problem string | solution string | candidates list | tags list | metadata dict |
|---|---|---|---|---|
Let $P(x) = x^2-1$ be a polynomial, and let $a$ be a positive real number satisfying $$ P(P(P(a))) = 99. $$ The value of $a^2$ can be written as $m+\sqrt{n}$ , where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$ .
*Proposed by **HrishiP*** | <details><summary>Solution</summary>First, we obtain $P(a)=a^2-1$ . Upon plugging in this value into the polynomial again, we obtain $$ P(P(a))=(a^2-1)^2-1=(a^2-1+1)(a^2-1-1)=a^2(a^2-2)=a^4-2a^2. $$ Finally, upon plugging in this value into the polynomial again, we obtain
\begin{align*}
P(P(P(a)))&=(a^4-2a^2)^2-1
&=(a^4-2a^2+1)(a^4-2a^2-1)
&=(a^2-1)^2((a^2-1)^2-2)
&=(a^2-1)^4-2(a^2-1)^2.
\end{align*}
Setting this equal to $99$ and letting $y=a^2-1$ , we get $y^4-2y^2=99$ . Adding $1$ to both sides of the equation gives us $$ y^4-2y^2+1=100 \implies (y^2-1)^2=100 \implies (a^4-2a^2)^2=100 \implies a^4-2a^2=\pm 10. $$ Next, by the quadratic formula, we obtain $$ a^2=\frac{2\pm \sqrt{4\pm40}}{2} = 1 \pm \sqrt{11}. $$ Since $a^2$ is positive, we have that $a^2=1 + \sqrt{11}$ , so the requested answer is $1+11=\boxed{012}$ .</details>**Remark.** This gives me DMC 10B #15 vibes, but I think that this one is more clever. | [
"alternatively you can just note that the answer is constant so we just need to find one value of $a$ that works\n\nwe have $P(P(a))=10$ so that $a^2-1=\\sqrt{11}\\rightarrow \\boxed{012}$ ",
"Wait yes, working backwards here is probably cleaner.",
"i did it a diff way $a^2=x$ so you get $x^4-4x^3+4x^2-1... | [
"origin:aops",
"2022 Contests",
"2022 DIME"
] | {
"answer_score": 1028,
"boxed": true,
"end_of_proof": false,
"n_reply": 7,
"path": "Contest Collections/2022 Contests/2022 DIME/2673239.json"
} |
In $\triangle ABC$ with $AC>AB$ , let $D$ be the foot of the altitude from $A$ to side $\overline{BC}$ , and let $M$ be the midpoint of side $\overline{AC}$ . Let lines $AB$ and $DM$ intersect at a point $E$ . If $AC=8$ , $AE=5$ , and $EM=6$ , find the square of the area of $\triangle ABC$ .
*Proposed by **DeToasty3*** | Interesting problem!
Trivially $AM=MC=DM=4$ when drawing the circumcircle of $\triangle ADC$ , so $D=2$ . Applying LoC on $\triangle AME$ we get $25=36+16-48 \cos(\angle AME) \implies \frac{9}{16} = \cos(\angle AME)$ Applying LoC now on $\triangle AMD$ we get $AD^2 = 32-32 (\tfrac{9}{16}) = 14 \implies AD=\sqrt{14}$ Doing Pythagoras now on $\triangle ADC$ we get $DC=5 \sqrt{2}$ For the final step in the solution, let $N$ be the midpoint of $AB$ , and $AN=BN=y$ . Stewart’s Theorem on $\triangle AME$ yields $$ 14(5-2y)+4(2y) = 10y(5-2y)+5(4y^2-14) \implies y=2 $$ $$ (2+2)^2 - (\sqrt{14})^2 = BD^2 \implies BD=\sqrt{2} $$ The final step is $$ \left(\frac{\sqrt{2} \cdot \sqrt{14}}{2} + \frac{\sqrt{14} \cdot 5 \sqrt{2}}{2} \right)^2 = (\sqrt{7}+5 \sqrt{7})^2 = \boxed{252} $$ | [
"MD=4, DE=2, then do some area ratios and heron bash to get the answer",
"oops I fakesolved by using the fact that aime answers are only integers. :blush: ",
"<blockquote>oops I fakesolved by using the fact that aime answers are only integers. :blush:</blockquote>\n\nHuh? Now I'm curious as to how you did it.... | [
"origin:aops",
"2022 Contests",
"2022 DIME"
] | {
"answer_score": 1038,
"boxed": false,
"end_of_proof": false,
"n_reply": 13,
"path": "Contest Collections/2022 Contests/2022 DIME/2673241.json"
} |
Let $a_1,a_2,\ldots,a_6$ be a sequence of integers such that for all $1 \le i \le 5$ , $$ a_{i+1}=\frac{a_i}{3} \quad \text{or} \quad a_{i+1}={-}2a_i. $$ Find the number of possible positive values of $a_1+a_2+\cdots+a_6$ less than $1000$ .
*Proposed by **stayhomedomath*** | AHHHH HOW IS EVERYONE SO SMART
For this problem I basically spent an hour bashing, first mapping a triangle that had all possible values of each set of $a_n$ terms for $1 \le n \le 6$ , of course setting the initial value to $x$ . Then, I spend EVEN LONGER finding all possible values of sums with making an EVEN LARGER triangle. In the end, I recognized all the numerators were multiples of $7$ , so the answer was simply $\lfloor \tfrac{1000}{7} \rfloor = \boxed{142}$ | [
"the sequence 27, -54, -18, 36, 12, 4 and all its multiples gives 7x => 142, but i cant figure out how to explicitly prove",
"Note that ${-}2 \\equiv \\tfrac{1}{3} \\equiv 5 \\pmod{7}$ , so $$ a_1+a_2+\\cdots + a_6 \\equiv a_1(1+5+5^2+\\cdots+5^5) \\equiv 0 \\pmod{7}, $$ which means that $a_1+a_2+\\cdots+a_... | [
"origin:aops",
"2022 Contests",
"2022 DIME"
] | {
"answer_score": 1010,
"boxed": true,
"end_of_proof": false,
"n_reply": 15,
"path": "Contest Collections/2022 Contests/2022 DIME/2673242.json"
} |
Given a parallelogram $ABCD$ , let $\mathcal{P}$ be a plane such that the distance from vertex $A$ to $\mathcal{P}$ is $49$ , the distance from vertex $B$ to $\mathcal{P}$ is $25$ , and the distance from vertex $C$ to $\mathcal{P}$ is $36$ . Find the sum of all possible distances from vertex $D$ to $\mathcal{P}$ .
*Proposed by **HrishiP*** | Since $ABCD$ must be a parallelogram, the plane $\mathcal{P}$ will always be fixed for some configuration of $ABCD$ . Thus, we fix the plane $\mathcal{P}$ and then proceed.
Let the notation $\{\mathcal{F} \}\{\mathcal{B} \} \in \mathcal{P}$ denote the set of all points relative to $\mathcal{P}$ such that $\mathcal{F}$ contains all points in the "front", and $\mathcal{B}$ has all points in the "back". There are multiple cases:**Case 1:** $\{A,B,C \}\{\} \in \mathcal{P}$
By parallelogram properties, the distance from $D$ to the plane is $(49-25)+36=60$ and is obvious by a diagram**Case 2:** $\{A,C \} \{B \} \in \mathcal{P}$ By parallelogram properties, the net distance between $A$ and $B$ is $49+25$ , so $D$ is $49+25+36=110$ units away from the plane**Case 3:** $\{A,B \} \{C \} \in \mathcal{P}$ By parallelogram properties, the net distance between $A$ and $B$ is $49-25$ , so the distance from $D$ to the plane is $36-(49-25)=12$ units away**Case 4:** $\{A \} \{B,C \} \in \mathcal{P}$ By parallelogram properties, the net difference between $C$ and $B$ is $36-25$ , so the distance from $D$ to the plane is $49-(36-25)=38$ units away
The answer is $$ 60+110+12+38=\boxed{220} $$ | [
"let the z-coordinate for $A$ be $z_A$ , and similarly for the other points\n\nthen we have $|z_A|=49,|z_B|=25,|z_C|=36$ .\n\nnotice that because $ABCD$ is a parallelogram, we have $z_A+z_D=z_B+z_C\\rightarrow z_D=z_B+z_C-z_A$ .\n\nthe possible values for $z_B+z_C$ are $-61,-11,11,61$ and the possible va... | [
"origin:aops",
"2022 Contests",
"2022 DIME"
] | {
"answer_score": 1062,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2022 Contests/2022 DIME/2673243.json"
} |
Richard has an infinite row of empty boxes labeled $1, 2, 3, \ldots$ and an infinite supply of balls. Each minute, Richard finds the smallest positive integer $k$ such that box $k$ is empty. Then, Richard puts a ball into box $k$ , and if $k \geq 3$ , he removes one ball from each of boxes $1,2,\ldots,k-2$ . Find the smallest positive integer $n$ such that after $n$ minutes, both boxes $9$ and $10$ have at least one ball in them.
*Proposed by **vvluo** & **richy*** | Simple recursion suffices. Let each marble be labeled $1,2,3 \cdots $ such that for every $n$ th minute, the marble with label $a$ is placed into a box such that there exists no marbles with labels $b<a$ . Let the function $f(c)$ denote the number of the marble that was placed into box $c$ first. We get $$ f(1)=1 $$ $$ f(2)=2 $$ $$ f(3)=3 $$ $$ f(4)=5 $$ $$ f(5)=8 $$ $$ f(6)=13 $$ It is clear $f(c)=f(c-1)+f(c-2)$ for $c>2$ . Therefore, $f(7)=21$ , $f(8)=34$ , $f(9)=55$ , and $f(10)=\boxed{89}$ . Note when marble $89$ is placed, box $9$ is guaranteed to have at least $1$ marble because boxes $1-8$ are only cleared. | [
"clearly the first time that box $10$ is filled is when we have $$ 1111111110\\rightarrow 0000000011 $$ which means that box $9$ will also be filled, so we just need to smallest positive integer $n$ such that after $n$ minutes box $10$ is filled.\n\nlet $f(n)$ be the smallest positive integer such t... | [
"origin:aops",
"2022 Contests",
"2022 DIME"
] | {
"answer_score": 1056,
"boxed": false,
"end_of_proof": false,
"n_reply": 8,
"path": "Contest Collections/2022 Contests/2022 DIME/2673244.json"
} |
Let $a$ and $b$ be real numbers such that $$ \left(8^a+2^{b+7}\right)\left(2^{a+3}+8^{b-2}\right)=4^{a+b+2}. $$ The value of the product $ab$ can be written as $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
*Proposed by **stayhomedomath*** | Everything was right until somehow I got $a=\tfrac{29}{4}$ ....
First off, make everything base $2$ to get $$ \left(2^{3a} + 2^{b+7} \right)\left(2^{a+3} + 2^{3b-6} \right) = 2^{2a+2b+4} $$ Clearly the RHS is a power of $2$ , which means both expressions in the LHS product are also powers of $2$ .
Now, suppose $2^c+2^d=2^e$ and WLOG let $2^c < 2^d$ . Clearly we can factor to get $2^c(1+2^{d-c}) = 2^e$ , so contradiction because $1+2^{d-c}$ is never a power of $2$ . Reverse the expression and we get the same result. Thus $2^c=2^d$ Now, back to the original problem, we get $2^{3a}=2^{b+7}$ and $2^{a+3} = 2^{3b-6}$ . Therefore, $$ 3a-b=7 $$ $$ 3a-9b=-27 $$ $$ 8b = 34 \implies b = \frac{17}{4} $$ $$ 3a=\frac{45}{4} \implies a = \frac{15}{4} $$ $$ ab=\frac{15 \cdot 17}{4 \cdot 4} = \frac{255}{16} \implies 255+16=\boxed{271} $$ If the following configuration had not worked, we would have ended up with $3$ systems which would have not been fun to solve... | [
"the LHS is $(2^{3a}+2^{b+7})(2^{a+3}+2^{3b-6})=4^{a+b+2}$ , this motivates us to set $3a=b+7$ and $a+3=3b-6$ giving $(a,b)=(\\frac{15}{4},\\frac{17}{4})$ . this works so the answer is $\\frac{255}{16}\\rightarrow \\boxed{271}$ .",
"wait i don't really understand the motivation for the above solution\n\nes... | [
"origin:aops",
"2022 Contests",
"2022 DIME"
] | {
"answer_score": 1050,
"boxed": false,
"end_of_proof": false,
"n_reply": 16,
"path": "Contest Collections/2022 Contests/2022 DIME/2673245.json"
} |
A positive integer $n$ is called $\textit{un-two}$ if there does not exist an ordered triple of integers $(a,b,c)$ such that exactly two of $$ \dfrac{7a+b}{n},\;\dfrac{7b+c}{n},\;\dfrac{7c+a}{n} $$ are integers. Find the sum of all un-two positive integers.
*Proposed by **stayhomedomath*** | If $n \equiv 0 \pmod{7}$ it's not hard to see that $n$ is not un-two. Otherwise, suppose we have $7a + b \equiv 0 \pmod{n}$ and $7b+c \equiv 0 \pmod{n}$ which means that $49a \equiv c \pmod{n}.$ We know $n$ is un-two if and only if $7c+a \equiv 344a \equiv 0 \pmod{n}$ for all $a$ so we just sum up all factors of $344$ to get $\boxed{660}.$ | [
"<details><summary>Solution</summary>Set $a=1,b=-7,c=49$ to get $$ \\frac{7a+b}{n}=0,\\frac{7b+c}{n}=0,\\frac{7c+a}{n}=\\frac{344}{n}. $$ Then clearly we must have $n\\mid 344$ . <insert explanation that i got during the test but forgot> it is sufficient so we have $(15)(44)=660$ .</details>"
] | [
"origin:aops",
"2022 Contests",
"2022 DIME"
] | {
"answer_score": 1020,
"boxed": true,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 DIME/2673246.json"
} |
A sequence of polynomials is defined by the recursion $P_1(x) = x+1$ and $$ P_{n}(x) = \frac{(P_{n-1}(x)+1)^5 - (P_{n-1}(-x)+1)^5}{2} $$ for all $n \geq 2$ . Find the remainder when $P_{2022}(1)$ is divided by $1000$ .
*Proposed by **treemath*** | <details><summary>Author's/Official Solution</summary>The recursion is applying the second roots of unity filter to $(P_n(x)+1)^5$ that only returns terms with odd degree. For the first step, $P_2(x)$ is the odd-degree terms of $(x+2)^5$ , which is $x^5+40x^3+80x$ . For all subsequent steps, we again apply the second roots of unity filter to $(P_n(x)+1)^5$ that only returns terms with odd degree.
In general, if a polynomial $P(x)$ has only terms of odd degree, then $(P(x))^n$ will only have terms of odd degree if $n$ is odd and will only have terms of even degree if $n$ is even. The proof for this is left as an exercise to the reader. Since $P_n(x)$ will have all odd coefficients for all $n \ge 2$ , we can treat $P_n(x)$ as a single term with odd degree. Thus, by the Binomial Theorem, we get that $$ P_{n+1}(x)=(P_n(x))^5+10(P_n(x))^3+5(P_n(x)) $$ for all $n \geq 2$ .
Next, we see that $P_2(1)=121$ , which is equivalent to $1$ modulo $8$ . We see that $$ P_3(1) \equiv 1^5+10(1)^3+5(1)\equiv 0 \pmod{8}, $$ so $P_n(1)\equiv 0 \pmod{8}$ for all $n \geq 3$ .
Next, we see that $121 \equiv {-}4 \pmod{125}$ . Computing, we get $$ P_3(1) \equiv ({-}4)^5+10({-}4)^3+5({-}4) \equiv 66 \pmod{125}. $$ We can rewrite $66$ as $50+16$ . From here, we get that
\begin{align*}
P_4(1) &\equiv (50+16)^5+10(50+16)^3+5(50+16)
&\equiv 16^5 + 10(16)^3 + 5(16)
&\equiv 116 \pmod{125}.
\end{align*}
Note that $116=100+16$ , so for any $n \geq 5$ , we see that
\begin{align*}
P_n(1) &\equiv (100+16)^5+10(100+16)^3+5(100+16)
&\equiv 16^5 + 10(16)^3 + 5(16)
&\equiv 116 \pmod{125}.
\end{align*}
Thus, we have that $P_{2022}(1)$ is equivalent to $0$ modulo $8$ and $116$ modulo $125$ . By the Chinese Remainder Theorem, the requested answer is $\boxed{616}$ .</details>**Remark (DeToasty3).** While I really like the <details><summary>idea</summary>keeping in only the terms with an odd degree</details> of this problem, unfortunately the big mod bash afterwards kind of downgrades the quality of the problem for me by a bit. | [
"the only thing i was able to get on the test was $P_{n-1}(x)=-P_{n-1}(-x)$ for all $n\\geq 2$ giving $$ P_{n+1}(x)=\\frac{(P_n(x)+1)^5+(P_n(x)-1)^5}{2} $$ for all $n\\geq 2$ ",
"Try showing that $$ P_{n+1}(x)=(P_n(x))^5+10(P_n(x))^3+5(P_n(x)) $$ for all $n \\geq 2$ .\n\nNow, use CRT with mod $125$... | [
"origin:aops",
"2022 Contests",
"2022 DIME"
] | {
"answer_score": 1070,
"boxed": true,
"end_of_proof": false,
"n_reply": 27,
"path": "Contest Collections/2022 Contests/2022 DIME/2673247.json"
} |
A spinner has five sectors numbered ${-}1.25$ , ${-}1$ , $0$ , $1$ , and $1.25$ , each of which are equally likely to be spun. Ryan starts by writing the number $1$ on a blank piece of paper. Each minute, Ryan spins the spinner randomly and overwrites the number currently on the paper with the number multiplied by the number the spinner lands on. The expected value of the largest number Ryan ever writes on the paper can be written as $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
*Proposed by **treemath*** | Cool problem!
<details><summary>Sol</summary>Let $a_k$ denote the $k^{\text{th}}$ term in our sequence of numbers, where the $0^{\text{th}}$ term is $1$ .
Now let's figure out the probability that $a_n$ is the last positive term in the sequence for a given $n$ . Following this positive we can have:
A $0$ . Probability: $\frac{1}{5}$ A negative followed by a $0$ . Probability: $\frac{2}{5} \cdot \frac{1}{5}$ A negative, a positive, and a $0$ . Probability: $\frac{2}{5} \cdot \frac{2}{5} \cdot \frac{1}{5}$ A negative, a positive, a positive, and a $0$ . Probability: $\frac{2}{5} \cdot \frac{2}{5} \cdot \frac{2}{5} \cdot \frac{1}{5}$
This sequence of a negative, positives, and $0$ continues infinitely. $\frac{\frac{1}{5}}{1-\frac{2}{5}} = \frac{1}{3}$ , so there's a $\frac{1}{3}$ probability that none of the terms following $a_n$ are positive.
We can compute the probability that $a_n$ is positive inductively.
Claim: The probability that $a_n$ is positive is $\frac{(\frac{4}{5})^n}{2}$ Base case: The probability that $a_1$ is positive is $\frac{\frac{4}{5}}{2} = \frac{2}{5}$ .
Inductive step: Given that the probability that $a_n$ is positive is $\frac{(\frac{4}{5})^n}{2}$ , we know that the probability that $a_n$ is negative is also $\frac{(\frac{4}{5})^n}{2}$ , since the probability that $a_n$ is not $0$ is $(\frac{4}{5})^n$ . From here, if $a_n$ is negative there is probability of $\frac{2}{5}$ that $a_{n+1}$ is positive, and if $a_n$ is positive there is probability of $\frac{2}{5}$ that $a_{n+1}$ is positive. Thus the probability that $a_{n+1}$ is positive is $\frac{2}{5} \cdot 2 \cdot \frac{(\frac{4}{5})^n}{2} = \frac{(\frac{4}{5})^{n+1}}{2}$ , completing our induction.
Combining the probability that $a_n$ is positive with the probability that all terms after $a_{n}$ are non-positive, we get that there is a probability of $\frac{(\frac{4}{5})^n}{2} \cdot \frac{1}{3}$ that $a_n$ is the last positive in our sequence. Note that this means that $a_n$ is also the greatest positive in our list, since magnitude cannot decrease unless we come across a $0$ .
Now we just need the expected value of $a_n$ , given that it is positive. Note that we can ignore if the spins leading up to $a_n$ are negative or positive, since we know that $a_n$ is positive. The $n$ terms that we take the product of to get $a_n$ have magnitude $\frac{5}{4}$ or $1$ with equal probability. The expected value of $a_n$ is thus $\frac{{n \choose n} \cdot (\frac{5}{4})^n + {n \choose {n-1}} \cdot (\frac{5}{4})^{n-1} + \ldots + {n \choose 0} \cdot (\frac{5}{4})^0}{2^n} = \frac{(\frac{5}{4} + 1)^n}{2^n} = (\frac{9}{8})^n$ .
Thus we can get our expected value by adding up the expected value of $a_n$ for all $n \geq 1$ : $\[ \sum_{n=1}^{\infty} (\frac{9}{8})^n \cdot \frac{(\frac{4}{5})^n}{2} \cdot \frac{1}{3} = \frac{3}{2}\]$ , but note that this discards the probability of our sequence being entirely non-positve after $a_0 = 1$ , which happens with a probability of $\frac{1}{3}$ , and thus contributes $1 \cdot \frac{1}{3}$ to the expected value. $\frac{3}{2} + \frac{1}{3} = \frac{11}{6} \implies \boxed{17}$</details> | [
"PROBABLY WRONG! but here goes\n\n<details><summary>eek</summary>Let $f(n)=\\text{E}[\\text{max}(\\text{largest number Ryan writes if he starts with the number }n,1)]$ . Then $$ f(1)=\\frac{f(-1.25)+f(-1)+f(0)+f(1)+f(1.25)}{5} $$ or $$ f(1)=\\frac{f(-1.25)+f(-1)+1+f(1.25)}{4}. $$ Notice that $f(1.25)=1.25... | [
"origin:aops",
"2022 Contests",
"2022 DIME"
] | {
"answer_score": 1120,
"boxed": false,
"end_of_proof": false,
"n_reply": 8,
"path": "Contest Collections/2022 Contests/2022 DIME/2673248.json"
} |
Let $\triangle ABC$ be acute with $\angle BAC = 45^{\circ}$ . Let $\overline{AD}$ be an altitude of $\triangle ABC$ , let $E$ be the midpoint of $\overline{BC}$ , and let $F$ be the midpoint of $\overline{AD}$ . Let $O$ be the center of the circumcircle of $\triangle ABC$ , let $K$ be the intersection of lines $DO$ and $EF$ , and let $L$ be the foot of the perpendicular from $O$ to line $AK$ . If $BL = 6$ and $CL = 8$ , find $AL^2$ .
