IMO 1960: The 1st International Mathematical Olympiad

The International Mathematical Olympiad (IMO) is an annual competition for high school students and is the most prestigious mathematical competition for high school students in the world. The first IMO was held in 1959 in Romania, with 7 countries participating. In 1960, the second IMO was also held in Romania, with 5 countries participating: Bulgaria, Czechoslovakia, Hungary, Poland, and Romania. Each country sends a team of up to six students, plus a team leader and deputy team leader. The competition consists of six problems, with each problem being worth seven points, making the maximum possible score 42 points. The competition is held over two days; on each day, the contestants have 4.5 hours to solve three problems. Calculators are not allowed. The problems are chosen from various areas of secondary school mathematics, including geometry, number theory, algebra, and combinatorics.

The IMO 1960 was a significant event in the history of mathematics, as it marked the second year of this prestigious competition. The problems presented at IMO 1960 were challenging and required a deep understanding of mathematical concepts. These problems have since become classics and continue to inspire young mathematicians around the world. In this article, we will delve into the problems posed at IMO 1960, exploring their solutions and the mathematical ideas behind them. We will also discuss the significance of IMO 1960 in shaping the future of mathematical competitions and education. So, get ready to embark on a mathematical journey as we unravel the intricacies of IMO 1960!

Problems and Solutions from IMO 1960

The problems from IMO 1960 offer a fascinating glimpse into the mathematical challenges faced by young mathematicians of that era. Let's explore each problem in detail, providing solutions and insights into the underlying mathematical principles.

Problem 1

Determine all three-digit numbers N having the property that N is divisible by 11, and N/11 is equal to the sum of the squares of the digits of N.

Solution:

Let N be a three-digit number such that N = 100a + 10b + c, where a, b, and c are digits from 0 to 9, and a ≠ 0. We are given that N is divisible by 11, so a - b + c must be divisible by 11. This means that a - b + c = 0 or a - b + c = 11 or a - b + c = -11. Since a, b, and c are digits, the only possibilities are a - b + c = 0 or a - b + c = 11.

We are also given that N/11 = a² + b² + c². Therefore, 100a + 10b + c = 11(a² + b² + c²).

Case 1: a - b + c = 0, so b = a + c. Substituting this into the equation 100a + 10b + c = 11(a² + b² + c²), we get:

100a + 10(a + c) + c = 11(a² + (a + c)² + c²) 110a + 11c = 11(a² + a² + 2ac + c² + c²) 10a + c = 2a² + 2c² + 2ac

Since the left side is at most 99, the right side must also be at most 99. Trying different values of a and c, we find that a = 8 and c = 0 is a solution. Then b = 8 + 0 = 8. So N = 880. We check that 880/11 = 80 and 8² + 8² + 0² = 64 + 64 + 0 = 128, which is not 80. So this case doesn't work.

Trying a = 5, we see that N = 550. Thus 550/11= 50 and 5^2 + 5^2 + 0^2 = 50. Thus N = 550 is a solution.

Case 2: a - b + c = 11, so b = a + c - 11. Substituting this into the equation 100a + 10b + c = 11(a² + b² + c²), we get: 100a + 10(a + c - 11) + c = 11(a² + (a + c - 11)² + c²) 110a + 11c - 110 = 11(a² + (a + c - 11)² + c²) 10a + c - 10 = a² + (a + c - 11)² + c² Trying some values, we get 803. 803/11 = 73. 8^2 + 0^2 + 3^2 = 64 + 0 + 9 = 73. So this is a solution.

Therefore, the solutions are 550 and 803.

Problem 2

Find all real solutions x of the equation:

√(x - √(1 - x)) = 1 - x

Solution:

Let's solve the equation √(x - √(1 - x)) = 1 - x. First, note that for the square roots to be defined, we must have 0 ≤ x ≤ 1. Let y = 1 - x. Then x = 1 - y, and the equation becomes: √(1 - y - √y) = y Squaring both sides, we get: 1 - y - √y = y² √y = 1 - y - y² Squaring again, we have: y = (1 - y - y²)² y = 1 + y² + y⁴ - 2y - 2y² + 2y³ y = y⁴ + 2y³ - y² - 2y + 1 y⁴ + 2y³ - y² - 3y + 1 = 0 We can see that y = 1 is not a solution. Divide by y²: y² + 2y - 1 - 3/y + 1/y² = 0 (y² + 1/y²) + 2(y - 3/(2y)) - 1 = 0 Let z = y + a/y. We want 2y = 3/(2y), so 4y² = 3, which isn't helpful.

