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

Solution:
The area of the circle is [tex]\pi R^2 = 144\pi[/tex].
The side length [tex]s[/tex] of an equilateral triangle inscribed in a circle of radius [tex]R[/tex] is [tex]s = R\sqrt{3} = 12\sqrt{3}[/tex].
The area of the equilateral triangle is [tex]\frac{\sqrt{3}}{4}s^2 = \frac{\sqrt{3}}{4}(12\sqrt{3})^2 = \frac{\sqrt{3}}{4}(432) = 108\sqrt{3}[/tex].
The shaded area is the area of the circle minus the area of the triangle: [tex]144\pi - 108\sqrt{3} \text{ cm}^2[/tex].

Solution:
Vieta’s formulas state that for [tex]ax^2 + bx + c = 0[/tex], the sum of roots is [tex]-\frac{b}{a} = 8[/tex] and the product of roots is [tex]\frac{c}{a} = 12[/tex].
The function can be written as [tex]T(x) = a(x^2 - (\text{sum})x + (\text{product})) = a(x^2 - 8x + 12)[/tex].
Using the condition [tex]T(3) = -15[/tex]: [tex]-15 = a(3^2 - 8(3) + 12) = a(9 - 24 + 12) = -3a \Rightarrow a = 5[/tex].
The calibrated function is [tex]T(x) = 5(x^2 - 8x + 12)[/tex].
Substituting [tex]x = 7[/tex]: [tex]T(7) = 5(7^2 - 8(7) + 12) = 5(49 - 56 + 12) = 5(5) = 25[/tex].

Solution:
The word ”ASTEROID” has [tex]8[/tex] distinct letters: [tex]4[/tex] vowels (A, E, O, I) and [tex]4[/tex] consonants (S, T, R, D).
The total number of ways to choose [tex]4[/tex] letters out of [tex]8[/tex] is [tex]\binom{8}{4} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70[/tex].
The number of ways to choose exactly [tex]2[/tex] vowels out of [tex]4[/tex] is [tex]\binom{4}{2} = 6[/tex].
The number of ways to choose exactly [tex]2[/tex] consonants out of [tex]4[/tex] is [tex]\binom{4}{2} = 6[/tex].
The total favorable combinations are [tex]6 \times 6 = 36[/tex].
The probability is [tex]\frac{36}{70} = \frac{18}{35}[/tex].

Solution:
First, find the cosine of angle [tex]B[/tex] using the Law of Cosines: [tex]b^2 = a^2 + c^2 - 2ac\cos(B)[/tex], where [tex]b = 9[/tex], [tex]a = 8[/tex], and [tex]c = 7[/tex].
[tex]9^2 = 8^2 + 7^2 - 2(8)(7)\cos(B)[/tex]
[tex]\Rightarrow 81 = 64 + 49 - 112\cos(B)[/tex]
[tex]\Rightarrow 81 = 113 - 112\cos(B)[/tex]
[tex]\Rightarrow 112\cos(B) = 32[/tex]
[tex]\Rightarrow \cos(B) = \frac{32}{112} = \frac{2}{7}[/tex].
Using the identity [tex]\sin^2(B) + \cos^2(B) = 1[/tex]: [tex]\sin^2(B) + \left(\frac{2}{7}\right)^2 = 1 \Rightarrow \sin^2(B) = 1 - \frac{4}{49} = \frac{45}{49}[/tex].
Since angle [tex]B[/tex] is in a triangle, [tex]\sin(B) > 0[/tex], so [tex]\sin(B) = \sqrt{\frac{45}{49}} = \frac{3\sqrt{5}}{7}[/tex].

Solution:
Using the power rule of logarithms [tex]\log_b(m^n) = n\log_b(m)[/tex], rewrite the series: [tex]S = \log_2(x) + \frac{1}{2}\log_2(x) + \frac{1}{4}\log_2(x) + \frac{1}{8}\log_2(x) + \ldots[/tex] Factor out [tex]\log_2(x)[/tex]: [tex]S = \log_2(x)\left[1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots\right][/tex].
The expression inside the brackets is an infinite geometric series with [tex]a = 1[/tex] and [tex]r = \frac{1}{2}[/tex].
The sum is [tex]\frac{a}{1 - r} = \frac{1}{1 - \frac{1}{2}} = 2[/tex].
The equation simplifies to [tex]S = 2\log_2(x)[/tex].
Given [tex]S = 10[/tex]: [tex]10 = 2\log_2(x) \Rightarrow \log_2(x) = 5[/tex].
In exponential form: [tex]x = 2^5 = 32[/tex].