*Proposed by **Awesome_guy*** | <details><summary>Solution?</summary>Reflect $O$ over side $\overline{BC}$ to a point $P$ . Due to reflections, we have that $E$ is the midpoint of $\overline{OP}$ , and we are given that $E$ is also the midpoint of $\overline{BC}$ , so $BOCP$ is a parallelogram. We also have that $\angle BAC = 45^{\circ}$ , so $\angle BOC = 90^{\circ}$ . Since $BO=CO$ , we have that $BOCP$ is a square.**<span style="color:#f00">Claim:</span>** Points $A$ , $K$ , and $P$ are collinear.
*<span style="color:#00f">Proof:</span>* We have that $\angle APB = \angle OEB = 90^{\circ}$ , so $AD \parallel OP$ . Thus, $\triangle PKF \sim \triangle OKE$ . We also have that $AF:DF = OE:PE = 1:1$ , so $\triangle AKD \sim \triangle PKO$ by spiral similarity. It then follows that points $A$ , $K$ , and $P$ are collinear, as desired.
Now, we see that $\angle ALO = \angle PLO = 90^{\circ}$ , so $L$ lies on the circumcircle of $BOCP$ . By the Pythagorean Theorem, $$ BC^2=BL^2+CL^2=64+36=100 \implies BC=10. $$ By Ptolemy's Theorem on quadrilateral $BCLO$ ,
\begin{align*}
BC \cdot LO + CL \cdot BO &= BL \cdot CO
\implies 10 \cdot LO + 6 \cdot 5\sqrt{2} &= 8 \cdot 5\sqrt{2}
\implies LO &= \sqrt{2}.
\end{align*}
Finally, we have that $AO=5\sqrt{2}$ (radius of the circumcircle of $\triangle ABC$ ), $LO=\sqrt{2}$ , and $\angle ALO = 90^{\circ}$ , so by the Pythagorean Theorem, $$ AL^2=AO^2-LO^2=(5\sqrt{2})^2-(\sqrt{2})^2=50-2=\boxed{048}, $$ as requested.**Remark.** The idea of reflecting $O$ over side $\overline{BC}$ and the claim were both due to **ApraTrip**, and the proof of the claim was due to **richy**. The rest was written by me.</details>
Alternative formation: Prove that $AL^2=BL \cdot CL$ . | [
"it suffices to show that B, O, L, C concyclic and i half guessed this during the test but basically BLC is right by 6-8-10 (very cringe) and then BOC is 90 cuz BAC = 45 so B O L C concyclic. now it's simple by pythag and ptolemy => 48",
"official sol?",
"Guessed $50$ :wallbash: ",
"<blockquote>official so... | [
"origin:aops",
"2022 Contests",
"2022 DIME"
] | {
"answer_score": 1076,
"boxed": false,
"end_of_proof": false,
"n_reply": 15,
"path": "Contest Collections/2022 Contests/2022 DIME/2673249.json"
} |
For positive integers $n$ , let $f(n)$ denote the number of integers $1 \leq a \leq 130$ for which there exists some integer $b$ such that $a^b-n$ is divisible by $131$ , and let $g(n)$ denote the sum of all such $a$ . Find the remainder when $$ \sum_{n = 1}^{130} [f(n) \cdot g(n)] $$ is divided by $131$ .
*Proposed by **ApraTrip*** | <details><summary>Author's Solution</summary>Let all sums in this solution be taken modulo $131$ , and let $\text{ord}_{131}(x)$ be the order of $x$ modulo $131$ .
We claim that there exists an integer $s$ such that $r^{s} \equiv x \pmod{131}$ if and only if $\text{ord}_{131}(x) \mid \text{ord}_{131}(r)$ . Let $g$ be a primitive root $\pmod{131}$ . Notice that when $\text{ord}_{131}(x) \mid \text{ord}_{131}(r)$ , then $\tfrac{130}{\text{ord}_{131}(r)} \mid \tfrac{130}{\text{ord}_{131}(x)}$ . Thus, since $r \equiv g^{\tfrac{130a}{\text{ord}_{131}(r)}}$ and $x \equiv g^{\tfrac{130b}{\text{ord}_{131}(x)}}$ for some $a$ and $b$ where $\gcd(a, 130) = 1$ and $\gcd(b, 130) = 1$ , if we let $c \equiv \tfrac{b}{a} \pmod{130}$ and $d = \frac{\tfrac{130}{\text{ord}_{131}(x)}}{\tfrac{130}{\text{ord}_{131}(r)}}$ , then $$ r^{cd} = g^{\frac{130ac}{\text{ord}_{131}(x)}} = g^{\frac{130b}{\text{ord}_{131}(x)}} = x. $$ Thus, there exists an integer $s$ such that $r^{s} \equiv x \pmod{131}$ if $\text{ord}_{131}(x) \mid \text{ord}_{131}(r)$ .
Now, notice that if $r^{s} \equiv x \pmod{131}$ , then $$ x^{\text{ord}_{131}(r)} \equiv r^{s \cdot \text{ord}_{131}(r)} \equiv 1 \pmod{131}, $$ which implies that $\text{ord}_{131}(x) \mid \gcd(\text{ord}_{131}(r), 130) = \text{ord}_{131}(r)$ , as $\text{ord}_{131}(r) \mid 130$ . Thus, our claim is true.
Thus, $f(x)$ is the number of integers $1 \le r \le 130$ such that $\text{ord}_{131}(x) \mid \text{ord}_{131}(r)$ , and $g(x)$ is the sum of all integers $1 \le r \le 130$ such that $\text{ord}_{131}(x) \mid \text{ord}_{131}(r)$ . We will deal with both functions separately, and put them together at the end. $f(x)$ : It's well known that there are $\varphi(n)$ numbers $\pmod{p}$ with order $n$ where $p$ is prime and $n \mid p-1.$ Thus,
\begin{align*}
f(x) &= \sum_{\substack{\text{ord}_{131}(x) \mid d, d \mid 130}} \varphi(d) = \sum_{d \mid \tfrac{130}{\text{ord}_{131}(x)}} [\varphi(\text{ord}_{131}(x)) \cdot \varphi(d)]
&= \varphi(\text{ord}_{131}(x)) \cdot \sum_{d \mid \tfrac{130}{\text{ord}_{131}(x)}} \varphi(d) = \frac{130\varphi(\text{ord}_{131}(x))}{\text{ord}_{131}(x)},
\end{align*}
where the second equality follows as $\varphi(x)$ is multiplicative and any two factors of $130$ which multiply to be a factor of $130$ are relatively prime (as $130$ isn't divisible by any perfect powers). $g(x)$ : It's known that for all primes $p$ such that that $p-1$ isn't divisible by any perfect powers, the sum of all numbers with order $d$ (for some $d \mid p-1$ ) is $(-1)^{\text{the number of prime factors of } d}$ (see 2021 DIME #14). Thus, in order to find $g(x)$ , we'll do casework on the number of factors of $\text{ord}_{131}(x)$ , while utilizing the fact that $130$ has three prime factors.**Case 1:** $\text{ord}_{131}(x)$ has $0$ prime factors.
In this case, there would be $1$ multiple of $\text{ord}_{131}(x)$ with $0$ factors that divides $130$ , $3$ multiples of $\text{ord}_{131}(x)$ with one factor that divide $130$ , $3$ multiples of $\text{ord}_{131}(x)$ with two factors that divide $130$ , and $1$ multiple of $\text{ord}_{131}(x)$ with three factors that divides $130$ . Thus, $g(x) = 1-3+3-1 = 0$ when $\text{ord}_{131}(x)$ has $0$ prime factors.**Case 2:** $\text{ord}_{131}(x)$ has $1$ prime factor.
In this case, there would be $1$ multiple of $\text{ord}_{131}(x)$ with one factor that divides $130$ , $2$ multiples of $\text{ord}_{131}(x)$ with two factors that divide $130$ , and $1$ multiple of $\text{ord}_{131}(x)$ with three factors that divides $130$ . Thus, $g(x) = -1+2-1 = 0$ when $\text{ord}_{131}(x)$ has $1$ prime factor.**Case 3:** $\text{ord}_{131}(x)$ has $2$ prime factors.
In this case, there would be $1$ multiple of $\text{ord}_{131}(x)$ with two factors that divides $130$ , and $1$ multiple of $\text{ord}_{131}(x)$ with three factors that divides $130$ . Thus, $g(x) = 1-1 = 0$ when $\text{ord}_{131}(x)$ has $2$ prime factors. At this point, it might seem that $g(x)$ is $0$ for all $x$ (which would make this a very trolly problem), but this is not true.**Case 4:** $\text{ord}_{131}(x)$ has $3$ prime factors.
In this case, there would be $1$ multiple of $\text{ord}_{131}(x)$ with three factors that divides $130$ . Thus, $g(x) = -1$ when $\text{ord}_{131}(x)$ has three prime factors.
Thus, $g(x) = 0$ when $x$ is not a primitive root, and $g(x) = -1$ when $x$ is a primitive root. Thus, our sum becomes
\begin{align*}
\sum_{\text{ord}_{131}(x) = 130} [f(x) \cdot g(x)] &= - \sum_{\text{ord}_{131}(x) = 130} f(x)
&= - \sum_{\text{ord}_{131} (x) = 130} \frac{130 \varphi(130)}{130}
&= -\varphi(130)^2
&= -2304.
\end{align*}
Thus, the requested remainder when our sum is divided by $131$ is $\boxed{054}$ .</details>
If you have any questions about this solution or if you find any issues with it, please send a private message to **ApraTrip** (with the subject "Regarding 2022 DIME #15"), as I do not know what is going on here... :maybe: | [
"any solutions to this?",
"i think its smt with primitive roots",
"sol pls?",
"neither do I @above\n\nyou guys rlly should've made a DJMO that included this problem",
"yeah thats a rly hard order problem\n\n",
"<blockquote>neither do I @above\n\nyou guys rlly should've made a DJMO that included this probl... | [
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Let $f:\mathbb{N}^*\rightarrow \mathbb{N}^*$ be a function such that $\frac{x^3+3x^2f(y)}{x+f(y)}+\frac{y^3+3y^2f(x)}{y+f(x)}=\frac{(x+y)^3}{f(x+y)},~(\forall)x,y\in\mathbb{N}^*.$ $a)$ Prove that $f(1)=1.$ $b)$ Find function $f.$ | Giả sử tồn tại hàm số thỏa mãn $$ \frac{x^3+3x^2f(y)}{x+f(y)}+\frac{y^3+3y^2f(x)}{y+f(x)}=\frac{(x+y)^3}{f(x+y)},\forall x,y\in \mathbb{Z^+}.(1) $$ Từ $(1)$ cho $x=y=1$ ta được $$ \frac{2(1+3f(1))}{1+f(1)}=\frac{8}{f(2)} $$ Hay ta được $$ f(2)=\frac{4(1+f(1))}{1+3f(1)}. $$ Suy ra $(1+f(1))|(4+4f(1))$ hay $(1+3f(1))|8$ suy ra $f(1)=1$ nên suy ra $f(2)=2.$ Ta sẽ chứng minh rằng $f(x)=x,\forall x\in \mathbb{Z^+}.$ Thật vật với $x=1,2$ thì khẳng định hiển nhiên đúng.Giả sử khẳng định đúng đến $x=k,k\in \mathbb{Z^+}.$ Thì ta chứng minh rằng $k+1$ cũng đúng.Từ $(1)$ thay $x=k,y=1$ ta được $$ \frac{k^3+3x^2}{k+1}+\frac{1+3k}{k+1}=\frac{(k+1)^3}{f(k+1)},\forall k\in \mathbb{Z^+}. $$ Hay ta được $$ f(k+1)=k+1,\forall k\in \mathbb{Z^+}. $$ Vậy theo quy nạp ta có tất cả hàm số cần tìm là $$ f(x)=x,\forall x\in \mathbb{Z^+}. $$ | [
"Let $a=f(1)$ . $$ (x,y)=(1,1) \\rightarrow \\dfrac{2(1+3a)}{1+a}=\\dfrac{8}{f(2)} \\rightarrow 1+3a\\; |\\; 4a+4 \\rightarrow f(1)=a=1. $$ Using induction with the relation $(x,1)$ we deduce that $f(x)=x$ ."
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$a)$ Prove that $2x^3-3x^2+1\geq 0,~(\forall)x\geq0.$ $b)$ Let $x,y,z\geq 0$ such that $\frac{2}{1+x^3}+\frac{2}{1+y^3}+\frac{2}{1+z^3}=3.$ Prove that $\frac{1-x}{1-x+x^2}+\frac{1-y}{1-y+y^2}+\frac{1-z}{1-z+z^2}\geq 0.$ | $a)$ The expression factorizes as $(x-1)^2(2x+1)$ which is clearly positive for $x\geq 0$ . Moreover, equality occurs for $x=1$ . $b)$ Using the result from point $a)$ , we get $1-x^3\leq \frac{3}{2}(1-x^2)$ , thus $$ 3=\sum_{cyc}\frac{2}{1+x^3}\implies 0=\sum_{cyc}\frac{1-x^3}{1+x^3}\leq \frac{3}{2}\sum_{cyc}\frac{1-x^2}{1+x^3}=\frac{3}{2}\sum_{cyc}\frac{1-x}{1-x+x^2} $$ which is exactly what we wanted to prove. Equality occurs for $x=y=z=1$ (which does verify the constraint over the variables). | [
"Let $x,y,z\\geq 0$ such that $\\frac{2}{1+x^3}+\\frac{2}{1+y^3}+\\frac{2}{1+z^3}=3.$ Prove that $$ \\frac{2-x}{2-x+x^2}+\\frac{2-y}{2-y+y^2}+\\frac{2-z}{2-z+z^2}\\leq \\frac{3}{2} $$ \n"
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$a)$ Solve over the positive integers $3^x=x+2.$ $b)$ Find pairs $(x,y)\in\mathbb{N}\times\mathbb{N}$ such that $(x+3^y)$ and $(y+3^x)$ are consecutive. | Part a) $x=1$ only solution. For $x>1, LHS>RHS$ Part b)
If $(x+3^y)$ and $(y+3^x)$ are consecutive $\Longrightarrow x+3^y+1= y+3^x$ Case $x=y$ $\Longrightarrow x+3^x+1= x+3^x \Longrightarrow 1= 0$ impossible
Case $y>x$ Let $y=x+n\Longrightarrow x+3^n3^x+1=x+n+3^x\Longrightarrow 3^x=\frac{n-1}{3^n-1}<1$ Contradiction
Case $x>y$ Let $x=y+n\Longrightarrow y+n+3^y+1=y+3^n3^y\Longrightarrow3^y=\frac{n+1}{3^n-1}$ $n=1\Longrightarrow 3^y=\frac{2}{2}\Longrightarrow 3^y=1\Rightarrow x=1, y=0$ is the only solution if $0\in Z$ .
If $n>1, 3^y=\frac{n+1}{3^n-1}<1$ . Contradiction. | [] | [
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Let $e$ be the identity of monoid $(M,\cdot)$ and $a\in M$ an invertible element. Prove that
[list=a]
[*]The set $M_a:=\{x\in M:ax^2a=e\}$ is nonempty;
[*]If $b\in M_a$ is invertible, then $b^{-1}\in M_a$ if and only if $a^4=e$ ;
[*]If $(M_a,\cdot)$ is a monoid, then $x^2=e$ for all $x\in M_a.$ [/list]
*Mathematical Gazette* | Notice that the set can be redefined as $M_a=\{x\in M : x^2=a^{-2}\}$ . $a)$ Observe that $a^{-1}\in M_a$ . $b)$ Since $b\in M_a$ , we have $b^2=a^{-2}$ , thus $$ b^{-1}\in M_a\iff b^{-2}=a^{-2}\iff e=b^2b^{-2}=a^{-2}a^{-2}=a^{-4}\iff a^4=e. $$ $c)$ Let $f$ be the identity of $M_a$ . $f\in M_a$ , thus $a^{-2}=f^2=f$ , implying that $a^{-4}=(a^{-2})^2=f^2=a^{-2}$ , and finally $a^{-2}=e$ . From the definition of $M_a$ , the result follows. | [] | [
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Let $(G,\cdot)$ be a group and $H\neq G$ be a subgroup so that $x^2=y^2$ for all $x,y\in G\setminus H.$ Show that $(H,\cdot)$ is an Abelian group. | Let $x \in G \setminus H.$ For any $a \in H,$ we have $ax \notin H$ and so $(ax)^2=x^2,$ which gives $axa=x$ hence $$ a=xa^{-1}x^{-1}, \ \ \ \ \ \forall a \in H. $$ So if $a,b \in H,$ then $$ xb^{-1}a^{-1}x^{-1}=x(ab)^{-1}x^{-1}=ab=(xa^{-1}x^{-1})(xb^{-1}x^{-1})=xa^{-1}b^{-1}x^{-1}, $$ which gives $b^{-1}a^{-1}=a^{-1}b^{-1}$ and so $ab=ba,$ i.e. $H$ is abelian.
The condition given in the problem is strong, and so I think we can say more about $G.$ For example, an obvious one is that $x^4=1$ for all $x \notin H$ because if $x \notin H,$ then $x^{-1} \notin H$ and so $x^2=x^{-2},$ which gives $x^4=1.$ | [] | [
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Find all values of $n\in\mathbb{N}^*$ for which \[I_n:=\int_0^\pi\cos(x)\cdot\cos(2x)\cdot\ldots\cdot\cos(nx) \ dx=0.\] | [
"[url=https://artofproblemsolving.com/community/c7h1646205p10391941] here [ \\url ]\n\nhttps://artofproblemsolving.com/community/c7h1646205p10391941"
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Let $I\subseteq \mathbb{R}$ be an open interval and $f:I\to\mathbb{R}$ a strictly monotonous function. Prove that for all $c\in I$ there exist $a,b\in I$ such that $c\in (a,b)$ and \[\int_a^bf(x) \ dx=f(c)\cdot (b-a).\] | Let $I=(k,l)$ and suppose $f$ is strictly decreasing (otherwise, look at the function $-f$ ). Consider some real number (which we will define later) $y_0\in (c,l)$ and let $g:(k,c)\rightarrow \mathbb{R}$ with $$ g(x)=\int_x^{y_0}f(t)dt - f(c)(y_0-x). $$ Observe that $g(c)<0$ since from the monotonicity of $f$ , $\int_c^{y_0}f(t)dt<f(c)(y_0-c)$ . Now $$ g(k)=\int_k^{y_0}f(t)dt - f(c)(y_0-k)=\int_k^c f(t)dt - f(c)(c-k)+\int_c^{y_0}f(t)dt - f(c)(y_0-c). $$ Letting $h(x)=\int_c^x f(t)dt - f(c)(x-c)$ , one can see that $\lim_{x\rightarrow c^{+}}h(x)=0$ , and since $\int_k^c f(t)dt - f(c)(c-k)>0$ (from the fact that $f$ is decreasing), there exists some $y_0\in(c,l)$ such that $g(k)>0$ . But now, $g$ is continuous (both the integral and the linear function are continuous), the Darboux property (MVT) implying that there exists some $r\in(k,c)$ such that $g(r)=0$ , which is exactly what we wanted to prove. | [
"Assume $f$ is strictly increasing (otherwise work with $-f$ ). Since $I$ is open, there exists $\\delta>0$ such that $(c-\\delta, c+\\delta) \\subseteq I$ . Fix such a $\\delta$ and consider the function (for $x>c$ )\n\\[\ng(x) = \\int_{c-\\delta}^{x} f(t)\\mathrm{d}t.\n\\]\nSince $f$ is strictly incr... | [
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Let $f,g:\mathbb{R}\to\mathbb{R}$ be functions which satisfy \[\inf_{x>a}f(x)=g(a)\text{ and }\sup_{x<a}g(x)=f(a),\]for all $a\in\mathbb{R}.$ Given that $f$ has Darboux's Property (intermediate value property), show that functions $f$ and $g$ are continuous and equal to each other.
*Mathematical Gazette* | <blockquote>Let $f,g:\mathbb{R}\to\mathbb{R}$ be functions which satisfy \[\inf_{x>a}f(x)=g(a)\text{ and }\sup_{x<a}g(x)=f(a),\]for all $a\in\mathbb{R}.$ Given that $f$ has Darboux's Property (intermediate value property), show that functions $f$ and $g$ are continuous and equal to each other.
*Mathematical Gazette*</blockquote>**<span style="color:#f00">Claim 01.</span>** $f,g$ are increasing
*Proof.* Indeed, suppose $a > b$ , then we have
\begin{align*}
f(a) &= \sup_{x < a} g(x) \ge \sup_{x < b} g(x) = f(b)
g(a) &= \inf_{x > a} f(x) \ge \inf_{x > b} f(x) = g(b)
\end{align*}
---------------------------
Indeed, as $f,g$ have Darboux's Property and are increasing, it follows that $f$ and $g$ are continuous.
Let $x > y$ . We know that $\inf_{x > y} f(x) = g(y)$ implies $f(x) \ge g(y)$ . Furthermore,
\begin{align*}
f(x) \ge f(y) \ \forall x > y &\implies g(y) = \inf_{x > y} f(x) \ge f(y)
f(x) \ge g(y) \ \forall y < x &\implies f(x) \ge \lim_{y \to x} g(y) = g(x)
\end{align*}
and thus we conclude that $f = g$ . | [
"https://ssmr.ro/",
"<blockquote>[ see here ](https://ssmr.ro/) </blockquote>\n\nRomanian maths problems are very nice",
"Just a question, what is the name of this competition, can you attach website or something because these questions are college level yet they are for grade 11, so I want to know. Also i sear... | [
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Let $A,B\in\mathcal{M}_3(\mathbb{R})$ de matrices such that $A^2+B^2=O_3.$ Prove that $\det(aA+bB)=0$ for any real numbers $a$ and $b.$ | Interesting problem $A^2=-B^2$ take the det then $det(A)^2=-det(B)^2$ ==> det(A)=det(B)=0 (1)
If a=0 or b=0 with (1) ==> det(aA+bB)=0
If a=/=0 and b=/=0 we must prove det(A+xB)=0 for any x in R $Q(x)=det(A+xB)=det(B)x^3 + px^2 + qx +det(A)= px^2 + qx$ where p,q reals numbers $Q(i)Q(-i)=|Q(i)|^2 = (-p+iq)(-p-iq) = p^2 +q^2$ (2)
Q(i)Q(-i)=det((A+iB)(A-iB))= $det(A^2+B^2+iBA-iAB)=i^3det(BA-AB)$ =-idet(BA-AB) (3)
(2) and (3) give p=q=det(BA-AB)=0
This prove for any x in R, det(A+xB)=0
Remark AB-BA is not invertible | [
"What does the $O_3$ mean?",
"zero matrix $ 3\\times 3$ ",
"<blockquote> $A^2=-B^2$ take the det then $det(A)^2=-det(B)^2$ ==> det(A)=det(B)=0 (1)\nIf a=0 or b=0 with (1) ==> det(aA+bB)=0\n\nIf a=/=0 and b=/=0 we must prove det(A+xB)=0 for any x in R $Q(x)=det(A+xB)=det(B)x^3 + px^2 + qx +det(A)= px^2 ... | [
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Let $(x_n)_{n\geq 1}$ be the sequence defined recursively as such: \[x_1=1, \ x_{n+1}=\frac{x_1}{n+1}+\frac{x_2}{n+2}+\cdots+\frac{x_n}{2n} \ \forall n\geq 1.\]Consider the sequence $(y_n)_{n\geq 1}$ such that $y_n=(x_1^2+x_2^2+\cdots x_n^2)/n$ for all $n\geq 1.$ Prove that
[list=a]
[*] $x_{n+1}^2<y_n/2$ and $y_{n+1}<(2n+1)/(2n+2)\cdot y_n$ for all $n\geq 1;$ [*] $\lim_{n\to\infty}x_n=0.$ [/list] | Here is yet another way of doing part b) without the help of the auxiliary sequence.
