Use Score Conservation and Standings
第11讲 逻辑推理（2）· 第4课
Competition results may vary, but the total number of points often stays fixed. Use that point budget together with rank inequalities and whole-number possibilities.
Count the games. Fix the point budget. Then make the standings fit.
Every match has a point budget
Under the 2-for-a-win, 1-each-for-a-draw, 0-for-a-loss chess rule used on this page, the players' shares can change while one match's total stays fixed.
Two possible ways to share 2 points
winner＋0
loser
draw＋1
draw
Invariant: a quantity that does not change when the match result changes.
Count every round-robin game once
In a single round robin, every pair of players meets exactly once.
Build the worked example standings problem
First create the complete point budget and translate the ranking words into inequalities.
Complete practice problem
Four students A, B, C, and D play a chess round robin. Every pair plays once. A win earns 2 points, a draw earns 1 point for each player, and a loss earns 0. A finishes first, B and C tie for second, and D finishes last. How many points does B earn?
| Standing clue | Mathematical meaning |
|---|---|
| A is first | A has more points than B and C |
| B and C tie for second | B = C = x |
| D is last | D has fewer points than x |
Test the tied score x
Scores must be whole numbers from 0 through 6. A and D must share the remaining point budget while staying on opposite sides of x.
Can A > x > D?
| x | A + D | Can A > x > D with 0–6 points? | Result |
|---|
Realize every valid score pattern
A “score vector” lists the four totals in A, B, C, D order; for example, 6–3–3–0 means A has 6, B and C each have 3, and D has 0. A score vector is not enough by itself; it should come from actual match results. All three surviving vectors below are realizable.
Match results
Why all three score vectors matter
The middle vector 5–3–3–1 is also possible. It occurs when A draws D, B draws C, A beats B and C, and B and C both beat D. The requested answer, B = 3, is unchanged.
Recover a missing player's score
When most scores are known, subtract them from the complete tournament budget.
Complete guided-practice problem
Five players play a chess round robin under the same 2-point-per-game rule. Four players have 16 points altogether. How many points does the fifth player have?
Complete budget
Subtract the known part
One-player point-budget machine
Use a different scoring invariant in volleyball
The shares change between 3–0 and 2–1, but every match still contributes 3 league points.
Complete guided-practice problem
Four volleyball teams play a single round robin. A match gives 3–0 league points when the set score is 3–0 or 3–1, and 2–1 league points when the set score is 3–2. The four final team scores are four consecutive natural numbers. What is the first-place score?
Fixed total
Combine conservation with systematic feasibility
A practice problem asks for the greatest possible number of draws under two standings restrictions.
Complete practice problem
Four chess players play one game against every other player. A win gives 2 points, a draw gives 1 point to each player, and a loss gives 0. Nobody wins all three games, and all four final point totals are different. What is the greatest possible number of drawn games?
All 36 result schedules checked
Each of the six games has three outcomes: first player wins, draw, or second player wins.
Below each bar: number of drawn games. Above it: number of valid schedules.
One schedule with the maximum
Prove the upper bound without checking 729 schedules
With all six games drawn, every player has 3 points. A decisive game changes the two players’ totals by +1 and −1.
With five draws, two players stay at 3. With four draws, the two decisive games either involve four different players (two scores become 4 and two become 2), or share a player. In the shared-player case, that person changes by +2, 0, or −2. If the change is +2 or −2, the two opponents tie; if it is 0, that player ties the untouched fourth player. Thus four, five, or six draws cannot give four different totals.
The displayed three-draw schedule has four distinct scores and nobody wins all three games, so three is achievable and is the exact maximum.
Optional challenges: take the method further
These four optional investigations have their own checkpoint and do not affect the ten-mission certificate. Try them with local hints, check your answers, then open worked review to compare full arguments.
Challenge 1 · Infer the number of events
A, B, and C take the top three places in every event. Their total scores are 8, 7, and 17 respectively. A wins exactly one event. Every event awards the same three positive whole-number scores: first place gets more than second, second more than third, and first place gets at least the sum of second and third. How many events are there, and what does A score in each?
Hint
The total point budget is 8 + 7 + 17 = 32. Each event awards the same integer budget. List the divisors of 32, remembering that three different positive scores sum to at least 6.
Study the solution
There can be at most five events, and the event count divides 32: test 1, 2, or 4. One event cannot work because A wins but C has more points. In two events, each budget is 16; the winner must get at least 8, so A’s win plus another positive score would exceed A’s total of 8.
Therefore there are four events, each awarding 8 points. The descending positive triples satisfying the winner condition are (5,2,1) and (4,3,1). With (4,3,1), A’s one win leaves 4 points for three scores drawn from 3 and 1, impossible: the sum of three odd numbers is odd. With (5,2,1), A scores 5,1,1,1.
