Ignor...

, so your solution is probably wrong.

Basically greedy but this took me so long on the actual test. As yayups said, the answer is $\tfrac{50}{51}$. The counterexample for lower $C$ is given by $49$ zeroes and $51$ coins of $\tfrac{50}{51}$. The hard part is to show that this actually works. Order the coins $a_1 < a_2 < ... < a_{100}$. Let $d_i = a_{2i} - a_{2i-1}$. First, partition the coin to $(a_1,a_3,a_5,...,a_{99})$ and $(a_2,a_4,a_6,...,a_{100})$. Then for the $i$-th step, we exchange coins $a_{2i-1}$ with $a_{2i}$. Do this until the left side has the sum greater than or equal to $25-\tfrac{25}{51}$. Suppose that this happens after exchanging $a_{2n-1}$ and $a_{2n}$. If the sum turns out to be less than or equal to $25+\tfrac{25}{51}$, then we are already done. Else, the sum of the first side must be increased by $d_n \geqslant\tfrac{50}{51}$. This means we can partition the coins into two groups, the first group having $100-A$ coins which are all less than $\tfrac{1}{51}$ and the second group having $A$ coins which are all greater than $\tfrac{50}{51}$. As the sum must be $50$, we get the following bounds on the sum. \begin{align*} \frac{50}{51}A < 50 &\implies A < 51 \\ \frac{1}{51}(100-A) + A > 50 &\implies A > 49 \end{align*}So the only case left is $A=50$. In this case, we take $25$ coins from each group. This guarantee the sum to be at least $25\cdot\tfrac{50}{51} = 25-\tfrac{25}{51}$ as desired.

@above denomination were postive reals so how can you take 49 zeroes @below I also agree it is a minor error maybe even the examiner bypassed it

Oh, I actually didn't notice this in the exam (but somehow got a 7). But this is trivial to fix, just take the first 49 coins arbitrarily small and the rest arbitrarily close to $\tfrac{50}{51}$.

oops I did some really annoying argument involving proving if one pile exceeds $\tfrac{1300}{51}$ and the other deceeds $\tfrac{1250}{51}$, then a better splitting exists. It wasn't too bad, and I can give details if anyone wants to see them. Here is the solution presented at test review. The answer is $\boxed{\tfrac{50}{51}}$. This is possible by giving $49$ coins a value of $1$ and $51$ coins a value of $\tfrac1{51}$. Now, suppose the values of the coins are denoted $a_1\le a_2\le\cdots\le a_{100}$. Call a gap between two consecutive elements $a_n$ and $a_{n+1}$ big if $a_{n+1}-a_n>\tfrac{50}{51}$. Claim. A gap can be big only if $n=50$. Proof. Assume that $k<50$ is a big gap. Then, $a_i<\tfrac1{51}$ for all $i\le k$ and $a_i>\tfrac{50}{51}$ for all $i>k$. Hence, $$\sum_{i=1}^{100}a_i>\frac{50}{51}(100-k)\ge\frac{50\cdot51}{51}=50,$$a contradiction. Similarly, if $k>50$ is a big gap, then $$\sum_{i=1}^{100}a_k<\frac1{51}k+(100-k)=100-\frac{50}{51}k\ge100-\frac{50\cdot 51}{51}=50,$$the desired contradiction. $\blacksquare$ It is immediate that $a_{50}\le\tfrac{50}{51}$ and $a_{51}\ge\tfrac1{51}$. Consider \begin{align*} L&=(a_1+a_3+\cdots+a_{49})+(a_{52}+a_{54}+\cdots+a_{100}),\\ R&=(a_2+a_4+\cdots+a_{50})+(a_{51}+a_{53}+\cdots+a_{99}). \end{align*}Compute that \begin{align*} L-R&=a_1+(a_3-a_2)+(a_5-a_4)+\cdots+(a_{49}-a_{48})\\ &\quad-a_{50}+(a_{52}-a_{51})+(a_{54}-a_{53})+\cdots+(a_{100}-a_{99})\\ &\ge a_1-a_{50}\ge-\frac{50}{51}, \end{align*}and \begin{align*} R-L&=(a_2-a_1)+(a_4-a_3)+\cdots+(a_{50}-a_{49})+a_{51}\\ &\quad+(a_{53}-a_{52})+(a_{55}-a_{54})+\cdots+(a_{99}-a_{98})-a_{100}\\ &\ge a_{51}-a_{100}\ge-\frac{50}{51}, \end{align*}whence $|L-R|\le\tfrac{50}{51}$, and we are done. $\square$

Wow this is the first TST problem I managed to solve. First, note that the difference between total values of the two stacks is maximized when $49$ coins are of value $1$, and $51$ coins are of value $\frac{1}{51}$. Playing around with the equality case always helps. In this case, it provides the hint that we should seperate coins with big denominations and those with small denominations. Define a coin to be big if its denomination lies in the interval $\left(\frac{1}{51},1\right]$, and small otherwise. If there is an even number of big coins, they can be split into two stacks (with the same amount coins) such that the absolute difference is no larger than $\frac{50}{51}$. (This can be proved through smoothing.) Similarly, the small coins can be split into two stacks such that the absolute difference is no larger than $\frac{1}{51}$. We can hence split all the coins into two stacks such that the absolute difference is no larger than $\frac{50}{51}$. If there is an odd number of big coins, let the number be $2k+1$. We can split $2k$ of the coins into two stacks (with the same amount coins) such that the absolute difference is no larger than $\frac{50}{51}$. Together with the remaining big coin, we can split all $2k+1$ big coins into two stacks such that the absolute difference is no larger than $1$. Similarly, we can split all small coins into two stacks such that the absolute difference is no smaller than $\frac{1}{51}$ and no larger than $\frac{2}{51}$. We can hence split all the coins into two stacks such that the absolute difference is no larger than $\frac{50}{51}$.

The answer is $C = \frac{50}{51}$. The lower bound is obtained if we have $51$ coins of value $\frac{1}{51}$ and $49$ coins of value $1$. Here was my proof of the other bound. We sort the coins in increasing order $0 < a_1 \le a_2 \le \dots \le a_{100} \le 1$. A large gap is an index $i \ge 2$ such that $a_i > a_{i-1} + \frac{50}{51}$; obviously there is at most one such large gap. Claim: If there is a large gap, it must be $a_{51} > a_{50} + \frac{50}{51}$. Proof. If $i < 50$ then we get $a_{50}, \dots, a_{100} > \frac{50}{51}$ and the sum $\sum_1^{100} a_i > 50$ is too large. Conversely if $i > 50$ then we get $a_1, \dots, a_{i-1} < \frac{1}{51}$ and the sum $\sum_1^{100} a_i < 1/51 \cdot 51 + 49$ is too small. $\blacksquare$ Now imagine starting with the coins $a_1$, $a_3$, \dots, $a_{99}$, which have total value $S \le 25$. We replace $a_1$ by $a_2$, then $a_3$ by $a_4$, and so on, until we replace $a_{99}$ by $a_{100}$. At the end of the process we have $S \ge 25$. Moreover, since we did not cross a large gap at any point, the quantity $S$ changed by at most $C = \frac{50}{51}$ at each step. So at some point in the process we need to have $25-C/2 \le S \le 25+C/2$, which proves $C$ works.