Firstly, we will prove that $(x_n)$ is decreasing. Notice that $$ x_{n+1}-x_n = \left(\sum_{k=1}^n \frac{x_k}{n+k}\right) - x_n = \left(\sum_{k=1}^{n-1} \frac{x_k}{n+k}\right) + \frac{x_n}{2n} - x_n = \sum_{k=1}^{n-1} x_k\left(\frac{1}{n+k} - \left(1-\frac{1}{2n}\right)\frac{1}{n-1+k}\right). $$ However, every term of the form $\frac{1}{n+k} - \left(1-\frac{1}{2n}\right)\frac{1}{n-1+k}$ is negative, proving our claim.
Next, we will use the well-known lemma, stating that if a function $f:[0,1]\rightarrow\mathbb{R}$ is Riemann integrable and $(x_n)$ is a convergent sequence with $\lim_{n\rightarrow \infty} x_n = x$ , then $$ \lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^n x_k f\left(\frac{k}{n}\right) = x\int_0^1 f(x)dx. $$ Returning to our problem, quite clearly all terms of the sequence are positive, and since the sequence is decreasing, by MCT the sequence has a real limit $x$ . Passing to the limit in the recurrence relation, we get that $$ x=\lim_{n\rightarrow\infty} \frac{1}{n}\sum_{k=1}^n x_k \frac{1}{1+\frac{k}{n}} =x\int_0^1 \frac{dx}{1+x} = x\ln 2, $$ finally implying that $x=0$ . | [
"<blockquote>\n[list=a]\n[*] $x_{n+1}^2<y_n/2$ \n[/list]</blockquote>\n\nThis fails even for $n=2$ . Check your definition of $y_n$ ...",
"I am very sorry, I made a typo. It's supposed to say $y_n=(x_1^2+x_2^2+\\cdots+x_n^2)/n.$ Fixing it.",
"You can do b without using the auxiliary sequence $(y_n)_n$ .\n... | [
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Let $A\in\mathcal{M}_n(\mathbb{C})$ where $n\geq 2.$ Prove that if $m=|\{\text{rank}(A^k)-\text{rank}(A^{k+1})":k\in\mathbb{N}^*\}|$ then $n+1\geq m(m+1)/2.$ | Let $m = \{ d_1 , d_2, \dots, d_m\}$ we have $$ \frac{m(m+1)}{2}=1+2+\cdots + m \leq d_1 +d_2 +\cdots d_m $$ $$ \leq \text{rank}(A^{i_1})-\text{rank}(A^{i_1+1})+\text{rank}(A^{i_2})-\text{rank}(A^{i_2+1})+\cdots +\text{rank}(A^{i_m})-\text{rank}(A^{i_m+1}) $$ $$ \leq \text{rank}(A^{i_1})-\text{rank}(A^{i_m+1})\leq n $$ That is assuming $i_1\leq i_2 \leq \cdots \leq i_n$ correspond to $d_1, d_2, \dots , d_m$ in some order and noting the fact that for each $j$ , $\text{rank}(A^{i_j+1})\geq \text{rank}(A^{i_{j+1}})$ .
It remains to show the equality case can not hold. But the equality of the last inequality holds when $\text{rank}(A^{i_1})=n$ and $\text{rank}(A^{i_m+1})=0$ that requires $A$ to be singular and full rank which is not possible. The conclusion follows. | [] | [
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We call a set of $6$ points in the plane *splittable* if we if can denote its elements by $A,B,C,D,E$ and $F$ in such a way that $\triangle ABC$ and $\triangle DEF$ have the same centroid.
[list=a]
[*]Construct a splittable set.
[*]Show that any set of $7$ points has a subset of $6$ points which is *not* splittable.
[/list] | a). Let $ABC$ and $DEF$ are equilateral triangles with the same circumcircle. Then the center is the centroid.
b). Let $O$ be the origin and $\vec{v}_{P} $ the position vector of the point $P$ . Now if $ABC$ and $DEF$ have the same centroid $G$ then $$ \vec{v}_{A}+\vec{v}_{B}+\vec{v}_{C}=3\vec{v}_{G}=\vec{v}_{D}+\vec{v}_{E}+\vec{v}_{F}. $$ For the set $\{A,B,C,D,E,F\}$ call this property a split.
Let $\mathcal{P}$ be set of $n$ points in the plane with $n\geq 7$ . Then fixing five of the points, we cannot have two distinct points with the same split. Thus say we have $\{P,Q,R,S,T,U\}$ and $\{P,Q,R,S,T,V\}$ then if we split the first set into $PQR$ and $STU$ , the second set can be split only into $PQS$ and $RTU$ or $PQU$ and $RST$ . After some algebra for the first split of the second set, we get $\vec{UV}=2\vec{RS}$ and for the second split, we get $\vec{UV}=2\vec{UR}$ .
Consider $\mathcal{S} \subseteq \mathcal{P}$ such that $|\mathcal{S}|=7$ for which we have a split. Now consider $X, Y\in\mathcal{S}$ such that the distance $XY$ is the smallest. However, applying the split with the remaining members of $\mathcal{S}$ with $X$ and $Y$ separately (not both in the same splitting set) we get a contradiction to the minimality of $XY$ . | [] | [
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} |
Determine all $x\in(0,3/4)$ which satisfy \[\log_x(1-x)+\log_2\frac{1-x}{x}=\frac{1}{(\log_2x)^2}.\] | I claim $x=1/2$ is the only possibility.
We have
\begin{align*}
&\frac{\log_2(1-x)}{\log_2 x} + \log_2(1-x) = \log_2 x + \frac{1}{(\log_2 x)^2}
&\iff \log_2(1-x)\left(\frac{1+\log_2 x}{\log_2 x}\right) = \frac{(\log_2 x+1)((\log_2 x)^2-\log_2 x+1)}{(\log_2 x)^2}.
\end{align*}
If $\log_2 x+1=0$ , that is if $x=1/2$ , we enjoy equality. Assuming this is not the case, we perform cancellations and get
\[
\log_2(1-x) = \frac{(\log_2 x)^2 -\log_2 x+1}{\log_2 x}.
\]
Now, set $x=2^t$ . We then get
\[
\log_2 (1-x) =\frac{t^2-t+1}{t} \implies 1-x = 2^{\frac{t^2-t+1}{t}} \iff f(t)\triangleq 2^t + 2^{\frac{t^2-t+1}{t}}=1.
\]
It is readily verified $t\mapsto (t^2-t+1)/t$ is increasing on $(-\infty,-1]$ and decreasing on $(-1,0)$ . Moreover, $f(-1)<1$ , it follows that such a $t$ , if exists, is in $(-1,0)$ . Notice, however, that as $x<3/4$ , we must have $2^{(t^2-t+1)/t}>1/4\iff (t^2-t+1)/t > -2$ . Notice, however, that setting $t=-s$ for $s>0$ and recalling $s+1/s\ge 2$ , we have
\[
\frac{t^2-t+1}{t} = t+\frac1t -1 = -1 - (s+1/s)\le -3,
\]
impossible. Hence, $x=1/2$ is the only solution. | [] | [
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} |
Let $z_1,z_2$ and $z_3$ be complex numbers of modulus $1,$ such that $|z_i-z_j|\geq\sqrt{2}$ for all $i\neq j\in\{1,2,3\}.$ Prove that \[|z_1+z_2|+|z_2+z_3|+|z_3+z_2|\leq 3.\]*Mathematical Gazette* | Joy Ma Tara! :bruce: Let's look at the geometric interpretation of the problem:
<details><summary>New problem</summary>Let $ABC$ be a non obtuse triangle such that all the interior angles are atleast $45^o$ each. Let $D,E,F$ be the midpoints of sides $BC,CA,AB$ respectively. If $O$ is the circumcenter prove that $2OD+2OE+2OF \leq 3$ .</details>
Observe that $2OD=|z_1+z_2|$ , $2OE=|z_3+z_2|$ , $2OF=|z_1+z_3|$ by parallelogram law of vector addition. Also $OD+OE+OF = \cos A+\cos B+\cos C \stackrel{Jensen}{\leq}3\cos(\frac{A+B+C}{3}) = \frac{3}{2}$ as desired. | [
"This is simple geo. Just use Jensen after introducing antipodes of these points.",
"<blockquote>This is simple geo. Just use Jensen after introducing antipodes of these points.</blockquote>\nExactly; but before doing so, prove that the triangle determined by $z_1,z_2$ and $z_3$ is either acute or right, so J... | [
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} |
A positive integer $n\geq 4$ is called *interesting* if there exists a complex number $z$ such that $|z|=1$ and \[1+z+z^2+z^{n-1}+z^n=0.\] Find how many interesting numbers are smaller than $2022.$ | <details><summary>storage</summary><span style="color:#f00">**Claim:-**</span>all $n$ of the form $5\rho-1$ , for some $\rho \in \mathbb{Z^{+}}$ works
<span style="color:#600">pf:-</span>since $|z|=1\implies z\cdot \overline{z}=1$
now taking conjugate of given equation we arise at $z^{n}+z^{n-1}+z^{n-2}+z+1=0$ subtract this with original equation to get: $z^{n-2}=z^2\implies z^{n-4}=1\implies z^{n-1}=z^3, z^{n}=z^{4}$ so plugging these back in original equation , equation re-transforms to $1+z+z^3+z^{4}+z^{5}$ , clearly we can observe $z\neq 1$ so we have $z^{5}=1\implies z$ is $5$ th root of unity
now this gives $n\equiv -1 \pmod 5$ hence all $n$ of form $5\rho-1$ works , for $4\le n\le 2022$ as claim follows $\blacksquare$ such number of n's are $\boxed{404}$</details> | [
"We observe: $1$ and $-1$ are not solutions of the equation.\nLet be $z_0=\\cos\\alpha+i\\sin\\alpha$ a solution of the equation, where $\\alpha\\in[0,2\\pi)$ .\nUsing $(\\cos\\alpha+i\\sin\\alpha)^k=\\cos (k\\alpha)+i\\sin (k\\alpha),\\;\\forall k\\in\\mathbb{N}$ , the equation becomes: $1+\\cos\\alpha+\\c... | [
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} |
Find all functions $f:\mathbb{R}\to\mathbb{R}$ which satisfy the following relationship for all $x,y\in\mathbb{R}:$ \[f(f(y-x)-xf(y))+f(x)=y\cdot(1-f(x)).\] | Let $P(x,y)$ denote the given assertion.
Claim: $f$ is surjective.
Proof: We note that \[f(f(y-x)-xf(y))=y(1-f(x))-f(x)\]
Noting that $f\equiv 1$ is not a solution, set $x$ such that $f(x)\ne1$ . Varying $y$ gives the desired result. $\blacksquare$ Claim: $f$ is injective.
Proof: $P(0,x): f(f(x))+f(0)=x(1-f(0))$ . If $f(a)=f(b)$ , then $a=b$ or $f(0)=1$ .
We will show $f(0)\ne 1$ . Suppose FTSOC $f(0)=1$ . Then $f(f(x))=-1$ for any $x$ . Since $f$ is surjective, $f\equiv -1$ , which is not a solution.
Thus, $a=b$ . $\blacksquare$ $P(x,-1): f(f(-x-1)-xf(-1))=1$ , so $f(-x-1)-xf(-1)=c$ for some constant $c$ . Set $x\to -x-1$ .
So $f(x)-(-x-1)f(-1)=c$ , which implies $f$ is linear.
Let $f(x)=ax+b$ . Note that $f(f(x))+b=x-xb\implies f(f(x))=x(1-b)-b$ . $f(f(x))=f(ax+b)=a^2x+(a+1)b$ . So $(a+1)b=-b$ .
Case 1: $a+1=-1$ .
Then $f(f(x))=4x-b$ . So $b=-3$ . Thus, $f(x)=-2x-3$ , which is not a solution.
Case 2: $b=0$ .
Then $f(x)=ax$ for some $a$ . But $f(f(x))=x\implies \boxed{f(x)=x}$ or $f(x)=-x$ . The former works but the latter doesn't. | [
"REDACTED",
"<blockquote>Denote $P(x,y)$ as usual be the assertion $P(0,x)\\leftrightarrow f(f(x))+f(0)=x(1-f(0)) \\leftrightarrow f(f(x)) = x(1-f(0)) - f(0)$ From here we get that $f(x)$ is bijective</blockquote>Not yet. What if $f(0)=1?$ ",
"Case 1: $f(0)\\neq 1$ $P(0,y)\\implies f(f(y)) = (1-f(0))y - ... | [
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"boxed": false,
"end_of_proof": false,
"n_reply": 13,
"path": "Contest Collections/2022 Contests/2022 District Olympiad/2811501.json"
} |
Charlotte is playing the hit new web number game, Primle. In this game, the objective is to guess a two-digit positive prime integer between $10$ and $99$ , called the *Primle*. For each guess, a digit is highlighted blue if it is in the *Primle*, but not in the correct place. A digit is highlighted orange if it is in the *Primle* and is in the correct place. Finally, a digit is left unhighlighted if it is not in the *Primle*. If Charlotte guesses $13$ and $47$ and is left with the following game board, what is the *Primle*? $$ \begin{array}{c}
\boxed{1} \,\, \boxed{3} [\smallskipamount]
\boxed{4}\,\, \fcolorbox{black}{blue}{\color{white}7}
\end{array} $$ *Proposed by Andrew Wu and Jason Wang* | <details><summary>Solution</summary>Because we are working with a two digit number and the digit $7$ is not the units digit, then it follows that the *Primle* has a ten's digit of $7$ . The prime numbers in the $70's$ are $71, 73,$ and $79$ . Because when Charlotte put the prime $13$ none of the digits were blue, $71$ and $73$ don't work. Therefore, the *Primle* is $\boxed{79}.$</details> | [
"<details><summary>Solution</summary>The blue $7$ tells us that the number is a prime with $7$ in the tens place. The uncolored numbers tell us that the units digit is not $1, 3$ , or $4$ . The only working prime number that satisfies these conditions can easily be checked as $\\fbox{79}$ .</details>",
"<d... | [
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"boxed": true,
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"n_reply": 6,
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} |
How many ways are there to fill in a $2\times 2$ square grid with the numbers $1,2,3,$ and $4$ such that the numbers in any two grid squares that share an edge have an absolute difference of at most $2$ ?
*Proposed by Andrew Wu* | <details><summary>Solution</summary>Note that the $1$ must be diagonally opposite the $4$ . Then, the $2$ and $3$ can be placed in any way.
Therefore, there are $4$ ways to place the $1$ , $1$ way to place the $4$ after that, and $2$ ways to place the $2$ and $3$ after that, for a total of $$ 4\cdot1\cdot2=\boxed{8}. $$</details> | [
"<details><summary>Solution</summary>By symmetry, we can place the $1$ in the top-left corner and then multiply our answer by $4$ for symmetry. This means that the $4$ must go diagonally across from the $1$ since the absolute difference is more than $2$ . Obviously, both orderings of the $2$ and $3$ wi... | [
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"n_reply": 3,
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} |
The **Collaptz function** is defined as $$ C(n) = \begin{cases} 3n - 1 & n\textrm{~odd}, \frac{n}{2} & n\textrm{~even}.\end{cases} $$
We obtain the **Collaptz sequence** of a number by repeatedly applying the Collaptz function to that number. For example, the Collaptz sequence of $13$ begins with $13, 38, 19, 56, 28, \cdots$ and so on. Find the sum of the three smallest positive integers $n$ whose Collaptz sequences do not contain $1,$ or in other words, do not **collaptzse**.
*Proposed by Andrew Wu and Jason Wang* | <details><summary>Solution</summary>Hoping that the three smallest positive integers that do not collaptzse are reasonably small, we use brute force:**1:** Obviously, the Collaptz sequence of $1$ contains a $1,$ so $1$ collaptzses**2:** The first two terms of the Collaptz sequence of $2$ are $2,1$ so $2$ also collaptzses.**3:** The first five terms of the Collaptz sequence of $3$ are $3,8,4,2,1$ so $3$ also collaptzses.**4:** The first three terms of the Collaptz sequeuce of $4$ are $4,2,1$ so $4$ also collaptzses**5:** The first six terms of the Collaptz sequence of $5$ are $5,14,7,20,10,5$ so it repeats with period $5$ without having a $1$ in the first $5$ terms. Hence, $5$ does not collaptzse.**6:** The first two terms of the Collaptz sequence of $6$ are $6,3.$ Since $3$ collaptzses, $6$ must also collaptzse.**7:** The first four terms of the Collaptz sequeuce of $7$ are $7,20,10,5.$ Since $5$ does not collaptzse, $7$ also does not collaptzse.**8:** The first two terms of the Collaptz sequeunce of $8$ are $8,4.$ Since $4$ collaptzses, $8$ must also collaptzse.**9:** The first nine terms of the Collaptz sequence of $9$ are $9,26,13,38,19,56,28,14,7.$ Since $7$ does not collaptzse, $9$ also does not collaptzse. We can stop here since we have already found $3$ numbers which do not collaptzse.
Hence, the three smallest positive integers $n$ that do not collaptzse are $5,7,9$ which have sum $$ 5+7+9=\boxed{21}. $$</details> | [
"<details><summary>Solution</summary>We use brute force with some quick observations. Obviously $1$ to $4$ all do not work by testing. $5$ works because its sequence is $5, 14, 7, 20, 10, 5, ...$ . $6$ doesn't work because it leads into the sequence for $3$ . $7$ works because its sequence leads into th... | [
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"boxed": true,
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} |
Kara rolls a six-sided die, and if on that first roll she rolls an $n$ , she rolls the die $n-1$ more times. She then computes that the product of all her rolls, including the first, is $8$ . How many distinct sequences of rolls could Kara have rolled?
*Proposed by Andrew Wu* | <details><summary>Solution</summary>Since the product of all the rolls is $8,$ the first roll must be a factor of $8$ which can be $1,2,$ or $4$ ( $8$ is out of range).
Obviously, $1$ wouldn't work because then she would have $0$ more rolls, so the product couldn't be $8$ .
If she rolls a $2,$ she has $1$ more roll, so her second roll must be a $4$ .
If she rolls a $4,$ the product of her next $3$ rolls must be $2.$ Obviously, they must be some permutation of $(1,1,2)$ of which there are $3$ of them.
Hence, the total number of distinct sequences of rolls Kara could have rolled is $$ 1+3=\boxed{4}. $$</details> | [
"<details><summary>sol</summary>We know it must start with a factor of $8$ , so it has to be $1,2,4$ . $1$ clearly doesn't work because we roll the die $0$ times. For $2$ , we have 1 roll so that has to be $4$ . If she rolls a $4$ , she has 3 rolls to get a product of $2$ . This has to be $2,1,1$ in som... | [
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"answer_score": 1036,
"boxed": true,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Girls in Math at Yale/2790586.json"
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Cat and Claire are having a conversation about Cat's favorite number.
Cat says, "My favorite number is a two-digit positive integer with distinct nonzero digits, $\overline{AB}$ , such that $A$ and $B$ are both factors of $\overline{AB}$ ."
Claire says, "I don't know your favorite number yet, but I do know that among four of the numbers that might be your favorite number, you could start with any one of them, add a second, subtract a third, and get the fourth!"
Cat says, "That's cool, and my favorite number is among those four numbers! Also, the square of my number is the product of two of the other numbers among the four you mentioned!"
Claire says, "Now I know your favorite number!"
What is Cat's favorite number?
*Proposed by Andrew Wu* | <details><summary>Solution</summary>All two-digit positive integers with distinct nonzero digits $\overline{AB}$ , such that $A$ and $B$ are both factors of $\overline{AB}$ can be listed pretty easily: $$ 12,15,24,36,48. $$ (For $A\geq5,$ for $A$ and $B$ to both divide $\overline{AB},$ we must have $A=B,$ contradiction.)
Then, note that $12+48-36=24$ or $12+48-24-36,$ so the set of four numbers mentioned is $\{12,24,36,48\}.$
Finally, one can easily find (by guess and check) that $$ 24^2 = 12 \cdot 48, $$ so Cat's favorite number is $\boxed{24}.$</details>
| [
"<details><summary>Solution</summary>We must have $A|10A+B$ , so $A|B$ . From here, we can see that the only possibilities are $12, 15, 24, 36, 48$ . We have $12 + 48 - 36 = 24$ , so our four numbers are $12, 48, 36, 24$ . We have $12 \\cdot 48 = 24^2$ , so the answer is $\\boxed{24}$ .</details>\n",
"<det... | [
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"boxed": true,
"end_of_proof": false,
"n_reply": 4,
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} |
Carissa is crossing a very, very, very wide street, and did not properly check both ways before doing so. (Don't be like Carissa!) She initially begins walking at $2$ feet per second. Suddenly, she hears a car approaching, and begins running, eventually making it safely to the other side, half a minute after she began crossing. Given that Carissa always runs $n$ times as fast as she walks and that she spent $n$ times as much time running as she did walking, and given that the street is $260$ feet wide, find Carissa's running speed, in feet per second.
*Proposed by Andrew Wu* | <details><summary>Sol</summary>Let $a$ be the number of seconds she spent walking. Then $a+an=30$ and $2a+2an^2=260$ or $a+an^2=130$ . From the first equation, we can divide by $n+1$ to get $a=\frac{30}{n+1}$ . We can substitute to get $\frac{30}{n+1}+\frac{30n^2}{n+1}=130$ . Multiplying by $\frac{n+1}{10}$ , we get $3n^2+3=13n+13$ . This becomes $3n^2-13n-10=0$ and factors to $(n-5)(3n+2)$ , so $n=5$ . We get $a=5$ , so her running speed is $2n =\boxed{10}$ .</details> | [
"<details><summary>Solution</summary>Let's say Carissa walks at $2$ ft/sec for $x$ seconds. She then travels a distance of $2x$ feet, and after that travels at $2n$ ft/sec for $nx$ seconds. She thus travels a total distance of $2x+2n^2x = 260$ , so $x+n^2x = 130$ . We also have $x + nx = 30$ . Dividing... | [
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"answer_score": 1028,
"boxed": true,
"end_of_proof": false,
"n_reply": 5,
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} |
Given that six-digit positive integer $\overline{ABCDEF}$ has distinct digits $A,$ $B,$ $C,$ $D,$ $E,$ $F$ between $1$ and $8$ , inclusive, and that it is divisible by $99$ , find the maximum possible value of $\overline{ABCDEF}$ .
