100 boxes contain fruits

by OronSH, Feb 1, 2024, 12:58 AM

I'm posting the (readable) solution here, hopefully others see the beauty of this problem outside just the ham sandwich sol.

Problem statement:
$100$ boxes contain apples, oranges and bananas. Prove that we can choose $51$ boxes in such a way that they contain at least half of all apples, and half of all oranges and half of all bananas.

First, isolate the box $\alpha$ with the most apples and the box $\omega$ with the most oranges (if these two are the same box, choose that box, designated both $\alpha$ and $\omega,$ and another arbitrary box $\varepsilon$ ). Among the rest, assign each a label, and write their labels in a row, in decreasing order of the number of apples they contain, from left to right (ties broken arbitrarily). Below this row, write their labels in decreasing order of number of oranges they contain. Draw a square around boxes in positions $2i,2i+1$ in both rows; this splits the $98$ boxes in each row into $49$ pairs. (In the example shown, we replace $98$ with $8$ for simplicity.)
$[asy] size(10cm); draw((0,0)--(2.5,0)--(2.5,3)--(0,3)--cycle); label("A", (1.25,1.5)); draw((4,0)--(6.5,0)--(6.5,3)--(4,3)--cycle); label("B", (5.25,1.5)); draw((8,0)--(10.5,0)--(10.5,3)--(8,3)--cycle); label("C", (9.25,1.5)); draw((12,0)--(14.5,0)--(14.5,3)--(12,3)--cycle); label("D", (13.25,1.5)); draw((16,0)--(18.5,0)--(18.5,3)--(16,3)--cycle); label("E", (17.25,1.5)); draw((20,0)--(22.5,0)--(22.5,3)--(20,3)--cycle); label("F", (21.25,1.5)); draw((24,0)--(26.5,0)--(26.5,3)--(24,3)--cycle); label("G", (25.25,1.5)); draw((28,0)--(30.5,0)--(30.5,3)--(28,3)--cycle); label("H", (29.25,1.5)); draw((0,-10)--(2.5,-10)--(2.5,-7)--(0,-7)--cycle); label("C", (1.25,-8.5)); draw((4,-10)--(6.5,-10)--(6.5,-7)--(4,-7)--cycle); label("A", (5.25,-8.5)); draw((8,-10)--(10.5,-10)--(10.5,-7)--(8,-7)--cycle); label("D", (9.25,-8.5)); draw((12,-10)--(14.5,-10)--(14.5,-7)--(12,-7)--cycle); label("E", (13.25,-8.5)); draw((16,-10)--(18.5,-10)--(18.5,-7)--(16,-7)--cycle); label("B", (17.25,-8.5)); draw((20,-10)--(22.5,-10)--(22.5,-7)--(20,-7)--cycle); label("F", (21.25,-8.5)); draw((24,-10)--(26.5,-10)--(26.5,-7)--(24,-7)--cycle); label("H", (25.25,-8.5)); draw((28,-10)--(30.5,-10)--(30.5,-7)--(28,-7)--cycle); label("G", (29.25,-8.5)); draw((-0.25,-0.25)--(6.75,-0.25)--(6.75,3.25)--(-0.25,3.25)--cycle,linewidth(1.5)); draw((7.75,-0.25)--(14.75,-0.25)--(14.75,3.25)--(7.75,3.25)--cycle,linewidth(1.5)); draw((15.75,-0.25)--(22.75,-0.25)--(22.75,3.25)--(15.75,3.25)--cycle,linewidth(1.5)); draw((23.75,-0.25)--(30.75,-0.25)--(30.75,3.25)--(23.75,3.25)--cycle,linewidth(1.5)); draw((-0.25,-10.25)--(6.75,-10.25)--(6.75,-6.75)--(-0.25,-6.75)--cycle,linewidth(1.5)); draw((7.75,-10.25)--(14.75,-10.25)--(14.75,-6.75)--(7.75,-6.75)--cycle,linewidth(1.5)); draw((15.75,-10.25)--(22.75,-10.25)--(22.75,-6.75)--(15.75,-6.75)--cycle,linewidth(1.5)); draw((23.75,-10.25)--(30.75,-10.25)--(30.75,-6.75)--(23.75,-6.75)--cycle,linewidth(1.5)); [/asy]$