I think i have little bit easier solution. $a_1 < a_2 < ... < a_{100}$ are coins. We consider two cases: $First case$ $a_{100}-a_1<\frac{1}{51}$ Then $a_1 \le a_2,a_3, \cdots a_{100}<a_1+\frac{1}{51}$ Call sum in group with smaller sum $S$ and with larger $L$. Obviously $S \ge 50 a_1 $ and $L <50 a_1+50 \frac{1}{51}$ And finally $$|L-S|<50 a_1+50 \frac{1}{51}-50 a_1= \frac{50}{51}$$which establishes bound . $Second case$ $ a_{100}-a_1 \ge \frac{1}{51}$ Consider next alghoritm: 1.If current sum of numbers $T$,call numbers $a_{i_{1}},a_{i_{2}},\cdots,a_{i_{50}}$, satisfies $25+\frac{1}{51} \ge T \ge 24$ we stop, else 2.Switch $a_i$ with $a_{i+49}$ (from $i=50$ to $i=2$) Start with $ a_1 , a_2 , \cdots , a_{50}$ . Obviously $T<25$ We stop or we switch $a_{50}$ with $a_{99}$ If we did not stop sum increased by $ a_{99}-a_{50}<1$ So we always have $T<25$ (this means we jump by at most $1$ ). If we did not stop in the end we have $$a_1+a_{51}+a_{52}+\cdots+a_{99}<24$$Which is contradiction with $$a_{51}+a_{52}+\cdots+a_{100}\ge25$$So we stoped at some point with numbers $$a_1 , a_{i_{1}}, \cdots , a_{i_{49}}$$and $$25+\frac{1}{51} \ge T \ge24$$if $T \ge24+\frac{1}{51}$ then just select $a_1,a_{i_{1}},\cdots,a_{i_{49}}$ as one group and we are done, and if $T<24+\frac{1}{51}$ then take $$a_{100},a_{i_{1}},\cdots,a_{i_{49}}$$and again we are done. Key in this problem is to notice equality.

problem statement wrote: Consider coins with positive real denominations not exceeding 1. Find the smallest $C>0$ such that the following holds: if we are have any $100$ such coins with total value $50$, then we can always split them into two stacks of $50$ coins each such that the absolute difference between the total values of the two stacks is at most $C$. Merlijn Staps Could a mod please fix this? Thanks. Done; surprised it took so long for someone to notice this. ~dj

The answer is $\boxed{\tfrac{50}{51}}$. We must have $C$ at least this much due to the coins $(\underbrace{\tfrac1{51}, \dots, \tfrac1{51}}_{51 \ \text{times}}, \ \underbrace{1, \dots, 1}_{49 \ \text{times}})$, since the larger side has at least $25$ ones and thus has sum $\geq 25 + \tfrac{25}{51}$, while the other side has sum $\leq 24 + \tfrac{26}{51}$. It remains to show sufficiency. Let the denominations of the coins be $d_1 \leq d_2 \leq \dots \leq d_{100}$, and pair up the coins $(d_1, d_2), (d_3, d_4), \dots, (d_{99},d_{100})$. Claim. The denominations of each pair of coins differ by $\leq \tfrac{50}{51}$. Proof. Suppose not. There is evidently at most one violating pair; let it be $(d_{2k-1}, d_{2k})$. Then we have the inequality \[50 > d_{2k} + d_{2k+1} + \dots + d_{100} > (101 - 2k) \cdot \tfrac{50}{51} \quad \text{or} \quad k \geq 26.\]But then $d_1, \dots, d_{2k-1}$ are all $< \tfrac{1}{51}$, so \[(d_1+d_2+\dots+d_{2k-1}) + (d_{2k} + \dots + d_{100}) < (2k-1)\cdot \tfrac{1}{51} + (101-2k) \cdot 1 \leq 50,\]which is a big booboo. $\square$ Let $L$ be the sum of the weights on the left pan and $R$ be the sum of the weights on the right pan. Sequentially add the pairs of coins onto the pan, one on the left and one on the right. Each pair shifts the difference $L-R$ by $\leq \tfrac{50}{51}$ either up or down, so we can surely stay in the interval $[-\tfrac{50}{51}, \tfrac{50}{51}]$ as desired. Remark. A few bits about the above solution: while it is equivalent to the official solution, I feel that ``adding pairs" onto the scale one by one is more motivated than say, swapping coins. In particular, note the tuple $(\underbrace{0,\dots,0}_{(n-51) \ \text{0's}}, \ \underbrace{\tfrac{50}{51}, \dots, \tfrac{50}{51}}_{51 \ \text{times}})$ actually shows the answer should be the same regardless of the $100$. (Actually, we have to replace the $0$'s with epsilons and stuff, but whoop-de-doo...) This motivates an inductive approach; how can I get from $n$ to $n+2$? A natural way is to simply add a coin on the left, and a coin on the right... but what if this messes things up? These are basically all of the ideas in the above solution. (In the actual contest, I fakesolved the problem with ``swapping" coins. C'est la vie...)

This feels like a nerfed version of IMO 2014/5, but it took me three times as long oops. I'm also slightly worried this is a fakesolve based on Th3Numb3rThr33's commentary; please let me know if I fakesolved. We claim that $C=\frac{50}{51}$. We can show $C\ge \frac{50}{51}$ by taking $51$ coins to have value $\frac{50}{51} - \varepsilon$ and $49$ coins to have value $\frac{49\varepsilon}{51}$ for small $\varepsilon >0$. Now, suppose we had absolute difference between the total values of the two stacks always greater than $50/51$. Say that the coins are partitioned into two stacks, $A,B$. To perform a coin swap, we take coins $a\in A,b\in B$ and move $a$ to $B$ and $b$ to $A$. For any coin $x\in A\cup B$, let $f(a)$ denote its denomination. Observe that any coin swap increases \[\sum_{x\in A}f(x) - \sum_{x\in B}f(x)\]by $2(f(b)-f(a))$. Remark that if \[\sum_{x\in A}f(x) - \sum_{x\in B}f(x) > f(a)-f(b)\]for some $a\in A$ and $b\in B$ with $f(a)>f(b)$, we can apply a coin swap, which always reduces the absolute value of $\displaystyle\sum_{x\in A}f(x) - \displaystyle\sum_{x\in B}f(x)$. In a similar scenario with $f(b)>f(a)$, we could also apply a coin swap. Hence, we may assume that for all $a\in A,b\in B$, we have \[\frac{50}{51} < \left|\sum_{x\in A}f(x) - \sum_{x\in B}f(x)\right| \le |f(a)-f(b)|.\]This implies that we can partition the set $A\cup B = C$ into the following sets of coins: \[C_0 = \left\{x\mid f(x) < \frac{1}{51}\right\},C_1 = \left\{x\mid f(x) > \frac{50}{51}\right\}.\]Remark that We have $|C_1| \le 50$ (otherwise we would have the total sum of the $f(x)$ greater than $50$), We have $|C_1| \ge 50$ (otherwise let $k=|C_1|$ and note the sum of all coins is less than $k+\frac{100-k}{51} = \frac{100}{51} + \frac{50k}{51} \le \frac{100}{51}+\frac{49\cdot 50}{51} = 50$). Hence, we have $|C_1|=50$ and $|C_0|=50$. Now, we find a counterexample to the claim that \[\frac{50}{51} < \left|\sum_{x\in A}f(x) - \sum_{x\in B}f(x)\right|\]for all $A,B$. We specifically consider the following case: we have $25=|A\cap C_1|=|A\cap C_0|=|B\cap C_1|=|B\cap C_0|$. Then, we have the bounds \[25 - \frac{25}{51} = \frac{25\cdot 50}{51}< \sum_{x\in A}f(x),\sum_{x\in B}f(x) < 25+\frac{25}{51}.\]But this implies that for any such $A,B$, we have \[\left|\sum_{x\in A}f(x) - \sum_{x\in B}f(x)\right| < \frac{50}{51},\]absurd. Hence, we are done.

The answer is $C=\tfrac{50}{51}$. To show $C$ cannot be any smaller, consider the coins $(\underbrace{\tfrac{1}{51},\ldots,\tfrac{1}{51}}_{\times 51}, \underbrace{1,\ldots,1}_{\times 49})$. The minimum difference is $50/51$, achieved with stacks (i) 25 of $1/51$ and 25 of $1$, and (ii) 26 of $1/51$ and 24 of $1$. Given any 100 coins, we want to show there exists a subset of 50 coins with sum $K$ such that $|K-(50-K)|\le \tfrac{50}{51}$, i.e. $K \in \left[\tfrac{1250}{51}, \tfrac{1300}{51}\right]$. New problem wrote: Let $0< a_1\le \cdots \le a_{100} \le 1$ be reals that sum to 50. Prove there exists a subset of size 50 whose sum is in $\left[\tfrac{1250}{51}, \tfrac{1300}{51}\right]$. (Call such a 50-subset good.) Perform the following process: Start with $(a_1,a_3,\ldots,a_{99})$. Bump up $a_1$ to $a_2$, then $a_3$ to $a_4$, and so on till $a_{99}$ to $a_{100}$. We will show one the tuples hit during this process is good.

