aops_id int64 2 3.62M | problem stringlengths 17 4.71k | best_solution stringlengths 7 38.7k | problem_vector listlengths 1.02k 1.02k | best_solution_vector listlengths 1.02k 1.02k | last_modified stringdate 2025-08-25 00:00:00 2025-08-25 00:00:00 |
|---|---|---|---|---|---|
1,000,044 | Justin has a 55% chance of winning any given point in a ping-pong game. To the nearest 0.1%, what is the probability that he wins exactly 7 out of the first 10 points? | So if he has a 55% chance of winning, he conversely has a 45% chance of losing. The problem calls for him winning 7 times and losing 3, so his win percentage will be multiplied by itself 7 times and his losing percentage will be multiplied by itself 3 times so your expression should look like this $ 0.55^7*0.45^3$. | [
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100,009 | Evaluate
\[
\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sqrt{\sin x}+\sqrt{\cos x}+3(\sqrt{\sin x}-\sqrt{\cos x})\cos 2x}{\sqrt{\sin 2x}}\;dx.
\] | You may be right. I made this problem by the differentiation of $\sin x \sqrt{\cos x}+\cos x\sqrt{\sin x}$ as shyong showed. | [
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-0.03054... | 2025-08-25 |
100,010 | Billy Bob has a pet snail called Larry. The wall is 37 feet tall. Larry can climb 3 feet in one day, but at night he slips down 2 feet. Larry starts the climb on Sunday, June 2, 2006. On what day (day of the week and date) will Larry finish the climb? | [quote="mtms5467"][hide]So basically Larry climbs 1ft/day. The day/date 37 days from June 2. (Oh wait...June 2, 2006 is a Friday...)
[hide="Assuming June 2, 2006 is a Sunday"]Saturday July 8[/hide]
[hide="Assuming he means July 2 (actually a Sun.!)"] Monday Aug 7[/hide][/hide][/quote]
It's a classic trick question. ... | [
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1,000,136 | Deriving the Quadratic Formula
Problem:
Derive the quadratic formula.
Solution:
Start with \(ax^2 + bx + c = 0\).
\(ax^2 + bx = -c\)
\(x^2 + \frac{b}{a}x = -\frac{c}{a}\)
\(x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} = -\frac{c}{a} + \frac{b^2}{4a^2}\)
\(\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}\)
\(x +... | Lol. I figured out how to do it this past year in 6th grade...
My math teacher never showed us, so one lonely lunch period I tried to figure out how to do it. I began with $ ax^2 \plus{} bx \plus{} c \equal{} 0$, and on a whim, I began to complete the square. I was shocked when, after 42 minutes and 39 seconds(yes, ... | [
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-0.... | 2025-08-25 |
1,000,141 | "[b]Coin Problems[/b]\r\n\r\n[i]Tony has 11 more nickels than quarters. If the total value of his co(...TRUNCATED) | "there's a few ways to do problems like the second one that work for all positive integer number of (...TRUNCATED) | [0.039213865995407104,0.019115909934043884,0.0004175684880465269,0.015549140982329845,-0.05275301262(...TRUNCATED) | [0.01666051521897316,0.011526957154273987,0.0007536772754974663,-0.015084879472851753,-0.04363686218(...TRUNCATED) | 2025-08-25 |
100,015 | "A 6-letter car plaque is to be made using the letters \\(A,\\dots,Z\\) such that the letters are in(...TRUNCATED) | "[hide]I get $\\binom{26}{6}$. Choose any 6 letters and there exists a unique alphabetical arrangeme(...TRUNCATED) | [0.03238161653280258,-0.018685823306441307,0.030574249103665352,-0.0035495604388415813,-0.0360959544(...TRUNCATED) | [0.005885143298655748,-0.005541427060961723,0.03213439881801605,0.023898232728242874,-0.049293328076(...TRUNCATED) | 2025-08-25 |
100,019 | "Billy Bob has a huge garden. He picks a few flowers from it. There is one red flower, one blue flow(...TRUNCATED) | "[hide]Or you can count the number of total ways $4!=24$ and then subtract the number of ways the re(...TRUNCATED) | [0.014576289802789688,0.0034330927301198244,0.02435494400560856,0.008475987240672112,-0.042872611433(...TRUNCATED) | [-0.006250646896660328,-0.0003476288402453065,0.001258213073015213,-0.029833216220140457,-0.04505798(...TRUNCATED) | 2025-08-25 |
100,023 | Simplify
\[
(1+x)(1+x^{2})(1+x^{4})(1+x^{8})\cdots
\]
for \(|x|<1\). | "[hide]When you multiply it out, you can see that the product is equal to\n$1+x+x^{2}+x^{3}\\dots$\n(...TRUNCATED) | [-0.011075973510742188,-0.007155919447541237,0.024705244228243828,-0.0047179763205349445,0.036557868(...TRUNCATED) | [0.023143166676163673,0.003017634619027376,0.011099766008555889,-0.03369978070259094,0.0238790772855(...TRUNCATED) | 2025-08-25 |
1,000,249 | "Two players (You and Ben) are each arrested and placed in separate jail cells with no communication(...TRUNCATED) | "If all four possibilities are equally likely, then confessing is better:\r\n\r\nMe Ben Number of(...TRUNCATED) | [0.04948459193110466,0.0268138125538826,-0.009900674223899841,0.037839602679014206,-0.03475932776927(...TRUNCATED) | [0.04233216494321823,-0.03452478349208832,-0.0022878588642925024,0.026167631149291992,-0.03814950957(...TRUNCATED) | 2025-08-25 |
100,026 | "Let r and s be the roots of\n\\[\nx^{2}-(a+d)x+(ad-bc)=0.\n\\]\nProve that \\(r^{3}\\) and \\(s^{3}(...TRUNCATED) | "From ?vietta's? sums $r+s=a+d$ and $rs=ad-bc$. Thus $r^{3}+s^{3}=(r+s)^{3}-3rs(r+s)=(a+d)^{3}-3(ad-(...TRUNCATED) | [0.0036721674259752035,0.023519733920693398,0.028071576729416847,-0.016635853797197342,0.01320050191(...TRUNCATED) | [-0.0007551855524070561,0.020630890503525734,0.029806489124894142,-0.03755967691540718,0.01932490989(...TRUNCATED) | 2025-08-25 |
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