*Proposed by Andrew Milas* | <details><summary>Sol</summary>It will always be divisible by $9$ because the sum of the digits is $36$ . For $11$ , we need alternate digits to have a sum that differs by a multiple of $11$ . We can try to set the first few digits in order from $8$ and find a way to place the rest of the digits to satisfy the $11$ divisibility condition. If we have $876543\_\_$ , then the sum is currently $18$ for odds and $15$ for evens, so there's no way to do this. If we have $87654\_\_\_$ , then the current sum for odd digits is $18$ and even digits have $12$ . We either need to close the gap of 6 or extend it to 11, but neither of these are possible. If we have $8765\_\_\_\_$ , then the current sum for odds is $14$ and evens is $12$ . Again, we either need to close the gap of $2$ or extend it to 11. To extend, we need a difference of $9$ but $1+2+3+4=10$ so this is not possible. To close the gap, we need even digits to have a sum 2 higher than the odd digits. The sum of the even digits will be $\frac{10+2}{2}=6$ and the odd digits will have $10-6=4$ . To maximize the number, the digit after $5$ is $3$ , because it can't be $4$ and the next digit is $4$ . Then we get $1$ and $2$ . Our answer is $\boxed{87653412}$ .</details> | [
"The problem asks for a six-digit positive integer. ",
"<details><summary>Solution</summary>Since we want to maximize $\\overline{ABCDEF}$ , I let $\\overline{ABCD}=9876$ . Since the sum of these numbers is $\\equiv 3 \\pmod 9$ , $E+F$ needs to be $\\equiv 6 \\mod 9$ . Since we have already used $9,8,7,$ ... | [
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"boxed": false,
"end_of_proof": false,
"n_reply": 8,
"path": "Contest Collections/2022 Contests/2022 Girls in Math at Yale/2790590.json"
} |
Triangle $ABC$ has sidelengths $AB=1$ , $BC=\sqrt{3}$ , and $AC=2$ . Points $D,E$ , and $F$ are chosen on $AB, BC$ , and $AC$ respectively, such that $\angle EDF = \angle DFA = 90^{\circ}$ . Given that the maximum possible value of $[DEF]^2$ can be expressed as $\frac{a}{b}$ for positive integers $a, b$ with $\gcd (a, b) = 1$ , find $a + b$ . (Here $[DEF]$ denotes the area of triangle $DEF$ .)
*Proposed by Vismay Sharan* |
Let $BD=x,ED=2x,AD=(1-x),DF=\frac{\sqrt{3}(1-x)}{2}$ The area will be noted by $\frac{\sqrt{3}}{2}x(1-x)$ , which is maximum value is got when $x=\frac{1}{2}$ , this time the area is $\frac{\sqrt{3}}{8}$ , the desired answer is $\frac{3}{64}$ leads to $\boxed{67}$ | [
"Diagram!\n\n(points not necessarily optimally chosen to satisfy problem statement.)\n\n[asy]\nunitsize(1.5inch);\npair A = dir(180);\npair B = dir(120);\npair C = dir(0);\ndraw(A--B--C--A);\npair D = (3A + 5B)/(3 + 5);\npair F = foot(D, A, C);\npair E = extension(D, rotate(90, D)*F, C, B); \ndraw(D--E--F--cycle, b... | [
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Suppose that $P(x)$ is a monic quadratic polynomial satisfying $aP(a) = 20P(20) = 22P(22)$ for some integer $a\neq 20, 22$ . Find the minimum possible positive value of $P(0)$ .
*Proposed by Andrew Wu*
(Note: wording changed from original to specify that $a \neq 20, 22$ .) | The conditions of the problem allow us to define
\[ XP(X) = (X-a)(X-20)(X-22) + C. \]
Since $X$ divides the RHS, then $C = 440a$ . So,
\begin{align*}
P(X) = \dfrac{1}{X} \left[(X-a)(X-20)(X-22) + 400a\right] = X^2 - (42+a)X + (440+42a).
\end{align*}
Henceforth, we want to minimize the value of $P(0) = 440+42a$ which is $\boxed{20}$ and occurs at $a = -10$ . | [
"Suppose $P(x) = x^2 + dx + e$ , then let $q(x) = xP(x)$ be a monic cubic of the form $x^3 + dx^2 + ex$ , then it is given that $q(20) = q(22) = q(a) = C$ for some $C$ . Translate down by $C$ , then we get that $q(x) - C$ has roots $20, 22, a$ meaning that $q(x) - C = (x-20)(x-22)(x-a)$ . \n\nThis beco... | [
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How many ways are there to choose distinct positive integers $a, b, c, d$ dividing $15^6$ such that none of $a, b, c,$ or $d$ divide each other? (Order does not matter.)
*Proposed by Miles Yamner and Andrew Wu*
(Note: wording changed from original to clarify) | We have $15^6 = 3^65^6$ . Let $a = 3^{x_1}5^{y_1}$ , $b = 3^{x_2}5^{y_2}$ , $c = 3^{x_3}5^{y_3}$ , $d = 3^{x_4}5^{y_4}$ .
Note that if any two $x_i$ or $y_i$ are equal, then the condition is violated (one number must divide the other). So they are all distinct, and the problem is equivalent to choosing $\{x_1, x_2, x_3, x_4, y_1, y_2, y_3, y_4\}$ , each nonnegative integers at most $6$ , such that that for all $i,j$ if $x_i < x_j$ then $y_i > y_j$ .
The number of ways to choose our variables so all $x_i$ and $y_i$ are distinct is just $(\binom{7}{4})^2 = 35^2$ . Given this set, since order does not matter, then exchange variables and assume WLOG that $x_1 < x_2 < x_3 < x_4$ . This uniquely forces $y_1 > y_2 > y_3 > y_4$ , and there is exactly $1$ way to rearrange the $y$ exponents in forming $a, b, c, d$ to satisfy this.
In total, the answer is just $35^2 * 1 = 1225$ . | [
"<blockquote>We have $15^6 = 3^65^6$ . Let $a = 3^{x_1}5^{y_1}$ , $b = 3^{x_2}5^{y_2}$ , $c = 3^{x_3}5^{y_3}$ , $d = 3^{x_4}5^{y_4}$ .\nIn total, the answer is just $35^2 * 1 = 1225$ .</blockquote>\n\nNot quite right. Keep in mind that ***<u>none of $a, b, c,$ or $d$ divide each other.</u>***\nThe follo... | [
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Georgina calls a $992$ -element subset $A$ of the set $S = \{1, 2, 3, \ldots , 1984\}$ a **halfthink set** if
- the sum of the elements in $A$ is equal to half of the sum of the elements in $S$ , and
- exactly one pair of elements in $A$ differs by $1$ .
She notices that for some values of $n$ , with $n$ a positive integer between $1$ and $1983$ , inclusive, there are no halfthink sets containing both $n$ and $n+1$ . Find the last three digits of the product of all possible values of $n$ .
*Proposed by Andrew Wu and Jason Wang*
(Note: wording changed from original to specify what $n$ can be.) | <blockquote>Georgina calls a $992$ -element subset $A$ of the set $S = \{1, 2, 3, \ldots , 1984\}$ a **halfthink set** if
- the sum of the elements in $A$ is equal to half of the sum of the elements in $S$ , and
- exactly one pair of elements in $A$ differs by $1$ .
She notices that for some values of $n$ , with $n$ a positive integer between $1$ and $1983$ , inclusive, there are no halfthink sets containing both $n$ and $n+1$ . Find the last three digits of the product of all possible values of $n$ .
*Proposed by Andrew Wu and Jason Wang*
(Note: wording changed from original to specify what $n$ can be.)</blockquote>
<details><summary>Click to expand</summary>All numbers except $989, 991, 993, 995$ have halfthink sets.
Compare with a halfthink set with no consecutive terms: $$ \{1, 3, 5, ..., 989, 991, 994, 996, ..., 1984\} $$ We can get four cases based on if $n$ is even or odd and if $n$ is $992$ or below, or greater than $992$ .
If $n$ is even and $n \leq 992$ , take our no consecutive term set, replace $n-1$ with $n$ , then replace $994$ with $993$ . All good. This even works for $n=992$ .
If $n$ is even and $n > 992$ , take our no consecutive term set, replace $n+2$ with $n+1$ , then replace $991$ with $992$ . All good.
If $n$ is odd and $n < 992$ , take our no consecutive term set, replace $n+2$ with $n+1$ , then replace $991$ with $992$ . However, these two replacements interfere when $n = 989, 991$ , we'll have to deal with these later. The rest are all good.
If $n$ is odd and $n > 992$ , take our no consecutive term set, replace $n-1$ with $n$ , then replace $994$ with $993$ . However, these two replacements interfere when $n = 993, 995$ , we'll have to deal with these later. The rest are all good.
Now, take a look at $n = 989$ . Then we must have $989$ and $990$ in our sequence. To get the maximum value to the left of $989$ , we must take $1, 3, 5, ..., 987$ . To get the maximum value to the right, we must take $992, 994, ..., 1984$ . However, there is one extra value, so we must remove one, and to satisfy the sum constraint, we must remove the value $991$ , which doesn't exist in our sequence.
We can use the same type of reasoning for $989, 991, 993, 995$ , which all don't work. Therefore our answer is $$ (989)(991)(993)(995) \equiv (11)(9)(7)(5) \equiv 465 \pmod {1000}. $$</details> | [
"I think this works? If yes, very funny problem. (Though highly doubting this does).\n\n<details><summary>Sol</summary>Consider $n=1000$ . Notice that if we have all the evens or odds below $1000$ , thats $499$ numbers each (we don't include 999 because only one pair of elements differ by 1), and all the evens ... | [
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Let $ABC$ be a triangle with $AB = 5$ , $BC = 7$ , and $CA = 8$ , and let $D$ be a point on arc $\widehat{BC}$ of its circumcircle $\Omega$ . Suppose that the angle bisectors of $\angle ADB$ and $\angle ADC$ meet $AB$ and $AC$ at $E$ and $F$ , respectively, and that $EF$ and $BC$ meet at $G$ . Line $GD$ meets $\Omega$ at $T$ . If the maximum possible value of $AT^2$ can be expressed as $\frac{a}{b}$ for positive integers $a, b$ with $\gcd (a,b) = 1$ , find $a + b$ .
*Proposed by Andrew Wu* | With **jeteagle.**
The main claim is that $T$ is fixed as the midpoint of arc $\widehat{BC}$ or equivalently $AT$ is the angle bisector of $\angle{BAC}$ . To prove this, we will show that $EF, TD, BC$ are concurrent at $G$ where $T$ is the midpoint of arc $\widehat{BC}$ and $D$ is a variable point on $\widehat{BC}$ .
Let $M$ and $N$ be the midpoints of arcs $\widehat{AB}$ and $\widehat{AC}$ respectively. Let the incenter of $\triangle{ABC}$ be $I$ . By Pascal's Theorem on $ABNDMC$ we see that $E$ , $I$ , $F$ are collinear. If $TD$ and $BC$ intersect at $G$ , then Pascal's Theorem on $TDMCBA$ yields that $G$ , $E$ , and $I$ are collinear. Since $E$ , $I$ , $F$ are collinear, it follows that $G$ lies on line $EF$ proving the claim.
Now for the computational part. Noting that $\angle{BAC} = 60^\circ$ , we have $\angle{BTC} = 120^\circ.$ Since $BC = 7$ and $BT = CT$ , we find that $BT = \frac{7}{\sqrt{3}}$ . By Ptolemy's Theorem, we have $5 \cdot BT + 8 \cdot BT = AT \cdot BC$ which yields $AT = \frac{13}{\sqrt{3}}$ . Thus, $AT^2 = \frac{169}{3}$ so the answer is $169+3 = \boxed{172}.$ | [
"I am being forced to learn asymptote for the first time :P so I decided I might as well practice on my own problems haha\n\n<details><summary>diagram (possible spoilers!)</summary>[asy]\nunitsize(1.5inch);\npair B = dir(210);\npair C = dir(330);\npair A = dir(134.5736);\nfilldraw(circumcircle(A,B,C), opacity(0.3)+... | [
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**p1** How many two-digit positive integers with distinct digits satisfy the conditions that
1) neither digit is $0$ , and
2) the units digit is a multiple of the tens digit?**p2** Mirabel has $47$ candies to pass out to a class with $n$ students, where $10\le n < 20$ . After distributing the candy as evenly as possible, she has some candies left over. Find the smallest integer $k$ such that Mirabel could have had $k$ leftover candies.**p3** Callie picks two distinct numbers from $\{1, 2, 3, 4, 5\}$ at random. The probability that the sum of the numbers she picked is greater than the sum of the numbers she didn’t pick is $p$ . $p$ can be expressed as $\frac{a}{b}$ for positive integers $a, b$ with $gcd (a, b) = 1$ . Find $a + b$ . | <details><summary>A2 (with sol)</summary>Essentially it means we divide $47-k$ candies to $n$ students evenly, or in other words the largest number less than or equal to $47$ which has a number between $10$ and $19$ inclusive as a factor.
We can try each $k$ starting from $0$ . $k=0$ doesn't work because $47$ is a prime $k=1$ doesn't work because the only factors are $1, 2, 23, 46$ , none of which are between $9$ and $20$ . $k=2$ works because $45$ candies can be divided to $15$ students (each student getting 3 candies), thus $\boxed{2}$ is our answer.</details> | [
"<details><summary>A1 (with sol)</summary>We can tell that there's not many integers (and it's the first problem) meaning we can directly do casework: **(i)** When the tens digit is 1, the units digit can be any integer from $2$ to $9$ , which is <u>$8$</u> numbers. **(ii)** When the tens digit is 2, the units d... | [
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**p4** Define the sequence ${a_n}$ as follows:
1) $a_1 = -1$ , and
2) for all $n \ge 2$ , $a_n = 1 + 2 + . . . + n - (n + 1)$ .
For example, $a_3 = 1+2+3-4 = 2$ . Find the largest possible value of $k$ such that $a_k+a_{k+1} = a_{k+2}$ .**p5** The taxicab distance between two points $(a, b)$ and $(c, d)$ on the coordinate plane is $|c-a|+|d-b|$ . Given that the taxicab distance between points $A$ and $B$ is $8$ and that the length of $AB$ is $k$ , find the minimum possible value of $k^2$ .**p6** For any two-digit positive integer $\overline{AB}$ , let $f(\overline{AB}) = \overline{AB}-A\cdot B$ , or in other words, the result of subtracting the product of its digits from the integer itself. For example, $f(\overline{72}) = 72-7\cdot 2 = 58$ . Find the maximum possible $n$ such that there exist distinct two-digit integers $ \overline{XY}$ and $\overline{WZ}$ such that $f(\overline{XY} ) = f(\overline{WZ}) = n$ . | <details><summary>A6 (with sol)</summary>$f(\overline{AB}) = 10A+B-A\cdot B = (A-1)(10-B) + 10$ , so essentially, we want $(A-1)(10-B) = (C-1)(10-D)$ where the two numbers are $\overline{AB}$ and $\overline{CD}$ If we want this to be large intuitively we should make A as large and B as small as possible . So we start from $A=9$ and $B=0$ . This doesnt work as there is no other way to form a product of 90. $91$ doesn't work because again, there is no other way to form this product. Similarly 92 doesn't work. However, 93, does - the product is 56 and we can also form it by C=8 D=2. So $n$ is $56+10 = \boxed{66}$ A=8 B=2 C=9 D=3</details> | [
"<details><summary>A4 (with sol)</summary>For all $n \\le 2$ , $a_n = \\frac{n(n+1)}{2}-(n+1) = \\frac{n^2-n-2}{2} = \\frac{(n+1)(n-2)}{2}$ . \nSo the problem wants the largest possible value of $k$ where $$ \\frac{(n+1)(n-2)}{2} + \\frac{(n+2)(n-1)}{2} = \\frac{(n+3)(n)}{2} $$ Solving this yields $\\boxed{... | [
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**p7** Cindy cuts regular hexagon $ABCDEF$ out of a sheet of paper. She folds $B$ over $AC$ , resulting in a pentagon. Then, she folds $A$ over $CF$ , resulting in a quadrilateral. The area of $ABCDEF$ is $k$ times the area of the resulting folded shape. Find $k$ .**p8** Call a sequence $\{a_n\} = a_1, a_2, a_3, . . .$ of positive integers *Fib-o’nacci* if it satisfies $a_n = a_{n-1}+a_{n-2}$ for all $n \ge 3$ . Suppose that $m$ is the largest even positive integer such that exactly one *Fib-o’nacci* sequence satisfies $a_5 = m$ , and suppose that $n$ is the largest odd positive integer such that exactly one *Fib-o’nacci* sequence satisfies $a_5 = n$ . Find $mn$ .**p9** Compute the number of ways there are to pick three non-empty subsets $A$ , $B$ , and $C$ of $\{1, 2, 3, 4, 5, 6\}$ , such that $|A| = |B| = |C|$ and the following property holds: $$ A \cap B \cap C = A \cap B = B \cap C = C \cap A. $$ | <details><summary>A8 (with sol)</summary>Note that $a_5$ represented in terms of $a_1$ and $a_2$ is $2a_1 + 3a_2$ . If there is exactly one, $m$ and $n$ can't be very large, so one quick sol is to just try each number. For example for $m$ we can start from... $8$ (only 2*1 + 3*2), which works. We don't need to try everything, we can jump to another number like.. $12$ , which also works (only 2*3 + 3*2). Let's go a bit further, to $14$ . This obvsly doesn't work- it an be $2*1 + 3*4$ or $2*4 + 3*2$ . Thus $m=12$ . Similar process finds $n=12$ , so our answer is $\boxed{109}$ .</details>
| [
"<details><summary>A7 (with sol)</summary>Very simple problem - if we fold B over AC the pentagon is $ACDEF$ , and if we fold A over CF the quadrilateral is $FEDC$ which is half of $ABCDEF$ , so $k=\\boxed{2}$ .</details>"
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**p10**Kathy has two positive real numbers, $a$ and $b$ . She mistakenly writes $$ \log (a + b) = \log (a) + \log( b), $$ but miraculously, she finds that for her combination of $a$ and $b$ , the equality holds. If $a = 2022b$ , then $b = \frac{p}{q}$ , for positive integers $p, q$ where $gcd(p, q) = 1$ . Find $p + q$ .**p11** Points $X$ and $Y$ lie on sides $AB$ and $BC$ of triangle $ABC$ , respectively. Ray $\overrightarrow{XY}$ is extended to point $Z$ such that $A, C$ , and $Z$ are collinear, in that order. If triangle $ ABZ$ is isosceles and triangle $CYZ$ is equilateral, then the possible values of $\angle ZXB$ lie in the interval $I = (a^o, b^o)$ , such that $0 \le a, b \le 360$ and such that $a$ is as large as possible and $b$ is as small as possible. Find $a + b$ .**p12** Let $S = \{(a, b) | 0 \le a, b \le 3, a$ and $b$ are integers $\}$ . In other words, $S$ is the set of points in the coordinate plane with integer coordinates between $0$ and $3$ , inclusive. Prair selects four distinct points in $S$ , for each selected point, she draws lines with slope $1$ and slope $-1$ passing through that point. Given that each point in $S$ lies on at least one line Prair drew, how many ways could she have selected those four points? | <details><summary>answer p10</summary>$\frac{p}{q} = \frac{2023}{2022}$ . Since they are relatively prime, $4045$ .</details> | [] | [
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**p13** Let $ABCD$ be a square. Points $E$ and $F$ lie outside of $ABCD$ such that $ABE$ and $CBF$ are equilateral triangles. If $G$ is the centroid of triangle $DEF$ , then find $\angle AGC$ , in degrees.**p14**The silent reading $s(n)$ of a positive integer $n$ is the number obtained by dropping the zeros not at the end of the number. For example, $s(1070030) = 1730$ . Find the largest $n < 10000$ such that $s(n)$ divides $n$ and $n\ne s(n)$ .**p15** Let $ABCDEFGH$ be a regular octagon with side length $12$ . There exists a region $R$ inside the octagon such that for each point $X$ in $R$ , exactly three of the rays $AX$ , $BX$ , $CX$ , $DX$ , $GX$ , and $HX$ intersect segment $EF$ . If the area of region $R$ can be expressed as $a -b\sqrt{c}$ for positive integers $a, b, c$ with $c$ squarefree, find $a + b + c$ . | <details><summary>13</summary>the answer should be $120^{\circ}$ Since $BE=BF;DE=DF$ and $\triangle{DEF}$ is equilateral triangle, it is clear that the centroid of the triangle lies on $BD$ $\angle{DAE}=150^{\circ},\angle{DAG}=75^{\circ}$ , it îs clear that $\triangle{AGD}\cong \triangle{CGD}$ , $\angle{AGC}=2\angle{AGD}=2*60^{\circ}=120^{\circ}$</details> | [] | [
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**p16** Madelyn is being paid $\$ 50 $/hour to find useful *Non-Functional Trios*, where a Non-Functional Trio is defined as an ordered triple of distinct real numbers $ (a, b, c) $, and a Non- Functional Trio is *useful* if $ (a, b) $, $ (b, c) $, and $ (c, a) $ are collinear in the Cartesian plane. Currently, she’s working on the case $ a+b+c = 2022 $. Find the number of useful Non-Functional Trios $ (a, b, c) $ such that $ a + b + c = 2022 $.**p17** Let $ p(x) = x^2 - k $, where $ k $ is an integer strictly less than $ 250 $. Find the largest possible value of k such that there exist distinct integers $ a, b $ with $ p(a) = b $ and $ p(b) = a $.**p18** Let $ ABC $ be a triangle with orthocenter $ H $ and circumcircle $ \Gamma $ such that $ AB = 13 $, $ BC = 14 $, and $ CA = 15 $. $ BH $ and $ CH $ meet $ \Gamma $ again at points $ D $ and $ E $, respectively, and $ DE $ meets $ AB $ and $ AC $ at $ F $ and $ G $, respectively. The circumcircles of triangles $ ABG $ and $ ACF $ meet BC again at points $ P $ and $ Q $. If $ PQ $ can be expressed as $ \frac{a}{b} $ for positive integers $ a, b $ with $ gcd (a, b) = 1 $, find $ a + b$. | [] | [
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**p1.** Suppose that $x$ and $y$ are positive real numbers such that $\log_2 x = \log_x y = \log_y 256$ . Find $xy$ .**p2.** Let the roots of $x^2 + 7x + 11$ be $r$ and $s$ . If f(x) is the monic polynomial with roots $rs + r + s$ and $r^2 + s^2$ , what is $f(3)$ ?**p3.** Call a positive three digit integer $\overline{ABC}$ fancy if $\overline{ABC} = (\overline{AB})^2 - 11 \cdot \overline{C}$ . Find the sum of all fancy integers.**p4.** In triangle $ABC$ , points $D$ and $E$ are on line segments $BC$ and $AC$ , respectively, such that $AD$ and $BE$ intersect at $H$ . Suppose that $AC = 12$ , $BC = 30$ , and $EC = 6$ . Triangle $BEC$ has area $45$ and triangle $ADC$ has area $72$ , and lines $CH$ and $AB$ meet at $F$ . If $BF^2$ can be expressed as $\frac{a-b\sqrt{c}}{d}$ for positive integers $a$ , $b$ , $c$ , $d$ with $c$ squarefree and $gcd(a, b, d) = 1$ , then find $a + b + c + d$ .**p5.** Find the minimum possible integer $y$ such that $y > 100$ and there exists a positive integer $x$ such that $x^2 + 18x + y$ is a perfect fourth power.**p6.** Let $ABCD$ be a quadrilateral such that $AB = 2$ , $CD = 4$ , $BC = AD$ , and $\angle ADC + \angle BCD = 120^o$ . If the sum of the maximum and minimum possible areas of quadrilateral $ABCD$ can be expressed as $a\sqrt{b}$ for positive integers $a, b$ with $b$ squarefree, then find $a + b$ .