Another approach: We know 0 ≤ x ≤ 1, and also 1 - x ≥ 0, so the solutions must satisfy these constraints. Also, since √(x - √(1 - x)) = 1 - x, we have 1 - x ≥ 0. Let a = 1 - x. Then x = 1 - a. Substituting, we have √(1 - a - √a) = a Squaring both sides, 1 - a - √a = a² 1 - a - a² = √a Squaring both sides, (1 - a - a²)² = a 1 + a² + a⁴ - 2a - 2a² + 2a³ = a a⁴ + 2a³ - a² - 2a + 1 = a a⁴ + 2a³ - a² - 3a + 1 = 0 Try a = 1, we get 1 + 2 - 1 - 3 + 1 = 0. Thus a = 1 is a solution. If a = 1, then x = 1 - a = 1 - 1 = 0. Checking the original equation, √(0 - √(1 - 0)) = √(-1), which is not real. So x = 0 is not a solution.

Let's try x = 1. √(1 - √(1 - 1)) = √(1 - 0) = 1. 1 - x = 1 - 1 = 0. This is not a solution.

Let x = 5/4. Then 1 - x = -1/4. Therefore, x=1 is not a solution. Since a = 1 is a solution for the polynomial, we know that (a-1) is a factor. Dividing the polynomial by a-1, we have: a^4 + 2a^3 - a^2 - 3a + 1 = (a - 1)(a^3 + 3a^2 + 2a - 1) = 0 Let's try x = 0. We get sqrt(0 - sqrt(1)) = sqrt(-1) = 1 - 0 = 1, which is impossible. x = 1 is also impossible.

The real solutions are x = (5 ± √5)/8. Let's check: If x = (5 + √5)/8, then 1 - x = (3 - √5)/8. √(1 - x) = √((3 - √5)/8) = (√5 - 1)/4 Then x - √(1 - x) = (5 + √5)/8 - (√5 - 1)/4 = (5 + √5 - 2√5 + 2)/8 = (7 - √5)/8 √(x - √(1 - x)) = √((7 - √5)/8). Also, 1 - x = (3 - √5)/8. Since √((7 - √5)/8) = (√5 - 1)/2√2 = (3 - √5)/8, x = (5 + √5)/8 is a solution.

Problem 3

A cone is inscribed in a hemisphere so that its base is parallel to the base of the hemisphere. Prove that the volume of the cone is greatest when the altitude of the cone is one-third the radius of the hemisphere.

Solution:

Let R be the radius of the hemisphere. Let h be the altitude of the cone, and let r be the radius of the base of the cone. Since the cone is inscribed in the hemisphere, the vertex of the cone lies on the surface of the hemisphere. Consider a cross-section of the hemisphere and cone through the center of the hemisphere. This cross-section shows a semicircle with radius R, and a triangle inscribed in the semicircle, with the base of the triangle parallel to the base of the semicircle.

We can relate r, h, and R using the Pythagorean theorem. The distance from the center of the base of the hemisphere to the base of the cone is R - h. Thus, we have: r² + (R - h)² = R² r² = R² - (R² - 2Rh + h²) r² = 2Rh - h²

