problem string | solution string | candidates list | tags list | metadata dict |
|---|---|---|---|---|
Fix a set of integers $S$ . An integer is *clean* if it is the sum of distinct elements of $S$ in exactly one way, and *dirty* otherwise. Prove that the set of dirty numbers is either empty or infinite.
*Note:* We consider the empty sum to equal \(0\).
*Proposed by Tony Wang and Ethan Tan* | [
"Any solution?",
"Check [https://icmathscomp.org/past-papers/2021-2022/ICMC_5.2_solutions.pdf](https://icmathscomp.org/past-papers/2021-2022/ICMC_5.2_solutions.pdf)."
] | [
"origin:aops",
"2022 Contests",
"ICMC 5"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/ICMC 5/2792549.json"
} | |
A robot on the number line starts at $1$ . During the first minute, the robot writes down the number $1$ . Each minute thereafter, it moves by one, either left or right, with equal probability. It then multiplies the last number it wrote by $n/t$ , where $n$ is the number it just moved to, and $t$ is the number... | [
"wrong so deleting",
"Bump :play_ball: \nAny ideas?",
"Bump again!",
"[https://icmathscomp.org/past-papers](https://icmathscomp.org/past-papers)"
] | [
"origin:aops",
"2022 Contests",
"ICMC 5"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/ICMC 5/2792553.json"
} | |
Let $P$ be a point on the circumcircle of square $ABCD$ such that $PA \cdot PC = 56$ and $PB \cdot PD = 90.$ What is the area of square $ABCD?$ | I'm super washed up so tried coordinates as the first thing - took me under five minutes, so figured that I'd write it up since no one has yet:
Let $A$ be at $(-x,x)$ , $B$ at $(-x, -x)$ , $C$ at $(x,-x)$ and $D$ at $(x,x)$ . The area is then $4x^2$ . Let $P$ have coordinates $(a,b)$ . We have $a^2+b... | [
"why is this problem so hard",
"why was question one so hard",
"this was too hard\ntook way too long",
"PTOLOMEY",
" $PA=14,PC=4,PB=9\\sqrt2,PD=5\\sqrt2$ so $106.$ This took me 45 minutes :(",
"Ez coordbash",
"Easy trig bash\nSet angle as x and other angle is 45-x\nDouble angle identities \nEzpz",
"... | [
"origin:aops",
"2023 AIME",
"2023 Contests"
] | {
"answer_score": 1080,
"boxed": false,
"end_of_proof": false,
"n_reply": 74,
"path": "Contest Collections/2023 Contests/2023 AIME/3011224.json"
} |
Five men and nine women stand equally spaced around a circle in random order. The probability that every man stands diametrically opposite a woman is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | <details><summary>Sol</summary>We have $14$ ways for a man to go in the circle, and the 2nd one can go in $12$ places, then $10$ , then $8$ , and finally $6$ . We can multiply these and divide by $120$ because of overcounting, resulting in $14\cdot{12}\cdot{10}\cdot{8}\cdot{6}$ . There is $24$ ways to place... | [
"anyone 191",
"191 confirmed",
"i got this one wrong because I forgot that it was 2^5 and only did x2\n",
"<blockquote>i got this one wrong because I forgot that it was 2^5 and only did x2</blockquote>\n\ni also got it wrong its not 2^5 its 14x12x10x8x6/14!",
"<blockquote>anyone 191</blockquote>\n\nYes can ... | [
"origin:aops",
"2023 AIME",
"2023 Contests"
] | {
"answer_score": 1032,
"boxed": true,
"end_of_proof": false,
"n_reply": 64,
"path": "Contest Collections/2023 Contests/2023 AIME/3011229.json"
} |
Let $\triangle ABC$ be an equilateral triangle with side length $55$ . Points $D$ , $E$ , and $F$ lie on sides $\overline{BC}$ , $\overline{CA}$ , and $\overline{AB}$ , respectively, with $BD=7$ , $CE=30$ , and $AF=40$ . A unique point $P$ inside $\triangle ABC$ has the property that \[\measuredangle A... | Solution that I think hasn't been posted yet:
<details><summary>Solution</summary>I claim that $\tan \angle AEP = 5\sqrt{3}$ , which would imply the requested answer is $\boxed{75}$ .**<span style="color:blue">Claim:</span>** The quadrilaterals $AEPF$ , $BFPD$ and $CDPE$ are cyclic.
*Proof.* This is since \[ \a... | [
"Drop the perpendiculars from $P$ to $\\overline{AB}$ , $\\overline{AC}$ , $\\overline{BC}$ , and call them $Q,R,$ and $S$ respectively. This gives us three similar right triangles $FQP$ , $ERP$ , and $DSP.$ \n\nThe sum of the perpendiculars to a point $P$ within an equilateral triangle is always con... | [
"origin:aops",
"2023 AIME",
"2023 Contests"
] | {
"answer_score": 1200,
"boxed": true,
"end_of_proof": true,
"n_reply": 51,
"path": "Contest Collections/2023 Contests/2023 AIME/3011230.json"
} |
There exists a unique positive integer $a$ for which the sum \[U=\sum_{n=1}^{2023}\left\lfloor\dfrac{n^{2}-na}{5}\right\rfloor\] is an integer strictly between $-1000$ and $1000$ . For that unique $a$ , find $a+U$ .
(Note that $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x$ ... | The key idea is that the summation without the floors should be close to $0$ because each floor can only change the sum by less than $1$ . The natural choice is a value of $a$ near the one that makes the aforementioned sum exactly $0$ . Luckily, we compute
\[ \sum_{n=1}^{2023}\frac{n^2-na}5 = \frac15 \left( \frac... | [
"I got $944$ can anyone confirm?",
"can confirm",
" $a=1349, U=-405$ ",
"arggggghhh I got $\\frac{2025}{5}=225$ I did that like one hundred times...",
"<blockquote>I got $944$ can anyone confirm?</blockquote>\n\nYea, 1349-405",
"no i did not btw",
"https://artofproblemsolving.com/community/c5h3011... | [
"origin:aops",
"2023 AIME",
"2023 Contests"
] | {
"answer_score": 1054,
"boxed": false,
"end_of_proof": false,
"n_reply": 31,
"path": "Contest Collections/2023 Contests/2023 AIME/3011232.json"
} |
Call a positive integer $n$ *extra-distinct* if the remainders when $n$ is divided by $2, 3, 4, 5,$ and $6$ are distinct. Find the number of extra-distinct positive integers less than $1000$ . | **Solution:** Firstly, we assume $n$ is even (0 mod 2). This must mean $n$ is $2 \pmod{4}$ and thus $4 \pmod{6}$ . $n \equiv 1 \pmod{3}, 3 \pmod{5}$ follows. This leads to $n \equiv 58 \pmod{60}$ .
Now, assume that the remainder when $n$ is divided by $2$ is $1$ . One subcase is where $n$ has remainde... | [
" $n\\equiv \\{35,58,59\\}\\pmod{60}$ so $17 + 16 + 16 = \\boxed{049}$ . ",
"16*3+1 = 49 ",
"Can anyone confirm 046?\nEdit: well rip",
"it's 049 my friend",
"I got 049",
"forgot that mod 2 and mod 3 determine mod 6 and got 16*32+30\n",
"I got 049 also",
"i sillied and missed a case lmao",
"missed ... | [
"origin:aops",
"2023 AIME",
"2023 Contests"
] | {
"answer_score": 1054,
"boxed": false,
"end_of_proof": false,
"n_reply": 45,
"path": "Contest Collections/2023 Contests/2023 AIME/3011233.json"
} |
The sum of all positive integers $m$ for which $\tfrac{13!}{m}$ is a perfect square can be written as $2^{a}3^{b}5^{c}7^{d}11^{e}13^{f}$ , where $a, b, c, d, e,$ and $f$ are positive integers. Find $a+b+c+d+e+f$ . | First note $13! = 2^{10} \cdot 3^{5} \cdot 5^2 \cdot 7\cdot 11\cdot 13$ .
This implies that $m = 2^{a_1} \cdot 3^{a_2}\cdot 5^{a_3} \cdot 7\cdot 11\cdot 13$ for nonnegative integers $a_1, a_2, a_3$ .
We must have $a_1$ is even and $\le 10$ , $a_2$ is odd and $\le 5$ , $a_3$ is even and $\le 2$ .
We ... | [
"1+2+1+3+1+4 = 12",
"012 confirmed",
"<blockquote>1+2+1+3+1+4 = 12</blockquote>\n\nafter i did that i realized i didnt even need to. Just needed to count the number of factors lol",
"More one-off errors I forgot a 13 and put 011.",
"ye the sum of all m is like\n13*11*7*3*(25+1)*(81+9+1)*(1024+256+64+16+4+1)... | [
"origin:aops",
"2023 AIME",
"2023 Contests"
] | {
"answer_score": 1022,
"boxed": false,
"end_of_proof": false,
"n_reply": 24,
"path": "Contest Collections/2023 Contests/2023 AIME/3011237.json"
} |
Find the largest prime number $p<1000$ for which there exists a complex number $z$ satisfying
- the real and imaginary part of $z$ are both integers;
- $|z|=\sqrt{p}$ , and
- there exists a triangle whose three side lengths are $p$ , the real part of $z^{3}$ , and the imaginary part of $z^{3}$ .
| <details><summary>Solution</summary>The answer is $\boxed{349}$ , which is achieved with $z = 18 + 5i$ .
First, let $z = a + bi$ for integers $a$ and $b$ ; then we have that $a^2 + b^2 = p$ and moreover, \[ z^3 = (a+bi)^3 = a^3 + 3a^2bi - 3ab^2 - b^3i \] Hence the real and imaginary parts are $a^3 - 3ab^2$ a... | [
"me when the guess 997 but oops that's 1 mod 4\n",
"Don’t know the solution, quick code bash gives the answer ",
"When you don't read the problem carefully and guess 025 :clown:",
"349 rite? (18+5i)",
"Yea confirmed, I got $349$ also",
"Confirm $349$ . The idea is that if $z=a+bi$ , we can take $a,b$... | [
"origin:aops",
"2023 AIME",
"2023 Contests"
] | {
"answer_score": 1224,
"boxed": true,
"end_of_proof": true,
"n_reply": 25,
"path": "Contest Collections/2023 Contests/2023 AIME/3011240.json"
} |
Rhombus $ABCD$ has $\angle BAD<90^{\circ}$ . There is a point $P$ on the incircle of the rhombus such that the distances from $P$ to lines $DA$ , $AB$ , and $BC$ are $9$ , $5$ , and $16$ , respectively. Find the perimeter of $ABCD$ .
| <details><summary>Subjective Quality/Difficulty</summary>4.75 out of 5. AIME P7 or P8.</details>
<details><summary>Solution 1</summary>Let $X$ , $Y$ , and $Z$ be the feet of the altitudes from $P$ to lines $DA$ , $AB$ , and $BC$ , respectively. Denote by $Q$ the intersection of height $\overline{XZ}$ with... | [
"wasted so much time on this but it was actually a good problem \n",
"<blockquote>wasted so much time on this but it was actually a good problem</blockquote>\n\n^^^^^^^",
"I guessed this one correctly! :thumbsup: ",
" sine addition in cyclic quadrilaterals ",
"4*(25+25/4) = 125\n\nbeautiful problem",
"yea... | [
"origin:aops",
"2023 AIME",
"2023 Contests"
] | {
"answer_score": 1210,
"boxed": true,
"end_of_proof": false,
"n_reply": 54,
"path": "Contest Collections/2023 Contests/2023 AIME/3011242.json"
} |
A plane contains $40$ lines, no $2$ of which are parallel. Suppose there are $3$ points where exactly $3$ lines intersect, $4$ points where exactly $4$ lines intersect, $5$ points where exactly $5$ lines intersect, $6$ points where exactly $6$ lines intersect, and no points where more than $6$ lin... | We have a total of $\binom{40}{2}$ pairs of intersections. Now note that some of these intersections overlap. We will count how many overlap. In places where $3$ lines intersect, note that we have $\binom{3}{2}$ intersection. Thus $3 \cdot \binom{3}{2}$ intersections there. Similarly for $4$ and $5$ for a f... | [
"607 just 40 choose 2 - 3*3choose 2 - 4*4choose 2 - 5*5choose 2 - 6*6choose 2",
"Almost the same as [https://artofproblemsolving.com/community/c3h2714268](https://artofproblemsolving.com/community/c3h2714268), but you're choosing pairs of lines instead of pairs of points. \n\nWas able to solve almost instantly be... | [
"origin:aops",
"2023 AIME",
"2023 Contests"
] | {
"answer_score": 1012,
"boxed": false,
"end_of_proof": false,
"n_reply": 32,
"path": "Contest Collections/2023 Contests/2023 AIME/3011245.json"
} |
Find the number of cubic polynomials $p(x) = x^3 + ax^2 + bx + c$ , where $a$ , $b$ , and $c$ are integers in $\{-20, -19,-18, \dots , 18, 19, 20\}$ , such that there is a unique integer $m \neq 2$ with $p(m) = p(2)$ . | this is good
Note that $p(x)-p(2)$ has roots $\{2,2,m\}$ or $\{2,m,m\}$ . These give the polynomials
\[x^3-(m+4)x^2+(4m+4)x-4m\]
\[x^3-(2m+2)x^2+(m^2+4m)x-2m^2\]
and since the constant is irrelevant (because $c$ can be anything) we have that
\[-6\le m\le 4\]
\[-6\le m\le 2\]
giving $11-1=10$ and $9-1=8$ sol... | [
"328 gang anyone? :wallbash_red: :wallbash_red: (correct answer is 738)",
"is the answer 287?\n",
"anyone get 369? im crying now",
"No, it should be 41*18 = 738.",
"I got $738$ .",
"FRICK 820 FORGOT THAT $m\\neq 2$ FRICKDKJFDHFKJSDHFJKDSHKJFGDSHJKLFDJKLSHFJKLDHSHJKLD",
"made 3 separate errors on thi... | [
"origin:aops",
"2023 AIME",
"2023 Contests"
] | {
"answer_score": 1118,
"boxed": false,
"end_of_proof": false,
"n_reply": 57,
"path": "Contest Collections/2023 Contests/2023 AIME/3011247.json"
} |
If $\sqrt{\log_bn}=\log_b\sqrt n$ and $b\log_bn=\log_bbn,$ then the value of $n$ is equal to $\frac jk,$ where $j$ and $k$ are relatively prime. What is $j+k$ ? | <details><summary>Literally the easiest problem on the test</summary>Just substitute $x$ for $\log_b n$ , so we get a systems of equations $\sqrt{x} = \frac{1}{2}x$ and $bx = 1 + x$ . Multiply both sides of the first equation by $2$ and then squaring both sides gives $x^2 = 4x$ . Solving for $x$ and we get ... | [
"625+256 = 881 yummy\n\nI initially thought the solution to $\\sqrt{x} = \\frac12 x$ is $x=\\frac14$ oops",
"Can confirm $881$ ",
"Just let $b^x=n$ and this is cheese",
"This was my favorite problem! <span style=\"font-size:50%\">/j dont hurt me</span>\n\nThe first equation gives $\\log_b n = \\frac{1... | [
"origin:aops",
"2023 AIME",
"2023 Contests"
] | {
"answer_score": 1032,
"boxed": false,
"end_of_proof": false,
"n_reply": 41,
"path": "Contest Collections/2023 Contests/2023 AIME/3011255.json"
} |
Each face of two noncongruent parallelepipeds is a rhombus whose diagonals have lengths $\sqrt{21}$ and $\sqrt{31}$ . The ratio of the volume of the larger of the two polyhedra to the volume of the smaller is $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . A parallelepiped i... | pov: an hour sunk because clearly the flat right triangles are 45-45-90! + why wasn't anyone who actually solved it synthetically nice enough to post a solution :'( + whoops it's already 11 pm, i guess i have to solve p12 in my sleep!**<span style="color:#f00">Part I: Visualization.</span>** Consider two adjacent faces... | [
"lol i tried to set the parallelopipeds as a cube and realized it was wrong and then cried and gave up.",
"This problem came with a diagram, I'm pretty sure.",
"yea i dont have the diagram rn oops..\n\nedit: should be there now, thanks to naman12",
"I didn't even understand the problem so :D ",
"<blockquot... | [
"origin:aops",
"2023 AIME",
"2023 Contests"
] | {
"answer_score": 1068,
"boxed": true,
"end_of_proof": false,
"n_reply": 44,
"path": "Contest Collections/2023 Contests/2023 AIME/3011264.json"
} |
Alice knows that $3$ red cards and $3$ black cards will be revealed to her one at a time in random order. Before each card is revealed, Alice must guess its color. If Alice plays optimally, the expected number of cards she will guess correctly is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive ... | There are $\frac{6!}{3! \cdot 3!}$ ways for the cards to show up. Then dividing by $2$ due to symmetry we have $10$ cases so it is feasible to bash.