Construction: A wins one event, with C second and B third. In the other three, C wins, B is second, A third. Totals: A = 5+1+1+1 = 8; B = 1+2+2+2 = 7; C = 2+5+5+5 = 17. Every clue is met.
Challenge 2 · Recover the shot-put score
Four athletes compete in long jump, 100-meter sprint, shot put, and high jump. Every event awards 5, 3, 2, and 1 points for first through fourth, with no ties within an event. The overall first-place athlete totals 17 points, and their high-jump score is lower than each of their other scores. The overall third-place athlete totals 11 points, and their high-jump score is higher than each of their other scores. Overall places are distinct. What is the overall second-place athlete’s shot-put score?
Hint
The overall winner’s other three event scores are at most 5 each. How can the four scores reach 17 while high jump is strictly lowest?
Study the solution
Name the athletes W (overall winner), R (runner-up), T (third), L (last). W must score 5,5,5,2: a high-jump score of 1 caps the total at 16, and a high-jump score of 3 would force all other scores to be 5, totaling 18. Thus W takes 5 in every non-high-jump event.
T cannot have a best score of 3 or less: the other three scores would be at most 2 each, totaling at most 9. Hence T gets 5 in high jump, leaving 6 across the other three events. Those scores are either 2,2,2 or a permutation of 1,2,3.
If T has 1,2,3, R can earn at most 3,3,2 in those events, and at most 3 in high jump: at most 11 overall, not above T. So T has 2,2,2. The remaining athletes receive 3 and 1 in every event. To beat 11, R must take all four 3s, totaling 12. L then totals 4. In particular, R earns 3 in shot put.
Challenge 3 · Maximize a gap in soccer standings
Twelve teams play every other team once. A win awards 3 points, a draw 1 to each team, and a loss 0. What is the greatest possible point gap between the teams in third and fourth place? Scores may tie; tied teams receive ordered places by a tie-break. We are maximizing the gap in points, not the tie-break value.
Hint
Separate the top three teams from the bottom nine. Bound the top three’s combined points from above and the bottom nine’s points from below.
Study the solution
The top three play 27 games against the bottom nine and 3 games among themselves. They can receive at most 27×3 + 3×3 = 90 points altogether. The third-place team has the fewest points of those three, so its score is at most 30.
The bottom nine play 9×8÷2 = 36 games among themselves. Every such game gives at least 2 total points. They therefore share at least 72 points, even before any points won from top teams. Fourth place has the most points among these nine, so it must have at least 72÷9 = 8 points. The gap is at most 30−8 = 22.
To reach 22, let all top three beat all bottom nine. Among the top three, arrange a cycle: A beats B, B beats C, C beats A. Each gets 30 points. All bottom-nine games are draws, so each bottom team gets 8. The stated tie-break convention orders tied teams, and the third/fourth gap is exactly 22.
This version explicitly permits equal point totals. Requiring every team’s score to be different would be a different optimization problem.
Challenge 4 · Tied first, then second
Teams A, B, C, and D play a single round robin. Each win gives 2 points, each draw 1 to each team, and a loss 0. A and B tie for first; C is second; D is third. Thus A = B > C > D. How many games are played, and what is C’s greatest possible score?
Hint
The total budget is fixed. If C scored 4 or more, how many points would A and B each need?
Study the solution
Four teams give 4×3÷2 = 6 games and 12 total points. If C had at least 4, A and B would each have at least 5; those three alone would require at least 14 points. So C ≤ 3.
Construction: A beats B and draws C and D; B beats C and D; C beats D. Totals are A = 4, B = 4, C = 3, D = 1. C reaches the upper bound while every ranking clue holds, so its maximum is 3.
Independent workshop
Solve at least 6 of 8 fresh problems. Each has a local hint.
Hint
Count unordered pairs.
Hint
Count matches, then multiply by the points per match.
Hint
Six players produce fifteen games.
Hint
Find the missing score and compare with four wins.
Hint
Write x + (x+1) + (x+2) + (x+3).
Hint
Compare 3+0 with 1+1.
Hint
Start with three per match, then remove one per draw.
Hint
A fixed total is necessary but not always sufficient.
Independent exit ticket
Solve all five fresh problems. The certificate also requires Missions 1–9.
Hint
Do not count A–B and B–A separately.
Hint
Compute the full point budget first.
Hint
Subtract the offsets 0+1+2+3 before dividing by four.
Hint
Compare the remainder with the number of games one person plays.
Hint
Separate a bound from an example.
Optional reflection — not automatically graded
Lesson checkpoints completed
Score-Conservation & Standings Architect
You completed the checkpoints on match counts, point budgets and feasible standings. Revisit any steps for which you needed solution help.
Lesson 11.4 · Grade 5 Mathematical Reasoning
Teaching notes
The original's printed discussion of the chess example omits the valid 5–3–3–1 score vector; this page explicitly includes it while preserving the original's correct requested answer, B = 3.Interactive graphs, complete enumeration, added practice, feedback, and assessment are new instructional scaffolds.