Now we color the boxes red and blue. Start with some box colored red:
$[asy] size(10cm); filldraw((0,0)--(2.5,0)--(2.5,3)--(0,3)--cycle,lightred); label("A", (1.25,1.5)); draw((4,0)--(6.5,0)--(6.5,3)--(4,3)--cycle); label("B", (5.25,1.5)); draw((8,0)--(10.5,0)--(10.5,3)--(8,3)--cycle); label("C", (9.25,1.5)); draw((12,0)--(14.5,0)--(14.5,3)--(12,3)--cycle); label("D", (13.25,1.5)); draw((16,0)--(18.5,0)--(18.5,3)--(16,3)--cycle); label("E", (17.25,1.5)); draw((20,0)--(22.5,0)--(22.5,3)--(20,3)--cycle); label("F", (21.25,1.5)); draw((24,0)--(26.5,0)--(26.5,3)--(24,3)--cycle); label("G", (25.25,1.5)); draw((28,0)--(30.5,0)--(30.5,3)--(28,3)--cycle); label("H", (29.25,1.5)); draw((0,-10)--(2.5,-10)--(2.5,-7)--(0,-7)--cycle); label("C", (1.25,-8.5)); draw((4,-10)--(6.5,-10)--(6.5,-7)--(4,-7)--cycle); label("A", (5.25,-8.5)); draw((8,-10)--(10.5,-10)--(10.5,-7)--(8,-7)--cycle); label("D", (9.25,-8.5)); draw((12,-10)--(14.5,-10)--(14.5,-7)--(12,-7)--cycle); label("E", (13.25,-8.5)); draw((16,-10)--(18.5,-10)--(18.5,-7)--(16,-7)--cycle); label("B", (17.25,-8.5)); draw((20,-10)--(22.5,-10)--(22.5,-7)--(20,-7)--cycle); label("F", (21.25,-8.5)); draw((24,-10)--(26.5,-10)--(26.5,-7)--(24,-7)--cycle); label("H", (25.25,-8.5)); draw((28,-10)--(30.5,-10)--(30.5,-7)--(28,-7)--cycle); label("G", (29.25,-8.5)); draw((-0.25,-0.25)--(6.75,-0.25)--(6.75,3.25)--(-0.25,3.25)--cycle,linewidth(1.5)); draw((7.75,-0.25)--(14.75,-0.25)--(14.75,3.25)--(7.75,3.25)--cycle,linewidth(1.5)); draw((15.75,-0.25)--(22.75,-0.25)--(22.75,3.25)--(15.75,3.25)--cycle,linewidth(1.5)); draw((23.75,-0.25)--(30.75,-0.25)--(30.75,3.25)--(23.75,3.25)--cycle,linewidth(1.5)); draw((-0.25,-10.25)--(6.75,-10.25)--(6.75,-6.75)--(-0.25,-6.75)--cycle,linewidth(1.5)); draw((7.75,-10.25)--(14.75,-10.25)--(14.75,-6.75)--(7.75,-6.75)--cycle,linewidth(1.5)); draw((15.75,-10.25)--(22.75,-10.25)--(22.75,-6.75)--(15.75,-6.75)--cycle,linewidth(1.5)); draw((23.75,-10.25)--(30.75,-10.25)--(30.75,-6.75)--(23.75,-6.75)--cycle,linewidth(1.5)); [/asy]$

Next, repeat the following steps:
1. Color the box in the other row with the same letter as the last box the same color as the last box.
2. Color the other box in the pair that the last-colored box is in with the opposite color.

