Math Olympiad Problems for Grades 11 and 12: Very Difficult Problems with Solutions

Solution:
Use the algebraic identity for the sum of cubes: [tex]a^3 + b^3 = (a + b)(a^2 - ab + b^2)[/tex].
Let [tex]a = \sin^2(15^\circ)[/tex] and [tex]b = \cos^2(15^\circ)[/tex].
[tex]E = (\sin^2(15^\circ) + \cos^2(15^\circ))(\sin^4(15^\circ) - \sin^2(15^\circ)\cos^2(15^\circ) + \cos^4(15^\circ))[/tex].
Since [tex]\sin^2(x) + \cos^2(x) = 1[/tex], the first term is [tex]1[/tex].
We know [tex]\sin^4(x) + \cos^4(x) = (\sin^2(x) + \cos^2(x))^2 - 2\sin^2(x)\cos^2(x) = 1 - 2\sin^2(x)\cos^2(x)[/tex].
Substituting this: [tex]E = 1 - 3\sin^2(15^\circ)\cos^2(15^\circ)[/tex].
Using the double angle identity [tex]\sin(2x) = 2\sin(x)\cos(x)[/tex], we have [tex]\sin^2(x)\cos^2(x) = \frac{1}{4}\sin^2(2x)[/tex].
Therefore, [tex]E = 1 - \frac{3}{4}\sin^2(30^\circ)[/tex].
Since [tex]\sin(30^\circ) = \frac{1}{2}[/tex], [tex]\sin^2(30^\circ) = \frac{1}{4}[/tex].
[tex]E = 1 - \frac{3}{4}\left(\frac{1}{4}\right) = 1 - \frac{3}{16} = \frac{13}{16}[/tex].

Solution:
Let [tex]Q(x) = P(x) - x[/tex].
Since [tex]P(x)[/tex] is a cubic polynomial, [tex]Q(x)[/tex] is also a cubic polynomial.
Based on the data, [tex]Q(1) = 0[/tex], [tex]Q(2) = 0[/tex], and [tex]Q(3) = 0[/tex].
This means [tex]1[/tex], [tex]2[/tex], and [tex]3[/tex] are the roots of [tex]Q(x)[/tex].
Thus, [tex]Q(x) = k(x - 1)(x - 2)(x - 3)[/tex].
Then [tex]P(x) = k(x - 1)(x - 2)(x - 3) + x[/tex].
Use the fourth data point [tex]P(4) = 5[/tex]: [tex]5 = k(4 - 1)(4 - 2)(4 - 3) + 4[/tex]
[tex]\Rightarrow 5 = k(3)(2)(1) + 4[/tex]
[tex]\Rightarrow 1 = 6k[/tex]
[tex]\Rightarrow k = \frac{1}{6}[/tex].
The full polynomial is [tex]P(x) = \frac{1}{6}(x - 1)(x - 2)(x - 3) + x[/tex].
To find [tex]P(5)[/tex]: [tex]P(5) = \frac{1}{6}(5 - 1)(5 - 2)(5 - 3) + 5 = \frac{1}{6}(4)(3)(2) + 5 = 4 + 5 = 9[/tex].

Solution:
Subtract the first equation from the second: [tex](x + yz) - (xy + z) = 161 - 160 \Rightarrow x - xy + yz - z = 1[/tex].
Factor by grouping: [tex]x(1 - y) - z(1 - y) = 1 \Rightarrow (x - z)(1 - y) = 1[/tex].
Since [tex]x, y, z[/tex] are integers, their differences are integers.
The only factors of [tex]1[/tex] are [tex](1)(1)[/tex] and [tex](-1)(-1)[/tex].
Case 1: [tex]1 - y = 1[/tex] gives [tex]y = 0[/tex], which is invalid because variables are strictly positive integers.
Case 2: [tex]1 - y = -1 \Rightarrow y = 2[/tex].
And [tex]x - z = -1 \Rightarrow z = x + 1[/tex].
Substitute [tex]y = 2[/tex] and [tex]z = x + 1[/tex] into the first equation: [tex]x(2) + (x + 1) = 160[/tex]
[tex]\Rightarrow 3x + 1 = 160[/tex]
[tex]\Rightarrow 3x = 159[/tex]
[tex]\Rightarrow x = 53[/tex].
Then [tex]z = 54[/tex].
The total weight sum is [tex]x + y + z = 53 + 2 + 54 = 109[/tex].

Solution:
Since [tex]A, B, P[/tex], and [tex]C[/tex] lie on a circle, [tex]ABPC[/tex] is a cyclic quadrilateral.
[tex]\Delta ABC[/tex] is equilateral, so [tex]\angle BAC = 60^\circ[/tex] and [tex]BC = 6[/tex].
By cyclic quadrilateral properties, opposite angles sum to [tex]180^\circ[/tex].
Thus, [tex]\angle BPC = 180^\circ - \angle BAC = 180^\circ - 60^\circ = 120^\circ[/tex].
Use the Law of Cosines on [tex]\Delta PBC[/tex]: [tex]BC^2 = PB^2 + PC^2 - 2(PB)(PC)\cos(120^\circ)[/tex].
Substitute knowns: [tex]6^2 = 2^2 + PC^2 - 2(2)(PC)(-0.5)[/tex]
[tex]\Rightarrow 36 = 4 + PC^2 + 2PC[/tex]
[tex]\Rightarrow PC^2 + 2PC - 32 = 0[/tex].
Use quadratic formula: [tex]PC = \frac{-2 \pm \sqrt{4 - 4(1)(-32)}}{2} = \frac{-2 \pm \sqrt{132}}{2} = \frac{-2 \pm 2\sqrt{33}}{2} = -1 \pm \sqrt{33}[/tex].
Since distance must be positive, [tex]PC = \sqrt{33} - 1[/tex] meters.

Solution:
This problem requires choosing specific items and finding the ”derangement” (a permutation where no element appears in its original position) for the remaining items.
Step 1: Choose the [tex]2[/tex] astronauts who get correct helmets.
Ways [tex]= \binom{5}{2} = 10[/tex].
Step 2: Calculate derangements for the remaining [tex]3[/tex] astronauts.
The formula is [tex]!n = n!\sum_{i=0}^{n}\frac{(-1)^i}{i!}[/tex].
For [tex]n = 3[/tex], [tex]!3 = 3!(\frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!}) = 6(1 - 1 + \frac{1}{2} - \frac{1}{6}) = 6(\frac{1}{3}) = 2[/tex].
So there are [tex]2[/tex] ways for the remaining [tex]3[/tex] to all get wrong helmets.
Total favorable outcomes [tex]= 10 \times 2 = 20[/tex].
Total possible outcomes for [tex]5[/tex] helmets [tex]= 5! = 120[/tex].
Probability [tex]= \frac{20}{120} = \frac{1}{6}[/tex].