The cases are given by, \[ \{ RRRBBB, RRBRBB, RRBBRB, RRBBBR, RBRRBB, RBRBRB, RBRBBR, RBBRRB, RBBRBR, RBBBRR\} \]
Now we take cases.**Case $1$** Expected number ... | [
"41/10 --> 051",
"can confirm",
"051 :smirk_cat:",
"did something dumb with tree diagrams and gave up \n",
"<blockquote>did something dumb with tree diagrams and gave up</blockquote>\n\nalso made a tree diagram but got the wrong answer probably made a calculation error somewhere",
"i spammed fractions\nac... | [
"origin:aops",
"2023 AIME",
"2023 Contests"
] | {
"answer_score": 1052,
"boxed": true,
"end_of_proof": false,
"n_reply": 43,
"path": "Contest Collections/2023 Contests/2023 AIME/3011266.json"
} |
Find the number of subsets of ${1,2,3,...,10}$ that contain exactly one pair of consecutive integers. Examples of such subsets are ${1,2,5}$ and ${1,3,6,7,10}$ . | We have the following crucial claim:**Claim.** The number of subsets of $\{k,k+1,\ldots,k+n-1\}$ without two consecutive numbers is $F_{n+2}$ , where $F_n$ is the $n$ th Fibonacci number.**Proof.** Note that this is true for $n=0,1$ , so it suffices to show that the sets satisfy the Fibonacci recurrence. Let the... | [
"Got 235. Anyone confirm?",
"Didn't solve this, but worked it out after the exam. The answer is $235$ .",
"bruh this just recursion",
"i wish i read this instead of panicking over p5 and p8 and spending way too long on those",
"I swear I've seen this question before somewhere\n\nLet $a_n$ denote the num... | [
"origin:aops",
"2023 AIME",
"2023 Contests"
] | {
"answer_score": 1134,
"boxed": false,
"end_of_proof": false,
"n_reply": 40,
"path": "Contest Collections/2023 Contests/2023 AIME/3011276.json"
} |
The following analog clock has two hands that can move independently of each other.
[asy]
unitsize(2cm);
draw(unitcircle,black+linewidth(2));
for (int i = 0; i < 12; ++i) {
draw(0.9*dir(30*i)--dir(30*i));
}
for (int i = 0; i < 4; ++i) {
... | Consider a $12\times 12$ grid (from $(0,0)$ to $(11,11)$ ) and call a point $(x,y)$ *good* if $x+y\equiv 0\pmod{12}$ . Note that if we plot the movements on the grid (and add portals along $x=12$ and $y=12$ to go to $x=0$ and $y=0$ , respectively), we get a path visiting every point exactly once. Further... | [
"clockblocked $\\binom{12}1+\\binom{12}5+\\binom{12}7+\\binom{12}{11}=1608$ ",
"clockblocked 2.0",
"my friend did some weird toroidal checkerboard thingy",
"I had this :sob:\n\nI HAD THIS\n\nI made two errors and only corrected one which made it wrong instead\n\ni would've gotten a 15",
"rip i spent an hour... | [
"origin:aops",
"2023 AIME",
"2023 Contests"
] | {
"answer_score": 1174,
"boxed": false,
"end_of_proof": false,
"n_reply": 11,
"path": "Contest Collections/2023 Contests/2023 AIME/3011300.json"
} |
Let $\omega=\cos\frac{2\pi}{7}+i\cdot\sin\frac{2\pi}{7}$ , where $i=\sqrt{-1}$ . Find $$ \prod_{k=0}^{6}(\omega^{3k}+\omega^k+1). $$ | <details><summary>solution</summary>Let's look at the factors of $1+\omega^k+\omega^{3k}$ and ignore the boring case where $k=0$ (in which case the factor is 3). Notice that
\begin{align*}(1+\omega^k+\omega^{3k})(1+\omega^{3-k}+\omega^{3(3-k)})&=(1+\omega^k+\omega^{3k})(1+\omega^{-k}+\omega^{-3k})
& =3+\omega^k+\om... | [
"Anyone else get 024? (a bit worried about this one)",
"Let me calculate while doing english",
"<blockquote>Anyone else get 024? (a bit worried about this one)</blockquote>\n\nYes someone confirmed 024. I didn’t solve this unfortunately :(",
"yep checked in wolfram alpha afterwards it is 024",
"24. Just p... | [
"origin:aops",
"2023 AIME",
"2023 Contests"
] | {
"answer_score": 1116,
"boxed": true,
"end_of_proof": false,
"n_reply": 51,
"path": "Contest Collections/2023 Contests/2023 AIME/3016170.json"
} |
Each vertex of a regular dodecagon (12-gon) is to be colored either red or blue, and thus there are $2^{12}$ possible colorings. Find the number of these colorings with the property that no four vertices colored the same color are the four vertices of a rectangle. | <details><summary>sol</summary>We can see that a rectangle of $4$ vertices of the same color is simply 2 pairs of opposite vertices of the same color.
There are $3$ cases $\textbf{Case 1: no opposite pairs of vertices share the same color}$ We have $2$ options for each of the opposite vertices thus we have $2^6... | [
" $928$ , rectangle is equivalent to two pairs of opposite points being the same color.",
"<details><summary>sol but i got sniped</summary>Notice that for every rectangle, the two diagonals must bisect each other, which is only possible when both diagonals connect opposite sides of the dodecagon.\n\nWe can do cas... | [
"origin:aops",
"2023 AIME",
"2023 Contests"
] | {
"answer_score": 1032,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2023 Contests/2023 AIME/3016173.json"
} |
Consider the L-shaped region formed by three unit squares joined at their sides, as shown below. Two points $A$ and $B$ are chosen independently and uniformly at random from inside this region. The probability that the midpoint of $\overline{AB}$ also lies inside this L-shaped region can be expressed as $\frac{m... | Bro everyone is overcomplicating this
<details><summary>Solution</summary>We use complementary counting. There is a $\frac{2!}{3^{2}} = \frac{2}{9}$ chance that the squares in the top left and bottom right are the places where $A$ and $B$ are. Let $A = (p, q+1), B = (r+1,s),$ then we need $p+r \ge 1,$ and $... | [
" $\\frac{17}{18}$ gives $35$ by a symmetry argument.",
"We calculate the probability it lies outside the shape, and subtract it from $1$ . \nLet the L shape be a 2x2 square with the top right 1x1 square missing.\nLet $A=(x_1, y_1)$ and $B=(x_2,y_2)$ and need $\\frac{x_1+x_2}{2}>1$ and $\\frac{y_1+y_2}... | [
"origin:aops",
"2023 AIME",
"2023 Contests"
] | {
"answer_score": 1020,
"boxed": true,
"end_of_proof": false,
"n_reply": 13,
"path": "Contest Collections/2023 Contests/2023 AIME/3016174.json"
} |
Circles $\omega_1$ and $\omega_2$ intersect at two points $P$ and $Q$ , and their common tangent line closer to $P$ intersects $\omega_1$ and $\omega_2$ at points $A$ and $B$ , respectively. The line parallel to line $AB$ that passes through $P$ intersects $\omega_1$ and $\omega_2$ for the secon... | <details><summary>Solution</summary>Firstly, dropping perpendiculars from the centers of the circles to the midpoints of their respective chords and then continuing the perpendiculars until they hit the tangency points of their comment tangent line, we find that
\begin{align*}
AB &= 5 + 7
&= 12.
\end{align*}
N... | [
"Best problem on the test imo. I used appolonius + power of point + dropping perpendiculars.",
"Tangent circles? Three given lengths? All the markings of a dj problem, I guess. (Also, should it say $PQ = 5$ ?)\n\n<details><summary>Solution</summary>Let $\\ell\\equiv AB$ be the common tangent and $m\\equiv ... | [
"origin:aops",
"2023 AIME",
"2023 Contests"
] | {
"answer_score": 1126,
"boxed": true,
"end_of_proof": false,
"n_reply": 34,
"path": "Contest Collections/2023 Contests/2023 AIME/3016185.json"
} |
Let $x$ , $y$ , and $z$ be real numbers satisfying the system of equations
\begin{align*}
xy+4z&=60
yz+4x&=60
zx+4y&=60.
\end{align*}
Let $S$ be the set of possible values of $x$ . Find the sum of the squares of the elements of $S$ . | Note, that $(xy+4z)-(yz+4x)=(x-z)(y-4)=0.$ Case 1: $x=z.$ If $x=z,$ then $x^2+4y=60,$ using the last equation, and $xy+4x=60,$ using the first equation. Now, note that $y=\frac{60-x^2}{4},$ so $xy+4x=\frac{60x-x^3}{4}+4x=60 \implies 60x-x^3+16x=240 \implies x^3-76x+240=(x-6)(x+10)(x-4)=0,$ so $x=(6,-10,4)$... | [
"<details><summary>Solution</summary>Saw something very similar just a day before the contest.\n\nFirst notice $60x - 4x^2 = 60y - 4y^2 = 60z - 4z^2 = xyz$ . Since any parabola and a horizontal line intersect at most twice, we can assume WLOG that up to cyclic rotation that $x = y$ . If $z = 15-x$ then we have ... | [
"origin:aops",
"2023 AIME",
"2023 Contests"
] | {
"answer_score": 1032,
"boxed": false,
"end_of_proof": false,
"n_reply": 15,
"path": "Contest Collections/2023 Contests/2023 AIME/3016186.json"
} |
In $\triangle ABC$ with side lengths $AB=13$ , $BC=14$ , and $CA=15$ , let $M$ be the midpoint of $\overline{BC}$ . Let $P$ be the point on the circumcircle of $\triangle ABC$ such that $M$ is on $\overline{AP}$ . There exists a unique point $Q$ on segment $\overline{AM}$ such that $\angle PBQ = \an... | <details><summary>In contest, apparently there is a kill</summary>Notice that since $(BPQ)$ and $(CPQ)$ must have the same circumradius, $\frac{\sin BQP}{\sin CQP} = \frac{BP}{CP} = \frac{\sin BAP}{\sin CAP}$ by ratio lemma, so it follows that $\frac{AB}{BQ} = \frac{AC}{CQ}$ , so $B, C$ must lie on the same ap... | [
"just a parallelogram lmao",
"Just construct a parallelogram and all the lengths are fixed. This took 2 mins on the test to solve lol. ",
"<blockquote>just a parallelogram lmao</blockquote>\n\nyup. bruh why was this the easiest problem on the test :skull:",
"One can also recall countless similar problems such... | [
"origin:aops",
"2023 AIME",
"2023 Contests"
] | {
"answer_score": 1044,
"boxed": true,
"end_of_proof": false,
"n_reply": 27,
"path": "Contest Collections/2023 Contests/2023 AIME/3016187.json"
} |
Find the number of collections of $16$ distinct subsets of $\{1, 2, 3, 4, 5\}$ with the property that for any two subsets $X$ and $Y$ in the collection, $X\cap Y \neq \emptyset$ . | Firstly, there cannot be two subsets with cardinality 1, or they will not intersect.
If there is one subset $A$ with cardinality $1$ ; let the element in $A$ be $a$ , then there are $2^4=16$ subsets that do not include $a$ so they do not work. Every remaining subsets $S$ will have $a$ as an element so $... | [
"<details><summary>Solution</summary>Hard! We perform casework. \nIf there is a subset of size $0$ you lose. If there is subset of size $1$ you get $5$ .\nElse subsets of size $4$ and size $5$ always work and you can only have one of each complementary pair of size $3$ and size $2$ subsets, so you can ... | [
"origin:aops",
"2023 AIME",
"2023 Contests"
] | {
"answer_score": 1094,
"boxed": false,
"end_of_proof": false,
"n_reply": 10,
"path": "Contest Collections/2023 Contests/2023 AIME/3016190.json"
} |
The number of apples growing on each of six apple trees form an arithmetic sequence where the greatest number of apples growing on any of the six trees is double the least number of apples growing on any of the six trees. The total number of apples growing on all six trees is $990$ . Find the greatest number of apples... | compare this to these #1's and laugh:
<blockquote>
Initially Alex, Betty, and Charlie had a total of $444$ peanuts. Charlie had the most peanuts, and Alex had the least. The three numbers of peanuts that each person had formed a geometric progression. Alex eats $5$ of his peanuts, Betty eats $9$ of her peanuts, a... | [
"Let the smallest tree have $x$ apples and the biggest tree have $2x$ apples. \n\nThis arithmetic sequence has sum $6(\\dfrac{3x}{2}) = 990$ so $x=110$ . So the biggest tree has $2x=220$ apples. Man that's a lot of apples.",
"yes 220 is right. I just solved it",
"Who at MAA thought making this an AIME... | [
"origin:aops",
"2023 AIME",
"2023 Contests"
] | {
"answer_score": 28,
"boxed": false,
"end_of_proof": false,
"n_reply": 34,
"path": "Contest Collections/2023 Contests/2023 AIME/3016191.json"
} |
Recall that a palindrome is a number that reads the same forward and backward. Find the greatest integer less than $1000$ that is a palindrome both when written in base ten and when written in base eight, such as $292=444_{\text{eight}}$ . | Such a number should have three digits in base ten and four digits in base 8. Write the base ten number as $aba$ and write the base eight number as $cddc$ . Then we have $101a + 10b = 513c + 72d$ . Clearly $c = 1$ , so we can rewrite as $101a+10b = 513+72d$ . We now want to maximize $a$ . If $a = 9$ , we have ... | [
"If the base-8 number had only 3 digits, the highest we could go is only $777_8 = 511$ .\nSo we assume it has 4 digits: $abba_8 = 512a + 64b + 8b + a = 513a + 72b$ a must be 1 because our number would already exceed 1000 with $a=2$ .\nBashing values of b from 1 to 6, we find that only b=1 works, resulting in $58... | [
"origin:aops",
"2023 AIME",
"2023 Contests"
] | {
"answer_score": 1036,
"boxed": false,
"end_of_proof": false,
"n_reply": 10,
"path": "Contest Collections/2023 Contests/2023 AIME/3016194.json"
} |
Let $S$ be the set of all positive rational numbers $r$ such that when the two numbers $r$ and $55r$ are written as fractions in lowest terms, the sum of the numerator and denominator of one fraction is the same as the sum of the numerator and denominator of the other fraction. The sum of all the elements of $... | Mine. (Also, darn, they replaced $91$ with something less trolly? What a shame.)