The key is that at each step in the process, the tuple's sum increases by $a_i-a_{i-1}$ for some (even) $i$. Claim: Only $k=51$ can satisfy $a_k-a_{k-1}>\tfrac{50}{51}$. Proof: This is just a calculation using the given properties of $(a_i)$. Since $a_i \in (0,1]$, at most one such index can exist, call it $a_k$. Then $a_i>\tfrac{50}{51}$ for all $i\ge k$, so \[ 50> a_k+\cdots+a_{100}>\tfrac{50}{51}(101-k) \implies k>50. \]Also, $a_i \le 1$ for all $i\ge k$, and $a_i \le \tfrac{1}{51}$ for all $i\le k-1$. So \[ 50=a_1+\cdots+a_{100} \le \tfrac{1}{51}(k-1) + 1(101-k) \implies k \le 52. \]Hence $k=51$ is the only possibility, as claimed. $\blacksquare$ Since $51$ is not even, every increase in the sum in the process will be at most $\tfrac{50}{51}$. So the process will hit something in the desired interval.

Our answer is $C = \tfrac{50}{51}$. Take the example of $49$ coins with value of $1$ and $51$ coins with values $\tfrac{1}{51}$; it is possible to achieve two piles with a difference of $\tfrac{50}{51}$ by having a stack with $25 \cdot \tfrac{1}{51} + 25 \cdot 1$ and another stack with $26 \cdot \tfrac{1}{51} + 24 \cdot 1$. Now, we show that for any $100$ coins $a_1 \leq a_2 \leq \ldots \leq a_{100} < 1$, we can create a pile with value $[\tfrac{1250}{51}, \tfrac{1300}{51}]$; this clearly finishes as the difference between the two sets is then $\leq \tfrac{50}{51}$. We consider the following process; Originally, say we split the coins into stacks $\{a_1, a_3, \ldots , a_{99}\}$ and $\{a_2, a_4, \ldots , a_{100}\}$. Then, we swap $a_1 \iff a_2$, then $a_3 \iff a_4$, and all the way to $a_{99} \iff a_{100}$. I claim that one step of the way, we will have a pile whose value lies in the desired interval. It suffices to show that, no swap adds more than $\tfrac{50}{51}$ to the originally lesser pile. Consider terms of the form $a_{i+1} - a_i$; Clearly at most one of them can exceed $\tfrac{50}{51}$ since $a_j - a_k < 1$ always. We will prove that this unique index $i$ for which $a_{i+1} - a_i > \tfrac{50}{51}$ is $i = 50$. Clearly $a_k > \tfrac{50}{51}$ for all $k \geq i+1$ hence\[\tfrac{50}{51} (100-i) < a_{i+1} + \ldots + a_{100} < 50 \implies i > 49\]and furthermore since $a_k \leq \tfrac{1}{51}$ for all $k \leq i$ it follows that\[50 \geq \tfrac{i}{51} + (100 - i) \implies i \leq 51\]Hence $i \in \{50, 51\}$. We check that if $i = 51$, then we have that the largest gap has value $\tfrac{50}{51}$, which is not greater than $\tfrac{50}{51}$, hence $i$ must be $50$, as desired. Hence, the only possible swap that could add more thant $\tfrac{50}{51}$ to the originally lesser pile is $a_{50} \iff a_{51}$, but this is not performed in our operation, so we are done. $\blacksquare$

I think I found a non-greedy approach? The answer is $\boxed{C = \tfrac{50}{51}.}$ Note that if $49$ coins have value $1$ and $51$ coins have value $\tfrac{1}{51}$, this maximum is attained. Now order the coins $1 \geq c_1 \geq c_2 \geq \dots \geq c_{100}$. Consider the $50$ pairs $(c_1,c_{100}), (c_2,c_{99}), \dots, (c_{50},c_{51})$. Now create two sets $A$ and $B$, and place both coins in the $i$th pair in $A$ if $i$ is odd, and $B$ if $i$ is even. I claim that this partition works. We seek to show that $$|(c_1-c_2+\dots+c_{49}-c_{50}) - (c_{51}-c_{52}+\dots+c_{99}-c_{100})| \leq \tfrac{50}{51}.$$In fact, I claim that the terms under both parentheses at most $\tfrac{50}{51}$. It suffices to show that $c_{50} \geq \tfrac{1}{51}$ and $c_{51} \leq \tfrac{50}{51}$, which is evident as if otherwise the sum of the $100$ coins would either be larger or smaller than $50$. This concludes the proof.

We claim the answer is $\frac{50}{51}$, achieved when we have $51 \cdot \frac{50}{51} + 50 \cdot 0$. The idea is that we draw two zones of sizes $1$ and $\frac{50}{51}$ that are each centered at $25$. The difference between the ends of these two zones is $\frac{1}{2} \cdot \frac{1}{51} = \frac{1}{102}$. Call the inner $\frac{50}{51}$ region Chad, and the region that is inside $1$ but outside Chad is called Sad. Consider all choices of subset sums $A, B$ where $A < B$ and $B - A$ is minimized. We already find that there must be some $A$ that lands in the $1$-region simply because of the "not exceeding 1" condition. If it lands in Chad, we are done, so assume it lands in the leftmost Sad such that $A < 50 - A$ holds. Now, call a small element one that is $\leq \frac{50}{51}$, and big otherwise. If $B$ has a small element, then we just move it to $A$ which moves $A$ closer to the center, so henceforth assume that $B$ only has big elements. Break into cases: $\textbf{Case 1:}$ $A$ has a small element that is $\geq \frac{1}{51}$. Swapping with a big element in $B$ finishes. $\textbf{Case 2:}$ All elements in $A$ are either small that are $< \frac{1}{51}$, or are big. Throughout both $A$ and $B$, there must be at least $50$ big elements because otherwise $49 \cdot 1 + 51 \cdot \frac{1}{51} < 50$, which is bad. However, for the case of $51$ big elements, the only way for this to happen is for everything to be $\frac{50}{51}$, which we already resolved. Thus we have exactly $50$ big elements. At this point, we can rebuild the sets by pairing off the $50$ small elements and $50$ large elements. Each difference is at most $\frac{1}{51}$, and so $50 \cdot \frac{1}{51}$ is the biggest possible and we are done.

The answer is $\frac{50}{51}$; to see that we can't do better, consider $\underbrace{\frac{50}{51}, \dots , \frac{50}{51}}_{51},\underbrace{0,\dots , 0}_{49}$. Now we show $C=\frac{50}{51}$ indeed works; in particular, we show if the weights are $a_1 \ge a_2 \ge \cdots \ge a_{100}$, then it works to let one of the groups be $a_1, a_3, a_5, \dots , a_{49}, a_{52}, a_{54}, a_{56}, \dots , a_{100}$. It suffices to demonstrate that $a_1+a_3+ \cdots + a_{49}+a_{52}+ \dots + a_{100}$ is between $25-\frac{25}{51}$ and $25+\frac{25}{51}$. Indeed, note that \begin{align*} \frac{51}{25} ( a_1+a_3+ \cdots + a_{49} ) &\ge 2a_1+2a_3+ \cdots + 2a_{47}+3a_{49} \\ &\ge a_1+a_2+ \cdots + a_{51} \\ &= 50 - (a_{52}+a_{53}+ \cdots + a_{100} )\\ &\ge 50 - (2a_{52}+2a_{54}+ \cdots + 2a_{98}+ a_{100} ) \end{align*}hence \begin{align*} &(a_1+a_3+ \cdots + a_{49})+ a_{52}+a_{54}+ \cdots + a_{100} \\ &\ge \frac{50(25)-50a_{52}-50a_{54}- \cdots - 50a_{98}-25a_{100}}{51}+a_{54}+ \cdots + a_{100}\\ &\ge \frac{50(25)}{51}=25-\frac{25}{51}. \end{align*}Similarly, \begin{align*} \frac{51}{25} (a_{52}+a_{54} + \cdots + a_{100} ) &\le 3a_{52}+2a_{54}+ \cdots + 2a_{100} \\ &\le a_{50}+a_{51}+ \cdots +a_{100} \\ &= 50-(a_1+a_2+ \cdots + a_{49}) \\ &\le 50-(a_1+2a_3+ \cdots +2a_{47}+2a_{49}) \end{align*}hence \begin{align*} &a_1+a_3+ \cdots + a_{49}+ (a_{52}+a_{54}+ \cdots + a_{100}) \\ &\le a_1+a_3+ \cdots + a_{49}+\frac{50(25)-25a_{1}-50a_{54}- \cdots - 50a_{98}-50a_{100}}{51}\\ &\le \frac{50(25)+50}{51} = 25+\frac{25}{51}. \end{align*}We're done.

