Take a square and four equal numbers {r_i} equal to half the area of the square divided by {6}.

First note that if the statement of the problem is not true for {n=5}, then it is not true for all {n \geq 5}. For {n= 5} my intuition said that we cannot find such points in the plane, and a well known problem came to mind:

If we have 5 points in the plane such that no three of them lie on the same line then there are four of them which form a convex quadrilateral.

The solution to this problem auxiliary problem led to the solution of the third problem in the following way: the convex hull of {5} non-collinear points is a triangle, a convex quadrilateral or a convex pentagon.

  • Case 1 The convex hull of the {5} points is a convex pentagon. Then using relation {(\star)} it is easy to see that the numbers {r_i} must be equal. But then the triangles {\Delta A_1A_2A_3, \Delta A_1A_2A_4,\Delta A_1A_2A_5} have the same area and {A_3,A_4,A_5} are collinear. Contradiction.
  • Case 2 The convex hull is a convex quadrilateral {A_1A_2A_3A_4}. Without loss of generality suppose that {A_5 \in \Delta A_2A_3A_4}. Using the same relation {(\star)} we find that {r_5=r_3} and therefore the areas of the triangles {\Delta A_2A_3A_4} and {\Delta A_2A_5A_4} are the same, which is impossible, since one of them is strictly contain in the other one.
  • Case 3 The convex hull is a triangle {A_1A_2A_3} and {A_4,A_5} are in the interior of this triangle. Without loss of generality suppose that the line {A_4A_5} intersects the lines {A_1A_2} and {A_1A_3} and {A_4} is closer to {A_1A_2}. Since we have

    \displaystyle Area(A_1A_2A_3) > Area(A_1A_2A_4)+Area(A_1A_3A_5)

    by the same relation {(\star)} we get

    \displaystyle r_1+r_2+r_3 > r_1+r_2+r_4+r_1+r_3+r_5,

    which is equivalent to

    \displaystyle 0 > r_1+r_4+r_5=Area(A_1A_4A_5)

    which is again a contradiction.

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