# Tournament's problem

Suppose that we have boys and girls who competes in a tournament where each of the person will play against each of the person exactly one time and if he wins then he gets $$1$$ point if he loses then he gets $$0$$ point and if the result of game is draw then both of players get $$0.5$$ point. We know that quantity of boys is three times bigger than quantity of girls. And we know that after the end of tournament overall points of boys are $$20\%$$ bigger than girls overall points.

We need to find how many people could participate in this tournament.

Let $$x$$ is quantity of girls and then $$3x$$ is quantity of boys, $$4x$$ is quantity of people.

I know that overall points of all people is $$\frac{4x(4x-1)}{2} = S$$

Boys got $$20\%$$ more points so if $$y$$ are points which were got by girls then we have $$y+1.2y=S \Rightarrow y = \frac{10x(4x-1)}{11}$$. At most girls can get $$x\cdot3x + \frac{x(x-1)}{2}$$ points, because they can win all boys (each girl wins against each boy) and they need to distribute other points beetwen them. So we get inequaltiy: $$\frac{10x(4x-1)}{11} \leq x\cdot3x + \frac{x(x-1)}{2}$$ and we get from it that $$x = 1$$ and overall quantity of people is $$4$$.

I think that there is an error (or errors) in my solution. Can somebody help me find them?

UPDATE.

From inequality $$\frac{10x(4x-1)}{11} \leq x\cdot3x + \frac{x(x-1)}{2}$$ we get that $$0 \leq x \leq 3$$.

So quantity of people can be $$4$$, $$8$$ or $$12$$.

Am i right now?

Thanks to Ross Millikan for pointing out:

• The obvious mistake of my confusing the total number of points scored with the total number of points scored by all of the girls.

• The more subtle mistake that the average score of all of the girls must not be greater than what their score would be, if every girl beat every boy.

If $$x$$ girls scored $$P$$ points, then $$3x$$ boys scored $$1.2P$$ points, so $$2.2P$$ points were scored in total.

Let $$~Q = 2.2P = ~$$ be the total number of points scored.

Then, the total score of all the boys is $$~\dfrac{6Q}{11}~$$ and the total score of all the girls is $$~\dfrac{5Q}{11},~$$ so $$~Q~$$ must be a multiple of $$~11.~$$

Also, each game generated $$1$$ point, distributed as either [1,0] or [1/2,1/2]. Therefore $$~Q~ = ~$$ the number of games played.

Since there were $$~x~$$ girls and $$~3x~$$ boys, you have that

$$Q = \binom{4x}{2} = (2x) \times (4x - 1).$$

So, $$~Q~$$ must be chosen so that there exists a positive integer $$x$$ such that
$$(2x) \times (4x-1) = Q.$$

So, either $$2x$$ is a multiple of $$11$$ or $$(4x - 1)$$ is a multiple of $$11$$.

The smallest positive integer $$x$$ that can serve as a $$\color{red}{\text{candidate value}}$$ is $$x = 3.$$

Exploring this $$\color{red}{\text{candidate value}}$$:

• The total number of games played is
$$\displaystyle \binom{12}{2}~ = 66 = Q.$$

• The total number of girls is $$~3~$$ and the total number of boys is $$~9.$$

• The total number of points collectively scored by all of the girls is $$~\dfrac{5Q}{11} = 30.$$
So, the average score of each girl was $$~10.$$

• The total number of points collectively scored by all of the boys is $$~\dfrac{6Q}{11} = 36.$$
So, the average score of each boy was $$~4.$$

$$\color{red}{\text{Does this work?}}$$

Suppose that every girl beat all 9 boys.

Further suppose that every girl:girl game and every boy:boy game ended in a draw.

Then each girl would score
$$\displaystyle 9 + \left[2 \times \frac{1}{2}\right] = 10,$$

and each boy would score
$$\displaystyle 0 + \left[8 \times \frac{1}{2}\right] = 4.$$

So, the $$\color{red}{\text{candidate value}}$$ of $$x = 3$$ works okay.

Responding to the comment/question of Arty.

First of all, your overall approach was $$\color{red}{\text{better}}$$ than my $$\color{red}{\text{eventual}}$$ approach, in one way, and flawed in another way.

That is:

• You elegantly confronted the issue that the total score of the girls, assuming that each girl beat each boy, had to be at least $$~\dfrac{5}{11}~$$ of the total games played.

Initially, I totally overlooked that issue. Then, after Ross Millikan commented, I edited my answer after the fact. However, I did not try for the elegant (and better) approach of establishing an inequality involving $$~x.~$$

Instead, I (inelegantly) used the notion of a candidate value to verify that my solution worked.

• You totally overlooked the constraint that the number of games played, $$~\dfrac{4x(4x-1)}{2}~$$ has to be a multiple of $$~11.~$$ This is because the total number of points scored by the girls is
$$\displaystyle y = \frac{5}{11} \times \frac{4x(4x-1)}{2}.$$

$$\displaystyle \frac{10x(4x - 1)}{11} \leq [x \times 3x] + \frac{x(x-1)}{2}.$$

I agree with your updated interpretation of the above inequality. That is you have that:

$$\displaystyle \frac{40x^2 - 10x}{11} \leq 3x^2 + \frac{x^2 - x}{2} = \frac{7x^2 - x}{2}.$$

Cross multiplying, this implies that

$$\displaystyle 80x^2 - 20x \leq 77x^2 - 11x.$$

Since $$x$$ must be positive, (so $$x$$ is non-zero), you can divide the above inequality through by $$x$$ to give

$$80x - 20 \leq 77x - 11 \implies 3x \leq 9 \implies x \leq 3.$$

This begs the question, which value of $$x$$ is satisfactory.

You need $$~\displaystyle \frac{4x(4x-1)}{2} = (2x) \times (4x-1)~$$ to be a multiple of $$11$$.

Of the candidate values given by $$~x \in \{1,2,3\}~$$, only $$~x=3~$$ satisfies this constraint.

• Thank you for the answer. I mostly try to understand is my proof correct or not. Can you check it? And only after I understand am i right or not i can look at your proof. Excuse me, maybe i was needed to add a tag "proof verification" to clarify my intentions. Commented Dec 30, 2022 at 7:29
• @Arty See the Addendum that I have just added to my answer. Commented Dec 30, 2022 at 15:19
• Thank you for your answer. Number of games played is needed to be divisible by $11$ because it is natural number and $5 \cdot 4\cdot x(4x-1)$ is not divisible by eleven if $x = 2$ or $x=1$. Did i get it right? Commented Dec 30, 2022 at 17:58
• @Arty $(2x) \times (4x-1)$ needs to be divisible by $11.$ When $~x=1~$ or $~x=2,~$ the product is not divisible by $11$. Commented Dec 30, 2022 at 21:51