For example, after $2$ turns, it will look like this:
$[asy] size(10cm); filldraw((0,0)--(2.5,0)--(2.5,3)--(0,3)--cycle,lightred); label("A", (1.25,1.5)); draw((4,0)--(6.5,0)--(6.5,3)--(4,3)--cycle); label("B", (5.25,1.5)); filldraw((8,0)--(10.5,0)--(10.5,3)--(8,3)--cycle,lightblue); label("C", (9.25,1.5)); filldraw((12,0)--(14.5,0)--(14.5,3)--(12,3)--cycle,lightred); label("D", (13.25,1.5)); draw((16,0)--(18.5,0)--(18.5,3)--(16,3)--cycle); label("E", (17.25,1.5)); draw((20,0)--(22.5,0)--(22.5,3)--(20,3)--cycle); label("F", (21.25,1.5)); draw((24,0)--(26.5,0)--(26.5,3)--(24,3)--cycle); label("G", (25.25,1.5)); draw((28,0)--(30.5,0)--(30.5,3)--(28,3)--cycle); label("H", (29.25,1.5)); filldraw((0,-10)--(2.5,-10)--(2.5,-7)--(0,-7)--cycle,lightblue); label("C", (1.25,-8.5)); filldraw((4,-10)--(6.5,-10)--(6.5,-7)--(4,-7)--cycle,lightred); label("A", (5.25,-8.5)); draw((8,-10)--(10.5,-10)--(10.5,-7)--(8,-7)--cycle); label("D", (9.25,-8.5)); draw((12,-10)--(14.5,-10)--(14.5,-7)--(12,-7)--cycle); label("E", (13.25,-8.5)); draw((16,-10)--(18.5,-10)--(18.5,-7)--(16,-7)--cycle); label("B", (17.25,-8.5)); draw((20,-10)--(22.5,-10)--(22.5,-7)--(20,-7)--cycle); label("F", (21.25,-8.5)); draw((24,-10)--(26.5,-10)--(26.5,-7)--(24,-7)--cycle); label("H", (25.25,-8.5)); draw((28,-10)--(30.5,-10)--(30.5,-7)--(28,-7)--cycle); label("G", (29.25,-8.5)); draw((-0.25,-0.25)--(6.75,-0.25)--(6.75,3.25)--(-0.25,3.25)--cycle,linewidth(1.5)); draw((7.75,-0.25)--(14.75,-0.25)--(14.75,3.25)--(7.75,3.25)--cycle,linewidth(1.5)); draw((15.75,-0.25)--(22.75,-0.25)--(22.75,3.25)--(15.75,3.25)--cycle,linewidth(1.5)); draw((23.75,-0.25)--(30.75,-0.25)--(30.75,3.25)--(23.75,3.25)--cycle,linewidth(1.5)); draw((-0.25,-10.25)--(6.75,-10.25)--(6.75,-6.75)--(-0.25,-6.75)--cycle,linewidth(1.5)); draw((7.75,-10.25)--(14.75,-10.25)--(14.75,-6.75)--(7.75,-6.75)--cycle,linewidth(1.5)); draw((15.75,-10.25)--(22.75,-10.25)--(22.75,-6.75)--(15.75,-6.75)--cycle,linewidth(1.5)); draw((23.75,-10.25)--(30.75,-10.25)--(30.75,-6.75)--(23.75,-6.75)--cycle,linewidth(1.5)); draw((1.25,-0.5)--(5.25,-6.5),Arrow); draw((1.25,-6.5)--(9.25,-0.5),Arrow); [/asy]$
Repeat this process until it reaches a box that is already colored:
$[asy] size(10cm); filldraw((0,0)--(2.5,0)--(2.5,3)--(0,3)--cycle,lightred); label("A", (1.25,1.5)); filldraw((4,0)--(6.5,0)--(6.5,3)--(4,3)--cycle,lightblue); label("B", (5.25,1.5)); filldraw((8,0)--(10.5,0)--(10.5,3)--(8,3)--cycle,lightblue); label("C", (9.25,1.5)); filldraw((12,0)--(14.5,0)--(14.5,3)--(12,3)--cycle,lightred); label("D", (13.25,1.5)); filldraw((16,0)--(18.5,0)--(18.5,3)--(16,3)--cycle,lightblue); label("E", (17.25,1.5)); filldraw((20,0)--(22.5,0)--(22.5,3)--(20,3)--cycle,lightred); label("F", (21.25,1.5)); draw((24,0)--(26.5,0)--(26.5,3)--(24,3)--cycle); label("G", (25.25,1.5)); draw((28,0)--(30.5,0)--(30.5,3)--(28,3)--cycle); label("H", (29.25,1.5)); filldraw((0,-10)--(2.5,-10)--(2.5,-7)--(0,-7)--cycle,lightblue); label("C", (1.25,-8.5)); filldraw((4,-10)--(6.5,-10)--(6.5,-7)--(4,-7)--cycle,lightred); label("A", (5.25,-8.5)); filldraw((8,-10)--(10.5,-10)--(10.5,-7)--(8,-7)--cycle,lightred); label("D", (9.25,-8.5)); filldraw((12,-10)--(14.5,-10)--(14.5,-7)--(12,-7)--cycle,lightblue); label("E", (13.25,-8.5)); filldraw((16,-10)--(18.5,-10)--(18.5,-7)--(16,-7)--cycle,lightblue); label("B", (17.25,-8.5)); filldraw((20,-10)--(22.5,-10)--(22.5,-7)--(20,-7)--cycle,lightred); label("F", (21.25,-8.5)); draw((24,-10)--(26.5,-10)--(26.5,-7)--(24,-7)--cycle); label("H", (25.25,-8.5)); draw((28,-10)--(30.5,-10)--(30.5,-7)--(28,-7)--cycle); label("G", (29.25,-8.5)); draw((-0.25,-0.25)--(6.75,-0.25)--(6.75,3.25)--(-0.25,3.25)--cycle,linewidth(1.5)); draw((7.75,-0.25)--(14.75,-0.25)--(14.75,3.25)--(7.75,3.25)--cycle,linewidth(1.5)); draw((15.75,-0.25)--(22.75,-0.25)--(22.75,3.25)--(15.75,3.25)--cycle,linewidth(1.5)); draw((23.75,-0.25)--(30.75,-0.25)--(30.75,3.25)--(23.75,3.25)--cycle,linewidth(1.5)); draw((-0.25,-10.25)--(6.75,-10.25)--(6.75,-6.75)--(-0.25,-6.75)--cycle,linewidth(1.5)); draw((7.75,-10.25)--(14.75,-10.25)--(14.75,-6.75)--(7.75,-6.75)--cycle,linewidth(1.5)); draw((15.75,-10.25)--(22.75,-10.25)--(22.75,-6.75)--(15.75,-6.75)--cycle,linewidth(1.5)); draw((23.75,-10.25)--(30.75,-10.25)--(30.75,-6.75)--(23.75,-6.75)--cycle,linewidth(1.5)); draw((1.25,-0.5)--(5.25,-6.5),Arrow); draw((1.25,-6.5)--(9.25,-0.5),Arrow); draw((13.25,-0.5)--(9.25,-6.5),Arrow); draw((13.25,-6.5)--(17.25,-0.5),Arrow); draw((21.25,-0.5)--(21.25,-6.5),Arrow); draw((17.25,-6.5)--(5.25,-0.5),Arrow); [/asy]$