<details><summary>Solution</summary>Write $r = \tfrac ab$ , where $a$ and $b$ are relatively prime, and suppose $r$ satisfies the condition in the problem statement. Set $d = \gcd(55a,b) = \gcd(55,b)$ ; then $d\mid 55 = 5\cd... | [
"This felt like a DJ problem :D",
"Nice. Let $r = \\tfrac{m}{n}$ for coprime positive integers $m,n$ . Do casework on $\\gcd (n, 55)$ .\n\nIf $\\gcd(n, 55) = 1$ , then $55r = \\tfrac{55m}{n}$ when written in simplest form, so $55m + n = m + n\\implies m = 0$ , impossible.\n\nIf $\\gcd(n, 55) = 5$ , then ... | [
"origin:aops",
"2023 AIME",
"2023 Contests"
] | {
"answer_score": 1044,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2023 Contests/2023 AIME/3016197.json"
} |
For each positive integer $n$ let $a_n$ be the least positive integer multiple of $23$ such that $a_n\equiv1\pmod{2^n}$ . Find the number of positive integers $n$ less than or equal to $1000$ that satisfy $a_n=a_{n+1}$ . | Non p-adic solution, although it is probably similar in spirit to #2 (not sure, I don't really know how p-adics work).
For all $n\geq 1$ , $a_n = r\cdot 2^n+1$ for some integer $0\leq r < 23$ (since if $r \geq 23$ worked, then $r-23$ would work, contradicting the minimality of $a_n$ ). Furthermore, there is ... | [
"Note that this question is equivalent to determining the number of $0$ s in the first $1001$ digits of the $2$ -adic representation of $23^{-1}$ .\nNotice $\\frac{1}{23} = 1 + \\frac{2^{11}-1 - 89}{1-2^{11}} = 1 + (11110100110_2)(1+2^{11}+2^{22}+2^{33}+\\ldots) = \\overline{01111010011}1_2$ .\nNotice $1000 ... | [
"origin:aops",
"2023 AIME",
"2023 Contests"
] | {
"answer_score": 1136,
"boxed": true,
"end_of_proof": false,
"n_reply": 17,
"path": "Contest Collections/2023 Contests/2023 AIME/3016199.json"
} |
Let $A$ be an acute angle such that $\tan A = 2\cos A$ . Find the number of positive integers $n$ less than or equal to $1000$ such that $\sec^n A + \tan^n A$ is a positive integer whose units digit is $9$ . | Very creative hybrid alg/nt problem.
First, notice obviously that if $x = \sin A$ then $2x^2 + x - 2 = 0$ so $x = \frac{\sqrt{17}-1}{4}$ .
Thus $\cos A = \sqrt{\frac{\sqrt{17}-1}{8}}$ The desired expression is equivalent to $\frac{1+\sin^n A}{\cos^n A}$ . Suppose the value is an integer.
Let $\mathbb{Q}(\sqrt{... | [
"Thank god I got this right",
"Official wording:\n\n<blockquote>Let $x$ and $y$ be real numbers such that $0<x<1$ , $x=\\sqrt{\\tfrac{y}{2}}$ , and $\\tfrac{1}{y}-y=\\tfrac{1}{2}$ . Define the sequence $a_{n}:=\\left(\\frac{1}{x}\\right)^{n}+\\left(\\frac{y}{x}\\right)^{n}$ for $n\\geq 1$ . Find the num... | [
"origin:aops",
"2023 AIME",
"2023 Contests"
] | {
"answer_score": 1054,
"boxed": true,
"end_of_proof": false,
"n_reply": 13,
"path": "Contest Collections/2023 Contests/2023 AIME/3016200.json"
} |
Let $\triangle{ABC}$ be an isoceles triangle with $\angle A=90^{\circ}$ . There exists a point $P$ inside $\triangle{ABC}$ such that $\angle PAB=\angle PBC=\angle PCA$ and $AP=10$ . Find the area of $\triangle{ABC}$ . | <details><summary>Solution</summary>From some angle chasing, one identifies $\angle CPA = 90^{\circ}$ and $\angle APB = \angle BPC = 135^{\circ}$ . This point $P$ can be identified by identifying the second intersection of the circle centered at $F$ (the midpoint of $AC$ ) passing through $A$ and $C$ and th... | [
"Wow, turns out $P$ is just the $B$ -Dumpty point of $ABC$ . Unfortunately, I spent 50 minutes on this problem before rotating everything $90^{\\circ}$ about $A$ so that $C’ = B$ , yielding $PC^2 - PB^2 = 200$ after some observations.",
"It is also a Brocard point, surprised that configuration was even... | [
"origin:aops",
"2023 AIME",
"2023 Contests"
] | {
"answer_score": 1046,
"boxed": false,
"end_of_proof": false,
"n_reply": 26,
"path": "Contest Collections/2023 Contests/2023 AIME/3016204.json"
} |
Let $N$ be the number of ways to place the integers $1$ through $12$ in the $12$ cells of a $2\times 6$ grid so that for any two cells sharing a side, the difference between the numbers in those cells is not divisible by $3$ . One way to do this is shown below. Find the number of positive integer divisors ... | Consider residues modulo 3. Each column must contain either $01,02,12$ in some order, and symmetry tells us we must have exactly two of each in our grid. Thus we have $\tfrac{6!}{2!2!2!}$ ways to rearrange these strings, two ways to assign the orientation of the first column, and all subsequent columns are fixed in... | [
"(redacted)",
"Mine. This one is a bit evil. (It's also not the most original, admittedly; see the remark.)\n\n<details><summary>Solution</summary>Replace each integer with its remainder upon division by $3$ . Then there are $4$ copies of each number from $0$ to $2$ and the given condition implies no tw... | [
"origin:aops",
"2023 AIME",
"2023 Contests"
] | {
"answer_score": 1104,
"boxed": true,
"end_of_proof": false,
"n_reply": 12,
"path": "Contest Collections/2023 Contests/2023 AIME/3016229.json"
} |
A cube-shaped container has vertices $A$ , $B$ , $C$ , and $D$ where $\overline{AB}$ and $\overline{CD}$ are parallel edges of the cube, and $\overline{AC}$ and $\overline{BD}$ are diagonals of the faces of the cube. Vertex $A$ of the cube is set on a horizontal plane $\mathcal P$ so that the plane of... | linear algebra solution, done with the help of geogebra
[diagram here](https://www.geogebra.org/calculator/b8mzhkpv)
Let $s$ be the side length of the cube. Let $A=(0, 0, 0)$ , $B=(s, 0, 0)$ , $C=(0, s, s)$ , and $D=(s, s, s)$ . Let the equation of plane $\mathcal P$ be $ax+by+cz=0$ , since it must pass throu... | [
"The hardest part is making the diagram.\n\nTo be fair, after we ran out of sphere tangency problems, there isn't much left in 3d geo it seems... Conics now?\nDrop $A_{\\perp}, B_{\\perp}, C_{\\perp}, D_{\\perp}$ as feets of altitudes of $A, B, C, D$ respectively.\nFirst, cut out the plane with $\\triangle ABC... | [
"origin:aops",
"2023 AIME",
"2023 Contests"
] | {
"answer_score": 1148,
"boxed": true,
"end_of_proof": false,
"n_reply": 17,
"path": "Contest Collections/2023 Contests/2023 AIME/3016733.json"
} |
Given is a positive integer $n$ . There are $2n$ mutually non-attacking rooks placed on a grid $2n \times 2n$ . The grid is splitted into two connected parts, symmetric with respect to the center of the grid. What is the largest number of rooks that could lie in the same part? | **Answer:** $2n-1$ **<span style="color:#0f0">Construction:</span>**
Below is a construction for $n=4$ which easily generalizes.
[asy]
filldraw(((0,0)--(0,1)--(1,1)--(1,0)--cycle), grey);
filldraw(((1,1)--(1,2)--(2,2)--(2,1)--cycle), grey);
filldraw(((2,2)--(2,3)--(3,3)--(3,2)--cycle), grey);
filldraw(((3,3)--(3,4)... | [
"<details><summary>Solution</summary>The answer is $2n-1$ . It is easy to produce an example with $2n-1$ rooks as follows: Start with the decomposition cut along the main diagonal from top left to lower right. Give the upper left half of the diagonal to the upper right piece and the other half to the other piece... | [
"origin:aops",
"2023 Contests",
"2023 All-Russian Olympiad Regional Round"
] | {
"answer_score": 16,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 All-Russian Olympiad Regional Round/3016010.json"
} |
Let $a, b, c$ be positive integers such that no number divides some other number. If $ab-b+1 \mid abc+1$ , prove that $c \geq b$ . | Can anyone check my solution:
Suppose $b>c$ .Since $ab-b+1 | abc+1$ , implies $ab-b+1 | abc+1-c(ab-b+1)=c(b-1)+1<b^2-b+1$ , so $b>a$ .Let $m= \frac{abc+1}{ab-b+1}, abc+1=mab-mb+m$ and $a(bc-mb)=m(1-b)-1, a= \frac{m(1-b)-1}{bc-mb}>0$ .Now $m(1-b)-1 \geq bc-mb$ , so $m \geq bc+1$ .
\begin{align}
abc+1 \geq (bc+1... | [
"Note that clearly $a,b,c>1$ and $ab-b+1\\mid abc+1-(ab-b+1) = b(ac-a+1)$ implies, together with $(b,ab-b+1)=1$ , that $ab-b+1\\mid ac-a+1$ . Now if $ac-a+1=ab-b+1$ then $a(c-1)=b(a-1)$ , so that $a\\mid b$ , a contradiction. Hence, $ac-a+1\\ge 2ab-2b+2$ yielding $ac+2b\\ge 2ab+a+1$ . Further, $2ab+a+... | [
"origin:aops",
"2023 Contests",
"2023 All-Russian Olympiad Regional Round"
] | {
"answer_score": 134,
"boxed": false,
"end_of_proof": true,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 All-Russian Olympiad Regional Round/3016011.json"
} |
Let $ABCD$ be a cyclic quadrilateral such that the circles with diameters $AB$ and $CD$ touch at $S$ . If $M, N$ are the midpoints of $AB, CD$ , prove that the perpendicular through $M$ to $MN$ meets $CS$ on the circumcircle of $ABCD$ . | Let line $CS$ and $DS$ intersect the line through $M$ at $X$ and $Y$ respectively. Also, let $I$ be the intersection of radical axes of three circles. Then as the circles are tangent at $S$ , we have the two circles are homothetic. Letting $CS$ and $DS$ intersect $(ASB)$ at $J$ and $K$ respective... | [
"You could find some solutions in Russian here [https://youtu.be/JV_JkzqBLPE](https://youtu.be/JV_JkzqBLPE)",
"Call the perpendicular line through M to MN \"l\". assume $SC\\cap l=X$ and $DS\\cap l=Y.$ With angle chasing it can be easily shown that DCYX is cyclic. Notice that XSY is a right triangle and SM is... | [
"origin:aops",
"2023 Contests",
"2023 All-Russian Olympiad Regional Round"
] | {
"answer_score": 58,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 All-Russian Olympiad Regional Round/3016012.json"
} |
Given are $50$ distinct sets of positive integers, each of size $30$ , such that every $30$ of them have a common element. Prove that all of them have a common element. | We'll show that we can find a set of $k$ sets with a common element , where $30 \leq k \leq 50$ .
To show that we can find such a set for $k=31$ , take a set $B$ , there are now $\binom{49}{29}>30$ ways to form a set of $30$ sets with $B$ in it , therfore some common element much have appeared at least twi... | [
"Some steps of my solution: Let all of them don't have a common element and sets are $A_1,A_2....A_{50}$ $1.$ For all $30$ sets from $A_1,A_2....A_{50}$ define $k$ is the least number of common element. (Where $1 \\leq k \\leq30$ ) $2.$ There isn't a $29$ sets with $k$ common element. $3.$ And we ca... | [
"origin:aops",
"2023 Contests",
"2023 All-Russian Olympiad Regional Round"
] | {
"answer_score": 70,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 All-Russian Olympiad Regional Round/3016013.json"
} |
In a triangle $ABC$ , let $BD$ be its altitude and let $H$ be its orthocenter. The perpendicular bisector of of $HD$ meets $(BCD)$ at $P, Q$ . Prove that $\angle APB+\angle AQB=180^{o}$ | Inverting about $(BC)$ , $H^*$ is the Orthocenter Miquel point ( $MH \cap (AH)$ ) and $P, Q$ are such that $\frac{PD}{PH^*} = \frac{QD}{QH^*} = \frac{MD}{MH^*}$ , thus $\frac{DP}{DQ} = \frac{H^*P}{H^*Q} \implies (AH)$ , which is centered on $PQ$ , must be the Apollonian circle containing the locus of points who... | [
"Let us consider the isogonal conjugate $\\mathcal{H}$ of $PQ$ -bisector in triangle $BPQ$ . From one hand it will be the set of points $X$ such that $\\angle XPB+\\angle XQB=0$ (all angles are oriented). From the other hand, $\\mathcal{H}$ is a rectangular hyperbola passing through $B$ , $P$ , $Q$ , ... | [
"origin:aops",
"2023 Contests",
"2023 All-Russian Olympiad Regional Round"
] | {
"answer_score": 148,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 All-Russian Olympiad Regional Round/3016016.json"
} |
We write pairs of integers on a blackboard. Initially, the pair $(1,2)$ is written. On a move, if $(a, b)$ is on the blackboard, we can add $(-a, -b)$ or $(-b, a+b)$ . In addition, if $(a, b)$ and $(c, d)$ are written on the blackboard, we can add $(a+c, b+d)$ . Can we reach $(2022, 2023)$ ? | [
"Note for any pair $(a,b)$ , $b\\equiv 2a \\pmod7$ , hence the answer is no."
] | [
"origin:aops",
"2023 Contests",
"2023 All-Russian Olympiad Regional Round"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 All-Russian Olympiad Regional Round/3016017.json"
} | |
Given is a triangle $ABC$ with altitude $AH$ and median $AM$ . The line $OH$ meets $AM$ at $D$ . Let $AB \cap CD=E, AC \cap BD=F$ . If $EH$ and $FH$ meet $(ABC)$ at $X, Y$ , prove that $BY, CX, AH$ are concurrent. | <blockquote><blockquote>It is really hard to write Tex on the cellphone!
Claim:
Let $B'$ and $C'$ be the diametrical points of $B$ and $C$ .Then we have $C',E,H$ and $B',F,H$ collinear.
Proof:
Let $C'H \cap AB=E'$ and $B'H \cap AC=F'$ .We are going to prove that in fact $B,D,F'$ collinear,then it is eas... | [
"bump this...",
"<details><summary>'Hint'</summary>Prove that $\\angle BYF=\\angle CXE=90^{\\circ}$ .</details>",
"It is really hard to write Tex on the cellphone!\nClaim:\nLet $B'$ and $C'$ be the diametrical points of $B$ and $C$ .Then we have $C',E,H$ and $B',F,H$ collinear.\nProof:\nLet $C'H \\... | [
"origin:aops",
"2023 Contests",
"2023 All-Russian Olympiad Regional Round"
] | {
"answer_score": 154,
"boxed": false,
"end_of_proof": false,
"n_reply": 8,
"path": "Contest Collections/2023 Contests/2023 All-Russian Olympiad Regional Round/3016018.json"
} |
Does there exist a positive integer $m$ , such that if $S_n$ denotes the lcm of $1,2, \ldots, n$ , then $S_{m+1}=4S_m$ ? | <span style="color:#f00">**<span style="font-size:150%">Quite a standard technique.</span>**</span>
Assume that, $m+1=p_1^{z_1}p_2^{z_2}\ldots p_k^{z_k}$ wherein $k \in \mathbb{N}$ . The rest of the sequences follow the typical definition we use.
