12st IMO 1970

Solution　We need an expression for r/q. There are two expressions, one in terms of angles and the other in terms of sides. The latter is a poor choice, because it is both harder to derive and less useful. So we derive the angle expression.
Let O be the center of the in-circle for ABC and X the center of the external circle for ABC. O is the intersection of the two angle bisectors from A and B, so c = r(cot A/2 + cot B/2). The X lies on the bisector of the external angle, so angle XAB is 90 - A/2. Similarly, angle XBA is 90 - B/2, so c = q(tan A/2 + tan B/2). Hence r/q = (tan A/2 + tan B/2)/(cot A/2 + cot B/2) = tan A/2 tan B/2.
Applying this to the other two triangles, we get r₁/q₁ = tan A/2 tan CMA/2, r₂/q₂ = tan B/2 tan CMB/2. But CMB/2 = 90 - CMA/2, so tan CMB/2 = 1/tan CMA/2. Hence result.

Solution　We have aⁿb^m > bⁿa^m for n > m. Hence aⁿB‘ > bⁿA’. Adding aⁿbⁿ to both sides gives aⁿB > bⁿA. Hence x_naⁿB > x_nbⁿA. But x_naⁿ = A - A‘ and x_nbⁿ = B - B’, so (A - A‘)B > (B - B’)A. Hence result.
Note that the only purpose of requiring x_n-1 > 0 is to prevent A‘ and B’ being zero.

Solution　(a) Each term of the sum is non-negative, so b_n is non-negative. Let c_k = Öa_k. Then the kth term = (1 - a_k-1/a_k)/Öa_k = c_k-1²/c_k (1/a_k-1 - 1/a_k) = c_k-1²/c_k (1/c_k-1 + 1/c_k)(1/c_k-1 - 1/c_k). But c_k-1²/c_k (1/c_k-1 + 1/c_k) <= 2, so the kth term <= 2(1/c_k-1 - 1/c_k). Hence b_n <= 2 - 2/c_n < 2.
(b) Let c_k = d^k, where d is a constant > 1, which we will choose later. Then the kth term is (1 - 1/d²)1/d^k, so b_n = (1 - 1/d²)(1 - 1/dⁿ⁺¹)/(1 - 1/d) = (1 + 1/d)(1 - 1/dⁿ⁺¹). Now take d sufficiently close to 1 that 1 + 1/d > c, and then take n sufficiently large so that (1 + 1/d)(1 - 1/dⁿ⁺¹) > c.

Solution　The only primes dividing numbers in the set can be 2, 3 or 5, because if any larger prime was a factor, then it would only divide one number in the set and hence only one product. Three of the numbers must be odd. At most one of the odd numbers can be a multiple of 3 and at most one can be a multiple of 5. The other odd number cannot have any prime factors. Contradiction. So there are no such sets.

Solution　The first step is to show that angles ADB and ADC are also 90. Let H be the intersection of the altitudes of ABC and let CH meet AB at X. Planes CED and ABC are perpendicular and AB is perpendicular to the line of intersection CE. Hence AB is perpendicular to the plane CDE and hence to ED. So BD² = DE² + BE². Also CB² = CE² + BE². Subtracting: CB² - BD² = CE² - DE². But CB² - BD² = CD², so CE² = CD² + DE², so angle CDE = 90. But angle CDB = 90, so CD is perpendicular to the plane DAB, and hence angle CDA = 90. Similarly, angle ADB = 90.
Hence AB² + BC² + CA² = 2(DA² + DB² + DC²). But now we are done, because Cauchy‘s inequality gives (AB + BC + CA)² <= 3(AB² + BC² + CA²).
We have equality iff we have equality in Cauchy’s inequality, which means AB = BC = CA.

Solution At most 3 of the triangles formed by 4 points can be acute. It follows that at most 7 out of the 10 triangles formed by any 5 points can be acute. For given 10 points, the maximum no. of acute triangles is: the no. of subsets of 4 points x 3/the no. of subsets of 4 points containing 3 given points. The total no. of triangles is the same expression with the first 3 replaced by 4. Hence at most 3/4 of the 10, or 7.5, can be acute, and hence at most 7 can be acute.
The same argument now extends the result to 100 points. The maximum number of acute triangles formed by 100 points is: the no. of subsets of 5 points x 7/the no. of subsets of 5 points containing 3 given points. The total no. of triangles is the same expression with 7 replaced by 10. Hence at most 7/10 of the triangles are acute.