Math Olympiad Problems for Grades 3 and 4: Very Difficult Problems with Solutions

Solution:
1. Analyzing Zog: If a Zog says “I lied yesterday” and is telling the truth today, then today must be a day of truth and yesterday a day of lies. This occurs on Thursday (Thursday tells the truth, Wednesday lied). If the Zog were lying today, it means the truth was told yesterday. This occurs on Monday (Monday lies, Sunday told the truth). So, for Zog, today can be either Monday or Thursday.
2. Analyzing Ym: If a Ym tells the truth today, a lie must have been told yesterday. This occurs on Sunday (Sunday is truth, Saturday lied). If the Ym is lying today, the truth must have been told yesterday. This occurs on Thursday (Thursday lies, Wednesday told the truth). So, for Ym, today can be either Sunday or Thursday.
3. The only day on which both can say that phrase at the same time is Thursday.

Solution:
1. Let’s begin by identifying the initial pieces. There are [tex]2[/tex] identical fabrics. Each is a rectangle with Length: [tex]10[/tex] cm and Width: [tex]4[/tex] cm.
2. The perimeter of a single rectangular piece of fabric is the sum of its four sides: [tex]10 + 4 + 10 + 4 = 28[/tex] cm.
3. If the two fabrics were separate, the sum of their perimeters would be [tex]28 + 28 = 56[/tex] cm.
4. The two fabrics are joined to form a single piece shaped like the letter ”L.” They overlap exactly in a perfect [tex]4 \times 4[/tex] cm square at one of the corners.
5. When the fabrics are joined, the edges where they overlap are hidden within the new shape and are no longer part of the outer outline.
6. Since the overlapping area is a [tex]4 \times 4[/tex] cm square, this means that from the horizontal fabric, two [tex]4[/tex] cm sides are hidden, and from the vertical fabric, two [tex]4[/tex] cm sides are hidden.
7. In total, the length of the edges that disappear from the outer perimeter is: [tex]4 \text{ cm} + 4 \text{ cm} + 4 \text{ cm} + 4 \text{ cm} = 16 \text{ cm}[/tex].
8. To obtain the total perimeter of the new shape, the length of the hidden edges is subtracted from the sum of the individual perimeters: [tex]56 \text{ cm} - 16 \text{ cm} = 40 \text{ cm}[/tex].

Solution:
1. To solve this, the problem is divided into two cases depending on the opposite sections (imagining the sections numbered [tex]1, 2, 3, 4[/tex] in circular order).
2. Case 1: Opposite sections ([tex]1[/tex] and [tex]3[/tex]) have the SAME color.
- Choosing the color for [tex]1[/tex] and [tex]3[/tex]: there are [tex]3[/tex] options (Red, Blue, Green).
- Since [tex]1[/tex] and [tex]3[/tex] are the same color, section [tex]2[/tex] can be any of the other [tex]2[/tex] colors ([tex]2[/tex] options).
- Section [tex]4[/tex] is also between two sections of the same color, so it has [tex]2[/tex] options.
- Total for Case 1 [tex]= 3 \times 2 \times 2 = 12[/tex] ways.
3. Case 2: Opposite sections ([tex]1[/tex] and [tex]3[/tex]) have DIFFERENT colors.
- Choosing the color for section [tex]1[/tex]: [tex]3[/tex] options.
- Choosing the color for section [tex]3[/tex] (must be different): [tex]2[/tex] options.
- Section [tex]2[/tex] is now trapped between two different colors, so it only has [tex]1[/tex] option left.
- Section [tex]4[/tex] is also trapped between two different colors, leaving [tex]1[/tex] option.
- Total for Case 2 [tex]= 3 \times 2 \times 1 = 6[/tex] ways.
4. Adding both cases together: [tex]12 + 6 = 18[/tex] ways in total.

Solution:
1. The total weight of everything on the scales is calculated. Adding the objects on the left side of the three scales gives: [tex]2[/tex] Triangles + [tex]2[/tex] Squares + [tex]2[/tex] Circles.
2. Adding the weights on the right side: [tex]12 \text{ kg} + 17 \text{ kg} + 15 \text{ kg} = 44 \text{ kg}[/tex].
3. This means that twice the weight of the entire collection is [tex]44[/tex] kg.
4. To find out how much a single collection weighs ([tex]1[/tex] Triangle + [tex]1[/tex] Square + [tex]1[/tex] Circle), the total is simply divided by [tex]2[/tex]: [tex]44 \div 2 = 22[/tex] kg.

Solution:
1. Calculate the total area to be shaded: The target is exactly half of the total area of the grid. [tex]A_{target} = \frac{64 \text{ cm}^2}{2} = 32 \text{ cm}^2[/tex].
2. Calculate the area already shaded: [tex]20[/tex] entire squares have been shaded. Each small square has an area of [tex]1 \text{ cm}^2[/tex]. [tex]A_{shaded} = 20 \times 1 \text{ cm}^2 = 20 \text{ cm}^2[/tex].
3. Calculate the remaining area that needs to be shaded: [tex]A_{remaining} = 32 \text{ cm}^2 - 20 \text{ cm}^2 = 12 \text{ cm}^2[/tex].
4. Determine the area of a single individual triangle: Each triangle is half of a square cut by the diagonal. [tex]A_{triangle} = \frac{1 \text{ cm}^2}{2} = 0.5 \text{ cm}^2[/tex].
5. Calculate exactly how many triangles are needed to cover the remaining area: Number of triangles [tex]= \frac{12 \text{ cm}^2}{0.5 \text{ cm}^2} = 24[/tex] triangles.