Let us take, $f(n)$ denotes the number of primes dividing n. So in o... | [
"Perhaps you meant $S_{m+1}=4S_m$ (clearly $S_{m+1}\\ge S_m$ ). Assuming this, the answer is no. Notice $m$ is odd and that $2+\\textstyle \\max_{1\\le i\\le m}v_2(i) = v_2(m+1)$ . But we also have $v_2(m+1) = v_2((m+1)/2)+1 \\le 1+\\textstyle \\max_{1\\le i\\le m}v_2(i)$ . So, a contradiction. ",
"Yeah, ... | [
"origin:aops",
"2023 Contests",
"2023 All-Russian Olympiad Regional Round"
] | {
"answer_score": 146,
"boxed": false,
"end_of_proof": false,
"n_reply": 8,
"path": "Contest Collections/2023 Contests/2023 All-Russian Olympiad Regional Round/3016019.json"
} |
In an acute triangle $ABC$ , let $M$ and $N$ be the midpoints of $AB$ and $AC$ and let $BH$ be its altitude from $B$ . Its incircle touches $AC$ at $K$ and the line through $K$ parallel to $MH$ meets $MN$ at $P$ . Prove that $AMPK$ has an incircle. | I'll show that AK+MP=AM+KP. Assume $$ \alpha=\frac{NK}{NH} $$ . We need to prove that $$ p-a+\frac{a}{2}.(1-\alpha)=\frac{c}{2}+\frac{c}{2}.\alpha $$ . Which is equivalent that we must prove that $$ \alpha=\frac{b}{a+c} $$ . $$ \alpha=\frac{NK}{NH}=\frac{b}{a+c}\Leftrightarrow\frac{b}{a+c}=\frac{a/2-c/2}{b/2... | [
"It suffices to show that the incircle of $\\triangle{AMN}$ is tangent to $KP$ . By 2x homothety at $A$ , it suffices to show that the tangent from the reflection of $A$ over $K$ to the incircle of $\\triangle{ABC}$ is the reflection of $AB$ over the perpendicular from incenter of $\\triangle{ABC}$ to... | [
"origin:aops",
"2023 Contests",
"2023 All-Russian Olympiad Regional Round"
] | {
"answer_score": 20,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 All-Russian Olympiad Regional Round/3016020.json"
} |
Find the largest real $m$ , such that for all positive real $a, b, c$ with sum $1$ , the inequality $\sqrt{\frac{ab} {ab+c}}+\sqrt{\frac{bc} {bc+a}}+\sqrt{\frac{ca} {ca+b}} \geq m$ is satisfied. | I will only prove that $m=1$ works. Let $\frac{a}{1-a}=x^2, \frac{b}{1-b^2}=y^2,\frac{c}{1-c^2}=z^2$ . Then we need to prove that $xy+yz+zx \geq 1$ if $\sum \frac{x^2}{x^2+1} = 1$ or $2(xyz)^2+\sum x^2y^2 = 1$ or if $2r^2+q^2-2pr = 1$ then $q \geq 1$ , there $q=xy+yz+zx>0,r=xyz>0,p=x+y+z>0$ . But if $q<1$... | [
"...............",
"No, the statement is for positive reals and the answer is <details><summary>answer</summary>$m=1$ .</details>",
"<details><summary>Solution</summary>The minimal $m$ is $1$ . Indeed, the triple $(\\varepsilon,\\varepsilon^2,1-\\varepsilon-\\varepsilon^2)$ achieves values arbitrarily clos... | [
"origin:aops",
"2023 Contests",
"2023 All-Russian Olympiad Regional Round"
] | {
"answer_score": 28,
"boxed": false,
"end_of_proof": false,
"n_reply": 7,
"path": "Contest Collections/2023 Contests/2023 All-Russian Olympiad Regional Round/3016021.json"
} |
A $100 \times 100 \times 100$ cube is divided into a million unit cubes and in each small cube there is a light bulb. Three faces $100 \times 100$ of the large cube having a common vertex are painted: one in red, one in blue and the other in green. Call a $\textit{column}$ a set of $100$ cubes forming a block ... | We suppose that the switches are on/off switches, which are all off initially. We will show the equivalent claim that any attainable configuration of lights with $k$ lights on can be achieved with at most $k/100$ red switches on.
Note that for any $100\times100\times1$ slice of the cube parallel to the blue face... | [] | [
"origin:aops",
"2023 Contests",
"2023 All-Russian Olympiad Regional Round"
] | {
"answer_score": 20,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 All-Russian Olympiad Regional Round/3016022.json"
} |
The bisector of $\angle BAD$ of a parallelogram $ABCD$ meets $BC$ at $K$ . The point $L$ lies on $AB$ such that $AL=CK$ . The lines $AK$ and $CL$ meet at $M$ . Let $(ALM)$ meet $AD$ after $D$ at $N$ . Prove that $\angle CNL=90^{o}$ | Let $P$ be the Miquel point of $ABCMLK$ ; then spiral similarity at $P$ takes $AL$ to $KC$ , thus $PA = PK$ and $\angle{APK} = \angle{BAD} \implies P$ is the $B$ -antipode wrt $(BAK)$ and thus also the $L$ -antipode wrt $(ALM)$ , which means it suffices to show that $C, P, N$ are collinear, AKA that ... | [
" $AK\\cap DC=P$ $\\angle KPC=\\angle KAB=\\angle DAK=\\angle AKC \\implies AL=KC=CP$ $\\implies LAM \\cong CPM \\implies LM=MC$ $\\angle NLM=\\angle NAM=\\angle MAB=\\angle LNM \\implies LM=MN$ $\\implies LM=MN=MC$ $\\implies \\angle CNL=90$ ",
"First, $ALMN$ is concyclic and $AM$ bisects $\\angle LAN... | [
"origin:aops",
"2023 Contests",
"2023 All-Russian Olympiad Regional Round"
] | {
"answer_score": 130,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 All-Russian Olympiad Regional Round/3016023.json"
} |
Given is a positive integer $k$ . There are $n$ points chosen on a line, such the distance between any two adjacent points is the same. The points are colored in $k$ colors. For each pair of monochromatic points such that there are no points of the same color between them, we record the distance between these two ... | Answer: $3k-1$ .
Construction: 1 2 3 $\cdots k$ $k \cdots$ 3 2 1 2 3 $\cdots k$ Bound: Note that there are $n-k$ gaps, hence the total gap length is $\ge 1+2+\cdots (n-k)=\frac{(n-k)(n-k+1)}{2}$ .
At the same time, if the first (resp. last) occurence of colour $i$ is at position $x_i$ (resp. $y_i$ ), th... | [] | [
"origin:aops",
"2023 Contests",
"2023 All-Russian Olympiad Regional Round"
] | {
"answer_score": 22,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 All-Russian Olympiad Regional Round/3016024.json"
} |
Prove that for all positive reals $x, y, z$ , the inequality $(x-y)\sqrt{3x^2+y^2}+(y-z)\sqrt{3y^2+z^2}+(z-x)\sqrt{3z^2+x^2} \geq 0$ is satisfied. | This is equivalent to show that $$ \sum x \sqrt{3x^2 + y^2} \geq \sum y \sqrt{3x^2 + y^2} $$ By $\text{AM-GM}$ we have $$ \sum x \sqrt{3x^2 + y^2} \geq \sum x(\frac{3}{2}x + \frac{1}{2}y) = \frac{3}{2} \sum x^2 + \frac{1}{2}\sum xy. $$ On the other hand, by Cauchy–Schwarz, we can write $$ \sum y \sqrt{3x^2 + y^2... | [
"............",
"@above $x-y, y-z, z-x$ can be negative.",
" $(x-y)(\\sqrt{3x^2+y^2}-(x+y))=\\frac{2x(x-y)^2}{\\sqrt{3x^2+y^2}+x+y} \\geq 0$ so $(x-y)\\sqrt{3x^2+y^2} \\geq x^2-y^2$ And so $(x-y)\\sqrt{3x^2+y^2}+(y-z)\\sqrt{3y^2+z^2}+(z-x)\\sqrt{3z^2+x^2} \\geq 0$ ",
"<blockquote> $(x-y)(\\sqrt{3x^2+y^2... | [
"origin:aops",
"2023 Contests",
"2023 All-Russian Olympiad Regional Round"
] | {
"answer_score": 142,
"boxed": false,
"end_of_proof": true,
"n_reply": 9,
"path": "Contest Collections/2023 Contests/2023 All-Russian Olympiad Regional Round/3016025.json"
} |
Given is a triangle $ABC$ with circumcenter $O$ . Points $D, E$ are chosen on the angle bisector of $\angle ABC$ such that $EA=EB, DB=DC$ . If $P, Q$ are the circumcenters of $(AOE), (COD)$ , prove that either the line $PQ$ coincides with $AC$ or $PQCA$ is cyclic. | [] | [
"origin:aops",
"2023 Contests",
"2023 All-Russian Olympiad Regional Round"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/2023 Contests/2023 All-Russian Olympiad Regional Round/3016026.json"
} | |
If $a, b, c$ are non-zero reals, prove that $|\frac{b} {a}-\frac{b} {c}|+|\frac{c} {a}-\frac{c}{b}|+|bc+1|>1$ . | Look at $|bc+1|$ ; If $bc>0$ or $bc<-2$ , then $|bc+1|>1$ and we are done. So assume that $-2 \leq bc <0$ . Thus, one of $b$ or $c$ is positive and the other is negative. The inequality is symmetric in $b$ and $c$ , so we can assume that $c<0<b$ . Considering $-c=c'>0$ , we should prove that $$ |\frac{... | [
"Note that everything is invariant if we change all the signs, so w.l.o.g. $a>0$ .\nMoreover, if $b$ and $c$ have the same sign, then $\\vert bc+1\\vert>1$ already.\nSince we also have the symmetry $b \\leftrightarrow c$ , we may thus w.l.o.g. assume that $b>0$ and $c<0$ .\nReplacing $c$ by $-c$ , it ... | [
"origin:aops",
"2023 Contests",
"2023 All-Russian Olympiad Regional Round"
] | {
"answer_score": 166,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 All-Russian Olympiad Regional Round/3016027.json"
} |
Given is a simple connected graph with $2n$ vertices. Prove that its vertices can be colored with two colors so that if there are $k$ edges connecting vertices with different colors and $m$ edges connecting vertices with the same color, then $k-m \geq n$ . | This is such a nice problem :-D . We prove the stronger statement that for any graph on $n$ vertices, $k - m \geq \left \lfloor\frac{n}{2}\right \rfloor$ . We proceed with induction on $n$ , the base case $n=2$ is trivial and now we look at a graph $G$ with $n + 1$ vertices. We say a graph is coloured ***amb... | [] | [
"origin:aops",
"2023 Contests",
"2023 All-Russian Olympiad Regional Round"
] | {
"answer_score": 170,
"boxed": false,
"end_of_proof": true,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 All-Russian Olympiad Regional Round/3016028.json"
} |
Let $n \geq 5$ be an integer. Consider $n$ squares with side lengths $1, 2, \dots , n$ , respectively. The squares are arranged in the plane with their sides parallel to the $x$ and $y$ axes. Suppose that no two squares touch, except possibly at their vertices. Show that it is possible to arrange these squares... | <blockquote>Let $n \geq 5$ be an integer. Consider $n$ squares with side lengths $1, 2, \dots , n$ , respectively. The squares are arranged in the plane with their sides parallel to the $x$ and $y$ axes. Suppose that no two squares touch, except possibly at their vertices. Show that it is possible to arrange t... | [
"Oh actually it is already available.",
"<blockquote>Where did you get this problem?</blockquote>\nThe official site. [https://www.apmo-official.org/static/problems/apmo2023_prb.pdf](https://www.apmo-official.org/static/problems/apmo2023_prb.pdf)",
"i have read this statement five times and i still have no clue... | [
"origin:aops",
"2023 APMO",
"2023 Contests"
] | {
"answer_score": 60,
"boxed": false,
"end_of_proof": false,
"n_reply": 17,
"path": "Contest Collections/2023 Contests/2023 APMO/3104922.json"
} |
Let $ABCD$ be a parallelogram. Let $W, X, Y,$ and $Z$ be points on sides $AB, BC, CD,$ and $DA$ , respectively, such that the incenters of triangles $AWZ, BXW, CYX,$ and $DZY$ form a parallelogram. Prove that $WXYZ$ is a parallelogram. | hahahahahahahahaha what
Let $I_A,\ldots,I_D$ be the incenters and $r_1,\ldots,r_4$ be the radii of $I_A,\ldots,I_D$ respectively. Since $\overline{I_AI_B}$ and $\overline{I_CI_D}$ have the same "slope" with respect to $AB \parallel CD$ and are the same length, it follows that $$ r_1-r_2=d(I_A,\overline{AB... | [
"wrong sol\n\ncan we use inversion perhaps (inspired by egmo)",
"how will u find incentres when circle is not unit?? :maybe:",
"If $I_A, I_B, I_C, I_D$ are their respective incenters then the midpoint of $I_AI_C$ lies on the line halfway between the bisectors of $\\angle{A}, \\angle{C}$ , and the midpoint ... | [
"origin:aops",
"2023 APMO",
"2023 Contests"
] | {
"answer_score": 204,
"boxed": false,
"end_of_proof": false,
"n_reply": 12,
"path": "Contest Collections/2023 Contests/2023 APMO/3104927.json"
} |
Find all integers $n$ satisfying $n \geq 2$ and $\dfrac{\sigma(n)}{p(n)-1} = n$ , in which $\sigma(n)$ denotes the sum of all positive divisors of $n$ , and $p(n)$ denotes the largest prime divisor of $n$ . | <details><summary>Solution</summary>The answer is $n = \boxed{6}$ only, which works since $$ \frac{\sigma(6)}{p(6)-1} = \frac{1+2+3+6}{3-1} = 6. $$ Now, we will show that this is the only solution.
Suppose that $n\geq 2$ was a solution. Note that
\begin{align*}
\frac{\sigma(n)}{n} &= \prod_{p\mid n}\frac{1 + ... | [
"Disappointing problem. Just use the bound $\\frac{\\sigma(n)} {n} \\leq \\omega(n)$ from [Baltic Way 2022 P17](https://artofproblemsolving.com/community/c6h2959324p26503920) and realize that you are done.",
"**Sketch :** The obvious way to deal with this type of problem is to take the generalized case of $n$ ... | [
"origin:aops",
"2023 APMO",
"2023 Contests"
] | {
"answer_score": 1220,
"boxed": false,
"end_of_proof": true,
"n_reply": 13,
"path": "Contest Collections/2023 Contests/2023 APMO/3104934.json"
} |
There are $n$ line segments on the plane, no three intersecting at a point, and each pair intersecting once in their respective interiors. Tony and his $2n - 1$ friends each stand at a distinct endpoint of a line segment. Tony wishes to send Christmas presents to each of his friends as follows:
First, he chooses an... | something something bijective process
Extend the segments to lines, then consider a large circle $\omega$ containing every intersection point. Now cut off the lines at their intersection points with $\omega$ and move everyone to the intersection points in the obvious way; clearly this doesn't change anything.