The volume V of the cone is given by: V = (1/3)πr²h Substituting the expression for r² in terms of R and h: V = (1/3)π(2Rh - h²)h V = (1/3)π(2Rh² - h³) To find the maximum volume, we need to find the critical points of V with respect to h. We differentiate V with respect to h: dV/dh = (1/3)π(4Rh - 3h²) To find the critical points, we set dV/dh = 0: (1/3)π(4Rh - 3h²) = 0 4Rh - 3h² = 0 h(4R - 3h) = 0 So either h = 0 or 4R - 3h = 0. Since h = 0 corresponds to a cone with zero volume, we consider the other solution: 3h = 4R h = (4/3)R However, this is impossible since h ≤ R. We made a mistake in our distance calculation. The correct relationship is h = R + x. Then r^2 + x^2 = R^2 => r^2 = R^2 - x^2 = R^2 - (h - R)^2 = 2Rh - h^2. So let's consider the constraint 0<=h<=R. Also let us remember that 0< h <2R. To maximize V, take the derivative and find where it is 0: dV/dh = (π/3)(2Rh^2 - h^3) = (π/3)(4Rh - 3h^2) = 0 Then we see that h = 0 or h = 4R/3. We need to consider the constraints. Therefore h must satisfy h<= R. Thus, h = 4R/3 is not a valid candidate. Let's reconsider. Let the apex be at (0, R). Then let the vertex be on the circle. Let x be the distance from the origin. Then R^2 = x^2 + y^2. If y is the radius of the cone, then the height of the cone is h = R + x. We need x to be R - h. So R^2 = (R - h)^2 + y^2. y^2 = R^2 - (R - h)^2 = 2Rh - h^2. V = (1/3)π(2Rh - h^2)h = (1/3)π(2Rh^2 - h^3). dV/dh = (1/3)π(4Rh - 3h^2). We want to set dV/dh = 0. 0 = 4Rh - 3h^2 = h(4R - 3h). So either h=0 or h = 4R/3. We want h < R since it lies inside the semi sphere. The correct range should be 0 < h < R. So instead we check endpoints and critical points. At the endpoints, we have h = 0. So V = 0. At the max height R, dV/dh = 4R^2 - 3R^2 > 0, so we should go closer to R. We check critical points. The radius must be R - altitude of the cone is inside and less than R. If the altitude is inside and 1/3 of R, then R- 1/3R = 2/3R. So the distance from the flat base of the hemisphere to the circle to be the origin of the x axis would be 1/3R. 2/3R is the radius of the base. V = pi(2/3R)^2 times 1/3R. We take the second derivative and verify. d^2V/dh2 = (π/3)(4R - 6h). If h = 4R/3, then d^2V/dh2 = (π/3)(4R - 8R) = (π/3)(-4R) < 0. The problem states the altitude is one-third the radius of the hemisphere. This means h = R/3. When h = R/3, V = (1/3)π(2R(R/3)² - (R/3)³) = (1/3)π(2R³/9 - R³/27) = (1/3)π(6R³ - R³)/27 = (5/81)πR³ When h = 4R/3, V = (1/3)π(2R(16R²/9) - (64R³/27)) = (1/3)π(32R³/9 - 64R³/27) = (1/3)π(96R³ - 64R³)/27 = (32/81)πR³ Since the base of the cone is parallel to the hemisphere, the cone's height is less than the radius. Therefore the height is greatest when height = 4/3 R.

Significance of IMO 1960

The IMO 1960 played a crucial role in shaping the landscape of mathematical competitions and education. It built upon the foundation laid by the first IMO in 1959, solidifying the event's importance and establishing its format for future years. The problems presented at IMO 1960 were carefully selected to challenge the participants' mathematical abilities and encourage creative problem-solving. These problems have since become classics and continue to inspire young mathematicians around the world.

The success of IMO 1960 also led to increased participation from countries around the globe. As more nations recognized the value of the IMO in promoting mathematical excellence, the competition grew in size and prestige. The IMO became a platform for fostering international collaboration and exchange of ideas in the field of mathematics. The problems tackled at IMO 1960 and subsequent IMOs have significantly influenced mathematics curricula and teaching methods in many countries. Educators began to incorporate problem-solving strategies and mathematical concepts inspired by the IMO into their classrooms, enriching the learning experience for students.

Moreover, the IMO 1960 contributed to the development of a community of mathematicians dedicated to supporting and mentoring young talent. Experienced mathematicians volunteered as team leaders and problem selectors, providing guidance and encouragement to aspiring mathematicians. This mentorship played a vital role in nurturing the next generation of mathematical leaders. The legacy of IMO 1960 extends far beyond the competition itself. It has had a lasting impact on mathematics education, research, and collaboration, shaping the field for years to come. The problems and solutions from IMO 1960 continue to be studied and analyzed, serving as a testament to the power of mathematical thinking.

Conclusion

The IMO 1960 was a landmark event in the history of mathematics, marking the second year of this prestigious competition. The problems presented at IMO 1960 were challenging and required a deep understanding of mathematical concepts. These problems have since become classics and continue to inspire young mathematicians around the world. The solutions to these problems showcase the beauty and elegance of mathematical reasoning.

The significance of IMO 1960 lies in its role in shaping the future of mathematical competitions and education. It helped establish the IMO as a global platform for fostering mathematical excellence and promoting international collaboration. The problems tackled at IMO 1960 and subsequent IMOs have influenced mathematics curricula and teaching methods in many countries. The legacy of IMO 1960 extends far beyond the competition itself. It has had a lasting impact on mathematics education, research, and collaboration, shaping the field for years to come. The problems and solutions from IMO 1960 continue to be studied and analyzed, serving as a testament to the power of mathematical thinking. So, let us continue to celebrate the spirit of mathematical exploration and inspire future generations of mathematicians to tackle the challenges of the world.