answer:First, we label the diagram: [asy] import olympiad; draw((0,0)--(sqrt(3),0)--(0,sqrt(3))--cycle); draw((0,0)--(-1,0)--(0,1)--cycle); label("6",(-1/2,1/2),NW); label("y",(sqrt(3)/2,sqrt(3)/2),NE); draw("45^{circ}",(1.5,0),NW); draw("45^{circ}",(-0.9,0),NE); draw(rightanglemark((0,1),(0,0),(sqrt(3),0),4)); label("A",(0,0),S); label("B",(-1,0),W); label("C",(sqrt(3),0),E); label("D",(0,1),N); [/asy] Triangle ABD is a 45-45-90 triangle, so: - AB = BD = 6 (given) - AD = 6sqrt{2} (from the properties of a 45-45-90 triangle) Triangle ACD is also a 45-45-90 triangle: - AC = AD = 6sqrt{2} - Therefore, CD = AC = 6sqrt{2} We conclude y = CD = boxed{6sqrt{2}}.

answer:There are 5 houses and 5 packages. We first choose 3 out of the 5 houses to receive the correct packages. This can be done in {5 choose 3} = 10 ways. For the remaining two houses, we need them to exchange their packages (which is essentially a derangement of two items). There are exactly one way for this to happen (each must receive the other's package). The probability calculation for the 5 total deliveries includes choosing the three correct deliveries and ensuring the last two are incorrect as follows: - Probability of selecting 3 correct out of 5 houses: {5 choose 3} = 10 - Probability of the remaining 2 being incorrect in the only possible incorrect way: frac{1}{2} (as there's only one way to arrange two items such that neither is in its original position). Thus, the probability is calculated as: [ frac{10}{5!} times 1 = frac{10}{120} = boxed{frac{1}{12}} ]

answer:1. We start by analyzing the layout of the pathways in the park. There are 5 longer pathways each of length (a) which form the sides of an equilateral triangle (ABC), and 6 shorter pathways each of length (b) which connect the centers of the triangle's sides to their vertices. 2. Label the vertices of the equilateral triangle as (A), (B), and (C), while the centers of sides opposite to these vertices are labeled as (D) (middle of (BC)), (E) (middle of (AC)), and (F) (middle of (AB)) respectively. 3. The goal is for a watering machine to travel along all these pathways and return to its starting point (A). To accomplish this optimally, note that every time the machine enters a vertex, it must also leave it (except at the starting/ending point). 4. To minimize the path, consider Eulerian paths and circuits. Each time we enter and exit a vertex (other than the starting point (A)), we must do this through an even total degree (paths and pathways connected). In our case: - ( A ) connects to ( 3 ) paths (to ( B ), ( F ), and ( E )), - Each of ( B, C, D, E, ) and ( F ) connects to ( 3 ) paths. 5. From graph theory, for an Eulerian Circuit to exist, each vertex must have an even degree. Since (A) starts and ends the path it has an odd degree (3 in -1 out = 2), the same holds for other vertices, which implies we must revisit the connecting paths (closed loops). 6. Design the path: - Starting at (A) going through (B, E, D, B, C, F, C, A, D, F ), finally returning to (A). 7. Calculate the total length of the path sequentially following each pathway. - The direct paths joined using (a) go through (AB), (BC), (CA), (AD), and revisit (CF), returning to (A) - The shorter paths joined (b): - (( BF )), (( EF )), ((DE)) - And revisit So total journey involves revisiting paths described by revised routes: [ 6a + 8b ] # Conclusion: Thus, the minimum path that covers all pathways and returns to the starting point (A) is: [ boxed{6a + 8b} ]

answer:To prove: A (2n times 2n) square board with one arbitrary removed cell can be tiled with (2 times 2) tetrominoes with one cell removed, provided (n) is not divisible by 3. 1. **Concept of Tilings and Reduced Board Size**: - We start by understanding that a (3 times 2) rectangle can be tiled using three (2 times 2) L-tetrominoes with one cell removed from each. - If the removed cell does not lie within a figure structured as a union of strips of width 6 cells attached to two adjacent sides of the square, we can proceed to tile the figure with (3 times 2) rectangles (Figure 130). 2. **Reduction to a Smaller Board**: - After ensuring the above tiling, the resulting problem reduces the size of the square to a (2(n-3) times 2(n-3)) square with one cell removed. - Because this operation can always be performed for (n geq 7), without loss of generality, it suffices to consider (n leq 5). 3. **Examine Cases for (n = 1, 2, 4, 5)**: - **Case (n = 1)**: A (2 times 2) square with one cell removed can be directly covered with one (2 times 2) L-tetromino (Fig. illustrating trimino not shown). - **Case (n = 2)**: A (4 times 4) square with one cell removed can be partitioned into four (2 times 2) squares. The one (2 times 2) square with the removed cell can be covered with one (2 times 2) L-tetromino, and the remaining cells by additional tetrominoes (Fig. 132). - **Case (n = 4)**: A (8 times 8) square with one cell removed uses the earlier principle for (n = 2), since it can be considered as smaller blocks, each of which can be tiled accordingly (Fig. 133). - **Case (n = 5)**: A (10 times 10) square with one cell removed can be divided into one (4 times 10) rectangle and two (3 times 10) rectangles such that the removed cell is in the (4 times 10). The (3 times 10) can be tiled using previously mentioned (2 times 2) tetrominoes. Subsequently, any (4 times 4) and (3 times 4) sections, as in previous cases, can be tiled (Fig. 132 and 133). 4. **Breakdown and Conclusion**: - For (n=3), we cannot satisfy the conditions as (3) divides (12), and thus, (n) not being divisible by 3 holds true to prevent tiling issues. - This confirms (2n times 2n) square with these properties always reduces and can be patterned with described tiling (2 times 2) removed L-tetrominoes. Conclusively, we affirm the tiling is possible for any (n leq 5), and for (n not equiv 0 mod 3). Hence, [ boxed{} ]