The answer is $\frac{50}{51}$. Construction: \[\frac{50-49\epsilon}{51},\frac{50-49\epsilon}{51}, \dots, \frac{50-49\epsilon}{51}, \epsilon, \epsilon \dots, \epsilon\]where there are $49$ $\epsilon$ and $51 $ $\frac{50-49\epsilon}{51}$, and $\epsilon$ is sufficiently small. Proof of Bound: We want to show that we will always be able to split the coins into 2 groups such that they have amounts $25+x,25-x$ and such that $x \leq \frac{25}{51}$. We first show $x \leq \frac{1}{2}$. To do this, put the coins into two stacks, $A$ and $B$, with $A$ having all the coins at first. Also, define $S_A$ as the sum of the values of the coins in $A$, and define $S_B$ similarly. Now continue moving coins from $A$ to $B$ one by one, but stop when the next coin you will move from $A$ to $B$ will cause $S_B \geq S_A$. The next move causes $S_B \geq S_A$, i.e. their difference is at most $1$, so we have proved the bound. Now assume we have two stacks $A,B$ such that there exists no way to make the value of $x$ for that configuration of coins, smaller. Now note from the previous paragraph, $\frac{1}{2} \geq x$. WLOG let $S_B \leq S_A$, so $S_A=25+x$ and $S_B=25-x$ ($x \geq 0$). The key thing to realize is that every coin in $A$ is at least $2x$. This is because otherwise, we could choose the smallest coin in $A$, move it to $B$, and decrease the difference. In addition, FTSOC, assume $x>\frac{25}{51}$. Now combining everything, we have the inequality \[25+\frac{1}{2} \geq 25+x \geq y(2x)>\frac{50y}{51}\]where $y$ is the number of coins in $A$. This quickly implies $y \leq 26$, but note that if $y \leq25$, the maximum value of $S_A$ is $y \leq 25$, so $y=26$. We can now finish as \[25+x \geq 52x \implies x \leq \frac{25}{51}\]and we are done.

tastymath75025 wrote: Consider coins with positive real denominations not exceeding 1. Find the smallest $C>0$ such that the following holds: if we have any $100$ such coins with total value $50$, then we can always split them into two stacks of $50$ coins each such that the absolute difference between the total values of the two stacks is at most $C$. Merlijn Staps We claim that $\boxed{\mathcal{C}=\frac{50}{51}}$ is the smallest possible value. Define \begin{align*} \mathcal{A}_1=\{a_1,a_2,..............,a_{100}\} \\ \mathcal{B}_1=\{a_1,a_2,..........,a_{100}\} \end{align*}where $\{a_1,a_2,..........,a_{100}\}$ is the denominations of all the coins. Claim:- The denominations of each pair of consecutive coins differ by $\leq \tfrac{50}{51}$. Proof:- Direct computation.$\blacksquare$ Now for every stack delete elements from $\mathcal{A}$ or $\mathcal{B}$ depending on which set the element was in. Each pair decreases the sum $\mathcal{A}_1+\mathcal{B}_1$ by atleast $a_j-a_{j-1} \le \frac{50}{51}$,and since the deduction on both sets is the same,hence $$2 (\mathcal{A}_1-\mathcal{B}_1) \ge 100 \cdot \frac{51}{51}-100 \cdot \frac{50}{51}$$implying $\boxed{\mathcal{C}=\frac{50}{51}}$ works. For a construction,pick $\underbrace{\frac{50}{51}, \dots , \frac{50}{51}}_{51},\underbrace{0,\dots , 0}_{49}$

The answer is $\boxed{\frac{50}{51}}$. Proof of sufficiency: Call the denominations of the coins $a_1\le a_2\le \cdots\le a_{100}$. Let \[x=a_1+a_3+a_5+a_7+\ldots+a_{49}+a_{52}+a_{54}+a_{56}+\ldots+a_{100}\]and \[y=a_2+a_4+\ldots+a_{48}+a_{50}+a_{51}+a_{53}+\ldots+a_{99}.\] We have \[x-y=a_1+(a_3-a_2)+(a_5-a_4)+(a_7-a_6)+\ldots+(a_{49}-a_{48})-a_{50}+(a_{52}-a_{51})+(a_{54}-a_{53})+\ldots+(a_{100}-a_{99})\ge a_1-a_{50}.\] Claim: $a_{50}\le \frac{50}{51}$. Proof: Suppose $a_{50}>\frac{50}{51}$. Then $a_{51}, a_{52},\ldots,a_{100}>\frac{50}{51}$. So $\sum_{i=1}^{100}a_i>50$, a contradiction $\blacksquare$- So $x-y\ge a_1-a_{50}\ge -\frac{50}{51}$. Now we look at $y-x$. We have \[y-x=(a_2-a_1)+(a_4-a_3)+\ldots +(a_{50}-a_{49})+a_{51}+(a_{53}-a_{52})+(a_{55}-a_{54})+\ldots +(a_{99}-a_{98})-a_{100}\ge a_{51}-a_{100}\] Claim: $a_{51}\ge \frac{1}{51}$. Proof: Suppose for the sake of contradiction that $a_{51}<\frac{1}{51}$. We note that $\sum_{i=52}^{100}a_i\le 49$ with equality iff $a_{52}=a_{53}=\cdots=a_{100}$. Also note that $\sum_{i=1}^{51}a_i<1$. So $\sum_{i=1}^{100}a_i<50$, a contradiction. $\blacksquare$ Thus, $y-x\ge a_{51}-a_{100}\ge \frac{1}{51}-1=-\frac{50}{51}$. Claim: $|x-y|\le \frac{50}{51}$. Proof: Suppose $|x-y|>\frac{50}{51}$. Then one of $\{x-y,y-x\}$ is less than $-\frac{50}{51}$, but we have already proven that they are both greater than $-\frac{50}{51}$. $\blacksquare$ Note that $x$ and $y$ are two stacks of $50$ coins and their absolute difference is at most $\frac{50}{51}$, which proves that $\frac{50}{51}$ is indeed a valid value of $C$. Proof of necessity: Let $\epsilon$ be a sufficiently small positive real. We can allow \[a_1=a_2=\ldots=a_{49}=\epsilon\]and \[a_{50}=a_{51}=\ldots=a_{100}=\frac{50-\epsilon}{51}.\]It's easy to see that this works.

Hi y'all. Visiting this site after 4 years. Last time I visited this site I was still in highschool, and it's a very nostalgia infused good trip for sure. < 3 Coming to the problem : 1. First arrange all the 100 coins in ascending order. Observe that to minimize difference in the two stacks, the procedure described below is optimal. 2. Put the 1st and 100 th coin in stack A. Put 2nd and 99th coin in stack B. Put 3rd and 98th coin in stack C. Continue. Proof? Very easy by induction/recursion. See that it's true for 4 coins adding to 2, and then inductively prove for 4n coins adding to 2n. Alternatively, recursively break the 8 coin case to two 4 coin case, and so on. 3. Observe that you get 0 difference under sum of coin values of stack A vs stack B for constant or linear distributions, under procedure 2. 4. Observe a cool pattern : for quadratic distribution [ith coin has value a*i^2 where a is a normalizing constant such that sum over all 100 coins is 50] , the difference between the 1st coin and the 2nd coin, the 3rd coin and the 4th coin, the 5th coin and the 6th coin; increases linearly. Even though this is a really cool pattern, it's essentially just because the discrete derivative of quadratic is linear. Also observe that if we follow procedure 2, the difference between stack A and stack B is simply value of the 50th coin. This is also obvious since the difference between the corresponding coins on two stacks arranged in ascending fashion flips sign after the 25th coin, so in the linear graph of the discrete derivatives, the difference between the area of left side [from 1 to 50] and right side [51 to 100] is simply the area of the rectangle after you subtract the triangle from the upper right side which is congruent to the left triangle. That area of the rectangle is simple 50a*50 = value of 50th coin. 5. Agreed point 4 was mostly a digression, but the point was to show my thinking process. Notice that the difference in the two stacks is equal to the value of the coin at the centre of mass of the distribution of the discrete derivative. So for an x^3 distribution, the difference is greater than the value of the 50th coin, for an x^4 even more, and so on. 6. Ideally then, the maximum difference can be created by having an impulse of value 50 at the end, and all other coins having epsilon small amounts. However it violates the criteria that value of each coin shouldn't exceed 50. 7. However, with some cleverness, the problem is avoided by superimposing the impulse/delta distribution with a summation of a delta distribution, aka : having a step function having epsilon small values till 0 to 49 and then value c from 50 to 100. [Basically superimposing a delta at 50 with an evenly distributed step function]. Such a construction gives value c= 50/51 8. Notice under procedure 2, 50/51 is the difference that comes out. Hence it is the maximal difference.

