7th Eötvös Competition Problems 1990

1.  d is not divisible by 5. For some integer n, a n³ + b n² + c n + d is divisible by 5. Show that for some integer m, a + b m + c m² + d m³ is divisible by 5.

2.  Construct the triangle ABC given c, r and r', where c = |AB|, r is the radius of the inscribed circle, and r' is the radius of the other circle tangent to the segment AB and the lines BC and CA.

3.  Two particles fall from rest 300 m under the influence of gravity alone. One particle leaves when the other has already fallen 1 μm. How far apart are they when the first particle reaches the end point (to the nearest 100 μm)?

Solutions

Problem 1

d is not divisible by 5. For some integer n, a n³ + b n² + c n + d is divisible by 5. Show that for some integer m, a + b m + c m² + d m³ is divisible by 5.

Solution

Evidently we need only consider n mod 5. Since d ≠ 0 mod 5, we must have n ≠ 0 mod 5. If n = 1 mod 5, then a+bn+cn²+dn³ = an³+bn²+cn+d mod 5, so can take m = 1. If n = 4 mod 5, then an³+bn²+cn+d = 4a+b+4c+d mod 5. If 4a+b+4c+d = 0 mod 5, then a+4b+c+4d = 0 mod 5, and since a+4b+4²c+4³d = a+4b+c+4d, we can take m = 4.

If n = 2 mod 5, then an³+bn²+cn+d = 3a+4b+2c+d mod 5. So if this is 0 mod 5, then so is 2(3a+4b+2c+d) = a+3b+4c+2d = a+3b+3²c+3³d mod 5, so we can take m = 3.

If n = 3 mod 5, then an³+bn²+cn+d = 2a+4b+3c+d mod 5. So if this is 0 mod 5, then so is 3(2a+4b+3c+d) = a+2b+4c+3d = a+2b+2²c+2³d mod 5, so we can take m = 2.

Problem 2

Construct the triangle ABC given c, r and r', where c = |AB|, r is the radius of the inscribed circle, and r' is the radius of the other circle tangent to the segment AB and the lines BC and CA.

Solution

Put a = BC, b = CA, s = (a+b+c)/2 as usual. We have AX = s-a, AY = s-b (just chase around using the fact that the two tangents from a given point to a circle are equal). So XY = c. Thus mark the points X and Y a distance c apart. Then draw the two circles. Then draw the other common tangents, thus getting A, B and C.

Problem 3

Two particles fall from rest 300 m under the influence of gravity alone. One particle leaves when the other has already fallen 1 μm. How far apart are they when the first particle reaches the end point (to the nearest 100 μm)?

Solution

Use s = ½gt². So fall time for lagging particle is √(600/g) - √(2·10^-6g), giving distance 300 + 10^-6 - √(12·10^-4). So the distance apart is 20√3 mm = 34.6 mm.