PS. You had better use hide for answers. Collected [here](https://artofproblemsolving.com/community/c5h2760506p24143309). | 3.I will assume $a\not=0$ . Rewriting $100a+10b+c=(10a+b)^2-11c$ gives $12c+25=100(a^2-a)+(b-5)^2+20ab$ .
Then $100(a^2-a)\le 108+25=133\implies a^2-a\le 1\implies a=1\implies 12c+25=(b+5)^2\implies 12c=b(b+10)$ .
Then $b$ is even and RHS is divisible by 3, which gives $100,122,168$ .
5. Note that $x=6,y=112$ gives a solution. Now suppose that $x^2+18x+y=k^4$ with $k>0$ and $y<112$ . Then $t:=y-81=k^4-(x+9)^2=(k^2+x+9)(k^2-x-9)$ and $t<31$ . Write $a,b$ for the factors, then $t=ab, a,b>0$ and $a-b=2(x+9)\ge 20\implies a\ge b+20 \implies 31>t\ge b(b+20)\implies b=1 \implies k^2=x+10\implies t=2k^2-1\implies 2k^2-1<31\implies k\le 3$ . But then $k^2=x+10$ is impossible. Hence 112 is minimal. | [
"P2:\n<details><summary>Click to expand</summary>$r + s = -7, rs = 11$ $f(x) = x^{2} - (rs + r + s)x + (r^{2} + s{2})$ $= x^{2} - (11 - 7)x + (r + s)^{2} - 2rs$ $ = x^{2} - 4x + 49 - 23$ $ = x^{2} - 4x + 27$ $f(3) = 9 - 12 + 27 = 24$</details>",
"P1:\n<details><summary>Click to expand</summary>Changing base... | [
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"path": "Contest Collections/2022 Contests/2022 Girls in Math at Yale/3171127.json"
} |
**p1.** Find the smallest positive integer $N$ such that $2N -1$ and $2N +1$ are both composite.**p2.** Compute the number of ordered pairs of integers $(a, b)$ with $1 \le a, b \le 5$ such that $ab - a - b$ is prime.**p3.** Given a semicircle $\Omega$ with diameter $AB$ , point $C$ is chosen on $\Omega$ such that $\angle CAB = 60^o$ . Point $D$ lies on ray $BA$ such that $DC$ is tangent to $\Omega$ . Find $\left(\frac{BD}{BC} \right)^2$ .**p4.** Let the roots of $x^2 + 7x + 11$ be $r$ and $s$ . If $f(x)$ is the monic polynomial with roots $rs + r + s$ and $r^2 + s^2$ , what is $f(3)$ ?**p5.** Regular hexagon $ABCDEF$ has side length $3$ . Circle $\omega$ is drawn with $AC$ as its diameter. $BC$ is extended to intersect $\omega$ at point $G$ . If the area of triangle $BEG$ can be expressed as $\frac{a\sqrt{b}}{c}$ for positive integers $a, b, c$ with $b$ squarefree and $gcd(a, c) = 1$ , find $a + b + c$ .**p6.** Suppose that $x$ and $y$ are positive real numbers such that $\log_2 x = \log_x y = \log_y 256$ . Find $xy$ .**p7.** Call a positive three digit integer $\overline{ABC}$ fancy if $\overline{ABC} = (\overline{AB})^2 - 11 \cdot \overline{C}$ . Find the sum of all fancy integers.**p8.** Let $\vartriangle ABC$ be an equilateral triangle. Isosceles triangles $\vartriangle DBC$ , $\vartriangle ECA$ , and $\vartriangle FAB$ , not overlapping $\vartriangle ABC$ , are constructed such that each has area seven times the area of $\vartriangle ABC$ . Compute the ratio of the area of $\vartriangle DEF$ to the area of $\vartriangle ABC$ .**p9.** Consider the sequence of polynomials an(x) with $a_0(x) = 0$ , $a_1(x) = 1$ , and $a_n(x) = a_{n-1}(x) + xa_{n-2}(x)$ for all $n \ge 2$ . Suppose that $p_k = a_k(-1) \cdot a_k(1)$ for all nonnegative integers $k$ . Find the number of positive integers $k$ between $10$ and $50$ , inclusive, such that $p_{k-2} + p_{k-1} = p_{k+1} - p_{k+2}$ .**p10.** In triangle $ABC$ , point $D$ and $E$ are on line segments $BC$ and $AC$ , respectively, such that $AD$ and $BE$ intersect at $H$ . Suppose that $AC = 12$ , $BC = 30$ , and $EC = 6$ . Triangle BEC has area 45 and triangle $ADC$ has area $72$ , and lines CH and AB meet at F. If $BF^2$ can be expressed as $\frac{a-b\sqrt{c}}{d}$ for positive integers $a$ , $b$ , $c$ , $d$ with c squarefree and $gcd(a, b, d) = 1$ , then find $a + b + c + d$ .**p11.** Find the minimum possible integer $y$ such that $y > 100$ and there exists a positive integer x such that $x^2 + 18x + y$ is a perfect fourth power.**p12.** Let $ABCD$ be a quadrilateral such that $AB = 2$ , $CD = 4$ , $BC = AD$ , and $\angle ADC + \angle BCD = 120^o$ . If the sum of the maximum and minimum possible areas of quadrilateral $ABCD$ can be expressed as $a\sqrt{b}$ for positive integers $a$ , $b$ with $b$ squarefree, then find $a + b$ .
PS. You had better use hide for answers. Collected [here](https://artofproblemsolving.com/community/c5h2760506p24143309). | <details><summary>Solution for p1</summary>Checking $N=1, 2, 3, \ldots$ going up, we see that the answer is $$ N = \boxed{13}. $$ $\square$</details> | [
"<details><summary>p6</summary>real solutions $$ (x=4,y=16) \\Longrightarrow xy=64 $$</details>",
"P4:\n<details><summary>Click to expand</summary>$r + s = -7, rs = 11$ $rs + r + s = 11 - 7 = 4$ , $r^2 + s^2 = (r + s)^2 - 2rs = 49 - 22 = 27$ $f(x) = x^2 - 4x + 27$ $f(3) = 9 - 12 + 27 = 24$</details>",
"P3:... | [
"origin:aops",
"2022 Contests",
"2022 Girls in Math at Yale"
] | {
"answer_score": 1108,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2022 Contests/2022 Girls in Math at Yale/3195043.json"
} |
(a) Find the value of the real number $k$ , for which the polynomial $P(x)=x^3-kx+2$ has the number $2$ as a root. In addition, for the value of $k$ you will find, write this polynomial as the product of two polynomials with integer coefficients.
(b) If the positive real numbers $a,b$ satisfy the equation $$ 2a+b+\frac{4}{ab}=10, $$ find the maximum possible value of $a$ . | <blockquote>(b) If the positive real numbers $a,b$ satisfy the equation $$ 2a+b+\frac{4}{ab}=10, $$ find the maximum possible value of $a$ .</blockquote>
We have $5-a=\frac b2+\frac 2{ab}>0$ and so $a<5$ Squaring, we have $(5-a)^2=\frac{b^2}4+\frac 2a+\frac 4{a^2b^2}$ $=\frac 4a+\left(\frac b2-\frac 2{ab}\right)^2\ge \frac 4a$ And so $a(a-5)^2\ge 4$ , which is $(a-4)(a^2-6a+1)\ge 0$ , which, associated to $a<5$ , implies $\boxed{a\le 4}$ , which indeed fits (with $b=1$ )
| [
"<blockquote>(a) Find the value of the real number $k$ , for which the polynomial $P(x)=x^3-kx+2$ has the number $2$ as a root. In addition, for the value of $k$ you will find, write this polynomial as the product of two polynomials with integer coefficients.</blockquote> $2^3-2k+2=0$ $\\implies$ $\\boxe... | [
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"answer_score": 1026,
"boxed": false,
"end_of_proof": false,
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"path": "Contest Collections/2022 Contests/2022 Greece Junior Math Olympiad/2790115.json"
} |
Let $ABC$ be an isosceles triangle, and point $D$ in its interior such that $$ D \hat{B} C=30^\circ, D \hat{B}A=50^\circ, D \hat{C}B=55^\circ $$
(a) Prove that $\hat B=\hat C=80^\circ$ .
(b) Find the measure of the angle $D \hat{A} C$ . | For the part (b) we denote by $\alpha$ measure of the $\angle DAC$ . By the Law of Sines in $\triangle ABD$ and $\triangle ACD$ we have that $$ 1 = \frac{BA}{AD}\cdot \frac{DA}{AC} = \frac{\sin (70^\circ - \alpha)}{\sin 50^\circ} \cdot \frac{\sin 25^\circ}{\sin (25^\circ + \alpha)} $$ **Claim.** $\alpha = 5^\circ$ works
*Proof.* We need to show that $\sin 30^\circ \cdot \sin 50^\circ = \sin 25^\circ \cdot \sin 65^\circ$ . Which is clear, since $2\sin 25^\circ \cdot \sin 65^\circ = 2\sin 25^\circ \cos 25^\circ = \sin 50^\circ$ . $\square$ Notice that measure of $\alpha$ is define in unique way by the condition of the problem, so we conclude that answer is $\boxed{5^\circ}$ | [
"If I am not wrong, Part a is trivial because \\[\\angle ABC=\\angle DBA+\\angle DBC=80\\] and since it's isoceles we're done.",
"Pretty sure this is the intended solution:\n\nLet $E$ be the point on the extension of $BD$ such that $AB=AE$ . Thus, $AE=AC$ .\n\nThen, $\\angle ABE=\\angle AEB=50$ , and $\\a... | [
"origin:aops",
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"answer_score": 1124,
"boxed": false,
"end_of_proof": false,
"n_reply": 8,
"path": "Contest Collections/2022 Contests/2022 Greece Junior Math Olympiad/2790118.json"
} |
On the board we write a series of $n$ numbers, where $n \geq 40$ , and each one of them is equal to either $1$ or $-1$ , such that the following conditions both hold:
(i) The sum of every $40$ consecutive numbers is equal to $0$ .
(ii) The sum of every $42$ consecutive numbers is not equal to $0$ .
We denote by $S_n$ the sum of the $n$ numbers of the board. Find the maximum possible value of $S_n$ for all possible values of $n$ . | Let out sequence is $a_1, a_2, ... , a_{40}, ...$ and notice that after $a_{40}$ we must have $a_1$ again, then $a_2$ , etc. So, our sequence is repeated by the first $40$ elements. By the second condition we have that $a_i + a_{i+1} \neq 0$ . Now we have $a_1=a_2=...=a_{20}=x$ and $a_{21}=a_{22}=...=a_{40}=-x$ , where $x\in \{-1,1\}$ . One can see that this type of sequence satisfies to the condition of the problem. Write $n=40m+k$ , where $m\geqslant 1$ and $40>k\geqslant 0$ . We don't care about sum of first $m$ $40$ tuples. So our problem asked us to find the maximum of the sum of first $k$ elements of the $a_1,a_2,...,a_{40}$ . Of course, its happen when $k=20$ and $x=1$ , so the answer is $\boxed{20}$ | [] | [
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"boxed": true,
"end_of_proof": false,
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"path": "Contest Collections/2022 Contests/2022 Greece Junior Math Olympiad/2790120.json"
} |
Find all couples of non-zero integers $(x,y)$ such that, $x^2+y^2$ is a common divisor of $x^5+y$ and $y^5+x$ . | Let $\gcd(x,y)=d,x=ad,y=bd$ . Obviously, $\gcd(a,b)=1$ .
Note that $d^2\mid d^2(a^2+b^2)\implies d^2\mid x^2+y^2$ .
But $x^5+y=d(a^5d^4+b)\implies d^2\mid d(a^5d^4+b)\implies d\mid b$ Similarly, $d\mid a$ . Thus, $d\mid \gcd(a,b)=1\implies d=1\implies\gcd(x,y)=1$ . $x^2+y^2\mid (x^5+y)(x^5-y)=(x^2)^5-y^2\implies (-y^2)^5-y^2\equiv (x^2)^5-y^2\equiv 0\pmod{x^2+y^2}$ . $x^2+y^2\mid y^{10}-y^2\implies x^2+y^2\mid y^8-1$ since $\gcd(x^2+y^2,y^2)=1$ .
So, $x^2+y^2\mid y^8-1-y^3(y^5+x)=-(xy^3+1)$ .
Similarly, $x^2+y^2\mid x^8-1\implies x^2+y^2\mid x^8-1-x^3(x^5+y)=-(x^3y+1)$ . $x^2+y^2\mid (x^3y+1)-(xy^3+1)=xy(x^2-y^2)$ .
Since $\gcd(x^2+y^2,x)=\gcd(x^2+y^2,y)=1$ , $x^2+y^2\mid x^2-y^2$ .
If $x^2-y^2\neq 0$ then $|x^2-y^2|\geq |x^2+y^2|\implies \max(x^2,y^2)\geq x^2+y^2>\max(x^2,y^2)$ which is absurd.
Thus, $x^2=y^2\implies x=\pm y$ . But since $\gcd(x,y)=1$ , $x,y\in\{-1,1\}$ .
It turn out that $(x,y)=(-1,1),(1,-1),(-1,-1),(1,1)$ all works. | [
"My solution as I solved it in the competition:\nFirst imply $\\gcd(x,y)=1$ as above. Then we have $x^2+y^2 \\mid x^2+y^2+x^2(x^3y-1)\\implies x^2+y^2 \\mid x^3y-1$ since $\\gcd(x^2,x^2+y^2)=1$ Same we get $x^2+y^2 \\mid y^3x-1$ . Combining these two we get $x^2+y^2 \\mid 2$ and we get the solutions $(x,y)... | [
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"answer_score": 40,
"boxed": false,
"end_of_proof": false,
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"path": "Contest Collections/2022 Contests/2022 Greece Junior Math Olympiad/2790121.json"
} |
Let $ABC$ be a triangle such that $AB<AC<BC$ . Let $D,E$ be points on the segment $BC$ such that $BD=BA$ and $CE=CA$ . If $K$ is the circumcenter of triangle $ADE$ , $F$ is the intersection of lines $AD,KC$ and $G$ is the intersection of lines $AE,KB$ , then prove that the circumcircle of triangle $KDE$ (let it be $c_1$ ), the circle with center the point $F$ and radius $FE$ (let it be $c_2$ ) and the circle with center $G$ and radius $GD$ (let it be $c_3$ ) concur on a point which lies on the line $AK$ . | Let $I$ be a incenter of $\triangle ABC$ . Denote by $2\alpha, 2\beta, 2\gamma$ angles $\angle A, \angle B, \angle C$ of the $\triangle ABC$ . Let $\odot (ABI)$ intersect side $\overline{BC}$ in $E'$ .**Claim.** $E' \equiv E$ *Proof.* We have that $\angle KE'C = 180^\circ - \angle AE'B = 180^\circ - \angle AIB = 90^\circ - \gamma \Longrightarrow \angle E'AC = 90^\circ - \gamma \Longrightarrow CE'=CA$ . $\square$ Analogously, for point $D$ .
[asy]
size(10cm);
import olympiad;
pair A,B,C,K,F,E,D,T, G, P;
A = dir(110);
B = dir(170);
C = dir(10);
K = incenter(A,B,C);
E = intersectionpoints(B--C, circle(C, 1.53))[0];
D = intersectionpoints(B--C, circle(B, 1))[0];
G = intersectionpoint(B--K, A--E);
F = intersectionpoint(C--K, A--D);
P = 11*K-10*A;
T = intersectionpoints(A--P, circumcircle(K,D,E))[1];
draw(A--B--C--A);
draw(A--D);
draw(A--E);
draw(circumcircle(K,D,E), dashed);
draw(circumcircle(A,B,K), dotted);
draw(circumcircle(A,C,K), dotted);
draw(A--T);
draw(F--E);
draw(F--T);
draw(G--D);
draw(G--T);
draw(T--E);
draw(T--D);
draw(F--G);
draw(B--K);
draw(C--K);
draw(K--E);
draw(K--D);
label(scale(0.8)*" $A$ ", A, dir(110), filltype = Fill(white));
label(scale(0.8)*" $B$ ", B, dir(200), filltype = Fill(white));
label(scale(0.8)*" $C$ ", C, dir(-20), filltype = Fill(white));
label(scale(0.8)*" $K$ ", K, dir(60), filltype = Fill(white));
label(scale(0.8)*" $T$ ", T, dir(-70), filltype = Fill(white));
label(scale(0.8)*" $D$ ", D, dir(-40), filltype = Fill(white));
label(scale(0.8)*" $E$ ", E, dir(200), filltype = Fill(white));
label(scale(0.8)*" $G$ ", G, dir(160), filltype = Fill(white));
label(scale(0.8)*" $F$ ", F, dir(50), filltype = Fill(white));
dot(A);
dot(B);
dot(C);
dot(E);
dot(D);
dot(K);
dot(G);
dot(F);
dot(T);
[/asy]
Observe that $\angle AGK = \angle BGA + \angle BAG = \beta + (2\alpha + \gamma - 90^\circ) = \alpha = \angle KAC = \angle KDC$ , so $G$ lie on the circumcircle of the $\triangle KDE$ . By the same argument we have that $F,G \in \odot (KDE)$ .**Claim.** $I \equiv K$ *Proof.* Just notice that $\angle AIE = 180^\circ - 2\beta$ and $\angle ADE = \angle ACE + \angle CAE = 2\gamma + (2\alpha - (90^\circ - \beta)) = 90^\circ - \beta$ , so $\angle AIE = 2\angle ADE$ . Similarly, $\angle AID =2\angle AED$ and the conclusion is follows. $\square$ Set $T = \odot (KDE) \cap \overline{AK}$ . To finish the proof we just notice that $\angle GTD = 180^\circ - \angle GKD = 180^\circ - \angle GKA = \angle GKT = \angle GDT$ , so $GT=GT$ and similarly $FT=FE$ . $\blacksquare$ | [
"<blockquote>Let $I$ be a incenter of $\\triangle ABC$ . Denote by $2\\alpha, 2\\beta, 2\\gamma$ angles $\\angle A, \\angle B, \\angle C$ of the $\\triangle ABC$ . Let $\\odot (ABI)$ intersect side $\\overline{BC}$ in $E'$ .**Claim.** $E' \\equiv E$ *Proof.* We have that $\\angle KE'C = 180^\\circ - ... | [
"origin:aops",
"2022 Greece National Olympiad",
"2022 Contests"
] | {
"answer_score": 172,
"boxed": false,
"end_of_proof": true,
"n_reply": 6,
"path": "Contest Collections/2022 Contests/2022 Greece National Olympiad/2790100.json"
} |
Let $n>4$ be a positive integer, which is divisible by $4$ . We denote by $A_n$ the sum of the odd positive divisors of $n$ . We also denote $B_n$ the sum of the even positive divisors of $n$ , excluding the number $n$ itself. Find the least possible value of the expression $$ f(n)=B_n-2A_n, $$ for all possible values of $n$ , as well as for which positive integers $n$ this minimum value is attained. | I claim the answer is $4$ , and is attained iff $n=4p$ , where $p>2$ is a prime or when $n=8$ . Check that indeed when $n=4p$ , then $A_n=1+p$ whereas $B_n=2+4+2p$ , yielding $B_n-2A_n =4$ .
Let $n=2^k\cdot m$ , where $m$ is odd and $k\ge 2$ . Notice that $A_n=\textstyle\sum_{d\mid m}d$ . On the other hand, $B_n = 2A_n+4A_n+\cdots+2^k A_n - n$ . Assume first that $k=2$ . Then, $B_n-2A_n = 4A_n - n =4(A_n-m)$ . Since $A_n\ge m+1$ (as $n>4$ ), it follows $B_n-2A_n\ge 4$ . Equality holds iff $A_n = m+1$ , that is when $m$ itself is an (odd) prime.
Next, assume $k=3$ . Then, $B_n=2A_n+4A_n+8A_n-n$ , yielding $B_n-2A_n = 4A_n + 8(A_n-m)$ . As $A_n\ge 1$ and $A_n-m\ge 0$ , we find $B_n-2A_n\ge 4$ with equality iff $k=3$ and $m=1$ , that is $n=8$ .