Ref... | [
"[wow, what a unique idea](https://artofproblemsolving.com/community/c6h1424941p26892279)",
"Is [Geoff](https://artofproblemsolving.com/community/c6h1271000p6642576) one of Tony's friends?",
"[lines are cool](https://artofproblemsolving.com/community/c6h1713439p11060031)",
"say $f(l)=1$ if the present is on... | [
"origin:aops",
"2023 APMO",
"2023 Contests"
] | {
"answer_score": 208,
"boxed": false,
"end_of_proof": true,
"n_reply": 8,
"path": "Contest Collections/2023 Contests/2023 APMO/3104935.json"
} |
Let $c>0$ be a given positive real and $\mathbb{R}_{>0}$ be the set of all positive reals. Find all functions $f \colon \mathbb{R}_{>0} \to \mathbb{R}_{>0}$ such that \[f((c+1)x+f(y))=f(x+2y)+2cx \quad \textrm{for all } x,y \in \mathbb{R}_{>0}.\] | <details><summary>Solution</summary>The answer is $f(x) = \boxed{2x}$ only, which works since $$ f((c+1)x + f(y)) = f(x+2y) + 2cx = (2c+2)x + 2y $$ for all $x, y\in \mathbb{R}_{>0}$ , for any given $c > 0$ . We will now show that this is the only solution.
Let $P(x, y)$ denote the assertion. First, we will s... | [
"The answer is $f(x)=2x$ only, which clearly works. Now we prove that this is the only solution.\n\nLet $P(x,y)$ denote the assertion. If $f(y)<2y$ for any $y$ , then from $P(x,\\tfrac{2y-f(y)}{c})$ we find that $2cx=0$ : contradiction. Therefore we may let $f(x)=g(x)+2x$ , where $g: \\mathbb{R}^+ \\to ... | [
"origin:aops",
"2023 APMO",
"2023 Contests"
] | {
"answer_score": 1364,
"boxed": false,
"end_of_proof": false,
"n_reply": 27,
"path": "Contest Collections/2023 Contests/2023 APMO/3104939.json"
} |
Find all functions $f\colon \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x,y \in \mathbb{R}$ ,
\[xf(x+f(y))=(y-x)f(f(x)).\]
*Proposed by Nikola Velov, Macedonia* | <details><summary>Solution</summary>The answer is $\boxed{f\equiv 0}$ and $\boxed{f(x) = -x+c}$ for some constant $c\in \mathbb{R}$ . $f\equiv 0$ obviously works, and $f(x) = -x+c$ works because $$ x(-(x+-y+c)+c) = x(y-x) = (y-x)(-(-x+c)+c). $$ Now, we show that these are the only functions that work. Let ... | [
"Let $P(x,y)$ denote the given assertion. $P(x,x) \\implies f(x+f(x))=0$ $P(0,y) \\implies f(f(0))=0$ $P(f(0),y) \\implies f(0)f(f(0)+f(y))=(y-f(0))f(0)$ So $f(0)=0$ or $ f(f(0)+f(y))=(y-f(0))$ If $ f(f(0)+f(y))=(y-f(0))$ , setting $y \\rightarrow f(y)+y \\implies f(y)=-y+f(0)$ and checking we see that $... | [
"origin:aops",
"2023 Contests",
"2023 Balkan MO"
] | {
"answer_score": 1218,
"boxed": false,
"end_of_proof": false,
"n_reply": 72,
"path": "Contest Collections/2023 Contests/2023 Balkan MO/3069477.json"
} |
In triangle $ABC$ , the incircle touches sides $BC,CA,AB$ at $D,E,F$ respectively. Assume there exists a point $X$ on the line $EF$ such that \[\angle{XBC} = \angle{XCB} = 45^{\circ}.\]
Let $M$ be the midpoint of the arc $BC$ on the circumcircle of $ABC$ not containing $A$ . Prove that the line $MD$ p... | <blockquote><details><summary>complex bash</summary>Identify the incircle of $\triangle ABC$ with the unit circle, so that
\begin{align*}
|d|=|e|=|f|&=1
a &= \frac{2ef}{e+f}
b &= \frac{2df}{d+f}
c &= \frac{2de}{d+e}
m &= \frac{2def}{(d+e)(d+f)}
\end{align*}
If point $X$ exists, it satisfies $$ \overline{x} = \... | [
"Very direct by the lemma that if $BI \\cap EF = U$ , then $\\angle BUC = 90^{\\circ}$ (this is since $I$ , $F$ , $U$ , $C$ are concyclic by angle chase). So $X$ must be one of the points $BI \\cap EF$ and $CI \\cap EF$ since $\\angle BXC = 90^{\\circ}$ and there are at most two intersection points ... | [
"origin:aops",
"2023 Contests",
"2023 Balkan MO"
] | {
"answer_score": 1214,
"boxed": false,
"end_of_proof": false,
"n_reply": 25,
"path": "Contest Collections/2023 Contests/2023 Balkan MO/3069479.json"
} |
For each positive integer $n$ , denote by $\omega(n)$ the number of distinct prime divisors of $n$ (for example, $\omega(1)=0$ and $\omega(12)=2$ ). Find all polynomials $P(x)$ with integer coefficients, such that whenever $n$ is a positive integer satisfying $\omega(n)>2023^{2023}$ , then $P(n)$ is also... | Let $k=2023^{2023}+1$ . We claim that only solutions are \[\boxed{P(x)=x^d} \text{ and } \boxed{P(x)=c} \text{ (where $d$ , $c \in \mathbb{N}$ and $\omega(c)<k$ )}\] Obviously these work, assume for now there is another $P$ .
Now by Schur's theorem; there exists $k+1$ distinct primes $p_1$ , $\dots$ , $p_{... | [
"@above $ -x^m$ isn’t a solution, because the problem statement requires $P(n)$ to be a positive integer whenever $n$ satisfies $\\omega(n)>2023^2023$ .",
"@2above I might be misunderstanding your solution, but are you sure you didn't solve the problem for $\\tau$ instead of $\\omega$ ? I don't see how ... | [
"origin:aops",
"2023 Contests",
"2023 Balkan MO"
] | {
"answer_score": 1038,
"boxed": true,
"end_of_proof": false,
"n_reply": 23,
"path": "Contest Collections/2023 Contests/2023 Balkan MO/3069481.json"
} |
Find the greatest integer $k\leq 2023$ for which the following holds: whenever Alice colours exactly $k$ numbers of the set $\{1,2,\dots, 2023\}$ in red, Bob can colour some of the remaining uncoloured numbers in blue, such that the sum of the red numbers is the same as the sum of the blue numbers.
Romania | Let $n=2023$ and $t=k_{\max}$ be the required maximum. The direct part is to feel that if Ana chooses $t$ highest numbers, meaning $(n-t)+1, (n-t)+2, \dots, (n-t)+t$ , their sum should be less than half of the sum of all numbers, which is $\dfrac{n(n+1)}{4}$ . Otherwise Bogdan, anyhow he chooses his numbers, th... | [
"If $k$ is too large, Alice could color the $k$ largest numbers in the set and there's no way for Bob to replicate the sum even if he chooses all of the remaining numbers. So this is the first constraint.\n\nThe sum of largest $k$ numbers is (letting $n=2023$ ) $kn - k(k-1)/2$ The sum of the first $n-k$ n... | [
"origin:aops",
"2023 Contests",
"2023 Balkan MO"
] | {
"answer_score": 140,
"boxed": false,
"end_of_proof": false,
"n_reply": 11,
"path": "Contest Collections/2023 Contests/2023 Balkan MO/3069485.json"
} |
Mr. Murgatroyd decides to throw his class a pizza party, but he's going to make them hunt for it first. He chooses eleven locations in the school, which we'll call $1, 2, \ldots, 11$ . His plan is to tell students to start at location $1$ , and at each location $n$ from $1$ to $10$ , they will find a message di... | Define the sequence $(a_1, a_2, \ldots, a_{11})$ where
1. $a_{11}=1$ 2. $a_i$ is the number on the message at location $i$ for $1 \le i \le 10$ .
Notice that $(a_1, a_2, \ldots, a_{11})$ is a permutation of $(1, 2, \ldots, 11)$ , and that $a_1 \neq 1$ . Construct the sequence $\pi$ as the set of all $a... | [
"[We refer to a location as a room]\n\nWe claim a room is never visited twice. It is obvious there is exactly one way to enter a room (other than room 1). Let the sequence of places you visit be $a_0=1, a_1, \\dots.$ If we have $a_i = a_j$ for $i<j$ , then by the fact above, $a_{i-1}=a_{j-1}, a_{i-2}=a_{j-2}$... | [
"origin:aops",
"2023 BAMO",
"2023 Contests"
] | {
"answer_score": 44,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 BAMO/3024919.json"
} |
Given a positive integer $N$ (written in base $10$ ), define its *integer substrings* to be integers that are equal to strings of one or more consecutive digits from $N$ , including $N$ itself. For example, the integer substrings of $3208$ are $3$ , $2$ , $0$ , $8$ , $32$ , $20$ , $320$ , $208$ , $3208... | Let $N = \overline{d_1d_2\dots d_n}$ . Define $p(k) = \sum_{\ell = 1}^k d_\ell$ for each $1\le k\le n$ and let $p(0) = 0$ . Then,
\[ \overline{d_id_{i+1}\dots d_j}\equiv \sum_{\ell = i}^jd_\ell = p(j) - p(i-1) \pmod{9}.\]
Hence, $9| \overline{d_id_{i+1}\dots d_j}$ if and only if $p(j)\equiv p(i-1)\pmod{9}$ , s... | [
"prefix sums on a math competition :gleam: \n\nThe largest is 8 eights. It is obvious this works. We can reduce the problem to proving N has less than 9 digits, since 8 eights is the largest 8 digit integer that works. Say $N$ has $D$ digits. Consider the sum of the first $i$ digits for $0\\leq i \\leq D$ .... | [
"origin:aops",
"2023 BAMO",
"2023 Contests"
] | {
"answer_score": 1034,
"boxed": true,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 BAMO/3024923.json"
} |
In the following figure---not drawn to scale!--- $E$ is the midpoint of $BC$ , triangle $FEC$ has area $7$ , and quadrilateral $DBEG$ has area $27$ . Triangles $ADG$ and $GEF$ have the same area, $x$ . Find $x$ .
[asy]
unitsize(2cm);
pair A = (0,38/16);
pair B = (0,0);
pair C = (38/16,0);
pair D = (0,25/1... | $\dagger$ <details><summary>lotus ratio lemma</summary>In a $\triangle{ABC}$ , a line from $C$ intersects $AB$ at $F$ and a line from $B$ intersects $AC$ at $E$ and $P$ be the intersection point of $BE$ and $CF$ , then $\frac{CP}{CF}=\frac{CE}{AE}\left(1+\frac{AF}{FB}\right)$ and $\frac{BP}{PE}... | [
" $F$ is useless. We see that $[DBE] = (27-[DGE]) = [DEC] = [DGE] + x+7$ , so we find that $[DGE] = 10 - \\frac x2$ . Similarly, $[ABE] = x+27 = [AEC] = [AGC] + x+7,$ so $[AGC] = 20$ . We compute $DG/GC$ in two ways. The first way is $[DGE]/[GEC]$ and the second is $[ADG]/[AGC]$ . This leads us to a quad... | [
"origin:aops",
"2023 BAMO",
"2023 Contests"
] | {
"answer_score": 1162,
"boxed": true,
"end_of_proof": true,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 BAMO/3024925.json"
} |
Zaineb makes a large necklace from beads labeled $290, 291, \ldots, 2023$ . She uses each bead exactly once, arranging the beads in the necklace any order she likes. Prove that no matter how the beads are arranged, there must be three beads in a row whose labels are the side lengths of a triangle. | To simplify calculations, let $k = 289$ . Let $a_1, a_2, \dots, a_{6k}$ be some arrangement of the beads, and suppose to the contrary that $a_i, a_{i + 1}, a_{i + 2}$ are not the sides of a triangle for $i = 1, 2, \dots, 6k$ (with indices taken mod $6k$ ). This implies that \[2\max(a_i, a_{i + 1}, a_{i + 2}) > ... | [
"Given values $x<y<z$ , they do not form a triangle iff $2z \\geq x+y+z$ by triangle inequality. Assume no consecutive triplet forms a triangle. Let $T$ be the set of the triplets. Then $$ 2\\cdot \\sum_{t \\in T} \\max(t) \\geq 3 \\sum_{i=290}^{2023}i , $$ where the right hand side comes from the fact tha... | [
"origin:aops",
"2023 BAMO",
"2023 Contests"
] | {
"answer_score": 126,
"boxed": false,
"end_of_proof": true,
"n_reply": 18,
"path": "Contest Collections/2023 Contests/2023 BAMO/3024927.json"
} |
A *lattice point* in the plane is a point with integer coordinates. Let $T$ be a triangle in the plane whose vertice are lattice points, but with no other lattice points on its sides. Furthermore, suppose $T$ contains exactly four lattice points in its interior. Prove that these four points lie on a straight line. | Let the triangle $T$ have vertices $A$ , $B$ , $C$ . Let $P$ be the point inside $T$ which is closest (in distance) to line $BC$ . Then, it follows that there are no lattice points lying inside or on $\triangle PBC$ . Consequently, by Pick's theorem, $\triangle PBC$ has area $1/2$ .
Now, consider the vect... | [
"<details><summary>Solution</summary>Shift $T$ so its vertices are $A = (0,0)$ , $B = (x_b, y_b)$ , and $C = (x_c, y_c)$ .\nThen repeatedly apply one of the following *shears*, which preserve all the properties of $T$ , and preserve the collinearity (or non-collinearity) of the four interior lattice points:\n... | [
"origin:aops",
"2023 BAMO",
"2023 Contests"
] | {
"answer_score": 270,
"boxed": false,
"end_of_proof": true,
"n_reply": 5,
"path": "Contest Collections/2023 Contests/2023 BAMO/3024929.json"
} |
Find all possible non-negative integer solution $(x,y)$ of the following equation- $$ x! + 2^y =(x+1)! $$ Note: $x!=x \cdot (x-1)!$ and $0!=1$ . For example, $5! = 5\times 4\times 3\times 2\times 1 = 120$ . | Note, that $(x+1)!-x!=(x+1)x!-x!=x\cdot x!=2^y.$ Hence, $x, x!$ , are powers of $2,$ if $x \ge 3$ , then $x!$ , is not a power of $2.$ If $x=0$ , no solution, if $x=1$ , we have $y=0,$ if $x=2, y=2$ hence, our solutions are $\boxed{(1,0),(2,2)}.$ | [
"I think the problem is wrong $(x+1)! \\ge x!$ so it may be $+2^y$ ",
"Ignore this",
"If this problem is true, this is way too easy for the national olympiad second round\n\nThe problem changed\nIgnore this",
"<blockquote>I think the problem is wrong $(x+1)! \\ge x!$ so it may be $+2^y$ </blockquote>\n\nSor... | [
"origin:aops",
"2023 Bangladesh Mathematical Olympiad",
"2023 Contests"
] | {
"answer_score": 1022,
"boxed": false,
"end_of_proof": false,
"n_reply": 17,
"path": "Contest Collections/2023 Contests/2023 Bangladesh Mathematical Olympiad/3013527.json"
} |
Let the points $A,B,C$ lie on a line in this order. $AB$ is the diameter of semicircle $\omega_1$ , $AC$ is the diameter of semicircle $\omega_2$ . Assume both $\omega_1$ and $\omega_2$ lie on the same side of $AC$ . $D$ is a point on $\omega_2$ such that $BD\perp AC$ . A circle centered at $B$ with... | $ABE$ is similar with $EBF$ $ABD$ is similar with $CBD$ Then $AB \times BF = BE^2 = BD^2 = BC \times BA$ Therefore, $BC = BF$ | [
"DB is Tangent to w1 at B therefore\n<DBE=<BAE=<BEF from parallel lines therefore by power of a point\nBD^2=BE^2=BF*BA\nFrom similarity in right triangle ADC we have\nBD^2=BC*BA\nthen BC=BF ",
"Huh? $\\triangle ABE \\sim \\triangle EBF, \\triangle ABD \\sim \\triangle CBD$ , now, note that $\\frac{AB}{BE}=\\fra... | [
"origin:aops",
"2023 Bangladesh Mathematical Olympiad",
"2023 Contests"
] | {
"answer_score": 12,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 Bangladesh Mathematical Olympiad/3013530.json"
} |
Solve the equation for the positive integers: $$ (x+2y)^2+2x+5y+9=(y+z)^2 $$ | If, $y\ge 3$ , we have, that $(x+2y)^2<(y+z)^2<(x^2+2y+2)$ , hence, we have $(y+z)^2=(x+2y+1)^2$ , so $y+z=x+2y+1$ , so $z=x+y+1,$ which yields, a contradiction.
