Math Olympiad Problems for Grades 9 and 10: Difficult Problems with Solutions

Solution:
The minute hand completes [tex]1[/tex] revolution every hour, meaning [tex]24[/tex] revolutions in a 24-hour day.
The special hour hand completes exactly [tex]1[/tex] revolution in a 24-hour day.
Since both move in the same continuous direction, the number of overlaps is the difference between their revolutions: [tex]24 - 1 = 23[/tex].

Solution:
Tie the Commander and Flight Engineer together into a single ”block”.
Now arrange [tex]5[/tex] units around the circle.
The formula for arranging [tex]n[/tex] distinct objects in a circle is [tex](n - 1)![/tex].
For [tex]n = 5[/tex], [tex](5 - 1)! = 4! = 24[/tex] ways.
Inside the block, the two individuals can swap seats ([tex]2! = 2[/tex] arrangements).
Total ways [tex]= 24 \times 2 = 48[/tex].

Solution:
The common difference between consecutive figures is [tex]11 - 6 = 5[/tex] and [tex]16 - 11 = 5[/tex].
This is an arithmetic progression with first term [tex]a_1 = 6[/tex] and common difference [tex]d = 5[/tex].
The formula for the [tex]n[/tex]-th term is [tex]a_n = a_1 + (n - 1)d[/tex].
Substitute the values: [tex]a_n = 6 + (n - 1)5 = 5n + 1[/tex].
For Figure 100, [tex]a_{100} = 5(100) + 1 = 501[/tex].

Solution:
Since the drone travels at a constant speed, the inspection time is directly proportional to the perimeter.
Because all frames are equilateral triangles, their perimeters are directly proportional to their base lengths.
Let the base lengths be [tex]3x[/tex], [tex]5x[/tex], and [tex]8x[/tex].
The total length of the beam is the sum of the bases: [tex]3x + 5x + 8x = 80 \Rightarrow 16x = 80 \Rightarrow x = 5[/tex].
The largest base is [tex]8x = 8(5) = 40[/tex] meters.

Solution:
For a triangle divided by concurrent cevians, the product of the areas of alternating smaller triangles is equal: [tex]A_1 \times A_3 \times A_5 = A_2 \times A_4 \times A_6[/tex].
Substituting the given values in clockwise order ([tex]A_1 = 4[/tex], [tex]A_2 = 6[/tex], [tex]A_3 = x[/tex], [tex]A_4 = 36[/tex], [tex]A_5 = 9[/tex], [tex]A_6 = 5[/tex]): [tex]4 \times x \times 9 = 6 \times 36 \times 5[/tex]
[tex]\Rightarrow 36x = 1080[/tex]
[tex]\Rightarrow x = 30 \text{ m}^2[/tex].

Solution:
Let [tex]S[/tex], [tex]C[/tex], and [tex]P[/tex] represent the mass of one Sphere, one Cube, and one Pyramid.
Equation 1: [tex]2C = 5S + 1P[/tex].
Equation 2: [tex]1P = 1C + 1S[/tex].
Substitute the expression for [tex]P[/tex] from Equation 2 into Equation 1: [tex]2C = 5S + (1C + 1S)[/tex]
[tex]\Rightarrow 2C = 6S + 1C[/tex]
[tex]\Rightarrow C = 6S[/tex].
Therefore, [tex]1[/tex] Cube equals the mass of [tex]6[/tex] Spheres.

Solution:
Let [tex]T[/tex] be Top, [tex]L[/tex] be Bottom-Left, and [tex]R[/tex] be Bottom-Right.
For Badge 1: [tex](T \times L) - R^2 = (8 \times 5) - 4^2 = 40 - 16 = 24[/tex].
Test on Badge 2: [tex](7 \times 6) - 3^2 = 42 - 9 = 33[/tex].
Applying the rule to Badge 3: [tex](9 \times 8) - 5^2 = 72 - 25 = 47[/tex].

Solution:
Area of the square [tex]= 12^2 = 144 \text{ cm}^2[/tex].
The [tex]4[/tex] semicircles have a diameter of [tex]12[/tex] cm, meaning a radius of [tex]r = 6[/tex] cm.
The sum of the areas of the [tex]4[/tex] semicircles is [tex]4 \times \left(\frac{\pi(6^2)}{2}\right) = 72\pi \text{ cm}^2[/tex].
The [tex]4[/tex] semicircles cover the entire square, but the shaded overlapping regions are counted twice.
Therefore, Area of [tex]4[/tex] Semicircles = Area of Square + Area of Overlap.
[tex]72\pi = 144 +[/tex] Area of Overlap [tex]\Rightarrow[/tex] Area of Overlap [tex]= 72\pi - 144 \text{ cm}^2[/tex].

Solution:
Let the side length of the inscribed square be [tex]x[/tex].
The square divides the large triangle into two smaller similar right-angled triangles.
The small triangle ”above” the square has a vertical leg of [tex](40 - x)[/tex] and a horizontal leg of [tex]x[/tex].
Since it is similar to the large triangle (ratio of vertical to horizontal leg is [tex]\frac{40}{60} = \frac{2}{3}[/tex]), we set up the proportion: [tex]\frac{40 - x}{x} = \frac{2}{3}[/tex]
[tex]\Rightarrow 3(40 - x) = 2x[/tex]
[tex]\Rightarrow 120 - 3x = 2x[/tex]
[tex]\Rightarrow 5x = 120[/tex]
[tex]\Rightarrow x = 24[/tex] cm.
Area [tex]= x^2 = 24^2 = 576 \text{ cm}^2[/tex].

Solution:
Using the Inscribed Angle Theorem, angles subtending the same arc are equal.
[tex]\angle BAC = 40^\circ[/tex] subtends arc [tex]BC[/tex], so [tex]\angle BDC = 40^\circ[/tex].
[tex]\angle CAD = 30^\circ[/tex] subtends arc [tex]CD[/tex], so [tex]\angle CBD = 30^\circ[/tex].
In [tex]\triangle BCD[/tex], the sum of angles is [tex]180^\circ[/tex].
Therefore,
[tex]\angle BDC + \angle CBD + \angle BCD = 180^\circ[/tex]
[tex]\Rightarrow 40^\circ + 30^\circ + \angle BCD = 180^\circ[/tex]
[tex]\Rightarrow \angle BCD = 110^\circ[/tex].