Stat Tracking

The first thing we should notice is that $q$ must be rational, since we are looking for points where $q=\frac{m}{n}$ for integers $m,n$ . Let $q=\frac{a}{b}$ for co-prime positive integers $a,b$ .

If $\frac{m}{n} \neq q$ at some point in the season, then we can say that there exists $m,n$ such that $\frac{m}{n}<q$ and $\frac{m+1}{n+1}>q$ . This leads to:

$m<qn \quad \text{and} \quad m+1>qn+q$ $m<qn \quad \text{and} \quad m+1-q>qn$ $m<qn<m+1-q$

Since $q=\frac{a}{b}$ , multiply the last inequality by $b$ :

$bm<an<bm+b-a$

Since $a,b,m,n$ are all integers, there will be no solutions to this inequality for $b-a=1$ . No integers lie between $bm$ and $bm+1$ .

On the contrary, solutions do exist for $b-a>1$ . let $an=bm+1$ . Then the inequality becomes $bm<bm+1<bm+b-a$ which is true. We are guaranteed to be able to find $m,n$ such that $an=bm+1$ if $a,b$ are co-prime, which we assumed to be the case.

Thus, all $q$ of the form $\frac{k}{k+1}$ will work, and the ones less than $.85$ are $\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}$ .

Their sum is $\boxed{3.55}$

If the mathlete gets the first question wrong and all the rest right, then the sequence of ratios is $\tfrac01\,,\,\tfrac12\,,\,\tfrac23\,,\,\tfrac34\,,\,\tfrac45\,,\,\tfrac56\,,\,\tfrac67\,,\,\tfrac89\,,\,...$ The first ratio is $0$ and all ratios after $\tfrac67$ are greater than $0.85$ . Thus the only possible values of $q$ are $\tfrac12\,,\,\tfrac23\,,\,\tfrac34\,,\,\tfrac45\,,\,\tfrac56$ .

If $k = 1,2,3,4,5$ , let $y_j$ be the number of correct answers, and $n_j$ the number of incorrect answers obtained, after $j$ questions have been asked. Then $y_j + n_j = j$ of course, and the ratio $r_j$ of correct answers is $\tfrac{y_j}{y_j + n_j}$ . It is easy to see that $r_j - \tfrac{k}{k+1} \; = \; \tfrac{y_j - kn_j}{j(k+1)}$ and so $r_j$ is less than/equal to/greater than $\tfrac{k}{k+1}$ precisely when $d_j = y_j - kn_j$ is negative/zero/positive respectively. Note now that $d_{j+1}$ is either equal to $d_j + 1$ or else to $d_j - k$ . Suppose that $d_u < 0$ for some integer $u$ , and that $d_v > 0$ for some integer $v > u$ . Since $d_j$ can only increase by $1$ at a time (although it can decrease more extravagantly), the weighted difference $d_j$ must take all integers between $d_u$ and $d_v$ while $u \le j \le v$ (and possibly some others as well). In particular, then, there must be $u < w < v$ with $d_w = 0$ , and hence $r_w = \tfrac{k}{k+1}$ .

Thus the sum of possible values of $q$ is $\tfrac12 + \tfrac23 + \tfrac34 + \tfrac45 + \tfrac56 = \boxed{3.55}$ .

(note: the below solution may be long but it is meant to be easy to understand and a quick read)

At first this problem seems like a lot, so a good first step is to see if we can reduce the number of solutions from infinity to a more manageable number. First realize that the order in which the mathlete wins and loses does not affect the answer, it is only the moment at which he passes the ratio q. So for the rest of this proof, I will assume he loses consecutively, then wins consecutively.

Consider if the mathlete got 1 problem incorrect. Then we would see the ratios: $0,1/2,2/3,3/4,4/5, 5/6 ...$

Notice that if a fraction is not on this list of fractions, then the statement must be false because the list of fractions would provide a counterexample, as it starts at 0 and later becomes >.85.

Now we have reduced it to the 5 possible solutions $\{1/2,2/3,3/4,4/5,5/6\}$ . Let's see if we can reduce it further by finding a counterexample. I consider 1/2 first, listing out the highest number strictly lower than 1/2 for some integer denominator:

$0/1,0/2,1/3,1/4,2/5,2/6 ...$ .

Realize that, if the mathlete ever wants to get an average greater than 1/2, he MUST reach a number on this list at some point. As you can see, starting a win streak at an odd denominator will cause the next number to be equal exactly to 1/2, and starting at an even denominator (or with a lower numerator) will only delay the problem. Next we consider 2/3. We again list out the highest number strictly lower than it for some integer denominator:

$2/4,3/5,3/6,4/7,5/8,5/9,6/10,7/11,7/12 ...$ .

Note again the consistency of the changes in the numerators, there is a repeating pattern of length 3 with 2 increasing numerators and then a numerator that stays the same because if it changed, a/b would equal exactly 2/3. This pattern appears for the other fractions of the form a/(a+1) as well. We conclude that all 5 possible solutions are valid, and the answer is $\boxed{3.55}$ .

Of course, you may want to prove that this pattern really exists and will continue. Let's imagine writing out the numbers for the fraction a/(a+1). Consider the fractions in this list with denominators of $(a+1) * N$ . (where N is some integer). The numerator will of course be $a*N - 1$ . The fractions with a denominator of $a+1)*N-1$ will also have this numerator, because $(a*N-1)/((a+1)*N-1 > a/(a+1) > (a*N-1)/((a+1)*N-1)$ . It is only at these points that the numbers will stay the same because we must have an average ratio of a/(a+1), which can only be achieved by increasing the numerator by 1 on every other fraction (unless of course the numerator increases by 2 somehow, but I think that is clearly impossible and will not prove it.)