Finally, let $k\ge 4$ . Then, $B_n = 2A_n + 4A_n + 8A_n +2^k A_n - n$ , yielding $B_n-2A_n\ge 4A_n+8A_n\ge 12$ , yielding a strictly worse value. | [] | [
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"2022 Greece National Olympiad",
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"answer_score": 64,
"boxed": false,
"end_of_proof": false,
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"path": "Contest Collections/2022 Contests/2022 Greece National Olympiad/2790101.json"
} |
The positive real numbers $a,b,c,d$ satisfy the equality $$ a+bc+cd+db+\frac{1}{ab^2c^2d^2}=18. $$ Find the maximum possible value of $a$ . | Note that we can write $a$ as $a=18-\left(bc+cd+db+\frac{1}{ab^2c^2d^2}\right)\leq18-\left(\frac{4}{\sqrt[4]{a}}\right)$
since we need to find $a_{\text{max}}$ , minimizing this part $\left(bc+cd+db+\frac{1}{ab^2c^2d^2}\right)$ will work, hence the above part which is true by AM-GM, we need to solve this $$ a+4a^{-\frac{1}{4}}-18=0 $$ which yields $a=16$ as the value. | [
"Using AM-GM on the last four terms, we obtain\n\\[\n18\\ge a+ 4a^{-\\frac14}\\iff t^4 +\\frac4t-18\\le 0 \\qquad\\text{where}\\qquad a=t^4.\n\\]\nFrom here, $t\\le 2$ and therefore $a\\le 16$ .\n\nWe now give an example attaining the equality. Note that if $b=c=d$ and $b^2 = (ab^6)^{-1}\\iff b^8 = \\frac1a$ ... | [
"origin:aops",
"2022 Greece National Olympiad",
"2022 Contests"
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"boxed": false,
"end_of_proof": false,
"n_reply": 7,
"path": "Contest Collections/2022 Contests/2022 Greece National Olympiad/2790106.json"
} |
Let $Q_n$ be the set of all $n$ -tuples $x=(x_1,\ldots,x_n)$ with $x_i \in \{0,1,2 \}$ , $i=1,2,\ldots,n$ . A triple $(x,y,z)$ (where $x=(x_1,x_2,\ldots,x_n)$ , $y=(y_1,y_2,\ldots,y_n)$ , $z=(z_1,z_2,\ldots,z_n)$ ) of distinct elements of $Q_n$ is called a *good* triple, if there exists at least one $i \in \{1,2, \ldots, n \}$ , for which $\{x_i,y_i,z_i \}=\{0,1,2 \}$ . A subset $A$ of $Q_n$ will be called a *good* subset, if any three elements of $A$ form a *good* triple. Prove that every *good* subset of $Q_n$ contains at most $2 \cdot \left(\frac{3}{2}\right)^n$ elements. | <details><summary>BRUH</summary>We proceed by induction on $n$ , with the base case, $n=1$ being clear.
For $t=0,1,2$ , let $J_t$ be the set of sequences with $x_n=t$ . We know that any three sequences in $J_t\cup J_u$ must have $x_j,y_j,z_j$ distinct, but $j\ne n$ . First observe $(x_1,\cdots,x_{n-1})$ cannot appear in $J_t$ and $J_u$ simultaneously. By inductive hypothesis, there are at most $2\cdot (\frac 32)^{n-1}$ possibilities for the first $n-1$ elements in $J_t\cup J_u$ . Since each of the possibilities can only appear in at most one of $J_t$ , $J_u$ , $|J_t|+|J_u|\le 2\cdot (\frac 32)^{n-1}$ sequences by inductive hypothesis. Thus, $2(|J_0|+|J_1|+|J_2|) \le 6\cdot (\frac 32)^{n-1}$ , completing the induction</details> | [
"<details><summary>Can we just?</summary>Note that total number of **unordered** triplets $(x,y,z) \\in Q^3 _{n} = \\frac{6! ^n}{6!} = 2^{n-1}3^{n-1}$ , suppose for the sake of contradiction assume that there exist a good subset with $ > 2 \\cdot (\\frac{3}{2})^n$ , then note that $\\binom{|A|}{3} \\le 2^{n... | [
"origin:aops",
"2022 Greece National Olympiad",
"2022 Contests"
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"answer_score": 36,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Greece National Olympiad/2790111.json"
} |
Positive integers $a$ , $b$ , and $c$ are all powers of $k$ for some positive integer $k$ . It is known that the equation $ax^2-bx+c=0$ has exactly one real solution $r$ , and this value $r$ is less than $100$ . Compute the maximum possible value of $r$ . | Since we must have $b^2=4ac$ , this implies $a$ , $b$ , and $c$ are powers of $2$ , so substitute $a=2^m$ , $b=-2^n$ , $c=2^\ell$ . Then $r=-\tfrac{b}{2a}$ , hence we want $r$ to be the largest power of $2$ less than $100$ , which just so happens to be $64$ .
Now we find a construction for this. $b^2=4ac$ translates to $2n=m+\ell + 2$ , and we want to make $n-m-1=6$ and $\ell-n +1=6$ . Taking $(m, n, \ell )=(1, 8, 13)$ satisfies this and indeed, the solution to $2x^2-256 x + 8192=0$ is $x=\boxed{64}$ . | [
"solution\n\n**Attachments:**\n\n[Hmmt 2022 P1.docx](https://cdn.artofproblemsolving.com/attachments/e/2/be004fd69ac8fcb133fa636153ecc729ddb017.docx)",
" $\\text{Discriminant }D=b^2-4ac=0 \\implies b \\text{ is even } \\implies k=2.$ It follows that $r$ is a power of $2$ . So max the power of $2$ smaller th... | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
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"answer_score": 1040,
"boxed": true,
"end_of_proof": false,
"n_reply": 7,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2798886.json"
} |
Compute the number of positive integers that divide at least two of the integers in the set $\{1^1,2^2,3^3,4^4,5^5,6^6,7^7,8^8,9^9,10^{10}\}$ . | First, notice that each of the integers from $1$ to $10$ (excluding $1$ , because it’s hard to define?) have at most two distinct prime factors. Therefore, the integers that divide *at least two* of these power things have only one distinct prime factor, since if it had two prime factors, there’s at most one integer with those factors and thus cannot exist. Then they are powers of $2$ , $3$ , or $5$ .
For powers of $2$ , we can have the exponent range from $0$ to $10$ ; powers of $3$ go from $0$ to $6$ ; powers of $5$ from $0$ to $5$ . Adding gives $(10+1)+(6+1)+(5+1)=24$ , but we must subtract off $2$ since we overcounted $1$ twice, ergo $\boxed{22}$ . | [
"22 is the answer ",
"Factorising the set will give you $$ {1^{1}, 2^2, 3^{3}, 2^{8}, 5^{5}, 2^{6} \\cdot 3^{6}, 7^7, 2^{24}, 3^{18}, 2^{10} \\cdot 5^{10}} $$ Let $x = 2^{a} \\cdot 3^{b} \\cdot 5^{c}$ , where $a$ , $b$ , and $c$ are nonnegative integers. (The case when $x = 1$ is obvious, so we will add ... | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 1038,
"boxed": true,
"end_of_proof": false,
"n_reply": 8,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2798887.json"
} |
Let $x_1, x_2, . . . , x_{2022}$ be nonzero real numbers. Suppose that $x_k + \frac{1}{x_{k+1}} < 0$ for each $1 \leq k \leq 2022$ , where $x_{2023}=x_1$ . Compute the maximum possible number of integers $1 \leq n \leq 2022$ such that $x_n > 0$ . | We note that if $x_{k}$ is positive, then $x_{k+1}$ must be negative. Therefore, $n$ must be less than 1012, as if we group our numbers by pairs of $(x_{k}, x_{k+1})$ then we would have 1011 groups (or at most 1011 positive integers).
Now, we will prove that $n = 1011$ is impossible through contradiction: First, let's suppose that the maximum possible number of integers that satisfy the conditions is 1011. We will then prove that the sequence $x_{1}, x_{2}, ..., x_{2021}, x_{2022}$ alternates: With $x_{2023} = x_{1}$ , the only possible sequences are $+, -, +, -, ..., +, -, +$ or $-, +, -, +, ..., -, +, -$ .
We can look at the case where the odd elements are positive, and apply the same arguments to the latter case. There are two subcases to the ordering of the elements:
[list=1]
[*] **Odd/even:** $x_{2n} +\frac{1}{x_{2n+1}} < 0$ and $x_{2n+1} \cdot x_{2n} < -1$ [*] **Even/odd:** $x_{2n+1} +\frac{1}{x_{2n}} < 0$ and $x_{2n+1} \cdot x_{2n} > -1$ [/list]
As the two subcases contradict each other, we have proved that $n = 1011$ is not possible.
Finally, we will prove that $n = 1010$ works with a simple example that ranges from $x_{1}$ to $x_{2023}$ : $1, -\frac{1}{2}, 3, -\frac{1}{4}, ..., 2019, -\frac{1}{2020}, \textrm{[any negative real number a]}, \textrm{[any negative fraction b such that the absolute value of b is less than 1]}, 1$ .
Therefore, $\boxed{n = 1010}$ . | [
"My answer is 1010.",
"This is fairly easy to do though prickly to actually prove; I'll only give an outline of how to prove $n \\geq 1011$ is impossible.\n\nFirst, observe as we cannot have consecutive $+$ signs, so $n$ is at most 1011. Consider the case when $n=1011$ . Now, assume without loss of general... | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 1038,
"boxed": true,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2798905.json"
} |
Compute the sum of all 2-digit prime numbers $p$ such that there exists a prime number $q$ for which $100q + p$ is a perfect square. | Note: $p\equiv11,31,41,61,71,19,29,59,79,89\pmod{100}
\Rightarrow 100q+p\equiv k^2\equiv 11,31,41,61,71,19,29,59,79,89\pmod{100}$ Now observe that the values $k^2$ are congruent to are nothing but the last $2$ digits of the perfect squares. So, we can reduce our calculations to just checking the last $2$ digits of the squares of 2 digit natural numbers as for three and for digits, we'll get the same value as we are working with $\pmod{100}$ Also, notice that $k^2\equiv 1,9\pmod{10}\Rightarrow k\equiv 1,3,7,9\pmod{10}$ So, our calculations are reduced to the numbers $11,21,31,41,51,61,71,81,91, $ $13,23,43,53,63,73,83,93,$ $ 17,27,37,47,57,67,87,97,$ $19,29,39,49,59,69,79,89,99$ Well, here (though I shouldn't have) I used a bit of java (coz I don't know python :P ) to find out what the last $2$ digits might be and the possibilites that came up were $41,61,29,89$ .
Now note that $41\equiv 1\pmod{8}$ and as $k^2$ is an odd perfect square, $100q+41\equiv 1\pmod {8}
\Rightarrow 100q+1\equiv 1\pmod {8}\Rightarrow 100q\equiv 0\pmod 8\Rightarrow q\equiv 0\pmod 8,
\Rightarrow\Leftarrow$ (a contradiction)
Hence the required answer is $29+61+89=\boxed{179}$ .
P.S1: Here's the code I used:
```
class congruence
{
public static void main()
{
for(int i=10;i<=99;i++)
{
if(i%10==1||i%10==3||i%10==7||i%10==9)
System.out.print((i*i)%100+" ,");
}//end for
}//end main()
}//end class
```
P.S2: My first solution to a problem in LATEX | [
"With mod 4 and mod 10 we obtain that p must be in the form 4k+1 and the last digit must be 1 or 9. Moreover, since a perfect square must have even number as its tens digit we got the only possibilities are 29, 41, 61, and 89. 89 obviously works when q=2 (17²). 61 when q=3 (19²) . 29 when q= 5 (23²). 41 is rejected... | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 1032,
"boxed": true,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2798908.json"
} |
Given a positive integer $k$ , let $||k||$ denote the absolute difference between $k$ and the nearest perfect square. For example, $||13||=3$ since the nearest perfect square to $13$ is $16$ . Compute the smallest positive integer $n$ such that $\frac{||1|| + ||2|| + ...+ ||n||}{n}=100$ . | Let $k = \lfloor \sqrt n \rfloor$ and let $n = k^2+r$ for some positive integers $r, k$ . Notice that $$ ||a^2|| + ||a^2+1|| + \cdots + ||(a+1)^2|| = a(a+1), $$ so we can write $$ 100n = 100k^2+100r = ||1||+||2||+ \cdots + ||n|| \leq \sum_{i=2}^k i(i-1) + \frac 12 r(r+1) \leq \frac 13 (k^3-k) + \frac 12 r(r+1). $$ Notice that $\frac 12 r(r+1) > 100r$ if and only if $r > 199$ , so we need the $\frac 13(k^3-k)$ term to be significantly larger than the $100k^2$ term. The inequality $$ \frac 13(k^3-k) \geq 100k^2 \iff k \geq 300 $$ under the integers. Thus, the smallest solutions should be in the neighborhood of 300; a solution with $k = 299$ should conspicuously exist as the margin for $k = 299$ is quite small. Indeed, we can rewrite this expression as $n = (k+1)^2 - r$ and plug in $k=299$ to obtain that $r = 200$ works. As a result, the smallest possible value is $300^2 - 200 = \boxed{89800}$ . | [
"I'm too lazy to make a mathematical solution xD.\nThe answer is 89800, checked by C++: \n```\n#include <bits/stdc++.h>\nusing namespace std;\n\nint f(int n) {\n int lo_sqrt_n = floor(sqrt(n));\n int up_sqrt_n = lo_sqrt_n + 1;\n int dif1 = n - lo_sqrt_n * lo_sqrt_n;\n int dif2 = up_sqrt_n * up_sqrt_n - ... | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 1038,
"boxed": true,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2798911.json"
} |
Let f be a function from $\{1, 2, . . . , 22\}$ to the positive integers such that $mn | f(m) + f(n)$ for all $m, n \in \{1, 2, . . . , 22\}$ . If $d$ is the number of positive divisors of $f(20)$ , compute the minimum possible value of $d$ . | The answer is $\boxed{2016}$ .
Claim: The function $f(n) = n\cdot \mathrm{lcm}(1,2,\ldots, 22)$ works.
Proof: Let $a = \mathrm{lcm}(1,2,\ldots, 22)$ . We must check that \[mn |a(m+n)\] for all $1\le m,n\le 22$ . Suppose for some fixed $m,n$ , and prime $p$ , we had $\nu_p(mn) >\nu_p(a(m+n))$ . Then, \[\nu_p(m) + \nu_p(n) > \nu_p(a) + \nu_p(m+n)\]WLOG that $\nu_p(m) \ge \nu_p(n)$ . Then we have \begin{align*}
\nu_p(a) + \nu_p(m+n)
\ge \nu_p(m) + \nu_p(n),
\end{align*}contradiction to our assumption. $\square$ Claim: $20a\mid f(20)$
Proof. Let $P(m,n)$ denote the given assertion. $P(m,m): m^2 \mid 2f(m)$ . If $m$ is odd, then $m^2 \mid f(m)$ , and if $m$ is even, $\frac{m^2}{2}\mid f(m)$ . Both imply $m\mid f(m)$ for all $m$ within range.
So $m\mid f(m) + f(n)\implies m\mid f(n)$ for all $m,n\in \{1,2,\ldots, 22\}$ . Thus $a$ divides $f(n)$ for each $n$ .
Now suppose there existed a prime $p$ such that $\nu_p(20a) > \nu_p(f(20))$ . Since $a\mid f(20)$ , we have $p=2$ or $p=5$ . Notice that $\nu_5(20a) = 2$ and $\nu_2(20a) = 6$ . $P(20,20): 200\mid f(20)$ , so $\nu_5(20a) \le \nu_p(f(20))$ .
This implies $\nu_2(f(20))<6$ . $P(16,16): \nu_2(f(16)) \ge 7$ . $P(16,20): \nu_2(f(16) + f(20)) \ge 6$ , which implies $\nu_2(f(20)) \ge 6$ , contradiction. $\square$ This implies the minimum number of divisors of $f(20)$ is when $f(20) = 20a$ , which is equal to \[2^6 \cdot 3^2 \cdot 5^2\cdot 7\cdot 11\cdot 13\cdot 17\cdot 19,\]which has \[7\cdot 3\cdot 3\cdot 2^5 = 2016\] divisors. | [
"Does $f(n)=n\\cdot \\mathrm{lcm}(1,2,\\ldots,22)$ work?",
"2016 is my answer .",
"For every prime $p$ , we can look at the minimum power of $p$ that must divide $f(20)$ . In general, we can use the following strategy: if $k$ is the number less than or equal to 22 with the maximum $\\nu_p$ , then we ca... | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 1178,
"boxed": true,
"end_of_proof": false,
"n_reply": 7,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2798917.json"
} |
Let $(x_1, y_1)$ , $(x_2, y_2)$ , $(x_3, y_3)$ , $(x_4, y_4)$ , and $(x_5, y_5)$ be the vertices of a regular pentagon centered at $(0, 0)$ . Compute the product of all positive integers k such that the equality $x_1^k+x_2^k+x_3^k+x_4^k+x_5^k=y_1^k+y_2^k+y_3^k+y_4^k+y_5^k$ must hold for all possible choices of the pentagon. | Much more difficult than #6, but a super interesting problem.
Throughout this solution, we drop all constants. By scaling, we may assume that the vertices of the pentagon lie on the unit circle. Now, let $(x_1, y_1) = \exp(\theta)$ for some angle $\theta$ . Then, $$ x_1^k + x_2^k + x_3^k + x_4^k + x_5^k = (\exp(\theta) + \exp(-\theta))^k + (\omega \cdot \exp(\theta) + \omega^{-1} \cdot \exp(-\theta))^k + \cdots + (\omega^4 \cdot \exp(\theta) + \omega^{-4} \cdot \exp(-\theta))^k. $$ Here $\omega$ is a fifth root of unity. Treating this as a homogeneous polynomial in $\exp(\theta)$ and $\exp(-\theta)$ , it expands as
\begin{align*}
&\phantom{+} \exp(\theta)^k (1+\omega^k + \omega^{2k} + \omega^{3k} + \omega^{4k})
&+ \exp(\theta)^{k-1} \exp(-\theta) \left({k \choose 1} + {k \choose 1}\omega^{k-2} + \cdots + {k \choose 1}\omega^{4k-4}\right)
&+ \cdots
&+ \exp(-\theta)^k \left(1+\omega^{-k} + \omega^{-2k} + \omega^{-3k} + \omega^{-4k}\right).
\end{align*}
Index these terms with $i=0, i=1, i=2, \cdots$ . Now, notice that because $\cos \theta = \cos(-\theta) = \sin(90^\circ + \theta)$ , the expression for the $y_i$ is simply the same polynomial with the value $\exp\left(\theta + \frac{\pi}2\right)$ . As this has to be true for all values of $\theta$ , this implies that all terms of $\exp(\theta)\exp(-\theta)$ with coefficients not subtracting to a multiple of 4 are each zero. Furthermore, observe that each of the terms is not equal to zero if and only if the exponent of the first term $k-2i$ is a multiple of five. Thus, if $k$ is odd, we need all terms to be zero, implying $k < 5$ . If $k$ is even, then the term with $i=5$ will fail, so we have $k < 10$ . Thus, the possible values multiply to $$ 1 \cdot 2 \cdot 3 \cdot 4 \cdot 6 \cdot 8 =\boxed{1152}. $$ | [
"Nice problem, my answer is 1×2×3×4×6×8=1152.",
"Here's a fast unrigorous solution: let $\\arg(x_i,y_i)=\\theta_i$ . Observe $\\textstyle\\sum\\cos k\\theta_i=\\sum\\sin k\\theta_i$ for $5\\nmid k$ ; expanding by Chevyshev gives the sum of degrees $k,k-2,\\dots$ , and so we can inductively build $1\\mapsto3... | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 1042,
"boxed": true,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2798920.json"
} |
Positive integers $a_1, a_2, ... , a_7, b_1, b_2, ... , b_7$ satisfy $2 \leq a_i \leq 166$ and $a_i^{b_i} \cong a_{i+1}^2$ (mod 167) for each $1 \leq i \leq 7$ (where $a_8=a_1$ ). Compute the minimum possible value of $b_1b_2 ... b_7(b_1 + b_2 + ...+ b_7)$ . | Note that $167$ is prime.
We have \[a_i^{2^{i-1}} \equiv a_1^{b_1 b_2 \cdots b_{i-1}}\pmod{167}\]for all $1<i\le 8$ .
This implies $a_1^{128}\equiv a_1^{b_1 b_2 \cdots b_7}\pmod{167}$ , so $a_1^{128 - b_1 b_2 \cdots b_7}\equiv 1\pmod{167}$ .
Case 1: $a_1 = 166$ .
If $a_i = 166$ , then since $-1$ isn't a QR mod $167$ , we get $b_i$ is even, then $a_{i+1} = 166$ . So $a_n = 166$ for all $1\le n\le 7 $ and $b_n$ is even. This gives a minimum possible value of $2^7 \cdot 7\cdot 2 = 1792$ .
Case 2: $a_1\ne 166$ .
Then the order of $a_1$ mod $167$ is either $83$ or $166$ . This implies $b_1 b_2 \cdots b_7\equiv 45\pmod{83}$ . Since we want to minimize $b_1 b_2 \cdots b_7(b_1 + b_2 + \cdots + b_7)$ , we can assume $b_1 b_2 \cdots b_7= 45$ since all other possible values are too big.