If, $y=3$ , we have no solutions, same for $y=2,$ but if $y=1,$ we have that $(x+2)^2+2x+14=(z+1)^2$ , so $x^2+6x+18=z^2+2z+1$ , hence, $x,z=1,4... | [
" $\\boxed{(x,y,z)=(1,1,4)}$ If $(x,y)\\ne (1,1)$ ,then $(x+2y+1)^2<(x+2y)^2+2x+5y+9<(x+2y+2)^2$ .Contradiction.",
"If $y \\ge 3$ $(x+2y)^2 < (y+z)^2 < (x+2y+2)^2$ So, $y+z = x + 2y + 1$ $z = x + y +1$ Contradiction\n\nIf $y <3$ $y = 1$ : $(x,y,z) = (1,1,4)$ $y = 2$ : $(z+2)^2 - (x+5)^2 = 10 \\equiv 2 ... | [
"origin:aops",
"2023 Bangladesh Mathematical Olympiad",
"2023 Contests"
] | {
"answer_score": 1024,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2023 Contests/2023 Bangladesh Mathematical Olympiad/3013532.json"
} |
$2023$ balls are divided into several buckets such that no bucket contains more than $99$ balls. We can remove balls from any bucket or remove an entire bucket, as many times as we want. Prove that we can remove them in such a way that each of the remaining buckets will have an equal number of balls and the total n... | If there are at least $100$ initially non-empty buckets, then we remove balls until exactly one remains in each of them and we are done. So we may assume that we have exactly $99$ buckets, some of which may be empty. Let $a_1 \geq a_2 \geq a_3 \geq \dots \geq a_{99}$ be their initial amounts of balls. If there ex... | [
"FTSOC, let the conclusion be false.\nThen there are 99 non empty buckets at most. \nSo there is 1 bucket with at least \\ceil{2023/99} = 21\nballs\n\nWe keep that bucket aside. So there are at most 2023-99 or 1924 balls remaining in at most 98 buckets. \n\nSo there is one bucket with at most \\ceil{1924/98} = 20 b... | [
"origin:aops",
"2023 Bangladesh Mathematical Olympiad",
"2023 Contests"
] | {
"answer_score": 58,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 Bangladesh Mathematical Olympiad/3013534.json"
} |
Let $m$ , $n$ and $p$ are real numbers such that $\left(m+n+p\right)\left(\frac 1m + \frac 1n + \frac1p\right) =1$ .
Find all possible values of $$ \frac 1{(m+n+p)^{2023}} -\frac 1{m^{2023}} -\frac 1{n^{2023}} -\frac 1{p^{2023}}. $$ | Claim: $\left(m+n+p\right)\left(\frac 1m + \frac 1n + \frac1p\right) =1 \implies (m+n)(n+p)(p+m)=0.$ Proof: $\left(m+n+p\right)\left(\frac 1m + \frac 1n + \frac1p\right)=3+\frac{m}{n}+\frac{m}{p}+\frac{n}{m}+\frac{n}{p}+\frac{p}{m}+\frac{p}{n}=1$ , note that this implies, $(m+n+p)(mn+mp+np)-mnp=(m+n)(m+p)(n+p)=0.$ N... | [
" $(m+n+p)(mn+np+pm)=mnp \\to (m+n)(n+p)(p+m)=0$ Let $m+n=0$ then $\\frac 1{(m+n+p)^{2023}} -\\frac 1{m^{2023}} -\\frac 1{n^{2023}} -\\frac 1{p^{2023}}=0$ ",
"After u simplify get $(m+n)(mn+p^2+pm+pn)=0$ Solve for two cases thus we get possible values to be $0$ :D",
"From the equiation we have that $$ (m... | [
"origin:aops",
"2023 Bangladesh Mathematical Olympiad",
"2023 Contests"
] | {
"answer_score": 1010,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2023 Contests/2023 Bangladesh Mathematical Olympiad/3013538.json"
} |
Let $\triangle ABC$ be an acute angle triangle and $\omega$ be its circumcircle. Let $N$ be a point on arc $AC$ not containing $B$ and $S$ be a point on line $AB$ . The line tangent to $\omega$ at $N$ intersects $BC$ at $T$ , $NS$ intersects $\omega$ at $K$ . Assume that $\angle NTC = \angle K... | The condition $\angle NTC=\angle KSB$ becomes $BNST{}$ cyclic. Hence
[list=disc]
[*] $\measuredangle NST=\measuredangle NBT=\measuredangle NBC=\measuredangle NKC\implies CK\parallel ST$ .
[*] $\measuredangle ABC=\measuredangle SBT=\measuredangle SNT=\measuredangle KAN$ , i.e. $ACKN{}$ is a CyclicISLscelesTrapez... | [
"<details><summary>hint</summary>(N.T.B.S concyclic) for obvious reasons. So $\\angle ABC = \\angle SNX$ let X be the point on the tangent of N and also on the other side of the point T. So, $\\angle KCN=\\angle SNX=\\angle ABC$ therefore, $KN=AC$ So, $AN$ , $CK$ parallel. Also, $\\angle CKN=\\angle CBN=\... | [
"origin:aops",
"2023 Bangladesh Mathematical Olympiad",
"2023 Contests"
] | {
"answer_score": 114,
"boxed": false,
"end_of_proof": true,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 Bangladesh Mathematical Olympiad/3013544.json"
} |
Prove that every positive integer can be represented in the form $$ 3^{m_1}\cdot 2^{n_1}+3^{m_2}\cdot 2^{n_2} + \dots + 3^{m_k}\cdot 2^{n_k} $$ where $m_1 > m_2 > \dots > m_k \geq 0$ and $0 \leq n_1 < n_2 < \dots < n_k$ are integers. | <blockquote>This is a classic. I prove it by strong induction.
Let $n=2k$ and assume all numbers $1,2,\dots,2k-1$ have such a representation. Letting $k=\textstyle \sum_{1\le i\le k}3^{m_i}2^{n_i}$ , we get $2k= \textstyle \sum_{1\le i\le k}3^{m_i}2^{n_i+1}$ . Assume now $n$ is odd. If $n$ is a power of $3... | [
"Induct, If it is even add one to every $n_1$ in the representation of $\\frac{n}{2}$ , if it is odd remove $3^{\\lfloor \\log_3(n)\\rfloor}$ and add one to every $n_1$ in $\\frac{n-3^{\\lfloor \\log_3(n)\\rfloor}}{2}$ .",
"<blockquote>Induct, If it is even add one to every $n_1$ in the representation o... | [
"origin:aops",
"2023 Bangladesh Mathematical Olympiad",
"2023 Contests"
] | {
"answer_score": 34,
"boxed": false,
"end_of_proof": false,
"n_reply": 8,
"path": "Contest Collections/2023 Contests/2023 Bangladesh Mathematical Olympiad/3013547.json"
} |
Find all functions $f\colon\mathbb{R}\to\mathbb{R}$ such that $(x-y)\bigl(f(x)+f(y)\bigr)\leqslant f\bigl(x^2-y^2\bigr)$ for all $x,y\in\mathbb{R}$ . | Let $P(x,y):=(x-y)(f(x)+f(y))\le f(x^2-y^2)$ $P(x,x)$ yields $f(0)\ge0$ $P(1,0)$ yields $f(0)\le0$ thus $f(0)=0$ $P(x,-x)$ yields $2x(f(x)+f(-x))\le0$ furthermore by $x\rightarrow -x$ we get that $2x(f(x)+f(-x))\ge0$ , thus $2x(f(x)+f(-x))=0\Longrightarrow -f(x)=f(-x)$ therefore $f$ is odd, also sin... | [
"<details><summary>Solution</summary>Let $P(x,y)$ be the assumption. Adding $P(x,y)$ and $P(y,x)$ we get that $f(t)+f(-t) \\ge 0$ for all $t$ .\nIn particular $f(0) \\ge 0$ . But $P(1,0)$ implies $f(0) \\le 0$ and hence $f(0)=0$ .\nPutting $y=-x$ , for $x>0$ we get $f(x)+f(-x) \\le 0$ and hence ... | [
"origin:aops",
"2023 Contests",
"2023 Benelux"
] | {
"answer_score": 1150,
"boxed": false,
"end_of_proof": false,
"n_reply": 10,
"path": "Contest Collections/2023 Contests/2023 Benelux/3067007.json"
} |
Determine all integers $k\geqslant 1$ with the following property: given $k$ different colours, if each integer is coloured in one of these $k$ colours, then there must exist integers $a_1<a_2<\cdots<a_{2023}$ of the same colour such that the differences $a_2-a_1,a_3-a_2,\dots,a_{2023}-a_{2022}$ are all power... | <details><summary>Answer</summary>$k = 1, 2$</details>
<details><summary>Proof everything above works</summary>For $k = 1$ take all powers of $2$ .
For $k = 2$ start at $a_1 = 1$ , and keep adding to the end of a sequence something that differs by a power of two, and is the same color. If at some point you can't,... | [
"Same as @above"
] | [
"origin:aops",
"2023 Contests",
"2023 Benelux"
] | {
"answer_score": 16,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Benelux/3067009.json"
} |
Let $ABC$ be a triangle with incentre $I$ and circumcircle $\omega$ . Let $N$ denote the second point of intersection of line $AI$ and $\omega$ . The line through $I$ perpendicular to $AI$ intersects line $BC$ , segment $[AB]$ , and segment $[AC]$ at the points $D$ , $E$ , and $F$ , respectively. T... | <blockquote>We redefines everything :)
Let $T$ be **mixtilinear Incircles touch point**.Note that circle pass though $E$ and $F$ Define $Y$ be antipode of $N$ then we know $\overline{Y-I-T}$ and $\angle YTN = 90$ Let $D = NT \cap BC$ and $M$ be midpoint of $BC$ . Redefine $Q = DM \cap YT$ Now note ... | [
"Hint : Prove that the intersection point is the point of tangency of the mixtilinear incircle.",
"This was quite classical. Let $M$ be the midpoint of major arc $BC$ . We split the proof into two Claims.**Claim 1:** Lines $MI$ and $DN$ meet on $\\omega$ .\n*Proof:* Let $DN$ intersect $\\omega$ at poi... | [
"origin:aops",
"2023 Contests",
"2023 Benelux"
] | {
"answer_score": 220,
"boxed": false,
"end_of_proof": true,
"n_reply": 14,
"path": "Contest Collections/2023 Contests/2023 Benelux/3067010.json"
} |
A positive integer $n$ is *friendly* if the difference of each pair of neighbouring digits of $n$ , written in base $10$ , is exactly $1$ . *For example, 6787 is friendly, but 211 and 901 are not.*
Find all odd natural numbers $m$ for which there exists a friendly integer divisible by $64m$ . | Call such a number $m$ *happy*. We claim that all happy integers are those for which $(m,5)=1$ . Note that $64 \mid 343232$ (this can be obtained by a finite casework on the possibilities of 6-digit strings divisible by $64$ that can be part of a friendly integer).
If $5\mid m$ and $m$ is happy, then if $... | [
"<blockquote>\nNow, suppose that $(m,5)=1$ . Let $n$ consist of $t$ consecutive strings of $343232,$ for some $t$ which we will pick later. Then, $n=343232 \\cdot \\dfrac{10^t-1}{9},$ and since $64 \\mid 343232,$ for $64m \\mid n$ to hold, we need $9m \\mid 10^t-1$ . However, since $(m,5)=1$ and $m... | [
"origin:aops",
"2023 Contests",
"2023 Benelux"
] | {
"answer_score": 68,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Benelux/3067011.json"
} |
Let $ABC$ be a triangle with $BA=BC$ and $\angle ABC=90^{\circ}$ . Let $D$ and $E$ be the midpoints of $CA$ and $BA$ respectively. The point $F$ is inside of $\triangle ABC$ such that $\triangle DEF$ is equilateral. Let $X=BF\cap AC$ and $Y=AF\cap DB$ . Prove that $DX=YD$ . | Since $D$ and $E$ are the midpoints of $BA$ and $CA$ , $\triangle{AED}\sim\triangle{ABC}$ and $AE=ED=DF=EF=EB$ . Since $\angle{DEF}=60$ , $\angle{FEB}=30$ . Since $\triangle{BEF}$ is isosceles, $\angle{EBF}=\angle{EFB}=75$ and since $\triangle{AEF}$ is isosceles with $\angle{AEF}=90+60=150$ , $\angl... | [
"Since $EA=EB=ED=EF$ , we have $\\widehat{BFA}=90^o$ Therefore, $Y$ is orthocenter of $\\Delta ABX$ . Hence $\\widehat{DXY}=45^o \\Rightarrow DX=DY$ ",
" First note that, we must have $\\angle BCA = \\angle CAB = 45^\\circ$ . Then, $\\angle BDA = 45^\\circ$ which gives us that $\\angle CDF = 180 - 45-60... | [
"origin:aops",
"2023 Contests",
"2023 Brazil EGMO TST -wrong source"
] | {
"answer_score": 44,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 Brazil EGMO TST -wrong source/2957866.json"
} |
Determine all the integers solutions $(x,y)$ of the following equation $$ \frac{x^2-4}{2x-1}+\frac{y^2-4}{2y-1}=x+y $$ | on re-writing the equation we get : $\frac{x^2-4}{2x-1}-x+\frac{y^2-4}{2y-1}-y=0$ which gives us that $\frac{-4x^2+4x-16}{2x-1}=\frac{4y^2-4y+16}{2y-1}$ which yields $2-2(x+y)=-15\left(\frac{2-2(x+y)}{(2x-1)(2y-1)}\right)$ <span style="color:#f0f">**Case1:-**</span> $x+y=1$ we set $y=1-x$ and plug back to observe t... | [
"By multiplying both sides with $4$ we get, $$ \\frac{-15}{2x-1}+\\frac{-15}{2y-1}+2x+1+2y+1=4x+4y $$ Let $(m,n)=(2x-1,2y-1)$ and note that $mn$ is not $0$ . $$ m+n+\\frac{15}{m}+\\frac{15}{n}=0\\implies (m+n)\\cdot (1+\\frac{15}{mn})=0 $$ Thus we have two cases; $$ m+n=0\\implies x=-y $$ $$ mn=-15\... | [
"origin:aops",
"2023 Contests",
"2023 Brazil EGMO TST -wrong source"
] | {
"answer_score": 1126,
"boxed": true,
"end_of_proof": true,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 Brazil EGMO TST -wrong source/2957868.json"
} |
There are $n$ cards. Max and Lewis play, alternately, the following game
Max starts the game, he removes exactly $1$ card, in each round the current player can remove any quantity of cards, from $1$ card to $t+1$ cards, which $t$ is the number of removed cards by the previous player, and the winner is the pla... | <details><summary>Solution</summary>I claim that Max wins for $n \equiv 1, 4 \pmod{5}$ and Lewis wins otherwise.