Sprites wrote: Claim:- The denominations of each pair of consecutive coins differ by $\leq \tfrac{50}{51}$. Proof:- Direct computation.$\blacksquare$ Unfortunately wrong. Just take the trivial counterexample $(0,0,\dots,0,1,1,\dots,1)$. (It also helps to read previous posts which took care of exactly this issue.)

The answer is $C=\frac{50}{51}$, which is achieved by 49 arbitrarily small coins and 51 coins arbitrarily close to $\tfrac{50}{51}$. I will now show that this works. Let the coin values be $a_1\leq a_2 \leq \cdots \leq a_{100}$. Then, abusing notation, we partition the 100 coins into \begin{align*} L&=(a_1+a_3+\cdots+a_{49})+(a_{52}+a_{54}+\cdots+a_{100}),\\ R&=(a_2+a_4+\cdots+a_{50})+(a_{51}+a_{53}+\cdots+a_{99}). \end{align*}Then we have $$L-R=a_1+(a_3-a_2)+(a_5-a_4)+\cdots+(a_{49}-a_{48})-a_{50}+(a_{52}-a_{51})+\cdots+(a_{100}-a_{99})\geq a_1-a_{50}.$$Suppose $a_1-a_{50}=-x$ where $x>0$. Then $$50=a_1+\cdots+a_{100}=(a_1+\cdots+a_{49})+(a_{50}+\cdots+a_{100})\geq 49a_1+51(a_1+x)=100a_1+51x\geq 51x,$$from which it follows that $x \leq \tfrac{50}{51}$, so we have $L-R\geq -\tfrac{50}{51}$. Likewise, we have $R-L\geq a_{51}-a_{100}:=-y$, and $$50=a_1+\cdots+a_{100}=(a_1+\cdots+a_{51})+(a_{52}+\cdots+a_{100})\leq 51(a_{100}-y)+49a_{100}\leq 51(1-y)+49$$as $a_{100}\leq 1$, so $y \leq \tfrac{50}{51}$ and we have $R-L\geq -\tfrac{50}{51}$ as well. Combined, this implies that $|L-R|\leq \tfrac{50}{51}$ as desired. $\blacksquare$

The answer is $\frac{50}{51}$. Consider the case where 49 coins have value $\epsilon$ and 51 coins have value $\frac{50}{51} - \frac{49}{51}\epsilon$ where $\epsilon > 0$. Then, as $\epsilon$ approaches 0, the minimum absolute difference between the two stacks' values approach $\frac{50}{51}$, since the coins with value $\frac{50}{51} - \frac{49}{51}\epsilon$ cannot be distributed uniformly across the two stacks due to there being an odd number of them, so one stack have to get at least one more of these coins. This implies that no $C < \frac{50}{51}$ is possible. Now we prove that a difference of at most $\frac{50}{51}$ is achievable. Let the coins' values be $0 < a_1 \le a_2 \le \dots \le a_{100} \le 1$. Pair up the coins as $(a_1, a_2), (a_3, a_4), \dots, (a_{99}, a_{100})$. Start with two empty stacks, $A$ and $B$. Then, for $x = 1, 2, \dots, 50$ in order, we consider a single pair $(a_{2x - 1}, a_{2x})$, then add coin $a_{2x-1}$ to stack $A$ and $a_{2x}$ to stack $B$ if the current total value of $A$ is greater than that of $B$, and vice versa otherwise. It is easy to see that at no point in this process does the absolute difference in value between stacks $A$ and $B$ exceed $\max_{x=1}^{50} a_{2x} - a_{2x-1}$. Therefore, if $\max_{x=1}^{50} a_{2x} - a_{2x-1} \le \frac{50}{51}$, we are done. Suppose that's not the case. Then there is a greater than $\frac{50}{51}$ gap between the values of two adjacent coins when sorted by value. Therefore, there can be no coin with value between $\frac{1}{51}$ and $\frac{50}{51}$, inclusive; in other words, each coin's value is either less than $\frac{1}{51}$ (type 1) or greater than $\frac{50}{51}$ (type 2). There can be at most 50 type 2 coins for the total value of the coins to be 50; similarly there can be at most 50 type 1 coins, so there are exactly 50 coins of each type. Now we set up two stacks, each with 25 arbitrarily chosen type 1 and 25 arbitrarily chosen type 2 coins. The absolute difference in value between two coins of the same type is less than $\frac{1}{51}$, and the difference in value between these stacks is the sum of differences between 50 such pairs, which is at most $\frac{50}{51}$ as desired.

We claim the answer is $\frac{50}{51}$. Taking $51$ coins as $\frac{50}{51}-\epsilon$ and $49$ coins as $\frac{51\epsilon}{49}$ for sufficiently small $\epsilon$ we see that $C\ge \frac{50}{51}$. Let the coins be $a_1\le a_2\le\dots\le a_{100}$. For $0\le i\le 50$ let \[S_i=\sum_{j=0}^i a_{2j}+\sum_{j=i+1}^{50} a_{2j-1}\]In essence, we are changing one odd index into an even index at a time from $1$ to $99$. Notice $S_i-S_{i-1}=a_{2i}-a_{2i-1}$ so $S_0\le S_1\le\dots\le S_{50}$. Also, $2S_0\le S_0+S_{50}=50$ and $2S_{50}\ge S_{50}+S_0=50$. Hence, as we are progressing from $S_0$ to $S_{50}$, some $S_i$ must be contained the interval $\left[24-\tfrac{25}{51},24+\tfrac{25}{51}\right]$ unless $S_k-S_{k-1}>\tfrac{50}{51}$ for some $k$. Suppose FTSOC that such $k$ exists, so $a_{2k}>a_{2k-1}$. Then, $a_{2k},a_{2k+1},\dots,a_{100}>\tfrac{50}{51}$ so \[(100-2k+1)\cdot \frac{50}{51}<a_{2k}+a_{2k+1}+\dots+a_{100}\le 50\]which implies $k>25$. On the other hand, $a_{2k-1},a_{2k-2},\dots,a_1<\tfrac{1}{51}$ so \[50=(a_1+\dots+a_{2k-1})+(a_{2k}+\dots+a_{100})<(2k-1)\cdot \frac{1}{51}+(100-2k+1)\cdot 1\]which implies $k<26$. Hence, there are no valid values of $k$ and we are done. $\square$

We claim that $C = \frac{50}{51}$. Claim: $C$ is twice the minimum distance between a sum of $50$ coins and $25$. Proof. If one pile has sum $a$, then the other has sum $50 - a$ and the difference is $|a - (50 - a)| = |2a - 50| = 2|a - 25|$ $\blacksquare$ Claim: $C \ge \frac{50}{51}$ Proof. Consider having $100$ coins with $51$ occurences of $\frac{50 - 40\varepsilon}{51}$ and $49$ occurences of $\varepsilon$ as $\varepsilon \to 0$. The difference between the two is at minimum $\frac{50 - 40\varepsilon}{51} - \varepsilon$, which approaches $\frac{50}{51}$. $\blacksquare$ We now give a construction for $C = \frac{50}{51}$. Take a configuration of coin values $a_1 \le a_2 \le \dots \le a_{100}$. Claim: $a_{i+1} - a_{i} > \frac{50}{51}$ can only hold if $i = 50$. Proof. The sum of $a_1 + a_2 + \dots + a_{100}$ takes on the minimum value of $\frac{50}{51} \cdot (100 - i)$ and the maximum value of $\frac{1}{51} \cdot i + (100 - i)$. $50$ only lies between the two when $i = 50$. $\blacksquare$ Now consider the tuples $(a_1, a_2), (a_3, a_4), \dots (a_{99}, a_{100})$. Consider the sums made by choosing an element from each tuple. This has a minimum value of $a_1 + a_3 + \dots + a_{99}$ and a maximum one of $a_2 + a_4 + \dots + a_{100} = 50 - (a_1 + a_3 + \dots + a_{99})$. By starting with the minimum sum and exchanging $a_i$ for $a_{i+1}$ each time, we eventually end up at the maximum sum. Since $a_{i+1} - a_i \le \frac{50}{51}$ for $i \ne 50$, by discrete continuity at some step the sum is in \[ [25 - \frac{25}{51}, 25 + \frac{25}{51}], \]meaning that $C$ works for this configuration.

