24th British Mathematical Olympiad 1988 Problems

24th British Mathematical Olympiad 1988 Problems

1.  ABC is an equilateral triangle. S is the circle diameter AB. P is a point on AC such that the circle center P radius PC touches S at T. Show that AP/AC = 4/5. Find AT/AC.

2.  Show that the number of ways of dividing {1, 2, ... , 2n} into n sets of 2 elements is 1.3.5 ... (2n-1). There are 5 married couples at a party. How many ways may the 10 people be divided into 5 pairs if no married couple may be paired together? For example, for 2 couples a, A, b, B the answer is 2: ab, AB; aB, bA.

3.  The real numbers a, b, c, x, y, z satisfy: x² - y² - z² = 2ayz, -x² + y² - z² = 2bzx, -x² - y² + z² = 2cxy, and xyz ≠ 0. Show that x²(1 - b²) = y²(1 - a²) = xy(ab - c) and hence find a² + b² + c² - 2abc (independently of x, y, z).

4.  Find all positive integer solutions to 1/a + 2/b - 3/c = 1.

5.  L and M are skew lines in space. A, B are points on L, M respectively such that AB is perpendicular to L and M. P, Q are variable points on L, M respectively such that PQ is of constant length. P does not coincide with A and Q does not coincide with B. Show that the center of the sphere through A, B, P, Q lies on a fixed circle whose center is the midpoint of AB.

6.  Show that if there are triangles with sides a, b, c, and A, B, C, then there is also a triangle with sides √(a² + A²), √(b² + B²), √(c² + C²).

Problem 1

ABC is an equilateral triangle. S is the circle diameter AB. P is a point on AC such that the circle center P radius PC touches S at T. Show that AP/AC = 4/5. Find AT/AC.

Answer

√(3/7)

Solution

 

Let O be the center of the circle. Let OA = R and PT = PC = r. Cosine rule applied to PAO gives PO² = PA² + AO² - 2PA·AO cos 60^o, or (R+r)² = (2R-r)² + R² - (2R-r)R. Hence r = 2R/5 and AP = 8R/5, so AP/AC = 4/5.

Applying cosine rule to APO and ATO, AP² = OP² + OA² - 2OPOA cos x, AT² = OT² + OA² - 2OTOA cos x, where x = ∠AOT. Hence AT² = (2 - 7/5 - 5/7 + 74/35)R² = 12R²/7. Hence AT/AC = √(3/7).

Problem 2

Show that the number of ways of dividing {1, 2, ... , 2n} into n sets of 2 elements is 1.3.5 ... (2n-1). There are 5 married couples at a party. How many ways may the 10 people be divided into 5 pairs if no married couple may be paired together? For example, for 2 couples a, A, b, B the answer is 2: ab, AB; aB, bA.

Answer

544

Solution

Any permutation gives a pairing. But 2ⁿn! permutations give the same pairing because the order of the two people in each of the n pairs is unimportant and the order of the n pairs is unimportant. Hence (2n)!/2ⁿn! = 1·3·5 ... (2n-1).

So there are 945 ways of pairing 10 people. If we pair a given married couple together, then there are 1·3·5·7 = 105 ways. If we pair two given married couples together, then there are 1·3·5 = 15 ways. If we pair three given married couples together, then there are 3 ways. If we pair 4 or 5 given couples together then there is just 1 way.

Hence by the principle of inclusion and exclusion, the number of ways or pairing so that no married couple is paired is 945 - 5·105 + 10·15 - 10·3 + 5·1 - 1 = 544.

 Problem 3

The real numbers a, b, c, x, y, z satisfy: x² - y² - z² = 2ayz, -x² + y² - z² = 2bzx, -x² - y² + z² = 2cxy, and xyz ≠ 0. Show that x²(1 - b²) = y²(1 - a²) = xy(ab - c) and hence find a² + b² + c² - 2abc (independently of x, y, z).

Answer

1

Solution

Adding the first two equations: -2z² = 2z(ay+bx). But z ≠ 0, so bx + ay + z = 0. Similarly, adding other pairs we get x + cy + bz = 0, cx + y + az = 0. Substituting z = -(bx+ay) in the second two equations gives x(1-b²) = y(ab-c) and y(1-a²) = x(ab-c). Hence x²(1-b²) = y²(1-a²) = xy(ab-c).

Multiplying x²(1-b²) = xy(ab-c) by y²(1-a²) = xy(ab-c) gives x²y²(1-a²)(1-b²) = x²y²(ab-c)². But x, y ≠ 0, so (1-a²)(1-b²) = (ab-c)². Expanding gives a²+b²+c²-2abc = 1.

Thanks to Suat Namli

Problem 4

Find all positive integer solutions to 1/a + 2/b - 3/c = 1.

Answer

(a,b,c) = (2,1,2), (2,3,18), (1,2n,3n), (n,2,3n)

Solution

If a ≥ 3, then 2/b = 1 + 3/c - 1/a > 2/3, so b < 3. So either a < 3 or b < 3. So we consider 4 cases: a = 1, a = 2, b = 1, b = 2.

If a = 1, then 2/b = 3/c, so 2c = 3b. Hence b must be even. Put b = 2n, then c = 3n, which gives a solution for any n.

If a = 2, then 2/b = 1/2 + 3/c > 1/2, so b ≤ 3. The cases b = 1, 2 we deal with below. If b = 3, then c = 18, which gives a solution.

If b = 1, then 3/c = 1 + 1/a > 1, so c = 1 or 2. c = 1 does not give a solution. c = 2 gives a = 2, which is a solution.

If b = 2, then 1/a = 3/c, so c = 3a. This gives a solution for any a.

Problem 5

L and M are skew lines in space. A, B are points on L, M respectively such that AB is perpendicular to L and M. P, Q are variable points on L, M respectively such that PQ is of constant length. P does not coincide with A and Q does not coincide with B. Show that the center of the sphere through A, B, P, Q lies on a fixed circle whose center is the midpoint of AB.

Solution

Let O be the midpoint of AB, and C be the center of the sphere. Then ∠COA = 90^o, so C lies in the plane p through O normal to AB. C also lies in the plane q normal to AP through its midpoint. Let P', Q' be the feet of the perpendiculars from P, Q to p. Then the plane q meets the plane p in the perpendicular bisector of OP'. So C must lie on this perpendicular bisector. Similarly, C must lie on the perpendicular bisector of OQ', so it must be the circumcenter of OP'Q'. But PP' = QQ' = OA, so PQ² = P'Q'² + AB², so P'Q' has the fixed length √(PQ²-AB²). The angle P'OQ' is also fixed, because if we take L' to be the line parallel to L through O and M' to be the line parallel to M through O, then L' and M' lie in the plane p, P lies on L', and Q lies on M'. Hence OC = P'Q'/(2 sin P'OQ'), which is fixed.

Problem 6

Show that if there are triangles with sides a, b, c, and A, B, C, then there is also a triangle with sides √(a² + A²), √(b² + B²), √(c² + C²).

Solution

We have to show that √(a²+A²) ≤ √(b²+B²) + √(c²+C²) (and two similar relations). Squaring, that is equivalent to a²+A² < b²+B²+c²+C² + 2√((b²+B²)(c²+C²)). By Cauchy-Scwartz we have bc + BC ≤ √((b²+B²)(c²+C²)), hence b²+B²+c²+C² + 2√((b²+B²)(c²+C²)) ≥ b²+B²+c²+C²+2bc+2BC = (b+c)² + (B+C)² > a² + A².