Let $s(n)$ denote the number of stones that the player whose turn it is can remove when there are $n$ stones left. Also, let a game state be called *winning* if the player whose turn it is can guaran... | [
"I think there might be some gaps in the problem such as;\n What is the winning condition?\n Is $t$ the total number of removed cards by previous player or are we considering just the last round?",
"I assume that one loses when he can't move.\nThe answer is :\nMax has a winning strategy when $n \\equiv 1,4 \\p... | [
"origin:aops",
"2023 Contests",
"2023 Brazil EGMO TST -wrong source"
] | {
"answer_score": 218,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2023 Contests/2023 Brazil EGMO TST -wrong source/2957879.json"
} |
The sequence of positive integers $a_1,a_2,a_3,\dots$ is *brazilian* if $a_1=1$ and $a_n$ is the least integer greater than $a_{n-1}$ and $a_n$ is **coprime** with at least half elements of the set $\{a_1,a_2,\dots, a_{n-1}\}$ . Is there any odd number which does **not** belong to the brazilian sequence? | <details><summary>Wrong Solution :( </summary>The answer is no.
Assume contrary. Let the first odd number which does not belong to sequence be $2l-1$ and let $a_k$ be the last term which is less than $2l-1$ .We are going to construct the sequence $S=\{a_1,a_2,\dots, a_{k}\}$ by using the following claims,**Clai... | [
"I am the creator of this problem and actually there are infinitely many odd numbers that do not belong to the sequence. We prove the following lemma: $\\textbf{Lemma:}$ Let $\\varphi(n)$ be the number of positive integers less than $n$ coprime with $n$ . Then, $$ \\liminf_{n\\to\\infty}\\frac{\\varphi(n)}{n... | [
"origin:aops",
"2023 Contests",
"2023 Brazil EGMO TST -wrong source"
] | {
"answer_score": 206,
"boxed": false,
"end_of_proof": true,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Brazil EGMO TST -wrong source/2957890.json"
} |
Let $ABC$ be an acute triangle with altitude $\overline{AH}$ , and let $P$ be a variable point such that the angle bisectors $k$ and $\ell$ of $\angle PBC$ and $\angle PCB$ , respectively, meet on $\overline{AH}$ . Let $k$ meet $\overline{AC}$ at $E$ , $\ell$ meet $\overline{AB}$ at $F$ , and $\... | **Tangent double angle formula:** We have $\tan 2\theta=\frac{2\tan \theta}{1-\tan^2\theta}$ . :coolspeak: :coolspeak: :coolspeak:
Let $H$ be on $\overline{BC}$ (WLOG). Use coordinates with $A=(1,0),B=(b,0),C=(c,0),H=(0,0)$ . Let $k \cap \ell \cap \overline{AH}:=I$ be the incenter of $\triangle PBC$ , and... | [
"Here is a short solution with harmonic quadrilaterals on hyperbolas :)\n\nRelabel $H$ as $D$ . First note since the incircle of $\\triangle PBC$ is tangent to $BC$ at $D$ we have $PB-PC=DB-DC$ , so the locus of $P$ is a hyperbola with foci $B,C$ passing through $D$ . Call this hyperbola $\\mathcal{... | [
"origin:aops",
"2023 Contests",
"2023 Brazil Team Selection Test"
] | {
"answer_score": 204,
"boxed": false,
"end_of_proof": false,
"n_reply": 30,
"path": "Contest Collections/2023 Contests/2023 Brazil Team Selection Test/3107320.json"
} |
Let $ABC$ be an acute-angled triangle with $AC > AB$ , let $O$ be its circumcentre, and let $D$ be a point on the segment $BC$ . The line through $D$ perpendicular to $BC$ intersects the lines $AO, AC,$ and $AB$ at $W, X,$ and $Y,$ respectively. The circumcircles of triangles $AXY$ and $ABC$ int... | <details><summary>complex bash</summary>Let $\triangle ABC$ be inscribed in the unit circle, and let the line through $D$ perpendicular to $\overleftrightarrow{BC}$ meet the unit circle at $U$ and $V$ . Then
\begin{align*}
|a|=|b|=|c|=|u|&=1
o &= 0
v &= -\frac{bc}u
d &= \frac12\left(b+c+u-\frac{bc}u\right) ... | [
"We first prove that $AZ\\parallel BC$ . We use Cartesian coordinates.\n\nLet $B=(-1,0)$ , $C=(1,0)$ , $A=(-a,b)$ , $D=(-d,0)$ . Let $P$ be the circumcenter of $\\triangle AXY$ . We require that the $OP\\perp BC$ , or that $P$ lies on the $y$ -axis.\n\nLine $AB$ passes through $(-1,0)$ and $(-a,b)$... | [
"origin:aops",
"2023 Contests",
"2023 Brazil Team Selection Test"
] | {
"answer_score": 1200,
"boxed": false,
"end_of_proof": false,
"n_reply": 44,
"path": "Contest Collections/2023 Contests/2023 Brazil Team Selection Test/3107324.json"
} |
Let $\mathbb R$ be the set of real numbers. We denote by $\mathcal F$ the set of all functions $f\colon\mathbb R\to\mathbb R$ such that $$ f(x + f(y)) = f(x) + f(y) $$ for every $x,y\in\mathbb R$ Find all rational numbers $q$ such that for every function $f\in\mathcal F$ , there exists some $z\in\mathbb R... | The answer is $\boxed{\frac{n+1}{n}} $ for any nonzero integer $n$ . First we show these work.
Claim: $f(nf(0)) = (n +1)f(0)$ for any integer $n$ .
Proof: If the result is true for $k$ , then $P(kf(0),0)$ and $P((k-1)f(0),0)$ give that the result is true for $k+1$ and $k - 1$ , respectively. Now, we ca... | [
"My favorite algebra in this shortlist, although quite easy for an A6.\n\n<details><summary>Solution</summary>The answer is $q=1+\\tfrac 1n$ for $n\\in\\mathbb Z\\setminus\\{0\\}$ .\n-----**Proof that such $q$ work**\n\nWe have $f(x+f(0)) - f(x) = f(0)$ , so induction $f(nf(0)) = (n+1)f(0)$ for all $n\\in... | [
"origin:aops",
"2023 Contests",
"2023 Brazil Team Selection Test"
] | {
"answer_score": 1238,
"boxed": false,
"end_of_proof": false,
"n_reply": 26,
"path": "Contest Collections/2023 Contests/2023 Brazil Team Selection Test/3107327.json"
} |
Let $n$ be a positive integer. We start with $n$ piles of pebbles, each initially containing a single pebble. One can perform moves of the following form: choose two piles, take an equal number of pebbles from each pile and form a new pile out of these pebbles. Find (in terms of $n$ ) the smallest number of nonemp... | For the rest of this proof, include $1$ as a power of $2$ .
The answer is
\[
\boxed{
\begin{cases}
1 & \text{if }n\text{ is a power of }2
2 & \text{otherwise}
\end{cases}
}.
\]
Rephrase the problem so that we have numbers on a blackboard(representing the sizes of the piles).**Construction.** We will first prove t... | [
"<details><summary>Solution</summary>The answer is $$ \\begin{cases}\n1 & n \\text{ is a power of two} \n2 & \\text{otherwise}.\n\\end{cases} $$ We split the solution into three parts.\n------**Part 1: construction for power of two**\n\nThis part is more or less obvious; from two piles of $x$ and $x$ pebbles,... | [
"origin:aops",
"2023 Contests",
"2023 Brazil Team Selection Test"
] | {
"answer_score": 1186,
"boxed": false,
"end_of_proof": false,
"n_reply": 29,
"path": "Contest Collections/2023 Contests/2023 Brazil Team Selection Test/3107333.json"
} |
A $\pm 1$ -*sequence* is a sequence of $2022$ numbers $a_1, \ldots, a_{2022},$ each equal to either $+1$ or $-1$ . Determine the largest $C$ so that, for any $\pm 1$ -sequence, there exists an integer $k$ and indices $1 \le t_1 < \ldots < t_k \le 2022$ so that $t_{i+1} - t_i \le 2$ for all $i$ , and ... | <details><summary>Solution</summary>The answer is $C = \boxed{506}$ .
To show that $C \leq 506$ , consider $$ a_i = \begin{cases} 1 & i\in \{0, 1\}\pmod{4} -1 & i\in \{2, 3\}\pmod{4} \end{cases}. $$ We will first prove a crucial lemma.**<span style="color:#0ff">Lemma:</span>** Consider some sequence of indices ... | [
"<details><summary>Answer</summary>$506$</details>\n<details><summary>Solution</summary>First consider the sequence $1,-1,-1,1,1,-1,-1,1,1,\\dots,-1,-1,1,1,-1$ . Suppose that $C \\ge 507$ . By symmetry, we may suppose that we have found a positive subsum $\\ge 507$ , and w.l.o.g. the first and last term are posi... | [
"origin:aops",
"2023 Contests",
"2023 Brazil Team Selection Test"
] | {
"answer_score": 1446,
"boxed": false,
"end_of_proof": false,
"n_reply": 40,
"path": "Contest Collections/2023 Contests/2023 Brazil Team Selection Test/3107334.json"
} |
Let $a > 1$ be a positive integer and $d > 1$ be a positive integer coprime to $a$ . Let $x_1=1$ , and for $k\geq 1$ , define $$ x_{k+1} = \begin{cases}
x_k + d &\text{if } a \text{ does not divide } x_k
x_k/a & \text{if } a \text{ divides } x_k
\end{cases} $$ Find, in terms of $a$ and $d$ , the greatest p... | <details><summary>Solution</summary>The answer is $\boxed{\lceil \log_a d \rceil}$ .
Say a term $a_k$ in this sequence is a starter if either $k = 1$ or $\frac{a_{k-1}}{a} = a_k$ and $a_k + d = a_{k+1}$ . Note that there are infinitely many such starters, since if $S$ is a starter (we know such an $S$ exis... | [
"<details><summary>Solution</summary>The answer is $n=\\lceil\\log_a d\\rceil$ . Before proving either part, let us show the following important claim of the sequence.\n-----\n<span style=\"color:red\">**Claim:**</span> $x_n < ad$ for all $n$ .\n\n*Proof.* We prove this by induction. By induction hypothesis, af... | [
"origin:aops",
"2023 Contests",
"2023 Brazil Team Selection Test"
] | {
"answer_score": 1444,
"boxed": false,
"end_of_proof": false,
"n_reply": 31,
"path": "Contest Collections/2023 Contests/2023 Brazil Team Selection Test/3107336.json"
} |
For each $1\leq i\leq 9$ and $T\in\mathbb N$ , define $d_i(T)$ to be the total number of times the digit $i$ appears when all the multiples of $1829$ between $1$ and $T$ inclusive are written out in base $10$ .
Show that there are infinitely many $T\in\mathbb N$ such that there are precisely two distin... | I found a solution for this that finds a $T$ such that all values of $d_i(T)$ are equal. From this we find what the problem asks for.
Let $n=1829$ . We first define $d_{i,j}(T)$ to be the amount of times the digit $i$ appears in the multiples of $n$ until $T$ , in the $j$ -th position (from right to left)... | [
"<details><summary>Solution</summary>Let $t$ be any positive integer divisible by $\\varphi(1829)$ , i.e., $1829\\mid 10^t-1$ , and we claim that $T=10^t$ works.\n\nThe key observation is that for any multiple $\\overline{a_ta_{t-1}\\dots a_1}$ of $1829$ , its cyclic rotation $\\overline{a_{t-1}a_{t-2}\\d... | [
"origin:aops",
"2023 Contests",
"2023 Brazil Team Selection Test"
] | {
"answer_score": 204,
"boxed": false,
"end_of_proof": false,
"n_reply": 14,
"path": "Contest Collections/2023 Contests/2023 Brazil Team Selection Test/3107337.json"
} |
Find all positive integers $n \geqslant 2$ for which there exist $n$ real numbers $a_1<\cdots<a_n$ and a real number $r>0$ such that the $\tfrac{1}{2}n(n-1)$ differences $a_j-a_i$ for $1 \leqslant i<j \leqslant n$ are equal, in some order, to the numbers $r^1,r^2,\ldots,r^{\frac{1}{2}n(n-1)}$ . | Solved with **pikapika007, OronSH**.
The answer is $\boxed{n \in \{2,3,4\}}$ .
To see that $n=2$ works, take $r = 1$ , $a_2 = 1, a_1 = 0$ .
To see that $n=3$ works, take $r = \phi = \frac{1 + \sqrt{5}}{2}$ , $a_1 = 0, a_2 = \phi, a_3 = \phi^2 + \phi$ . This works because $a_2 - a_1 = \phi, a_3 - a_2 = \... | [
"Such a configuration only exists for $\\fbox{n = 2,3,4}$ .**Construction**\nFor $n=2$ , just take $r=1$ and $a_1, a_2$ such that $a_2-a_1=1$ .\nFor $n=3$ , we take $r = \\varphi$ , so that $r^2 = 1+r$ . Let $a_2 - a_1 = r$ and $a_3-a_2 = r^2$ , so that $a_3 - a_1 = r+r^2 = r(1+r) = r^3$ .\nFor $n=4$... | [
"origin:aops",
"2023 Contests",
"2023 Brazil Team Selection Test"
] | {
"answer_score": 1286,
"boxed": false,
"end_of_proof": false,
"n_reply": 12,
"path": "Contest Collections/2023 Contests/2023 Brazil Team Selection Test/3107362.json"
} |
Let $Q$ be a set of prime numbers, not necessarily finite. For a positive integer $n$ consider its prime factorization: define $p(n)$ to be the sum of all the exponents and $q(n)$ to be the sum of the exponents corresponding only to primes in $Q$ . A positive integer $n$ is called *special* if $p(n)+p(n+1)$... | <blockquote> you need to show there is one special integer in $[1, 101]$ .</blockquote>
Here is a solution, essentially the same as the one of **TheUltimate123**, which also handles this issue.
<details><summary>Solution</summary>We will prove that $\boxed{c=1550}$ works.
-----First observe that the functions $p,... | [
"any ideas?",
"For each positive integer \\(N\\), consider \\[S=\\{5040N,\\;5040N+70,\\;5040N+72,\\;5040N+75,\\;5040N+80\\}.\\] There are 4 possible values of \\(\\{p(n)\\bmod2,q(n)\\bmod2\\}\\), so by Pigeonhole, there are distinct \\(a\\) and \\(b\\) in \\(S\\) with \\(p(a)\\equiv p(b)\\pmod2\\) and \\(q(a)\\eq... | [
"origin:aops",
"2023 Contests",
"2023 Brazil Team Selection Test"
] | {
"answer_score": 1110,
"boxed": false,
"end_of_proof": false,
"n_reply": 10,
"path": "Contest Collections/2023 Contests/2023 Brazil Team Selection Test/3107366.json"
} |
Lucy starts by writing $s$ integer-valued $2022$ -tuples on a blackboard. After doing that, she can take any two (not necessarily distinct) tuples $\mathbf{v}=(v_1,\ldots,v_{2022})$ and $\mathbf{w}=(w_1,\ldots,w_{2022})$ that she has already written, and apply one of the following operations to obtain a new tupl... | Assume for the sake of contradiction that two tuples is possible. Firstly, note that if $v_i\ge cv_j$ and $w_i\ge cw_j$ for some nonnegative $c$ then $v_i+w_i\ge c(v_j+w_j)$ and \[\max(v_i,w_i)\ge\max(cv_j,cw_j)=c\max(v_j,w_j)\] so if we start with only two tuples $\mathbf{v}$ and $\mathbf{w}$ , and there ex... | [
"<details><summary>Sol</summary>The answer is 3.\n\nProof that two fail: take two entries where on the first vector both are positive and second both are negative. Show there exists a positive constant c such that one entry is>=c times the other (equality occurs if two vectors are linearly dependent on these two po... | [
"origin:aops",
"2023 Contests",
"2023 Brazil Team Selection Test"
] | {
"answer_score": 104,
"boxed": false,
"end_of_proof": false,
"n_reply": 10,
"path": "Contest Collections/2023 Contests/2023 Brazil Team Selection Test/3107374.json"
} |
In a school there are $1200$ students. Each student is part of exactly $k$ clubs. For any $23$ students, they are part of a common club. Finally, there is no club to which all students belong. Find the smallest possible value of $k$ . | The answer is **$k=23$ .****Example**: Take $24$ clubs( $c_1,c_2,\dots,c_{24}$ ) and students $S_1,S_2,\dots,S_{1200}$ such that $S_i$ belongs to all the clubs except $c_j$ (with $i\equiv j$ mod 24). For any union of $23$ students at least one club can not be withdrawn.
