# A Lot of 90% Proportions

In the land of Puzzovania Sock Enrage:

Show that among the puzzles tagged with both and , more than 90% are also tagged .

Notes:

1. This is harder than it looks!
2. Based on a puzzle from the Russian Tournament of the Towns.

Let $A_{RM}$ be puzzles tagged riddle and math, but not logical-deduction, and define $A_{LM}$, $A_{RL}$ and $A_{RLM}$ similarly.

From 1: $A_{RM} + A_{RLM} > 9(A_{RL} + A_{LM})$ (a)
From 2: $A_{RL} + A_{RLM} > 9(A_{RM} + A_{LM})$ (b)

$$2A_{RLM} > 8A_{RL} + 8A_{RM} + 18A_{LM}$$ $$A_{RLM} > 4A_{RL} + 4A_{RM} + 9A_{LM}$$

Since all $A$s are non-negative, this implies $A_{RLM} > 9A_{LM}$, as desired.

• You make it look easy :-) Note that inequalities (a) and (b) aren't the most natural things to conclude from the given conditions: in fact $A_{RM}+A_{RLM}>9($all the rest$)$, and you have to see which terms to chuck out and which to keep. Did a Venn diagram help at all, or did you just do this with algebra? – Rand al'Thor May 29 '18 at 21:53
• A bit of both - I drew a Venn diagram to help me write the inequalities, then used algebra to finish. Noticing which terms to keep is made easier if you notice that in the worst case every puzzle is either tagged riddle or tagged both logical-deduction and mathematics, so the other terms shouldn't matter. – ffao May 29 '18 at 22:05

Here's the Venn diagram: $k$ = total number of puzzles
$a,b$ = the amounts added to $9k/10$ to make sure the ratios are greater
$x$ = the number of questions tagged with mathematics and logical-deduction, but not riddle

$b+x$ and $a+x <= k/10$, so $x<k/10$, which proves it.

• How are $k,a,b,x$ defined? – Rand al'Thor May 30 '18 at 6:27
• Added explanations to the answer. – Nautilus May 30 '18 at 6:49