Consider when this process stops. Notice that for all but the first box to be colored, its color and position uniquely determine the preceding box to be colored: red boxes on top are only colored immediately after the blue box in the same pair as them, blue boxes on top are only colored immediately after the blue box on the bottom with the same letter, and similarly for the bottom row. Thus if the process first ends at a box other than the initial box, that is, if the first box to be colored twice has a unique preceding box, then this preceding box must have been colored twice as well, contradicting our assumption. Thus, this process can only end at the initial box.

Furthermore, each pair either has neither or both of its boxes colored; this is easy to see, since if one box gets colored, the other must be colored immediately after. Additionally, notice that the difference between the number of colored boxes in the top row and the bottom row never exceeds $1,$ thus once this process stops, it must be $0,$ and an equal number of pairs have both boxes colored.

Now, color another box red, and repeat this entire process until all boxes are colored.
$[asy] size(10cm); filldraw((0,0)--(2.5,0)--(2.5,3)--(0,3)--cycle,lightred); label("A", (1.25,1.5)); filldraw((4,0)--(6.5,0)--(6.5,3)--(4,3)--cycle,lightblue); label("B", (5.25,1.5)); filldraw((8,0)--(10.5,0)--(10.5,3)--(8,3)--cycle,lightblue); label("C", (9.25,1.5)); filldraw((12,0)--(14.5,0)--(14.5,3)--(12,3)--cycle,lightred); label("D", (13.25,1.5)); filldraw((16,0)--(18.5,0)--(18.5,3)--(16,3)--cycle,lightblue); label("E", (17.25,1.5)); filldraw((20,0)--(22.5,0)--(22.5,3)--(20,3)--cycle,lightred); label("F", (21.25,1.5)); filldraw((24,0)--(26.5,0)--(26.5,3)--(24,3)--cycle,lightred); label("G", (25.25,1.5)); filldraw((28,0)--(30.5,0)--(30.5,3)--(28,3)--cycle,lightblue); label("H", (29.25,1.5)); filldraw((0,-10)--(2.5,-10)--(2.5,-7)--(0,-7)--cycle,lightblue); label("C", (1.25,-8.5)); filldraw((4,-10)--(6.5,-10)--(6.5,-7)--(4,-7)--cycle,lightred); label("A", (5.25,-8.5)); filldraw((8,-10)--(10.5,-10)--(10.5,-7)--(8,-7)--cycle,lightred); label("D", (9.25,-8.5)); filldraw((12,-10)--(14.5,-10)--(14.5,-7)--(12,-7)--cycle,lightblue); label("E", (13.25,-8.5)); filldraw((16,-10)--(18.5,-10)--(18.5,-7)--(16,-7)--cycle,lightblue); label("B", (17.25,-8.5)); filldraw((20,-10)--(22.5,-10)--(22.5,-7)--(20,-7)--cycle,lightred); label("F", (21.25,-8.5)); filldraw((24,-10)--(26.5,-10)--(26.5,-7)--(24,-7)--cycle,lightblue); label("H", (25.25,-8.5)); filldraw((28,-10)--(30.5,-10)--(30.5,-7)--(28,-7)--cycle,lightred); label("G", (29.25,-8.5)); draw((-0.25,-0.25)--(6.75,-0.25)--(6.75,3.25)--(-0.25,3.25)--cycle,linewidth(1.5)); draw((7.75,-0.25)--(14.75,-0.25)--(14.75,3.25)--(7.75,3.25)--cycle,linewidth(1.5)); draw((15.75,-0.25)--(22.75,-0.25)--(22.75,3.25)--(15.75,3.25)--cycle,linewidth(1.5)); draw((23.75,-0.25)--(30.75,-0.25)--(30.75,3.25)--(23.75,3.25)--cycle,linewidth(1.5)); draw((-0.25,-10.25)--(6.75,-10.25)--(6.75,-6.75)--(-0.25,-6.75)--cycle,linewidth(1.5)); draw((7.75,-10.25)--(14.75,-10.25)--(14.75,-6.75)--(7.75,-6.75)--cycle,linewidth(1.5)); draw((15.75,-10.25)--(22.75,-10.25)--(22.75,-6.75)--(15.75,-6.75)--cycle,linewidth(1.5)); draw((23.75,-10.25)--(30.75,-10.25)--(30.75,-6.75)--(23.75,-6.75)--cycle,linewidth(1.5)); draw((1.25,-0.5)--(5.25,-6.5),Arrow); draw((1.25,-6.5)--(9.25,-0.5),Arrow); draw((13.25,-0.5)--(9.25,-6.5),Arrow); draw((13.25,-6.5)--(17.25,-0.5),Arrow); draw((21.25,-0.5)--(21.25,-6.5),Arrow); draw((17.25,-6.5)--(5.25,-0.5),Arrow); draw((25.25,-0.5)--(29.25,-6.5),Arrow); draw((25.25,-6.5)--(29.25,-0.5),Arrow); [/asy]$

It is not hard to see that exactly one box in each pair is colored red, and exactly one is colored blue.

Now consider the $51$ boxes $\alpha,\omega$ (and $\varepsilon$ if necessary) together with the $49$ red boxes (this is well-defined since two boxes with the same letter have the same color, by construction). Consider the pairs of boxes in the top row. We see that $\alpha$ has at least as many apples as the blue box in the first (leftmost) pair, and for all $1\le i\le 48$ the red box in pair $i$ has at least as many apples as the blue box in pair $i+1.$ This shows that together, the blue boxes have at most as many apples as the union of $\alpha$ and the red boxes; thus, this choice of $51$ boxes has at least half of the total number of apples. By the same logic, it has at least half of the total number of oranges.