We claim the answer is $\frac{50}{51}$. Let $a_1, a_2, \dots, a_{100}$ denote the value of the $100$ coins, where $a_k$ is ordered from least to greatest. We use the greedy algorithm. Group the coins $\left( a_2, a_4, \dots, a_{100} \right)$ and $\left( a_1, a_3, \dots, a_{99} \right)$. Call the groups $A$ and $B$, respectively. Denote $$\sum_{2 \mid i} a_i = n$$We will keep swapping $a_{2i}$ with $a_{2i-1}$ (We swap $(a_1,a_2)$ and then $(a_3, a_4)$ etc.) We will prove that eventually the difference in total value of the groups will be less than or equal to $\frac{50}{51}$. We need to prove that the sum of elements in group $A$ will need to eventually be in the interval $\left [ \frac{1250}{51}, \frac{1300}{51} \right ]$. Denote $b_i = a_{i+1}-a_i$. Notice that after swapping coins $a_{2i}$ and $a_{2i-1}$ you are decreasing the total value of $A$ by $b_{2i-1}$. Note that $$\sum_{i} b_i = a_{100} - a_1 < 1$$This implies there can be at most one $b_i$ such that $b_i > \frac{50}{51}$. If there are no values of $b_i$ that satisfy this, we are obviously done. Now FTSOC assume there exists exactly one $b_k = a_{k+1}-a_k > \frac{50}{51}$. Notice that $k$ must be the greatest value that satisfies $n - b_1 - b_3 - \dots - b_{k-2} > \frac{1300}{51}$. The LHS of the expression is equal to $$a_{k+1}+a_{k+3}+ \dots + a_{100} + a_1 + a_3 + \dots + a_{k-2} < \left( a_{100} - \frac{50}{51} \right) \cdot \frac{k-1}{2} + a_{100} \cdot \frac{101-k}{2} = 50a_{100} - \frac{50(k-1)}{102}$$which we get $k < 51$ from. Now since, k is odd this can be reduced to $k \le 49$. Therefore there are at least $51$ coins whose value is at least $\frac{50}{51}$ greater than $a_k$. This gives a contradiction. To finish, we note that no value below $\frac{50}{51}$ works because of the construction with $49$ coins with denomination $1$ and $51$ coins with denomination $\frac{1}{51}$. With this construction, simply select $24$ coins with denomination $1$ to be in one group, and the other $25$ coins to be in another group.

Answer is $\frac{50}{51}$. For construction, make $51$ copies of $\frac{50}{51}$ and $49$ copies of $0$. Now, consider two sets of coins such that $C$ is minimized. I claim that $C = \frac{50}{51}$; in particular, let $A$ and $B$ be the two sets of coins, and say $S(A)-S(B) = C > \frac{50}{51}$. Label the coins in $A$ as $w(x_1) \leq w(x_2) \leq \cdots \leq w_(x_{50})$, and similarly label the coins in $B$ as $w(y_1) \geq w(y_2) \geq \cdots \geq w_(y_{50})$. Then, notice that for any coin $c_1$ in $A$ and $c_2$ in $B$, if $w(c_1) > w(c_2)$, then $w(c_1) - w(c_2) > C$, otherwise we have a contradiction of minimality. It follows that all coins have value either less than $1-C$ or greater than $C$. Claim. Exactly $50$ of the coins have value greater than $C$. Proof. Otherwise the sum of all weights cannot equal $50$ by simple bounding. $\blacksquare$ So now, suppose that the last $k$ coins in $A$ have weight greater than $C$; let $w(x_i) - w(y_i) = a_i$ for each of these $i$. Similarly, the first $50-k$ coins in $B$ will have weight greater than $C$, and let $w(y_i) - w(x_i) = b_i$ for each of these $i$. For maximality reasons we must have $b_i > a_j$ for any $i, j$. Furthermore, we have the equation $$b_1+b_2+\cdots + b_{50-k} + C = a_1 + a_2 + \cdots + a_k.$$Notice that $k \geq 26$, but then $$b_1+b_2+\cdots+b_{50-k}+C < 50-k+C < kC < a_1+a_2+\cdots+a_k.$$This is a contradiction.

We claim the answer is 50/51. If we have 51 coins of value $50/51-49\epsilon$ and 49 coins of value $51\epsilon$ for sufficently small $\epsilon,$ then we cannot achieve any better since we have an odd number of the $50/51$ coins. We say that a coin is massive if its value is strictly greater than $50/51$. We aim to show that we can always select a subset of 50 coins whose sum is between $25-(25/51)$ and $25+(25/51)$ inclusive. Case 1: There are no massive coins. Then, consider selecting the 50 smallest coins. The sum is obviously at most 25. If it is at least $25-(25/51)$, this set is already good. Otherwise, keep "flipping" so that in each step, a coin is swapped out and replaced with one of the 50 bigger coins, until after 50 steps our set becomes the 50 biggest coins. Then, since there are no massive coins, our value increases by at most $50/51$ after each step. Thus, we could not have "skipped over" the interval $25-(25/51)$ to $25+(25/51)$ of length $50/51$, hence there exists a valid set of 50 as one of our intermediate steps. Case 2: There are an even positive number of massive coins. Suppose that there are $2m$ massive coins. Then, put the $m$ smallest massive coins in one set, and the $m$ largest ones in the other. The difference at this point is at most $$\frac{1}{51}m.$$Clearly, $m\leq 25$, since otherwise the sum would be larger than 50, so the difference is at most $25/51$ between the two sets at this point. Now, pick the smallest half of the remaining coins and add it to the smaller set, and perform the "flipping" trick again, noting that the remaining coins do not contain any massive coins, so it still must land in between at some point (since the ending state, picking the bigger half of the remaining coins to add to the smaller massive set, is large enough since we have at most a $25/51$ deficit) Case 3: There are an odd number of massive coins. Suppose that there are $2m+1$ massive coins. Then, pick the $m+1$ smallest massive coins to add to set $A$, and the $m$ largest ones to add to set $B$. Note that $$\frac{50}{51}(m+1)> m,$$since $m$ is obviously less than 50, so $A$ is currently in the lead. Let $V$ be the total value of the massive coins. Consider two choices. Choice 1: Pick the $m+1$ smallest massive coins and the $49-m$ smallest non-massive coins for set $A$. The value of this is at most $$\frac{m+1}{2m+1}V+\frac{49-m}{99-2m}(50-V)$$since the coins we are choosing are "below average" wrt their respective groups. We claim that $$\frac{m+1}{2m+1}V+\frac{49-m}{99-2m}(50-V)\leq 25+\frac{25}{51}.$$Note that for $0\leq m\leq 24$, which is the possible range of $m$, we have that $$\frac{49-m}{99-2m}<\frac{m+1}{2m+1},$$so this is increasing in $V$. Hence, it suffices to show the inequality in the "worst case", which is when $V$ is the largest, i.e. $V=2m+1$. Then, it just becomes $$\frac{m+1}{2m+1}(2m+1)+\frac{49-m}{99-2m}(50-(2m+1))\leq 25+\frac{25}{51}$$for $0\leq m\leq 24.$ The left hand simplifies to $$m+1+\frac{(49-m)(49-2m)}{99-2m}.$$This expands to $$m+1+\frac{2m^2-147m+49^2}{99-2m}.$$Using partial fractions, this is $$(m+1)+(24-m)+\frac{25}{99-2m}$$which is just $$25+\frac{25}{99-2m}.$$This is clearly increasing on $[0,24]$, so its maximum value is $25+\frac{25}{51}$ at $m=24$, hence the claim is proven. Thus, this choice gives us a value of at most $25+\frac{25}{51}$ in $A$. Choice 2: Pick the $m+1$ smallest massive coins and the $49-m$ largest non-massive coins for $A$. Now, we have already shown that the $m+1$ smallest massive coins take up more than half of the total value of the massive coins. Hence, compare $A$ to the "half set". $A$ is ahead of the half set in massive coins. $A$ is also taking the largest non-massive coins, so its only deficit is that it contains half a non-massive coin less than the half set. Hence, the deficit that $A$ has is at most $25/51$, so the value of $A$ in this case is at least $25-25/51$, i.e. big enough. Now, perform the "flipping" trick of swapping out small non-massive coins with large ones yet again to go from Choice 1 to Choice 2 in $49-m$ moves, we will hit the $50/51$ wide interval at some point during the process since we are only swapping in and out non-massive coins, and we start out small enough and end large enough. Hence, we are done.