For $k\leq 22$ take two students $... | [] | [
"origin:aops",
"2023 Contests",
"2023 Brazil Team Selection Test"
] | {
"answer_score": 144,
"boxed": false,
"end_of_proof": true,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Brazil Team Selection Test/3144197.json"
} |
Let $ABCD$ be a parallelogram. The tangent to the circumcircle of triangle $BCD$ at $C$ intersects $AB$ at $P$ and intersects $AD$ at $Q$ . The tangents to the circumcircle of triangle $APQ$ at $P$ and $Q$ meet at $R$ . Show that points $A$ , $C$ , and $R$ are collinear. | By some straightforward angle chasing we have $\angle AQP = \angle BCP = \angle BDC = \angle ABD$ so $\overline{BD}$ and $\overline{PQ}$ are antiparallel in $\angle A$ . Thus, since $AR$ is the $A$ -symmedian in $\triangle APQ$ , it must be the $A$ -median in $\triangle ABD;$ but $C$ lies on the $A$ -m... | [
"Let $RQ \\cap BC=E,RP \\cap DC=F$ We have $\\angle DQC=\\angle BCP=\\angle BDC=\\angle DBA$ so $Q,D,B,P$ are cyclic. $\\angle PBE=\\angle PQE $ so $E,Q,B,P $ are cyclic. $\\angle FPQ= \\angle FDQ $ so $F,P,D,Q$ are cyclic. $\\implies E,Q,D,B,P,F$ are cyclic.\nBy Pascal at $EQDFPB,$ we get $R,A,C$ a... | [
"origin:aops",
"2023 Contests",
"2023 Brazil Team Selection Test"
] | {
"answer_score": 24,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Brazil Team Selection Test/3144201.json"
} |
Show that for all positive real numbers $a, b, c$ , we have that $$ \frac{a+b+c}{3}-\sqrt[3]{abc} \leq \max\{(\sqrt{a}-\sqrt{b})^2, (\sqrt{b}-\sqrt{c})^2, (\sqrt{c}-\sqrt{a})^2\} $$ | Suppose $a\geq b\geq c$ .
\[f(a,b,c)=3(\sqrt{a}-\sqrt{c})^2+3\sqrt[3]{abc}-a-b-c=2a+2c-b-6\sqrt{ac}+3\sqrt[3]{abc}\]
We want to show that $f(a,b,c)\geq 0$ . Let $a,b,c$ be the triplet where $f_{min}$ is achieved. If $ac\geq b^2$ , then
\[f(a,b,c)-f(\frac{ac}{b},b,b)=2a+2c-b-6\sqrt{ac}+3\sqrt[3]{abc}-\frac{2ac}{... | [
"<blockquote>Show that for all positive real numbers $a, b, c$ , we have that $$ \\frac{a+b+c}{3}-\\sqrt[3]{abc} \\leq \\max\\{(\\sqrt{a}-\\sqrt{b})^2, (\\sqrt{b}-\\sqrt{c})^2, (\\sqrt{c}-\\sqrt{a})^2\\} $$ </blockquote> $$ \\frac{a+b+c}{3}-\\sqrt[3]{abc} \\leq \\frac23\\cdot\\max\\{(\\sqrt{a}-\\sqrt{b})^2, (\... | [
"origin:aops",
"2023 Contests",
"2023 Brazil Team Selection Test"
] | {
"answer_score": 118,
"boxed": false,
"end_of_proof": true,
"n_reply": 7,
"path": "Contest Collections/2023 Contests/2023 Brazil Team Selection Test/3144205.json"
} |
Find all positive integers $n$ with the following property: There are only a finite number of positive multiples of $n$ that have exactly $n$ positive divisors. | <details><summary>Solution</summary>We claim that this happens iff $n$ is squarefree or $n=4$ .
Let $n=p_1^{e_1}\dots p_k^{e_k}$ .
If $e_i \ge 2$ for some odd $p_i$ , then note that $p_i^{e_i-1} \ge e_i+1$ (since $p_i \ge 3$ ) and $p_j^{e_j} \ge e_j+1$ for all the others, so $p_i^{p_i^{e_i-1}-1} \cdot \pro... | [
"We uploaded our solution [https://calimath.org/pdf/BrazilTST2023-4-4.pdf](https://calimath.org/pdf/BrazilTST2023-4-4.pdf) on youtube [https://youtu.be/7DdZOcjMMos](https://youtu.be/7DdZOcjMMos)."
] | [
"origin:aops",
"2023 Contests",
"2023 Brazil Team Selection Test"
] | {
"answer_score": 54,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Brazil Team Selection Test/3144208.json"
} |
Let $G$ be a graph on $n\geq 6$ vertices and every vertex is of degree at least 3. If $C_{1}, C_{2}, \dots, C_{k}$ are all the cycles in $G$ , determine all possible values of $\gcd(|C_{1}|, |C_{2}|, \dots, |C_{k}|)$ where $|C|$ denotes the number of vertices in the cycle $C$ . | We claim the possible values are $1$ or $2$ . $K_n$ and $K_{ \left\lfloor \frac{n}{2} \right\rfloor , \left\lfloor \frac{n+1}{2} \right\rfloor}$ are constructions for both. Let $A_1, \dots , A_k$ be the longest path in $G$ . By maximality, all of $A_1$ 's neighbours must be in this path, and by the problem ... | [
"Nicely generalizing \nhttps://artofproblemsolving.com/community/c6h514287_degrees_gt_3_implies_cycle_not_div_by_3\n\n",
"solved with @Chiaquinha.\n\nConsideer the maximal (in vertices) path $P$ , with vertices $T_1, T_2, \\dots, T_k$ . Note that $T_1$ doesn't have connections with any vertices out of the pat... | [
"origin:aops",
"2023 Contests",
"2023 Bulgaria National Olympiad"
] | {
"answer_score": 48,
"boxed": false,
"end_of_proof": false,
"n_reply": 10,
"path": "Contest Collections/2023 Contests/2023 Bulgaria National Olympiad/3048313.json"
} |
Let $ABC$ be an acute triangle and $A_{1}, B_{1}, C_{1}$ be the touchpoints of the excircles with the segments $BC, CA, AB$ respectively. Let $O_{A}, O_{B}, O_{C}$ be the circumcenters of $\triangle AB_{1}C_{1}, \triangle BC_{1}A_{1}, \triangle CA_{1}B_{1}$ respectively. Prove that the lines through $O_{A}, ... | Let $I_A$ , $I_B$ , $I_C$ be the excenters of triangle $ABC$ , and let $V$ be the Bevan point of triangle $ABC$ .
[asy]
import geometry;
unitsize(1 cm);
pair[] A, B, C, I, O;
pair X;
A[0] = (1,3);
B[0] = (0,0);
C[0] = (4,0);
I[1] = excenter(B[0],C[0],A[0]);
I[2] = excenter(C[0],A[0],B[0]);
I[3] = excenter(A... | [
"Let $O$ be circumcenter, $(I)$ tangent to sides at $D,E,F$ . Reflects $I$ over $O$ to get $B_e$ . Since $O$ midpoint $IB_e$ , $D,E,F$ are projections of $I$ on $BC,CA,AB$ so clearly $A_1,B_1,C_1$ are projections of $B_e$ on $BC,CA,AB$ . \nWe can easily see $O_B,O_C$ are midpoints of $LB,L... | [
"origin:aops",
"2023 Contests",
"2023 Bulgaria National Olympiad"
] | {
"answer_score": 86,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 Bulgaria National Olympiad/3048317.json"
} |
Let $f(x)$ be a polynomial with positive integer coefficients. For every $n\in\mathbb{N}$ , let $a_{1}^{(n)}, a_{2}^{(n)}, \dots , a_{n}^{(n)}$ be fixed positive integers that give pairwise different residues modulo $n$ and let
\[g(n) = \sum\limits_{i=1}^{n} f(a_{i}^{(n)}) = f(a_{1}^{(n)}) + f(a_{2}^{(n)}) + \do... | Way too easy for the #3 spot. Anyway, here is my <details><summary>solution.</summary>Let $f(x) = c_{d}x^{d}+c_{d-1}x^{d-1}+\dots +c_{1}x+c_{0}$ . The residue condition on $a_{i}^{(n)}$ combined with the fact that $b-c\mid f(b)-f(c)$ for all integers $b,c$ implies:
\[\gcd(m, g(m)) = \gcd(m, f(1)+f(2)+\dots +f(m)... | [
"It turned out it is enough to use that $\\sum_{i=1}^n i^k$ is a polynomial $f(n)$ in $n$ with rational coefficients and $f(0) = 0$ , i.e. (after multiplication by a sufficiently large constant) it is divisible by $n$ .",
"@above: [Indeed, it is!](https://dgrozev.wordpress.com/2023/04/18/bulgarian-nmo-202... | [
"origin:aops",
"2023 Contests",
"2023 Bulgaria National Olympiad"
] | {
"answer_score": 24,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 Bulgaria National Olympiad/3048321.json"
} |
Prove that there exists a unique point $M$ on the side $AD$ of a convex quadrilateral $ABCD$ such that
\[\sqrt{S_{ABM}}+\sqrt{S_{CDM}} = \sqrt{S_{ABCD}}\]
if and only if $AB\parallel CD$ . | There's actually a surprisingly neat solution to this question.
Square the equality to obtain
\[S_{ABM}+S_{CDM}+2\sqrt{S_{ABM}S_{CDM}}=S_{ABCD}\]
This is equivalent to
\[2\sqrt{S_{ABM}S_{CDM}}=S_{BCM}\]
Firstly, let us consider the case when $AB\parallel CD$ .
[asy]
size(7cm);
pair A = dir(100);
pair D = dir(200);
p... | [
"Here's my in-contest solution (excluding the case $AB\\parallel CD$ as it's the easier part and my approach was pretty much the same as in #2). <details><summary>Bash</summary>Assume there exists a unique such point $M$ . Let $AB=a$ , $CD=b$ , $\\angle A = \\alpha$ , $\\angle B = \\beta$ , $AD = x$ , $AM ... | [
"origin:aops",
"2023 Contests",
"2023 Bulgaria National Olympiad"
] | {
"answer_score": 92,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 Bulgaria National Olympiad/3049016.json"
} |
For every positive integer $n$ determine the least possible value of the expression
\[|x_{1}|+|x_{1}-x_{2}|+|x_{1}+x_{2}-x_{3}|+\dots +|x_{1}+x_{2}+\dots +x_{n-1}-x_{n}|\]
given that $x_{1}, x_{2}, \dots , x_{n}$ are real numbers satisfying $|x_{1}|+|x_{2}|+\dots+|x_{n}| = 1$ . | Solved with $\textit{carvilesp7}$ The answer is $2^{1-n}$ , achieved when $x_1=2^{1-n}$ and $x_i=2^{i-n-1}$ for every other $i$ .
It's clear this is true for $n=1$ . We will induct on $n$ .
Let's solve the following problem: we want to find the minimum of \[\frac{|x_{1}|+|x_{1}-x_{2}|+|x_{1}+x_{2}-x_{3}|+\do... | [
"[https://dgrozev.wordpress.com/2023/04/09/bulgarian-nmo-2023-problem-5/](https://dgrozev.wordpress.com/2023/04/09/bulgarian-nmo-2023-problem-5/)",
"The answer and equivalency condition is \n<blockquote>\n\nThe answer is $2^{1-n}$ , achieved when $x_1=2^{1-n}$ and $x_i=2^{i-n-1}$ for every other $i$ .\n\n</... | [
"origin:aops",
"2023 Contests",
"2023 Bulgaria National Olympiad"
] | {
"answer_score": 66,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 Bulgaria National Olympiad/3049023.json"
} |
In a class of $26$ students, everyone is being graded on five subjects with one of three possible marks. Prove that if $25$ of these students have received their marks, then we can grade the last one in such a way that their marks differ from these of any other student on at least two subjects. | Assume that each pair of the $25$ students is given different marks in at least one subject, otherwise we can remove one of these students.
I claim that we can assign the last student marks in each of the first three subjects such that they differ in at least two subjects in all except at most $6$ other students, ... | [
"Does anybody have a solution?",
"<blockquote>\n... \nNote that for the first subject there must be some mark that is achieved by at most $8$ of the $25$ students. Assign this to the last student, <span style=\"color:#f00\">and note that at most $2$ of the $17$ other students can differ from the last stud... | [
"origin:aops",
"2023 Contests",
"2023 Bulgaria National Olympiad"
] | {
"answer_score": 26,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 Bulgaria National Olympiad/3049041.json"
} |
Given is a set $A$ of $n$ elements and positive integers $k, m$ such that $4 \leq k <n$ and $m \leq \min \{k-3, \frac {n} {2}\}$ . Let $A_1, A_2, \ldots, A_l$ be subsets of $A$ , all with size $k$ , such that $|A_i \cap A_j| \leq m$ for all $i \neq j$ . Prove that there exists a subset $B$ of $A$ wi... | [] | [
"origin:aops",
"2023 Contests",
"2023 Bulgarian Spring Mathematical Competition"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/2023 Contests/2023 Bulgarian Spring Mathematical Competition/3039497.json"
} | |
Given is a polynomial $f$ of degree $m$ with integer coefficients and positive leading coefficient. A positive integer $n$ is $\textit {good for f(x)}$ if there exists a positive integer $k_n$ , such that $n!+1=f(n)^{k_n}$ . Prove that there exist only finitely many integers good for $f$ . | Suppose there are infinitely many integers good for $f$ , obviously we have $k_n<n$ and $f(n)$ is odd, using the LTE lemma we know that $v_2(f(n)^{k_n}-1) \le v_2(f(n)-1)+v_2(f(n)+1)+v_2(k_n)$ , so at least one of $\{v_2(f(n)-1), v_2(f(n)+1),v_2(k_n)\}$ is greater than $\frac{1}{3}v_2(n!)$ for good $n$ , as ... | [
"Write $n! + 1 = P(n)^{k_n}$ . Due to the left-hand side and $P(x) \\geq x$ , we have $k_n<n$ and $P(n)$ must be odd for $n\\geq 2$ . Hence, using the Lifting the Exponent Lemma, we obtain \\[\\nu_2(n!) = \\nu_2\\left(P(n)^{k_n}-1\\right) \\le \\nu_2(P(n)-1)+\\nu_2(P(n)+1)+\\nu_2(k_n).\\] \nSince $\\nu_2(x)... | [
"origin:aops",
"2023 Contests",
"2023 Bulgarian Spring Mathematical Competition"
] | {
"answer_score": 48,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Bulgarian Spring Mathematical Competition/3039500.json"
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.