Challenge 1 · physics
One speed, two directions
A launcher is 1.25 m above level ground. Fired horizontally, a ball lands 5.0 m from the point directly below the launcher. Ignore air resistance and use g = 10 m/s^2. The same launcher speed is then aimed vertically upward. How high above the launcher does the ball rise?
Reveal the checked result
5 m. In the interactive lab, the result stays hidden until the numeric gate passes.
3 checked solution families
Time → speed → vertical rise
From 1.25 = (1/2)(10)t^2, obtain t = 0.5 s. From R = ut, obtain u = 5.0/0.5 = 10 m/s. At the top of the vertical launch, 0 = u^2 - 2gH, so H = 5.0 m.
Trajectory / direct elimination
For the horizontal launch, h = gR^2/(2u^2). For the vertical launch, H = u^2/(2g). Eliminating u and g gives hH = R^2/4, hence H = R^2/(4h) = 5.0 m.
Recover speed, then use energy
The horizontal fall gives t = 0.5 s and u = 10 m/s. For the vertical launch, (1/2)mu^2 = mgH. Cancel m: H = u^2/(2g) = 5.0 m.
Challenge 2 · mathematics
Five separate ceiling tests
The same training-ball test is run separately in five test chambers—one launch per chamber. Without a ceiling, the ball's height in every trial would follow y = -0.5(x - 4)^2 + 9, where x and y are measured in metres and y is height above the floor. The five chambers have ceiling heights 7.0 m, 8.5 m, 9.0 m, 10.0 m, and 12.0 m. A sensor triggers if the ball would touch or pass that trial's ceiling. In how many of the five separate trials would the sensor trigger?
Reveal the checked result
3 trials. In the interactive lab, the result stays hidden until the numeric gate passes.
3 checked solution families
Read the vertex, then test thresholds
Read the vertex (4, 9) from y = -0.5(x - 4)^2 + 9. The negative coefficient makes 9 m the maximum height. Count 7.0, 8.5, and 9.0 m because touching also triggers the sensor: 3 trials.
Differentiate, classify, then compare
Differentiate: y' = -(x - 4), so the stationary point occurs at x = 4. Since y'' = -1, the point is a maximum, and y(4) = 9 m. Three ceiling heights are at or below the maximum, so 3 sensors trigger.
Test horizontal-line intersections
For ceiling height h, solve -0.5(x - 4)^2 + 9 = h, giving (x - 4)^2 = 2(9 - h). A real intersection exists exactly when the right-hand side is non-negative, so h <= 9 m. The feasible thresholds are 7.0, 8.5, and 9.0 m, including equality: 3 trials.
Challenge 3 · mathematics
When applying twice changes nothing
Let S = {1, 2, 3, 4, 5, 6}. A function f: S → S satisfies both of the following conditions: (1) f(f(x)) = f(x) for every x in S; (2) for every y in S, at most two elements x in S satisfy f(x) = y. How many such functions f are there?
Reveal the checked result
331 functions. In the interactive lab, the result stays hidden until the numeric gate passes.
2 checked solution families
Count oriented pair components
If y is in the image, write y = f(x). Then f(y) = f(f(x)) = f(x) = y, so every image point is fixed. Each fixed point is already one of its own preimages. The cap of two allows at most one additional element to map to it, so the functional graph consists only of fixed singletons and oriented pairs. With p oriented pairs, choose their 2p elements, partition them into pairs, and choose the fixed endpoint of every pair. The orientation choices cancel the 2^p denominator from unordered pairing, leaving 6!/((6 - 2p)!p!). For p = 0, 1, 2, 3 the counts are 1, 30, 180, and 120. Their sum is 331.
Build a distinguished-element recurrence
Use the structural lemma above and let Aₙ count decompositions of n labelled elements into fixed singletons and oriented pairs. A distinguished element can be a singleton, leaving Aₙ₋₁ possibilities. Otherwise choose its partner in n - 1 ways and choose the fixed endpoint in 2 ways, leaving Aₙ₋₂ possibilities. Hence Aₙ = Aₙ₋₁ + 2(n - 1)Aₙ₋₂, with A₀ = A₁ = 1. Successive values are 3, 7, 25, 81, and finally A₆ = 81 + 10·25 = 331.
Challenge 4 · mathematics
The camera lost the labels
Eight maintenance beacons stand along a test railway. Their IDs and positions are: beacon 1 at 2 km, beacon 2 at 7 km, beacon 3 at 13 km, beacon 4 at 18 km, beacon 5 at 24 km, beacon 6 at 31 km, beacon 7 at 35 km, and beacon 8 at 47 km. Only three beacons were armed during a night run. The train travelled in the direction of increasing kilometre marks. Its camera recorded the three flashes, but a blackout erased their labels. The odometer shows 17 km from the first flash to the second, then 11 km from the second flash to the third. Which three beacon IDs flashed? Enter their IDs in travel order as one three-digit code.
Reveal the checked result
257 code. In the interactive lab, the result stays hidden until the numeric gate passes.
2 checked solution families
Scan the first gap, then extend
The 17 km pairs are 7 to 24 and 18 to 35. Only 7 to 24 can continue another 11 km, because 35 is listed while 46 is not. The positions 7, 24, and 35 map to IDs 2, 5, and 7: code 257.
Find endpoints, then reconstruct the middle
The endpoint span is 17 + 11 = 28 km. Only 7 and 35 among the listed positions are 28 km apart. The middle is at 7 + 17 = 24 km, so the IDs are 2, 5, and 7: code 257.