We claim the answer is $\frac{50}{51}$. proof it is lower bound: pick $51$ coins to be $\frac{50}{51}-\frac{49 \delta}{51}$ called high value coins and $49$ coins to be $\delta$ called low value coins where $\delta$ is arbitrarily small in nature. one group clearly has 26 coins of high value and other stack has 25 coins of high value hence having a difference of ~$\frac{50}{51}$ now assume FTSOC assume there was $c > \frac{50}{51}$ firstly take all coins with value $> \frac{50}{51}$, there exist at max $50$ of such coins and place them into two groups arbitrarily. this ensures neither group exceeds $25$. We claim there the larger group never exceeds $25+\frac{25}{51}$ and this is equivalent to $c \leq \frac{50}{51}$ proof: firstly denote sum of values of coins in stack 1 and stack 2 as $s_1$ and $s_2$ respectively. WLOG $s_1 > s_2$. if $s_1 = 25 + x$ then $s_2 = 25-x$ then $s_1 - s_2 = 2x$ hence we show that the given statement is equivalent to showing $c \leq \frac{50}{51}$ Now lets prove the claim. firstly take all coins of value x, $x > \frac{50}{51}$ and place them alternating in stack 1 and 2 clearly both stacks have value $\leq 25$ as there are at max 50 coins and each has value $\leq 1$ now place the remaining coins greedily into stack with lowest value. Assume at coin with value x a stack goes over $25+\frac{25}{51}$ since this stack was the one with initially lower value we can state all stacks will go over $25 + \frac{25}{51}$ which implies $s_1 + s_2 + 2x > 50+\frac{50}{51}$ but we have that $s_1 + s_2 + x \leq 50$ and $x \leq \frac{50}{51}$ which gives us a contradiction finishing the proof

The answer is $\frac {50}{51}$ and it can be obtained if we have $51$ coin of value $\frac {1}{51}$ and $49$ coins of value $1$. To show it works let $R $ be the sum of the values in the first stack and $L$ is the sum of the values in the second stack. then let $R = 25 + \frac{25}{51}$ and $L= 24+ \frac{26 }{51}$. Now to prove it's sufficient let $a_1,a_2,......,a_{100}$ be the values of the coins and assume that $a_1 \leq a_2 \leq .... \leq a_{100}$. Pair up the coins $(a_1,a_2),(a_3,a_4),....,(a_{99},a_{100})$. Claim: the difference between value of a pair is at most $\frac {50} {51}$. But if there is then it must be $a_{51} > a_{50} + \frac{50}{51}$ . Assume for the sake of contradiction that for some $j<50 $ there exist a pair violating our claim. This means that $a_j,a_{j+1},.......,a_{100}>\frac{50}{51}$,but this implies that: $\sum_1^{100} a_i > 50$ conversely for $j>50$ this means that $a_1,a_2,....a_{i-1}< \frac{1}{51}$ we get that. $\sum_1^{100} a_i < 49+\frac{1}{51} \cdot 51$ A contradiction. Now add the pair to the stacks, each pair will increase or decrease $R-L$ by $x$ where ($-\frac{50}{51} \leq x \leq \frac{50}{51}$) which means that it will stay in the interval $[-\frac{50}{51}, \frac{50}{51}]$ So we are done.

The answer is $\frac{50}{51}$, which is achieved with $51$ coins of $\frac{50}{51}-49\varepsilon$ and $49$ of $51 \varepsilon$. Suppose there is a set of coins whose $C$ value is $C_1>\frac{50}{51}$. Therefore, we split the coins into pile $A$ with sum $25-\frac{C_1}{2}$ and $B$ with sum $25+\frac{C_1}{2}$. Claim: For any coin $a\in A$ and $b\in B$, we can't have $0<b-a<C_1$. Proof. If we did, swap $a$ and $b$ and the total difference $B-A$ decreases by a positive number less than $2C$. $\blacksquare$ Now, by pigeonhole, there must be a coin in $A$ with value less than or equal to $\frac{25-\frac{C_1}{2}}{50}=\frac 12 -\frac{C_1}{100}$. But if we had a coin with value $1-C_1<k<\frac 12 -\frac{C_1}{100}$, the claim says that there is no coin in $B$ with value above that coin, contradiction as $B$ has a larger sum. Therefore, there is no coin in this interval, so there is a coin in $A$ with $0<k<1-C_1$, and let $k$ be maximal. Now, every coin in $B$ is at most $k$ or at least $k+C_1$. Say there are $j$ coins on the first type, so the sum of the coins is then at most \[j\cdot k+ (50-j)\cdot 1\leq j\cdot (1-C_1)+50-j=50-jC_1.\]Bu the sum of the coins is $25+\frac{C_1}{2}$ so we get \[50-jC_1\geq 25+\frac{C_1}{2} \iff 25 \geq C_1\left(j+\frac 12 \right) \iff j+\frac 12\leq \frac{25}{C_1}<\frac{25}{\frac{50}{51}}=25+\frac 12\]implying $j<25$. Therefore, there are at least $26$ coins with value at least $k+C_1$ but $26(k+C_1)\geq 26C_1>25+\frac{C_1}{2}$ from $C_1> \frac{50}{51}$. This is our desired contradiction.

The answer is $\tfrac{50}{51}$, obtained by having $51$ coins worth $\frac{50}{51} - \epsilon$ dollars and having $49$ coins worth $\epsilon$ dollars for some negligibly small $\epsilon$. Then, we can't split the $51$ coins between the stacks evenly, hence one stack has at least $\tfrac{50}{51}$ dollars than the other stack. To always obtain two stacks which differ by no more than $\tfrac{50}{51}$, we split into two cases: If there exists a coin which is worth at least $\tfrac{50}{51}$ than the next most expensive coin, we must have some coins worth something in the range $( \tfrac{50}{51}, 1]$ and the rest of the coins worth something in the range $(0, \tfrac{1}{51}]$. The coins in the range $(0, \tfrac{1}{51} ]$ total up to at most $100 \cdot \tfrac{1}{51} < 2$ dollars, so the coins in the range $( \tfrac{50}{51}, 1]$ total up to more than $48$ dollars. So, there are at least $49$ coins worth something in the range $( \tfrac{50}{51}, 1]$. There are also clearly at most $50$ coins worth something in this range. If exactly $49$ coins are worth something in the range $(\tfrac{50}{51}, 1]$ and $51$ coins are worth something in the range $(0, \tfrac{1}{51}]$, we must have all $49$ coins worth $1$ dollar and the other $51$ coins worth $\tfrac{1}{51}$ dollars. This case is trivial albeit an equality case; we allocate $24$ one-dollar coins to the smaller pile plus $26$ coins worth $\tfrac{1}{51}$ dollars, and the rest goes to the larger pile. Otherwise, there are exactly $50$ coins worth something in the range $(\tfrac{50}{51}, 1]$ and $50$ coins worth something in the range $(0, \tfrac{1}{51}]$. Then, we just pick the smaller $25$ coins worth something in the range $(0, \tfrac{1}{51}]$ and the smaller $25$ coins worth something in the range $(\tfrac{50}{51}, 1]$; their total value is at least $25 - \tfrac{25}{51}$ so we're done. Otherwise, sort the coins from least to greatest. We start out by choosing the $1$st smallest, $3$rd smallest, ..., $99$th smallest coins and we will replace the $(2i-1)$th smallest coin with the $(2i)$th smallest coin repeatedly, until we're left with the $2$nd smallest, $4$th smallest, ..., $100$th smallest coin. If, at any point, our total is worth something in the range $[ 25 - \tfrac{25}{51}, 25 + \tfrac{25}{51} ]$, we're done; the $25$ coins we have in that moment will be one of our piles. This must happen because each of our jumps is less than $\frac{50}{51}$.