Now we want to minimize the value of $b_1 + b_2 + \cdots + b_7$ . We can replace a $(1,9)$ with $(3,3)$ , a $(1,15)$ with $(3,5)$ , and a $(1,45)$ with $(3,15)$ . If we want our construction to be optimal, then we must not have any 9's, 15's 45's. This leaves us with $(1,1,1,1,3,3,5)$ , so the ans is \[45\cdot (1+1+1+1+3+3+5) = \boxed{675}\]
| [
"The official solutions are here:\n[https://hmmt-archive.s3.amazonaws.com/tournaments/2022/feb/algnt/solutions.pdf](https://hmmt-archive.s3.amazonaws.com/tournaments/2022/feb/algnt/solutions.pdf).",
"Let $p = b_1b_2\\dots b_7$ . Note that\n\n\\[a_1^p = a_1^{b_1b_2\\dots b_7} \\equiv a_1^{2b_2b_3\\dots} \\dots \\... | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 1060,
"boxed": true,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2798921.json"
} |
Suppose $P(x)$ is a monic polynomial of degree $2023$ such that $P(k) = k^{2023}P(1-\frac{1}{k})$ for every positive integer $1 \leq k \leq 2023$ . Then $P(-1) = \frac{a}{b}$ where $a$ and $b$ are relatively prime integers. Compute the unique integer $0 \leq n < 2027$ such that $bn-a$ is divisible by the prime $2027$ . | bruh i'm not gonna lie this is probably the hardest algebra problem i've ever done :skull:
Firstly, we note that $P(k)-k^{2023}P\left(1-\frac{1}{k}\right)$ evidently has roots $1,2,3, \cdots 2023$ . Therefore, we can write $P(k)-k^{2023}P\left(1-\frac{1}{k}\right)=k(x-1)(x-2) \cdots (x-2023)$ for some $k$ . Let us write $P(x)=x^{2023}+a_{2022}x^{2022}+a_{2021}x^{2021}+\cdots +a_0$ , thus $x^{2023}P\left(1-\frac{1}{x}\right)=(x-1)^{2023}+a_{2022}(x-1)^{2022}x+a_{n-2}(x-1)^{2021}x^2+ \cdots a_0x^n$ . We now compare coefficients: the absolute difference of $x^{2023}$ in our polynomials should be $k$ . Therefore, $1-P(1)=k$ . Now we compare constant terms and receive $P(0)+1=-(1-P(1))2023!$ . Also, since $P(1)=P(0)$ from the given problem condition, we solve for $P(1)=\frac{2023!+1}{2023!-1}$ and $k=-\frac{2}{2023!-1}$ . Let us now plug in $P(2), P\left(\frac{1}{2}\right),$ and $P(-1)$ to solve for $P(-1)$ . We receive $P(2)=2^{2023}P\left(\frac{1}{2}\right)$ , $P\left(\frac{1}{2}\right)-\frac{P(-1)}{2^{2023}}=\frac{2 \cdot 4045!!}{(2023!-1) \cdot 2^{2023}} \implies 2^{2023}P\left(\frac{1}{2}\right)=P(-1)+\frac{2 \cdot 4045!!}{2023!-1}$ , and $P(-1)+P(2)=2024! \cdot \frac{2}{2023!-1}$ . We solve for $P(-1)=\frac{2024!-4045!!}{2023!-1}$ . Therefore, we want $n(2023!-1) \equiv 2024!-4045!! \pmod{2027}$ . We proceed with the extremely inelegant modular arithmetic bash. Note that $4045!!\equiv 0 \pmod{2027}$ so $n(2023!-1) \equiv 2024! \pmod{2027}$ . By Wilson, we have $2024! \equiv \frac{-1}{-1 \cdot -2} \equiv (-2)^{-1} \equiv 1013 \pmod{2027}$ and $2023! \equiv \frac{-1}{-1 \cdot -2 \cdot -3} \equiv 6^{-1} \implies337 \pmod{2027}$ . We wish to find $n$ such that $337n \equiv 1013 \pmod{2027}$ . We notice that $337 \cdot 6 \equiv -5 \pmod{2027}$ . Thus, we can express $n=6a+b$ . and therefore we want to find $-5a+337b \equiv 1013 \equiv -1014 \pmod{2027}$ . We find that $a=68$ and $b=-2$ works, therefore $n=6 \cdot 68 -2 = \boxed{406}$ . | [
"Here's a version of the official solution that I think is a bit easier (or at the very least more time-efficient) for a non-olympiad like the HMMT:\n\nWe begin by constructing the following equation: $P(x) - x^{2023}P(1-\\frac{1}{x}) = c(x-1)(x-2)\\ldots(x-2023)$ ( $c$ is some constant)\n\nWe first evaluate the ... | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 1066,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2798928.json"
} |
Compute the smallest positive integer $n$ for which there are at least two odd primes $p$ such that $\sum_{k=1}^{n} (-1)^{v_p(k!)} < 0$ . Note: for a prime $p$ and a positive integer $m$ , $v_p(m)$ is the exponent of the largest power of $p$ that divides $m$ ; for example, $v_3(18) = 2$ . | [
"It was very tough one in the individual round, though I have solved and found the answer 229"
] | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2798933.json"
} | |
Let $(a_1, a_2, ..., a_8)$ be a permutation of $(1, 2, ... , 8)$ . Find, with proof, the maximum possible number of elements of the set $$ \{a_1, a_1 + a_2, ... , a_1 + a_2 + ... + a_8\} $$ that can be perfect squares. | The answer is $\boxed{5}$ , achievable with $(1,3,5,7,2,4,6,8)$ .
For the bound, begin by noticing that $1+2+\dots+8 = 36$ , so we can have at most $6$ squares among the set. Thus, it suffices to show we cannot have all $6$ of them. Obviously, the squares must appear in increasing order, and also must be consecutive in order to keep the $a_i$ small enough. However, the difference between consecutive squares is odd, so we require $5$ odd numbers to have a construction with $6$ squares, a contradiction because we only have $4$ odd numbers. | [
"<blockquote>I submitted this and got full credit</blockquote>\n\ni like ur handwriting"
] | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 1018,
"boxed": true,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2801545.json"
} |
Find, with proof, the maximum positive integer $k$ for which it is possible to color $6k$ cells of $6 \times 6$ grid such that, for any choice of three distinct rows $R_1$ , $R_2$ , $R_3$ and three distinct columns $C_1$ , $C_2$ , $C_3$ , there exists an uncolored cell $c$ and integers $1 \le i, j \le 3$ so that $c$ lies in $R_i$ and $C_j$ | The answer is $\boxed{4}$ , and a construction can easily be found (too lazy to use asy).
Obviously, $k=6$ isn't possible. Consider $k=5$ . There are $6$ uncolored squares; pick three of them and report their respective rows as $r_1, r_2, r_3$ . Let the columns of the other three uncolored squares be $c_1, c_2, c_3$ . Note that picking $R_i$ out of the set $\{1,2,3,4,5,6\} \setminus \{r_1,r_2,r_3\}$ and picking $C_i$ out of the set $\{1,2,3,4,5,6\} \setminus \{c_1, c_2, c_3\}$ suffices as a contradiction. | [] | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 1020,
"boxed": true,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2801546.json"
} |
Let triangle $ABC$ be an acute triangle with circumcircle $\Gamma$ . Let $X$ and $Y$ be the midpoints of minor arcs $AB$ and $AC$ of $\Gamma$ , respectively. If line $XY$ is tangent to the incircle of triangle $ABC$ and the radius of $\Gamma$ is $R$ , find, with proof, the value of $XY$ in terms of $R$ . | Let the incenter be $I = \overline{CX} \cap \overline{BY}$ . Since $\overline{XY}$ is tangent to the incircle, we have $\triangle IBC \cong \triangle IXY$ . Thus, $BI = XI = XB$ , meaning that $\angle BXI = \angle BXC = \angle BAC = 60^\circ$ . Hence, $\angle OBC = \angle OXY = 120^\circ$ , giving $XY = \boxed{R\sqrt{3}}$ . | [
"Rsqrt3=XY"
] | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 1014,
"boxed": true,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2801547.json"
} |
Let $ABC$ be a triangle with centroid $G$ , and let $E$ and $F$ be points on side $BC$ such that $BE = EF = F C$ . Points $X$ and $Y$ lie on lines $AB$ and $AC$ , respectively, so that $X$ , $Y$ , and $G$ are not collinear. If the line through $E$ parallel to $XG$ and the line through $F$ parallel to $Y G$ intersect at $P\ne G$ , prove that $GP$ passes through the midpoint of $XY$ . | Trivial. Equivalent to $S_{GPX} = S_{GPY}$ and the parallel lines rewrite the latter into $S_{GEX} = S_{GFY}$ . But $GE \parallel AB$ (consider $M = CG \cap AB$ and note $BE/EM = CG/GM$ ) and similarly $GF \parallel AC$ , so we reduce to $S_{GEB} = S_{GFC}$ , which is clear. | [] | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 14,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2801549.json"
} |
Let $P(x) = x^4 + ax^3 + bx^2 + x$ be a polynomial with four distinct roots that lie on a circle in the complex plane. Prove that $ab\ne 9$ . | This is an incredibly deep problem. BTW, I don't think any teams solved this problem at the contest.
One of the roots is $0$ . Let the other roots be $z_1, z_2, z_3$ . By Vieta's, $z_1+z_2+z_3=-a\implies \frac{z_1+z_2+z_3}{3}=-\frac{a}{3}$ . Also by Vieta's, $\frac{\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}}{3}=-\frac{b}{3}$ . Now they key is to **invert said circle about the unit circle**. Because the circle passes through $0$ , it's image under this inversion is a line! Now the key is that we want to show that $-\frac{3}{a}\ne -\frac{b}{3}$ . It is enough to show that **the image of the centroid of $z_1, z_2, z_3$ is not on the line** from the Vieta's equations above and the fact that the centroid of $\frac{1}{z_1}, \frac{1}{z_2}, \frac{1}{z_3}$ is on the line (which is obvious). Therefore it is enough to show **the centroid of three points on a circle does not also lie on this circle.** This is true. | [
"Suppose by way of contradiction that some $a,b$ work. Let $0,p,q,r$ be the roots, which lie on the circle in that order. From Vieta, we have $pqr = -1$ and\n\\[ab = -(p + q + r)(pq + qr + rp) = (p + q + r)\\left(\\frac1p + \\frac1q + \\frac1r\\right) = 9\\text{.}\\]\nNow we deal with the circle condition. By... | [
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Find, with proof, all functions $f : R - \{0\} \to R$ such that $$ f(x)^2 - f(y)f(z) = x(x + y + z)(f(x) + f(y) + f(z)) $$ for all real $x, y, z$ such that $xyz = 1$ . | <details><summary>good game!</summary>Let $P(x,y,z)$ be the assertion $P(1,1,1)\implies f(1)^2-f(1)^2=9f(1)$ , i.e $\boxed{f(1)=0}$ $P(x,\frac{1}{x},1) \implies f(x)^2=x(x+\frac{1}{x}+1)(f(x)+f(\frac{1}{x}))$ $P(\frac{1}{x},x,1)\implies f(\frac{1}{x})^2=\frac{1}{x}(\frac{1}{x}+x+1)f(\frac{1}{x})+f(x))$ which gives $(f(x)-f(\frac{1}{x}))(f(x)+f(\frac{1}{x}))=(f(x)+f(\frac{1}{x}))(x^2-\frac{1}{x^2}+x-\frac{1}{x})$ **<u>case1:-</u>** $f(x)+f(\frac{1}{x})\neq 0$ then $f(x)-f(\frac{1}{x})=(x-\frac{1}{x})(x+\frac{1}{x}+1)$ $P(1,x,\frac{1}{x})\implies f(x)f(\frac{1}{x})=-(1+x+\frac{1}{x})(f(x)+f(\frac{1}{x}))$ set $a=f(x) , b=f(\frac{1}{x})$ gives $a-b=(x-\frac{1}{x})(x+\frac{1}{x}+1)$ and $ab=-(1+x+\frac{1}{x})(a+b)$ solving for $b$ we get $b=\pm \frac{1}{2}\cdot (x^2+\frac{1}{x^2}+x+\frac{1}{x}+2)$ **<u>subcase 1.1:-</u>** solving for "+" case we get $b=\frac{1}{x^2}-x$ , which gives $a=x^2-\frac{1}{x}$ **<u>subcase 1.2:-</u>** solving for "-" case we get $b=-x^2-2x-\frac{1}{x}-2$ which actually don't gives $f(1)=0$ , hence subcase 1.2 don't holds**<u>Case2:-</u>** $f(x)+f(\frac{1}{x})=0$ we get then $f(x)\cdot f(\frac{1}{x})=0$ which gives $f(x)=0$ on concluding we get $f(x)\equiv \boxed{x^2-\frac{1}{x},0}$ as our functions $\blacksquare$</details> | [
"Bump this",
"plug 1,1,1 to get f(1)=0 and then x,1/x,1 and 1/x,x,1 and solve for f(x), f(1/x)",
"Find, with proof, all functions $f : R - \\{0\\} \\to R$ such that $$ f(x)^2 - f(y)f(z) = x(x + y + z)(f(x) + f(y) + f(z)) $$ for all real $x, y, z$ such that $xyz = 1$ The only solutions are $\\boxed{f\\eq... | [
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Let $P_1P_2...P_n$ be a regular $n$ -gon in the plane and $a_1, . . . , a_n$ be nonnegative integers. It is possible to draw $m$ circles so that for each $1 \le i \le n$ , there are exactly $a_i$ circles that contain $P_i$ on their interior. Find, with proof, the minimum possible value of $m$ in terms of the $a_i$ .
. | [] | [
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Let $\Gamma_1$ and $\Gamma_2$ be two circles externally tangent to each other at $N$ that are both internally tangent to $\Gamma$ at points $U$ and $V$ , respectively. A common external tangent of $\Gamma_1$ and $\Gamma_2$ is tangent to $\Gamma_1$ and $\Gamma_2$ at $P$ and $Q$ , respectively, and intersects $\Gamma$ at points $X$ and $Y$ . Let $M$ be the midpoint of the arc $XY$ that does not contain $U$ and $V$ . Let $Z$ be on $\Gamma$ such $MZ \perp NZ$ , and suppose the circumcircles of $QVZ$ and $PUZ$ intersect at $T\ne Z$ . Find, with proof, the value of $T U + T V$ , in terms of $R$ , $r_1$ , and $r_2$ , the radii of $\Gamma$ , $\Gamma_1$ , and $\Gamma_2$ , respectively. | [] | [
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On a board the following six vectors are written: $$ (1, 0, 0), (-1, 0, 0), (0, 1, 0), (0, -1, 0), (0, 0, 1), (0, 0, -1). $$ Given two vectors $v$ and $w$ on the board, a move consists of erasing $v$ and $w$ and replacing them with $\frac{1}{\sqrt2} (v + w)$ and $\frac{1}{\sqrt2} (v - w)$ . After some number of moves, the sum of the six vectors on the board is $u$ . Find, with proof, the maximum possible length of $u$ . | [] | [
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Suppose $n \ge 3$ is a positive integer. Let $a_1 < a_2 < ... < a_n$ be an increasing sequence of positive real numbers, and let $a_{n+1} = a_1$ . Prove that $$ \sum_{k=1}^{n}\frac{a_k}{a_{k+1}}>\sum_{k=1}^{n}\frac{a_{k+1}}{a_k} $$ | The LaTeX is supposed to be
\[ \sum^{n}_{k=1} \frac{a_k}{a_{k+1}} > \sum^{n}_{k=1} \frac{a_{k+1}}{a_k}. \]
Anyway here's my solution. It's quite different to the official solutions.
<details><summary>Alternate Solution</summary>We'll show it for $n=3$ first. Notice that we want to show
\[ \frac{a_1}{a_2} + \frac{a_2}{a_3} + \frac{a_3}{a_1} > \frac{a_2}{a_1} + \frac{a_3}{a_2} + \frac{a_1}{a_3}. \]
Multiplying by $-1$ and adding $\frac{a_3}{a_1} + \frac{a_3}{a_2} + \frac{a_3}{a_3}$ , we obtain:
\begin{align*}
\frac{a_3 - a_1}{a_2} + \frac{a_3 - a_2}{a_3} + \frac{a_3 - a_3}{a_1} < \frac{a_3 - a_2}{a_1} + \frac{a_3 - a_3}{a_2} + \frac{a_3 - a_1}{a_3}
\end{align*}
which is true, since $\frac{a_3 - a_2}{a_3} < \frac{a_3 - a_1}{a_3}$ . Hence we need to show that $\frac{a_3 - a_1}{a_2} < \frac{a_3 - a_2}{a_1} \iff \frac{a_3 - a_2 - a_1}{a_2} < \frac{a_3 - a_2 - a_1}{a_1}$ which is equivalent to $a_1 < a_2$ , so it's true.
Now we'll induct. Suppose $\sum^{n}_{k=1} \frac{a_k}{a_{k+1}} > \sum^{n}_{k=1} \frac{a_{k+1}}{a_k}$ for some natural number $n \geq 3$ . Then,
\[ \sum^{n+1}_{k=1} \frac{a_k}{a_{k+1}} = \sum^{n}_{k=1} \frac{a_k}{a_{k+1}} - \frac{a_n}{a_1} + \frac{a_n}{a_{n+1}} + \frac{a_{n+1}}{a_1} \]
and \[ \sum^{n+1}_{k=1} \frac{a_{k+1}}{a_k} = \sum^{n}_{k=1} \frac{a_{k+1}}{a_k} - \frac{a_1}{a_n} + \frac{a_{n+1}}{a_n} + \frac{a_1}{a_{n+1}}. \]
By our induction hypothesis it suffices to show that:
\[ - \frac{a_n}{a_1} + \frac{a_n}{a_{n+1}} + \frac{a_{n+1}}{a_1} > - \frac{a_1}{a_n} + \frac{a_{n+1}}{a_n} + \frac{a_1}{a_{n+1}} \]
which, similar to the base case proof, is equivalent to
\[ \frac{a_{n+1} - a_n}{a_1} + \frac{a_{n+1} - a_{n+1}}{a_n} + \frac{a_{n+1} - a_1}{a_{n+1}} > \frac{a_{n+1} - a_1}{a_n} + \frac{a_{n+1} - a_n}{a_{n+1}} + \frac{a_{n+1} - a_{n+1}}{a_1}. \]
Since $\frac{a_{n+1} - a_1}{a_{n+1}} > \frac{a_{n+1} - a_n}{a_{n+1}}$ , we need to show that
\begin{align*}
\frac{a_{n+1} - a_n}{a_1} &> \frac{a_{n+1} - a_1}{a_n}
\iff \frac{a_{n+1} - a_n - a_1}{a_1} &> \frac{a_{n+1} - a_n - a_1}{a_n},
\end{align*}
which is true.</details> | [
"On the other hand, you can also solve this by smoothing $a_i$ between $a_{i-1}$ and $a_{i+1}$ , and using calculus to note that the inequality is closer when $a_i$ is very near $a_{i-1}$ "
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Let $ABC$ be a triangle with $\angle A = 60^o$ . Line $\ell$ intersects segments $AB$ and $AC$ and splits triangle $ABC$ into an equilateral triangle and a quadrilateral. Let $X$ and $Y$ be on $\ell$ such that lines $BX$ and $CY$ are perpendicular to ℓ. Given that $AB = 20$ and $AC = 22$ , compute $XY$ . | <details><summary>Solution</summary>Let the intersection of $\ell$ with $\overline{AB}$ and $\overline{AC}$ be $M$ and $N,$ respectively. Then, let the side length of $\triangle{AMN}$ be $x.$ Then, it follows that $AM=AN=x,$ so we get $BM=20-x$ and $CN=22-x.$ Since $\triangle{AMN}$ is equilateral, we know that $\angle{AMN}=\angle{ANM}=60^\circ,$ so by vertical angles, it follows that $$ \angle{BMX}=\angle{CNY}=60^\circ. $$ Therefore, $\triangle{BMX}$ and $\triangle{CNY}$ are $30-60-90$ triangles, implying that $$ MX = \frac{20-x}{2}=10-\frac{x}{2}, NY=\frac{22-x}{2} = 11-\frac{x}{2}. $$ Hence, the answer is
\begin{align*}
XY &= XM + MN + NY
&= 10-\frac{x}{2}+x+11-\frac{x}{2}
&= \boxed{21}.
\end{align*} $\square$</details> | [
"Let line l intersect AB and AC at P and Q. From 30-60-90 triangle properties, XP = (20-x)/2 and QY = (22-x)/2, where x is the length of the equilateral triangle with angle A of 60 degrees. XY = XP + PQ + QY = (20-x)/2 + (22-x)/2 + x = **21**.",
"Perfect for problem one: clean and simple.\n\nLet the side length o... | [
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Let $ABCD$ and $AEF G$ be unit squares such that the area of their intersection is $\frac{20}{21}$ . Given that $\angle BAE < 45^o$ , $\tan \angle BAE$ can be expressed as $\frac{a}{b}$ for relatively prime positive integers $a$ and $b$ . Compute $100a + b$ . | <details><summary>Solution</summary>Without loss of generality, let $G$ lie inside square $ABCD.$ Also, let the second intersection of the squares be $X \neq A.$ Furthermore, let $\angle{BAX}=\angle{GAX}=\theta.$ We wish to find $\tan(90^\circ-2\theta).$ Clearly, the area of the overlapping region between the two squares is $\tan(\theta),$ so it follows that $\tan(\theta)=\tfrac{20}{21},$ from which it follows that
\begin{align*}
\tan(2\theta) &= \frac{2\tan(\theta)}{1-\tan^2(\theta)}
&= \frac{2\cdot\tfrac{20}{21}}{1-\left(\tfrac{20}{21}\right)^2}
&= \frac{\tfrac{40}{21}}{1 - \tfrac{400}{441}}
&= \frac{\tfrac{40}{21}}{\tfrac{41}{441}}
&= \frac{840}{41}.
\end{align*}
Therefore, we get
\begin{align*}
\tan(90^\circ-2\theta) &= \frac{1}{\tan(2\theta)}
&= \frac{1}{\tfrac{840}{41}}
&= \frac{41}{840},
\end{align*}
yielding an answer of
\begin{align*}
100\cdot41+840 &= 4100+840
&= \boxed{4940}
\end{align*} $\square$</details> | [
"Again, trivial problem. Note the area of the intersection of the two unit squares is the sum of two triangles, each of which has a base of 1 and another side x, x < 1. Hence the intersection area = (1*x/2)*2 = x = 20/21. Let the two angles of these triangles, centered at point A, be alpha and beta. The tangent of ... | [
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Parallel lines $\ell_1$ , $\ell_2$ , $\ell_3$ , $\ell_4$ are evenly spaced in the plane, in that order. Square $ABCD$ has the property that $A$ lies on $\ell_1$ and $C$ lies on $\ell_4$ . Let $P$ be a uniformly random point in the interior of $ABCD$ and let $Q$ be a uniformly random point on the perimeter of $ABCD$ . Given that the probability that $P$ lies between $\ell_2$ and $\ell_3$ is $\frac{53}{100}$ , the probability that $Q$ lies between $\ell_2$ and $\ell_3$ can be expressed as $\frac{a}{b}$ , where $a$ and $b$ are relatively prime positive integers. Compute $100a + b$ .
| Here's a surprisingly clean trigonometric solution.
First, check that $B$ and $D$ must lie between $\ell_2$ and $\ell_3$ ; to prove this is the case, simply notice that the border case yields a probability of $\frac 12$ and the more "tilted" square $ABCD$ is, the greater portion of it will be contained between $\ell_2$ and $\ell_3$ .
Now, let $\ell$ be a fixed line perpendicular to the $\ell_i$ , and define $\theta = \measuredangle(\ell, \overline{AD})$ . Furthermore, assume that the distances between consecutive $\ell_i$ 's is 1. Then, notice that $$ AB^2 = \frac 32 \cdot \frac 1{\cos^2(45^\circ - \theta)} $$ and the area bounded by the square between $\ell_1$ and $\ell_2$ is given by $\tan \theta + \cot \theta$ . Thus,
\begin{align*}
\frac 32 \cdot \frac 1{\cos^2(45^\circ - \theta)} \cdot \frac{47}{100} &= \tan \theta + \cot \theta
\frac 32 \cdot \frac 1{(\frac{\sqrt 2}2 \sin \theta + \frac{\sqrt 2}2 \cos \theta)^2} \cdot \frac{47}{100} &= \tan \theta + \cot \theta
\frac 9{(\sin \theta + \cos \theta)^2} \cdot \frac{47}{100} &= \frac 1{\sin \theta \cos \theta}
\frac 9{1+\sin(2\theta)} \cdot \frac{47}{100} &= \frac 2{\sin(2\theta)}.