However, by the same logic, the set of $51$ boxes given by the union of $\alpha,\omega,$ (possibly $\varepsilon$ ) and the $49$ blue boxes also contains at least half of the total number of apples and at least half of the total number of oranges. Now, if both of these sets contain less than half of the total number of bananas, then their complements must both contain at least half. However, their complements are the sets of blue and red boxes, respectively, and the number of bananas in their union is at most the number of bananas in all $100$ boxes together, which is a contradiction.

Thus one of these subsets satisfies the conditions.

This post has been edited 2 times. Last edited by OronSH, Feb 1, 2024, 6:57 PM

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8 Comments

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HOW LONG DID THAT ASY AND LATEX TAKE YOU BROSKI
ORZ :omighty:

like 3 hours skull

This post has been edited 1 time. Last edited by OronSH, Feb 1, 2024, 1:08 AM

by CC_chloe, Feb 1, 2024, 1:06 AM

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is there a generalization to this problem lol yeah but its trivialized by ham sandwich

This post has been edited 1 time. Last edited by OronSH, Feb 1, 2024, 1:27 AM

by dolphinday, Feb 1, 2024, 1:23 AM

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I find the name ham sandwich to be misleading, since infinite slices of ham and convex bread slices are both incredibly rare.

by aryabhata000, Feb 1, 2024, 1:54 AM

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Wait haha skullers I was under the impression that the ham sandwich theorem is something it's not. Apparently it has nothing to do with separating convex bodies, and instead applies to general objects

by aryabhata000, Feb 1, 2024, 1:55 AM

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isn't ham sandwich theorem the first part of the claim in 18brmo6 (combigeo)

by Leo.Euler, Feb 1, 2024, 2:06 AM

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"What are math contests like in Russia?" A asked B.

"Russian math contests are a lot better than American ones," replied B.

"Hey have you done RMO fruits" said Oron.

"What's that?" asked B.

And in that moment, through the reflection in the bowl of soup, Oron saw his destiny, bright and clear. The duty had been enlisted upon him: from that moment onwards, he would dedicate his life to spreading the tale of "RMO fruits" far and wide, leaving not a single crevice of mankind behind.

Many, many years later, some two archeologists would be the visitors to a graveyard burrowed amongst the Ruins of New Jersey, silent for the past millenium. The first archeologist, in a fit of coughs (New Jersey had never been renowned for its air quality), picked up one of the hundreds of stones around him. He squinted in vein, but only hieroglyphics covered the front:
$~$

Oron Wang
2008-2108
100 boxes contain apples, oranges and bananas. Prove that we can choose 51 boxes in such a way that they contain at least half of all apples, and half of all oranges and half of all bananas.

by ihatemath123, Feb 1, 2024, 6:41 AM

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how long did it take for you to solve this the first time i think like 10 hours total

also how many mohs (is it even worth it for me to bother trying this lmao) probably 40

This post has been edited 1 time. Last edited by OronSH, Feb 1, 2024, 2:27 PM

by ihatemath123, Feb 1, 2024, 6:43 AM

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omg it prob took so long w the latex orz :omighty:

by Pdapda, Feb 2, 2024, 2:36 PM

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admits admits admits
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how so xor oren W
by peelybonehead, May 17, 2023, 2:49 AM
xook $\text{[math]}$ $\text{[math]}$
by OronSH, May 17, 2023, 2:44 AM
Hihihihihihihihi
by maxamc, May 17, 2023, 2:27 AM
hi orons !
by megarnie, May 17, 2023, 2:12 AM
ooks . $~~~~~~~$
by pi271828, May 17, 2023, 2:10 AM
how so orz
by UnearthedCyclone, May 17, 2023, 2:09 AM
so orz

by zhoujef000, May 17, 2023, 1:45 AM
imagine so xor
by pikapika007, May 17, 2023, 1:45 AM
XORON HOWSOXOR
by GoodMorning, May 17, 2023, 1:45 AM
HOW SO XOR
by TotallyNotSimon, May 17, 2023, 1:44 AM
orz $\text{[math]}$
by aa1024, May 17, 2023, 1:44 AM

415 shouts

xooks

100 boxes contain fruits

by OronSH, Feb 1, 2024, 12:58 AM

8 Comments