Kvant Math Problem 1021

Verified: no
Verdicts: SKIP + SKIP
Solve time: 7m06s
Source on kvant.digital

Problem

A mountaineer wants to climb a 1000 m cliff. After spending the night in a camp at the foot of the cliff, he can ascend while fixing a rope at a rate of 40 m/h, whereas after a cold overnight stay on the cliff he can do so at 30 m/h. He climbs a pre-installed rope at a rate of 400 m/h. In how many days can he reach the summit if he works on the cliff (including climbing the rope) for 6 hours per day? (Neglect the time required for descent and other operations.)

Exploration

Consider how the mountaineer’s progress depends on the day’s starting point. If he starts from the foot of the cliff after a night in a camp, he ascends at 40 m/h while fixing the rope. If he spends the night on the cliff, his rate drops to 30 m/h. Once the rope is installed, he can climb it at 400 m/h, which is far faster. The mountaineer can work six hours per day, but part of that time may be spent ascending newly fixed rope, and part climbing pre-installed rope.

Attempting a few small cases suggests that the optimal strategy is to leave part of the rope for subsequent rapid climbing, rather than fixing the entire cliff in one day. For example, if he fixes 240 m of rope in six hours on day one (6 h × 40 m/h), he can ascend all 240 m at 400 m/h on the next day almost instantaneously relative to fixing. If he were cold overnight at height, his fixing rate drops to 30 m/h, reducing daily progress. Therefore, the critical step is correctly accounting for the interaction between rope fixing, rope climbing, and daily limits, especially when only part of the rope is fixed each day.

Problem Understanding

A mountaineer seeks to climb a 1000 m cliff by working six hours per day. After sleeping at the base, he fixes new rope at 40 m/h. After sleeping on the cliff, the fixing rate drops to 30 m/h. Climbing pre-installed rope always occurs at 400 m/h. The question asks how many days he requires to reach the summit under these constraints, neglecting descent and other operations.

This is a Type C problem, as it asks to determine a minimum number of days. The core difficulty lies in balancing the time spent fixing new rope with the rapid climbing along pre-installed rope, accounting for daily working hours and the drop in efficiency after cold nights on the cliff. Intuitively, the fastest strategy is to fix the maximum rope possible each day while minimizing the penalty from cold overnight stays, then climb the fixed rope quickly the following day.

Proof Architecture

Lemma 1: If the mountaineer fixes $x$ meters of rope in a day, he can climb it the next day at a negligible fraction of the six hours due to 400 m/h climbing speed. This is true because 400 m/h is much faster than 30–40 m/h fixing rates.

Lemma 2: The optimal daily strategy is to sleep at the base every night until the rope reaches a point where sleeping on the cliff is unavoidable. This holds because 40 m/h > 30 m/h, so starting from the base is more efficient.

Lemma 3: The total number of days is determined by summing the distances fixed each day divided by the daily rate and including rapid climbs along pre-fixed rope. The hardest step is correctly computing cumulative progress and converting it into whole days while respecting the six-hour work limit.

Solution

Denote the total cliff height as $H = 1000$ m. Let $D$ be the number of days required. On the first day, the mountaineer starts from the base. He works six hours fixing rope at 40 m/h, so he fixes

$$h_1 = 6 \times 40 = 240\ \text{m}.$$

At the end of the first day, he can sleep at the top of the fixed rope and climb it immediately the next morning at 400 m/h. The time to climb 240 m at 400 m/h is

$$t_1 = \frac{240}{400} = 0.6\ \text{h}.$$

On the second day, he starts at 240 m. He fixes rope for six hours at 40 m/h (he spent the night at the base for the first day, so no cold penalty yet), adding

$$h_2 = 6 \times 40 = 240\ \text{m},$$

reaching 480 m total height. He can climb the newly fixed 240 m in $0.6$ h the next morning. Repeating this reasoning, each day adds 240 m of fixed rope and negligible time for climbing pre-fixed rope. The sequence of cumulative heights after each day of fixing is

$$H_1 = 240,\quad H_2 = 480,\quad H_3 = 720,\quad H_4 = 960.$$

On the fifth day, he starts at 960 m. The remaining distance is $1000 - 960 = 40$ m. He fixes this 40 m of rope in

$$t = \frac{40}{40} = 1\ \text{h},$$

climbing the 40 m at 400 m/h in $0.1$ h. Thus, the fifth day suffices to reach the summit. Summing the days, he requires five days in total.

This completes the proof.

∎

Verification of Key Steps

The delicate step is the calculation of daily progress, ensuring climbing the pre-fixed rope consumes negligible time relative to the six-hour workday. For 240 m of rope, climbing at 400 m/h takes 0.6 h. If one mistakenly assumed climbing took six hours, the estimated number of days would increase erroneously. Checking the last day, climbing 40 m at 400 m/h requires 0.1 h, well under the six-hour limit, confirming no additional day is necessary. The sequence of heights 240, 480, 720, 960, 1000 ensures all steps are covered without gaps.

Alternative Approaches

One could attempt a continuous model, treating progress as a differential equation with daily time constraints, then rounding up to whole days. Another approach is a backward induction from the summit, calculating how far he must be the previous day to reach the top in six hours. The main approach is preferable because it directly uses the given rates and daily hours, avoids unnecessary abstraction, and transparently accounts for the climbing and fixing split.