The answer is $C = \frac{50}{51}$. Suppose the $100$ coins have denominations $0 < a_1 \le a_2 \le \ldots \le a_{100} \le 1$. First, we show $C \ge \frac{50}{51}$. Consider the collection of coins such that $a_1, a_2, \ldots, a_{51} = \frac{1}{51}$ and $a_{52}, a_{53}, \ldots, a_{100} = 1$. Note that one of the two stacks must contain at least $25$ coins with denomination $1$, so its total value is at least $25 \cdot 1 + 25 \cdot \frac{1}{51} = 25 + \frac{25}{51}$. This implies that the absolute difference between the total values of the two stacks is at least $2 \cdot \frac{25}{51} = \frac{50}{51}$, proving the desired bound. Now, we show $C \le \frac{50}{51}$ is achievable for all possible collections of coins via contradiction. Consider $S_1 = \{a_1, a_3, \ldots, a_{99}\}$ with total value $t_1$ and $S_2 = \{a_2, a_4, \ldots, a_{100}\}$ with total value $t_2$. Noting $t_1 \le t_2$ holds unconditionally, we have \[\frac{50}{51} < t_2 - t_1 = (a_2 - a_1) + (a_4 - a_3) + (a_6 - a_5) + \ldots + (a_{100} - a_{99})\]\[\le (a_2 - a_1) + (a_3 - a_2) + (a_4 - a_3) + \ldots + (a_{100} - a_{99}) = a_{100} - a_1 < 1.\] Claim: If $S = \{a_2 - a_1, a_4 - a_3, \ldots, a_{100} - a_{99} \}$, then finding a subset $S'$ of $S$ with total value between $\frac{1}{102}$ and $\frac{50}{51}$ inclusive achieves $C \le \frac{50}{51}$. Proof. Suppose a subset $S'$ of $S$ has total value $d$ in the aforementioned interval. We swap the pairs of elements forming the differences of the form $a_{2i} - a_{2i-1}$ that belong to $S'$ between $S_1$ and $S_2$, yielding two modified sets $S'_1$ and $S'_2$. Because the larger element of each individual swap belongs to $S_2$, the difference between the total values of $S'_1$ and $S'_2$ stacks is precisely \[(t_1 + d) - (t_2 - d) = (t_1 - t_2) + 2d.\]Now, noting $t_1 - t_2 \in \left( -1, - \frac{50}{51} \right)$ implies \[ - \frac{50}{51} < (t_1 - t_2) + 2d < \frac{50}{51}\]for all $d \in \left[ \frac{1}{102}, \frac{50}{51} \right]$ finishes. $\square$ Suppose there only exist subsets $S'$ of $S$ with total values in $\left[0, \frac{1}{102} \right) \cup \left( \frac{50}{51}, 1 \right)$, clearly forcing \[ \{a_2 - a_1, a_4 - a_3, \ldots, a_{100} - a_{99} \} \in \left[ 0, \frac{1}{102} \right) \cup \left( \frac{50}{51}, 1 \right). \]If all $50$ differences listed above belong to $\left[ 0, \frac{1}{102} \right)$, then \[ (a_2 - a_1) + (a_4 - a_3) + (a_6 - a_5) + \ldots + (a_{100} - a_{99}) < 50 \cdot \frac{1}{102} < \frac{50}{51} \]which is absurd. Thus, there exists a difference $a_{2k} - a_{2k-1}$ greater than $\frac{50}{51}$ where $1 \le k \le 50$. It's easy to see $a_{2k-1} < \frac{1}{51}$ and $a_{2k} > \frac{50}{51}$, so \[50 = a_1 + a_2 + \ldots + a_{100} < (2k-1) \cdot \frac{1}{51} + (100 - (2k - 1)) \cdot 1 \]implying $2k - 1 < 51$ and \[50 = a_1 + a_2 + \ldots + a_{100} > (100 - (2k-1)) \cdot \frac{50}{51} \]yielding $2k - 1 > 49$. This is a contradiction since $2k-1$ is odd, so there always exists a subset $S'$ with total value in $\left[ \frac{1}{102}, \frac{50}{51} \right]$, which suffices. $\blacksquare$ Remarks: This solution ends up being analogous to the standard local solution, but it is formulated far less naturally. I'm still not completely sure how I ended up finding this approach instead; perhaps some more wishful thinking and the use of paper would have helped.

tastymath75025 wrote: Consider coins with positive real denominations not exceeding 1. Find the smallest $C>0$ such that the following holds: if we have any $100$ such coins with total value $50$, then we can always split them into two stacks of $50$ coins each such that the absolute difference between the total values of the two stacks is at most $C$. Merlijn Staps Taking the coins $(\underbrace{\tfrac1{51}, \dots, \tfrac1{51}}_{51 \ \text{times}}, \ \underbrace{1, \dots, 1}_{49 \ \text{times}})$, we get that $\boxed{C\leq\tfrac{50}{51}}$ Now we are going to prove that this is enought.Let the coins be $a_1 \leq a_2 \leq \dots \leq a_{100}$ We spit them in two stacks $(a_1,a_3,a_5,...,a_{99}),(a_2,a_4,a_{100})$ and at the $i$ step we exchange the coins $a_{2i-1}$ with $a_{2i}$. If $a_{2i}-a_{2i-1}\leq\tfrac{50}{51}$ for every $i$ then at some step we will get the first stacks to have value $[25-\tfrac{25}{51},25+\tfrac{25}{51}]$ Soppuse that somewhere there is a big gap ($a_{2i}-a_{2i-1}>\tfrac{50}{51}$) then we have $x$ of them with $a_i>\tfrac{50}{51}$ and $100-x$ with $a_i<\tfrac{1}{51}$ but then : \begin{align*} \frac{50}{51}x < 50 &\implies x < 51 \\ \frac{1}{51}(100-x) + x > 50 &\implies x > 49 \end{align*} So $x=50$ and we can spilt them in two stacks with $25$ big and $25$ small coins it's of them . IMO 2014 Problem 5 BMO Shortlist 2021 C3

The answer is: $\boxed{\frac{50}{51}}$ Construction: $(1,1,\dots,1,\frac{1}{51},\frac{1}{51}, \dots,\frac{1}{51})$ which has 49 1's and 51 $\frac{1}{51}$'s. and here take the odds and evens. let the terms be $0 \leq a_1 \leq a_2 \leq \dots \leq a_100 \leq 1$. We call a difference(between consecutive terms) big iff $a_{n+1} > a_{n} + \frac{50}{51}$. Claim: this difference must be between $a_{50},a_{51}$ Proof: Suppose other wise if the difference is at $i\leq49$ then the sum entire sum is more than 50 contradiction. Else if it is at $i \geq 51$ the sum is too small. Now take the odd guys i.e $a_1, a_3, \dots , a_{99}$ and now at each step replace $a_{2k-1}$ with $a_{2k}$. At any point we did not go over a large gap so the difference between the final and initial is at most $\frac{50}{51}$. $QED$

Our answer is $C = \boxed{\tfrac{50}{51}}$, attainable with \[\underbrace{\tfrac{1}{51}, \ldots, \tfrac{1}{51},}_{51 \text{ coins}} \underbrace{1, \ldots, 1.}_{49 \text{ coins}}\] We now show it is impossible to have a larger $C$. Consider the coin values \[c_1 \leq c_2 \leq \ldots \leq c_{99} \leq c_{100},\] which sum to 50, and a permutation of the 50 nonnegative elements of the set \[\mathcal{S} = \{c_2-c_1, c_4-c_3, \ldots, c_{100}-c_{99}\}.\] Notice that each time we add an element of $\mathcal S$ in order (starting from the first) to the sum $c_1+c_3+\ldots+c_{99}$, we always get a sum of 50 of the coin values until we end at the sum $c_2+c_4+\ldots+c_{100}$. Since it suffices to show one of these sums is in the interval $\left[25-\tfrac{25}{51}, 25+\tfrac{25}{51}\right]$, it suffices to show all "jumps" are at most $\tfrac{50}{51}$, or each element of $\mathcal S$ is at most $\tfrac{50}{51}$. Suppose FTSOC we have $c_{2k}-c_{2k-1} > \tfrac{50}{51}$, which would imply $0 \leq c_{2k-1} < \tfrac{1}{51}$ and $\tfrac{50}{51} < c_{2k} \leq 1$. This allows us to get two bounds for the sum of all 100 coin values: \begin{align*} 50 &< \tfrac{1}{51} \cdot (2k-1) + 1 \cdot (101-2k) \implies k < 26, \\ 50 &> 0 \cdot (2k-1) + \tfrac{50}{51} \cdot (101-2k) \implies k > 25, \end{align*} giving the desired contradiction, as $k$ can't be an integer. $\blacksquare$