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$QR = 10$

The solutions of the equation are x = 1 and x = 29.

There are 15 possible ways to color the five squares: Red-Yellow-Blue-Red-Yellow Red-Yellow-Blue-Yellow-Red Red-Blue-Yellow-Red-Yellow Red-Blue-Yellow-Yellow-Red Red-Yellow-Red-Blue-Yellow Red-Yellow-Red-Yellow-Blue Red-Blue-Red-Yellow-Yellow Red-Blue-Red-Yellow-Blue Yellow-Red-Blue-Red-Yellow Yellow-Red-Blue-Yellow-Red Yellow-Blue-Red-Red-Yellow Yellow-Blue-Red-Yellow-Red Blue-Red-Yellow-Red-Yellow Blue-Red-Yellow-Yellow-Red Blue-Yellow-Red-Red-Yellow Blue-Yellow-Red-Yellow-Red

The probability that the distance between the two points is at most 1.5 is 0.5. This is because the maximum distance between two points on a circle of radius 1 is 2, and 1.5 is half of that.

The probability that the first card Grok draws has an even number and is blue is 1/4. The probability that the second card Grok draws has an odd number and is red is also 1/4. Therefore, the probability that the first card Grok draws has an even number and is blue, and the second card Grok draws has an odd number and is red is 1/16.

Marvin can reach his home in 8 different ways. He can go right 4 times and up 6 times, right 3 times and up 7 times, right 2 times and up 8 times, right 1 time and up 9 times, left 1 time and up 9 times, left 2 times and up 8 times, left 3 times and up 7 times, and left 4 times and up 6 times. He cannot go through (2,4) because of the frog.

(a) Proof: Let x = a^2 + b^2, y = c^4 + d^4, and z = e^4 + f^4. By the AM-GM inequality, we have xyz >= 8xyz. Expanding this, we get (a^2 + b^2)^2 (c^4 + d^4)(e^4 + f^4) >= 8(a^2 + b^2)(c^4 + d^4)(e^4 + f^4). Simplifying, we get (a^2 + b^2)^2 (c^4 + d^4)(e^4 + f^4) >= 8abcdef. Since 8abcdef >= 32abcdef, we have (a^2 + b^2)^2 (c^4 + d^4)(e^4 + f^4) >= 32abcdef. (b) Proof: Let x = a^2 + b^2, y = c^2 + d^2, and z = e^2 + f^2. By the AM-GM inequality, we have xyz >= 4xyz. Expanding this, we get (a^2 + b^2)(c^2 + d^2)(e^2 + f^2) >= 4(a^2 + b^2)(c^2 + d^2)(e^2 + f^2). Simplifying, we get (a^2 + b^2)(c^2 + d^2)(e^2 + f^2) >= 8abcdef.