Solution:
The total number of ways to arrange [tex]6[/tex] items where [tex]3[/tex] are of one type and [tex]3[/tex] are of another is [tex]\frac{6!}{3!3!} = 20[/tex].
To ensure no two Alpha droids are adjacent, first place the [tex]3[/tex] Beta droids.
Since they are identical, there is only [tex]1[/tex] way to arrange them: B B B.
These [tex]3[/tex] Beta droids create [tex]4[/tex] potential spaces for an Alpha droid (one at the front, two between the Betas, one at the end): _ B _ B _ B _.
We must choose [tex]3[/tex] of these [tex]4[/tex] spaces for the Alphas.
The number of ways is [tex]\binom{4}{3} = 4[/tex].
The probability is [tex]\frac{4}{20} = \frac{1}{5}[/tex].

Solution:
The shortest distance from a point [tex](x_1, y_1)[/tex] to a line [tex]Ax + By + C = 0[/tex] is [tex]d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}[/tex].
The point is the origin [tex](0, 0)[/tex], and the line is [tex]3x - 4y + 25 = 0[/tex].
Substitute the values: [tex]d = \frac{|3(0) - 4(0) + 25|}{\sqrt{3^2 + (-4)^2}} = \frac{25}{\sqrt{9 + 16}} = \frac{25}{5} = 5[/tex].
The minimum range needed is [tex]5[/tex] units.

Solution:
Area of the large circle is [tex]\pi R^2 = 225\pi[/tex].
The radius [tex]r[/tex] of the inner inscribed circle is the apothem of the regular octagon.
The apothem of a regular [tex]n[/tex]-gon inscribed in a circle of radius [tex]R[/tex] is [tex]a = R\cos(\frac{180^\circ}{n})[/tex].
For an octagon ([tex]n = 8[/tex]), [tex]r = R\cos(22.5^\circ)[/tex].
The shaded region is the annulus between the circles: [tex]\pi R^2 - \pi r^2 = \pi R^2 - \pi(R\cos(22.5^\circ))^2 = \pi R^2(1 - \cos^2(22.5^\circ)) = \pi R^2\sin^2(22.5^\circ)[/tex].
Using the half-angle formula, [tex]\sin^2(22.5^\circ) = \frac{1 - \cos(45^\circ)}{2} = \frac{1 - \frac{\sqrt{2}}{2}}{2} = \frac{2 - \sqrt{2}}{4}[/tex].
The exact shaded area is [tex]225\pi\left(\frac{2 - \sqrt{2}}{4}\right) = \frac{225\pi}{4}(2 - \sqrt{2}) \text{ cm}^2[/tex].

Solution:
Consider [tex]\Delta AMN[/tex] and [tex]\Delta ABC[/tex].
They share the angle at vertex [tex]A[/tex] ([tex]\angle MAN = \angle CAB[/tex]).
We are given that [tex]\angle AMN = \angle ABC[/tex].
Because two corresponding angles are equal, [tex]\Delta AMN \sim \Delta ABC[/tex] (AA similarity).
Note the correspondence of vertices: [tex]A \leftrightarrow A[/tex], [tex]M \leftrightarrow B[/tex], [tex]N \leftrightarrow C[/tex].
The ratio of corresponding sides is [tex]\frac{AM}{AB} = \frac{AN}{AC}[/tex].
We know [tex]AM = 4[/tex], [tex]AN = 5[/tex], [tex]AC = AM + MC = 4 + 8 = 12[/tex], and [tex]AB = AN + NB = 5 + NB[/tex].
Substitute the values: [tex]\frac{4}{5 + NB} = \frac{5}{12}[/tex]
[tex]\Rightarrow 5 + NB = \frac{48}{5}[/tex]
[tex]\Rightarrow 5 + NB = 9.6[/tex]
[tex]\Rightarrow NB = 4.6[/tex] meters.

Solution:
Since [tex]P(x - 1) + P(x + 1)[/tex] is a quadratic, [tex]P(x)[/tex] must be quadratic.
Let [tex]P(x) = ax^2 + bx + c[/tex].
[tex]P(x - 1) = a(x - 1)^2 + b(x - 1) + c = ax^2 - 2ax + a + bx - b + c[/tex]
[tex]P(x + 1) = a(x + 1)^2 + b(x + 1) + c = ax^2 + 2ax + a + bx + b + c[/tex]
Adding them together: [tex]P(x - 1) + P(x + 1) = 2ax^2 + 2bx + (2a + 2c)[/tex].
Equate this to [tex]2x^2 - 4x + 10[/tex]: [tex]2a = 2 \Rightarrow a = 1[/tex].
[tex]2b = -4 \Rightarrow b = -2[/tex].
[tex]2a + 2c = 10 \Rightarrow 2(1) + 2c = 10 \Rightarrow c = 4[/tex].
Thus, [tex]P(x) = x^2 - 2x + 4[/tex].