Math Olympiad Problems for Grades 7 and 8: Difficult Problems with Solutions

Solution:
The central figure formed by the intersection of the diagonals of a regular pentagon is also a regular pentagon.
All regular polygons of the same type have the same interior angle measure, regardless of their size.
The formula for the sum of the interior angles of an [tex]n[/tex]-sided polygon is [tex](n - 2) \times 180^\circ[/tex].
For a pentagon, [tex]n = 5[/tex], so the sum is [tex](5 - 2) \times 180^\circ = 540^\circ[/tex].
Since it is regular, the measure of each interior angle is [tex]\frac{540^\circ}{5} = 108^\circ[/tex].

Solution:
Divisibility by [tex]4[/tex] requires the last two digits ([tex]B4[/tex]) to be divisible by [tex]4[/tex].
Possible single-digit values for [tex]B[/tex] are [tex]0, 2, 4, 6, 8[/tex] (yielding [tex]04, 24, 44, 64, 84[/tex]).
Divisibility by [tex]9[/tex] requires the sum of the digits to be a multiple of [tex]9[/tex].
The sum is [tex]2 + A + 3 + B + 4 = 9 + A + B[/tex].
For this to be a multiple of [tex]9[/tex], the remaining part ([tex]A + B[/tex]) must be a multiple of [tex]9[/tex] (so [tex]A + B = 0, 9[/tex], or [tex]18[/tex]).
If [tex]A + B = 0[/tex], then [tex]B = 0[/tex] and [tex]A = 0[/tex].
If [tex]A + B = 9[/tex], testing valid [tex]B[/tex]’s: [tex]B = 0 \Rightarrow A = 9[/tex]; [tex]B = 2 \Rightarrow A = 7[/tex]; [tex]B = 4 \Rightarrow A = 5[/tex]; [tex]B = 6 \Rightarrow A = 3[/tex]; [tex]B = 8 \Rightarrow A = 1[/tex].
If [tex]A + B = 18[/tex], [tex]B = 9[/tex] (invalid, since [tex]B[/tex] must be even).
The possible single-digit values for [tex]A[/tex] are [tex]0, 1, 3, 5, 7, 9[/tex].
Their sum is [tex]0 + 1 + 3 + 5 + 7 + 9 = 25[/tex].

Solution:
Let [tex]a[/tex] be the number of Alpha crates and [tex]b[/tex] be the number of Beta crates.
Equation 1: [tex]a + b = 10 \Rightarrow b = 10 - a[/tex].
Equation 2: [tex]15a + 25b = 210[/tex].
Substituting [tex]b[/tex]: [tex]15a + 25(10 - a) = 210[/tex]
[tex]\Rightarrow 15a + 250 - 25a = 210[/tex]
[tex]\Rightarrow -10a + 250 = 210[/tex]
[tex]\Rightarrow 10a = 40[/tex]
[tex]\Rightarrow a = 4[/tex].
There are [tex]4[/tex] crates containing Alpha modules.

Solution:
Let [tex]O[/tex] be the center of the circle.
We are given that chord [tex]PQ = r[/tex].
We also know that [tex]OP = r[/tex] and [tex]OQ = r[/tex].
Since all three sides are equal, triangle [tex]OPQ[/tex] is equilateral, meaning the central angle [tex]\angle POQ = 60^\circ[/tex].
The inscribed angle theorem states that an inscribed angle is exactly half the central angle that subtends the same arc.
Both angles subtend the minor arc [tex]PQ[/tex].
Therefore, [tex]\angle PRQ = \frac{1}{2} \times 60^\circ = 30^\circ[/tex].

Solution:
This problem asks for the smallest amount of time that is a common multiple of both Drone A’s period ([tex]18[/tex] hours) and Drone B’s period ([tex]24[/tex] hours).
This requires the Least Common Multiple (LCM).
The prime factorization is [tex]18 = 2 \times 3^2[/tex] and [tex]24 = 2^3 \times 3[/tex].
Taking the highest power of each prime factor gives the LCM: [tex]2^3 \times 3^2 = 8 \times 9 = 72[/tex].
Both drones will return simultaneously after [tex]72[/tex] hours.

Solution:
Step 1: Area of the Regular Hexagon.
A regular hexagon consists of six equilateral triangles.
Area [tex]= 6 \times \left(\frac{\sqrt{3}}{4} \times 6^2\right) = 6 \times 9\sqrt{3} = 54\sqrt{3} \text{ m}^2[/tex].
Step 2: Area of the Inscribed Circle.
The radius [tex]r[/tex] is the height of one of the equilateral triangles: [tex]r = \frac{\sqrt{3}}{2} \times 6 = 3\sqrt{3}[/tex] m.
Using [tex]\pi \approx 3[/tex], Area [tex]= \pi r^2 \approx 3 \times (3\sqrt{3})^2 = 3 \times 27 = 81 \text{ m}^2[/tex].
Step 3: The remaining area is [tex]54\sqrt{3} - 81 \text{ m}^2[/tex].

Solution:
We solve the quadratic equation to find [tex]t[/tex] when [tex]h(t) = 0[/tex].
So, [tex]-t^2 + 8t + 20 = 0[/tex].
Multiplying by [tex]-1[/tex] gives [tex]t^2 - 8t - 20 = 0[/tex].
We factor this by finding two numbers that multiply to [tex]-20[/tex] and add to [tex]-8[/tex], which are [tex]-10[/tex] and [tex]2[/tex].
The equation factors to [tex](t - 10)(t + 2) = 0[/tex].
The solutions are [tex]t = 10[/tex] or [tex]t = -2[/tex].
Since negative time is not physically meaningful, the correct positive value is [tex]10[/tex] seconds.

Solution:
There are [tex]5[/tex] slots to fill.
Slot 1 is constrained to ”Initialize” ([tex]1[/tex] option).
Slot 5 is constrained to ”Test” ([tex]1[/tex] option).
We must fill the remaining [tex]3[/tex] middle slots using the remaining [tex]3[/tex] commands (”Scan,” ”Align,” ”Calibrate”), using each exactly once.
This requires finding the number of permutations of [tex]3[/tex] distinct items in [tex]3[/tex] spots, calculated as [tex]3! = 3 \times 2 \times 1 = 6[/tex].

Solution:
Analyzing the sequence, we notice that each term is one less than a power of [tex]2[/tex]: [tex]3 = 2^2 - 1[/tex], [tex]7 = 2^3 - 1[/tex], [tex]15 = 2^4 - 1[/tex], [tex]31 = 2^5 - 1[/tex], and [tex]63 = 2^6 - 1[/tex].
The general formula for the [tex]n[/tex]-th term is [tex]2^{n+1} - 1[/tex].
For the 8th term ([tex]n = 8[/tex]), the value is [tex]2^{8+1} - 1 = 2^9 - 1 = 512 - 1 = 511[/tex].

Solution:
Let [tex]L[/tex] be the length and [tex]W[/tex] be the width of the original rectangle.
Perimeter [tex]= 2L + 2W = 48 \Rightarrow L + W = 24[/tex].
Folding the rectangle parallel to the shorter side halves the longer side ([tex]\frac{L}{2}[/tex]).
For the new figure to be a square, its dimensions must be equal, so [tex]W = \frac{L}{2} \Rightarrow L = 2W[/tex].
Substituting into the perimeter equation: [tex]2W + W = 24 \Rightarrow 3W = 24 \Rightarrow W = 8[/tex] cm.
The side length of the square is [tex]W[/tex], which is [tex]8[/tex] cm.