\end{align*}
This is linear in $\sin(2\theta)$ and yields $\sin(2\theta) = \frac{200}{223}$ . Then, the portion of perimeter not contained between $\ell_2$ and $\ell_3$ is $$ \frac{2\left(\frac 1{\cos \theta} + \frac 1{\sin \theta}\right)}{4 \cdot \frac 3{\sin \theta + \cos \theta}} = \frac{\frac{\sin \theta + \cos \theta}{\sin(2\theta)}}{\frac 3{\sin \theta + \cos \theta}} = \frac{1+\sin(2\theta)}{3 \sin(2\theta)} = \frac{141}{200}, $$ so the answer is $\frac{59}{200}$ which yields an answer of $\boxed{6100}$ . | [
"Different clean trig solution.\n\nFix a unit square $ABCD$ and let $l_1$ be incident from $A$ with angle $\\theta.$ By inspection, $\\theta < 45^\\circ$ <details><summary>equivalent to</summary><blockquote>First, check that $B$ and $D$ must lie between $\\ell_2$ and $\\ell_3$ ; to prove this is th... | [
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Let triangle $ABC$ be such that $AB = AC = 22$ and $BC = 11$ . Point $D$ is chosen in the interior of the triangle such that $AD = 19$ and $\angle ABD + \angle ACD = 90^o$ . The value of $BD^2 + CD^2$ can be expressed as $\frac{a}{b}$ , where $a$ and $b$ are relatively prime positive integers. Compute $100a + b$ . | Let $C' \neq B$ be a point such that $\triangle ABC \cong \triangle ACC'$ . In addition, let $D'$ be a point in $\triangle ACC'$ such that $\angle DAD' \cong \angle BAC$ .
Notice, $BD^2 + CD^2 = CD^2 + CD'^2 = DD'^2$ , but $\angle DCD' = 90^{\circ}$ . As $\triangle DAD' \sim \triangle BAC$ , then
\[ DD' = AD \cdot \tfrac{BC}{BA} = 19 \cdot \tfrac{11}{22} = \tfrac{19}{2} \quad \longrightarrow \quad DD'^2 = \tfrac{361}{4} \]
Hence, the answer is $\boxed{36104}$ . | [
"Rotate $\\triangle{ADC}$ and let $AC$ meet $AB$ , after rotation, we call the triangle be $\\triangle{ABD'}$ with $\\angle{ABD'}=\\angle{ACD}, \\angle AD'B=90^{\\circ}$ We can see that $\\angle{D'AB}=\\angle{DAC}, \\triangle{AD'D}\\sim \\triangle{ABC}, D'D=\\frac{19}{2}$ $BD^2+CD^2=BD^2+D'B^2=DD'^2=\\f... | [
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Let $ABCD$ be a rectangle inscribed in circle $\Gamma$ , and let $P$ be a point on minor arc $AB$ of $\Gamma$ . Suppose that $P A \cdot P B = 2$ , $P C \cdot P D = 18$ , and $P B \cdot P C = 9$ . The area of rectangle $ABCD$ can be expressed as $\frac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively prime positive integers and $b$ is a squarefree positive integer. Compute $100a + 10b + c$ . | Note that we have $PA \cdot PB=2, PC \cdot PD = 18, PB \cdot PC = 9$ , which implies that $PA \cdot PD=4$ . Then, by applying the Law of Cosines on $\triangle PAD, \triangle PDC, \triangle PBC,$ and $\triangle PAB$ , and setting the minor arcs subtended by lines $AB$ and $AD$ to be $2 \beta$ and $2\theta$ , we get that $$ y^2=PA^2+PB^2+4\cos \beta = PC^2+PD^2-36 \cos \beta $$ and $$ x^2=PA^2+PD^2-8 \cos \theta=PB^2+PC^2-18 \cos \theta $$ Let $PA^2+PB^2+PC^2+PD^2=S$ . Since $PA^2+PC^2=PB^2+PD^2=x^2+y^2$ as $AC$ and $BD$ are diameters of the circle, we note that $S=2x^2+2y^2$ . Also, $2x^2=S-26\cos\theta$ and $2y^2=S-32\cos\beta$ . Looking at triangle $ADC$ , we see that $\cos\beta=\frac{x}{\sqrt{x^2+y^2}}$ and $\cos\theta=\frac{y}{\sqrt{x^2+y^2}}$ . Therefore, $\cos\beta=\frac{x\cos\theta}{y}$ and $\cos\theta=\frac{y\cos\beta}{x}$ . We plug in our earlier value of $S$ into the equations with terms $2x^2$ and $2y^2$ to receive that $y^2=13\cos\theta$ and $x^2=16\cos\beta$ . Substituting our earlier values for $\cos\beta$ and $\cos\theta$ in terms of the other cosine, we see that $x=\frac{16\cos\theta}{y} \implies xy=16\cos\theta$ and $y=\frac{13\cos\beta}{x} \implies xy=13\cos\beta$ . Thus $16\cos\theta=13\cos\beta$ . We also know that $\cos(\beta+\theta)=\cos\beta\cos\theta-\sin\beta\sin\theta=0$ . Now it is simply a matter of solving for $\cos\beta$ and $\cos\theta$ . We get $\cos\theta=\frac{13}{5\sqrt{17}}$ . But $xy$ is the area, and we derived earlier that $xy=16\cos\theta$ . Then the answer is $16\cdot\frac{13}{5\sqrt{17}}=\frac{208\sqrt{17}}{85}$ , and the requested sum is $\boxed{21055}$ . | [
"Firstly we can observe that $\\frac{PD}{PB}=2; \\frac{PC}{PA}=\\frac{9}{2}$ Then, since rectangle $ABCD$ is inscribed in the circle, so $\\angle{APC}=\\angle{PBD}=90^{\\circ}; AC=BD$ We denote that $PB=a,PD=2a; PA=2m, PC=9m$ , now with the application of PT, $5a^2=85m^2, a=\\sqrt{17}m$ , the diameter of the ... | [
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Point $P$ is located inside a square $ABCD$ of side length $10$ . Let $O_1$ , $O_2$ , $O_3$ , $O_4$ be the circumcenters of $P AB$ , $P BC$ , $P CD$ , and $P DA$ , respectively. Given that $P A+P B +P C +P D = 23\sqrt2$ and the area of $O_1O_2O_3O_4$ is $50$ , the second largest of the lengths $O_1O_2$ , $O_2O_3$ , $O_3O_4$ , $O_4O_1$ can be written as $\sqrt{\frac{a}{b}}$ , where $a$ and $b$ are relatively prime positive integers. Compute $100a + b$ . | Just a quick note:
A shortcut to the official sol is scaling $\square{O_1O_2O_3O_4}$ by a factor of $2$ about $P$ to obtain a cyclic quadrilateral with circumcenter $P$ (This reduces the solution length to 4-5 lines) | [] | [
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Let $E$ be an ellipse with foci $A$ and $B$ . Suppose there exists a parabola $P$ such that $\bullet$ $P$ passes through $A$ and $B$ , $\bullet$ the focus $F$ of $P$ lies on $E$ , $\bullet$ the orthocenter $H$ of $\vartriangle F AB$ lies on the directrix of $P$ .
If the major and minor axes of $E$ have lengths $50$ and $14$ , respectively, compute $AH^2 + BH^2$ . | $AH=BH$ and the distance from $A$ to the directrix of parabola is equivalent to $AF$ which is half of the major axis, $AF=25$ .
Let $D$ be the projection of $A$ onto the directrix, $AD=25$ . Due to symmetry, $DH=\frac{AB}{2}=24$ , $AH^2=25^2+24^2=1201$ . As $AH=BH, AH^2+BH^2=2402$ | [
" $AH^{2} + BH^{2}=2402$ "
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Let $A_1B_1C_1$ , $A_2B_2C_2$ , and $A_3B_3C_3$ be three triangles in the plane. For $1 \le i \le3$ , let $D_i $ , $E_i$ , and $F_i$ be the midpoints of $B_iC_i$ , $A_iC_i$ , and $A_iB_i$ , respectively. Furthermore, for $1 \le i \le 3$ let $G_i$ be the centroid of $A_iB_iC_i$ .
Suppose that the areas of the triangles $A_1A_2A_3$ , $B_1B_2B_3$ , $C_1C_2C_3$ , $D_1D_2D_3$ , $E_1E_2E_3$ , and $F_1F_2F_3$ are $2$ , $3$ , $4$ , $20$ , $21$ , and $2020$ , respectively. Compute the largest possible area of $G_1G_2G_3$ . | Let $a_i, b_i, c_i, d_i, e_i, f_i, g_i$ be the points' vectors. It suffices to maximize $\frac{1}{2} |(g_2 - g_1) \times (g_3 - g_1)|.$ Observe that,
\begin{align*}
\frac{1}{2} |(g_2 - g_1) \times (g_3 - g_1)| &= \frac{1}{2} |(\frac{(a_2 - a_1) + (b_2 -b_1) + (c_2 - c_1)}{3}) \times (\frac{(a_3 - a_1) + (b_3 -b_1) + (c_3 - c_1)}{3})|
&= \frac{1}{18} |\sum_{cyc} (a_2 - a_1) \times (a_3 - a_1) + \sum_{cyc} (b_2 - b_1)(c_3 - c_1) + (b_3 - b_1)(c_2 - c_1)|
&= \frac{1}{18} |\sum_{cyc} (b_2 - b_1 + c_2 - c_1) \times (b_3 - b_1 + c_3 - c_1) - \sum_{cyc} (a_2 - a_1) \times (a_3 - a_1)|
\end{align*}
We are given $\frac{1}{2} |(a_2 - a_1) \times (a_3 - a_1)|, \dots$ and $\frac{1}{2}|(\frac{b_2 - b_1 + c_2 - c_1}{2}) \times (\frac{b_3 - b_1 + c_3 - c_1}{2})|, \dots$ from the areas of $A_1A_2A_3, \dots$ and $D_1D_2D_3, \dots$ respectively. To maximize $[G_1G_2G_3]$ let the signed areas of the vertex triangles be negative and the midpoint ones positive. Then we have, $[G_1G_2G_3] = \frac{1}{18}(8 \cdot (20 + 21 + 2020) + 2 \cdot (2 + 3 + 4)) = \boxed{917}.$ | [] | [
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Suppose $\omega$ is a circle centered at $O$ with radius $8$ . Let $AC$ and $BD$ be perpendicular chords of $\omega$ . Let $P$ be a point inside quadrilateral $ABCD$ such that the circumcircles of triangles $ABP$ and $CDP$ are tangent, and the circumcircles of triangles $ADP$ and $BCP$ are tangent. If $AC = 2\sqrt{61}$ and $BD = 6\sqrt7$ ,then $OP$ can be expressed as $\sqrt{a}-\sqrt{b}$ for positive integers $a$ and $b$ . Compute $100a + b$ . | Let $X=AB \cap CD$ and $Y=AD \cap BC$ . By radical axis theorem on $\omega$ , $(ABP)$ , and $(CDP)$ followed by PoP, $XP^2=XO^2-8^2$ . By similar logic, $YP^2=YO^2-8^2$ . By the perpendicularity lemma, $\overline{OP}$ is perpendicular to $\overline{XY}$ . Let $P' = OP \cap XY$ , and note that $P'$ is the projection of $O$ onto $\overline{XY}$ . By the Pythagorean theorem we have $OP \cdot (2OP'-OP)=OP'^2-(OP'-OP)^2=OP'^2-PP'^2=8^2$ . Now one can coordinate bash and compute $OP'=32$ which returns $OP=32-8\sqrt{15}$ , for a final answer of $1024 \cdot 100+960=\boxed{103360}$ .
<details><summary>Idea for potential synthetic solution</summary>Brokard's theorem on $ABCD$ implies that the triangle whose vertices are $Z=AC \cap BD$ , $X$ , and $Y$ has orthocenter $O$ . Then the problem reduces by orthocenter calculations to finding $ZX^2+ZY^2-XY^2$ . Bash using trig and similar triangles induced by the cyclic quadrilateral $ABCD$ .</details> | [] | [
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Rectangle $R_0$ has sides of lengths $3$ and $4$ . Rectangles $R_1$ , $R_2$ , and $R_3$ are formed such that: $\bullet$ all four rectangles share a common vertex $P$ , $\bullet$ for each $n = 1, 2, 3$ , one side of $R_n$ is a diagonal of $R_{n-1}$ , $\bullet$ for each $n = 1, 2, 3$ , the opposite side of $R_n$ passes through a vertex of $R_{n-1}$ such that the center of $R_n$ is located counterclockwise of the center of $R_{n-1}$ with respect to $P$ .
Compute the total area covered by the union of the four rectangles.
 | <details><summary>Solution</summary>Note that the overlapping region of any two consecutive triangles is a triangle which makes up half of the area of either triangle, so all triangles have the same area. Therefore, the answer is
\begin{align*}
12 + \frac{12}{2} + \frac{12}{2} + \frac{12}{2} &= 12+6+6+6
&= \boxed{30},
\end{align*}
by simply looking at the amount of area added by each new rectangle. $\square$</details> | [
"Trivial, can eyeball this problem. Note that R0, R1, R2, and R3 have equal areas, all which equal to 3*4 = 12. The incremental area added by each R_n+1 is simple 12/2 = 6. Hence the total area of the entire figure is 12+6*3 = **30**",
"Notice that the three rectangles have equal areas, which we will call $S$ . ... | [
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Sets $A, B$ , and $C$ satisfy $|A| = 92$ , $|B| = 35$ , $|C| = 63$ , $|A\cap B| = 16$ , $|A\cap C| = 51$ , $|B\cap C| = 19$ . Compute the number of possible values of $ |A \cap B \cap C|$ . | @above that's incorrect
<details><summary>Solution</summary>Let $|A\cap B\cap C|=x$ . Then $|A\cap B|=16-x$ , so we have $x\leq 16$ . Also, $|C\backslash \{A\cup B\}|=x-7$ , so we have $x\geq 7$ . We can verify that these are the only constraints on $x$ , so the answer is $\boxed{10}$ .</details> | [
"<details><summary>solution</summary>$|A \\cup B \\cup C| = |A| + |B| + |C| - |A \\cap B| - |A \\cap C| - |B \\cap C| + |A \\cap B \\cap C|$ . Plugging in the above constants, $|A \\cap B \\cap C| = |A \\cup B \\cup C| - 104$ . The intersection of the three sets is at least 0 and at most equal to the cardinality o... | [
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Compute the number of ways to color $3$ cells in a $3\times 3$ grid so that no two colored cells share an edge. | <details><summary>Solution</summary>If one of the cells is the center cell, then the number of possible colorings is $\tbinom{4}{2}=6,$ and if no cells are the center cell, then it is not difficult to see that there are $4$ possible configuration, each of which yields $4$ possible colorings. Hence, the answer is
\begin{align*}
6+4\cdot4 &= 6+16
&= \boxed{22}.
\end{align*} $\square$</details> | [
"There are three cases we need to consider for a configuration that violates the rule (in other words, a grid where two colored cells share an edge)\n1. We have three consecutive colored cells in the same row or column. \n2. We have two consecutive colored squares in the same row or column, and one colored cell tha... | [
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Michel starts with the string HMMT. An operation consists of either replacing an occurrence of H with HM, replacing an occurrence of MM with MOM, or replacing an occurrence of T with MT. For example, the two strings that can be reached after one operation are HMMMT and HMOMT. Compute the number of distinct strings Michel can obtain after exactly $10$ operations. | The final string must be $14$ characters long, and of the form $HM----------MT$ , where each $-$ is either an $O$ or an $M$ such that no two $O$ 's are adjacent. The number of $O$ 's is an integer from $0$ to $5$ inclusive, so we simply do casework on the number of $O$ 's to get $\tbinom{6}{5}+\tbinom{7}{4}+\tbinom{8}{3}+\tbinom{9}{2}+\tbinom{10}{1}+1=\boxed{144}$ . | [
"HMMT is the first string\n\n2nd string: HMMMT, HMOMT\n\n3rd string: HMMMMT, HMOMMT, HMMOMT\n\n4th string: HMMMMMT, HMOMMMT, HMMOMMT, HMMMOMT, HMOMOMT\n\nengineering induction tells us its just a Fibonacci sequence, 2-3-5-8-13-21-34-55-89-144, $\\boxed{144}$ is the answer "
] | [
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Compute the number of nonempty subsets $S \subseteq\{-10,-9,-8, . . . , 8, 9, 10\}$ that satisfy $$ |S| +\ min(S) \cdot \max (S) = 0. $$ | <details><summary>Solution</summary>Note that either $\min(S)$ or $\max(S)$ has an absolute value of $1$ or $(\min(S), \max(S))$ $\in$ $\{(-3,2),$ $(-2,2),$ $(-2,3)\}.$ The former case yields
\begin{align*}
2\binom{2}{2}+2\binom{3}{2}+\ldots+2\binom{10}{2} &=2 \binom{11}{3}
&= 2\cdot 165
&= 330,
\end{align*}
and the latter case yields $5,$ for a total of $330+5=\boxed{335}.$ $\square$</details> | [
"<details><summary>Solution</summary>Note that we must have $\\min S <0, \\max S>0.$ Let $-a= \\min S$ and $b=\\max S$ , then we get $|S|\\leq a+b+1$ and so $ab\\leq a+b+1.$ This shows that one of $a,b$ must be equal to $1$ or $(a,b)$ must be one of the pairs $(2,2), (2,3)$ or $(3,2).$ We can now... | [
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Five cards labeled $1, 3, 5, 7, 9$ are laid in a row in that order, forming the five-digit number $13579$ when read from left to right. A swap consists of picking two distinct cards, and then swapping them. After three swaps, the cards form a new five-digit number n when read from left to right. Compute the expected value of $n$ . | By linearity of expectation, we can compute the expected value of each digit.
In general, the probability that the digit remains the same after three swaps is $$ \left(\frac 35\right)^3 + {3 \choose 2} \cdot \frac 35 \cdot \frac 25 \cdot \frac 1{10} + \frac 25 \cdot \left(\frac 25 - \frac 1{10}\right) \cdot \frac 1{10} = \frac 3{10} $$ using casework on how many times that position engages in a swap. Thus, we can compute the expected value of the first digit as $$ \frac 3{10} \cdot 1 + \frac 7{10} \cdot \frac{3+5+7+9}4 = \frac 92 $$ as each of the other four digits are equally likely to be swapped in. Similarly, we may compute the expected values of the other digits to be $\frac{19}4$ , 5, $\frac{21}4$ , $\frac {11}2$ in that order. Thus, the answer is $$ 10000 \cdot \frac 92 + 1000 \cdot \frac {19}4 + 100 \cdot 5 + 10 \cdot \frac{21}4 + \frac{11}2 = \boxed{50308}. $$ | [
"Find the sum of $1+3+5+7+9=25$ . Find the probability that it will be the same digit. Then, we can have $\\frac{6^3+3\\cdot 6\\cdot 4+4\\cdot 3}{10^3}=\\frac{3}{10}$ . Thus, for each one its essentially the same. By LOE, we have $10000\\cdot \\left(\\frac{3}{10}+\\frac{7}{10}\\cdot \\frac{25-1}{4}\\right)+1000\... | [
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The numbers $1, 2, . . . , 10$ are randomly arranged in a circle. Let $p$ be the probability that for every positive integer $k < 10$ , there exists an integer $k' > k$ such that there is at most one number between $k$ and $k'$ in the circle. If $p$ can be expressed as $\frac{a}{b}$ for relatively prime positive integers $a$ and $b$ , compute $100a + b$ . | The idea is to use recursion. Let $a_n$ be the number of *ways* to permute $1, 2, 3, \cdots, n$ cyclically such that the condition holds. Call a number $a$ *close enough* to $b$ if there is at most one number between $a$ and $b$ .
To develop a recursive formula for $a_n$ , we consider two cases. In the first case, say that $n$ and $n-1$ are adjacent. Then, by removing $n$ from the circle of numbers, any number that satisfied the condition because $n$ was close enough to it now has $n-1$ the same distance from it, so every one of these numbers still works. Thus, the remaining circular permutation is a valid permutation under $a_{n-1}$ . Thus, every permutation with $n$ numbers and the two numbers adjacent can be obtained by inserting $n$ into one of the positions consecutive to $n-1$ , which yields $2a_{n-1}$ cases.
In the second case, say that $n$ and $n-1$ have one number between them, say $k$ . Then, consider removing both $n$ and $k$ . Any number that was close enough to $n$ is close enough to $n-1$ . Furthermore, $k$ is still adjacent to $n-1$ , so it satisfies the condition as well. Thus, we are left with a permutation of $n-2$ numbers that satisfies the given condition, for $a_{n-2}$ ways. Finally, every permutation with $n$ and $n-1$ not adjacent can be obtained by taking one of the $a_{n-2}$ smaller circular permutations, inserting $n$ consecutive to $n-1$ (the largest number in this permutation) and choosing one of $n-2$ numbers to omit from the outside permutation and instead insert between $n$ and $n-1$ .
From our previous discussion, we then have the recursive relation $$ a_n = 2a_{n-1} + 2(n-2) a_{n-2}. $$ Letting $p_n = \frac{a_n}{(n-1)!}$ , $$ p_n(n-1)! = 2p_{n-1}(n-2)! + 2p_{n-2} (n-2)! \iff p_n = \frac 2{n-1}(p_{n-1} + p_{n-2}). $$ Because $p_4 = p_5 = 1$ , we have $p_6 = \frac 45$ , $p_7 = \frac 35$ , $p_8 = \frac 25$ , $p_9 = \frac 14$ , and $p_{10} = \frac{13}{90},$ so the answer is $\boxed{1390}$ . | [
"<details><summary>solution</summary>There are at least 3 numbers less than or equal to 10 that are greater than numbers 1-7, so they will never be directly adjacent to all the numbers that are greater than them. The condition is only not satisfied when 9 and 10 are adjacent or 8 is between 9 and 10. Since reflecti... | [
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Let $S = \{(x, y) \in Z^2 | 0 \le x \le 11, 0\le y \le 9\}$ . Compute the number of sequences $(s_0, s_1, . . . , s_n)$ of elements in $S$ (for any positive integer $n \ge 2$ ) that satisfy the following conditions: $\bullet$ $s_0 = (0, 0)$ and $s_1 = (1, 0)$ , $\bullet$ $s_0, s_1, . . . , s_n$ are distinct, $\bullet$ for all integers $2 \le i \le n$ , $s_i$ is obtained by rotating $s_{i-2}$ about $s_{i-1}$ by either $90^o$ or $180^o$ in the
clockwise direction. | Observe that consecutive moves are either one forward move or one left move. Define a *run* as a maximal subsequence of $s$ in a line. Then $s$ can be segmented into alternating runs horizontally and vertically.
Given a set of numbers, we can determine the direction of the runs in the sequence due to the non-intersecting property. E.g for $1, 2, 4, 6$ the is a $x$ -runs span $1 \to 6, 6 \to 2, 2 \to 4$ in $x$ -coordinates respectively. With this property, it suffices to choose subsets $A \in \{0, 1, \dots, 11\}$ (starting $0$ doesn't count) and $B\in \{0, 1, \dots, 9\}$ ( $0 \in B$ is guaranteed) to determine the path $s.$ However, since the runs alternate, either $|A| = |B|, |B + 1|.$
We can enumerate the sequence as (note $n \ge 2$ ),
\begin{align*}
& \left ( \left (\binom{12}{1} - 2 \right ) \binom{9}{0}+ \left (\binom{12}{1} - 1 \right) \cdot \binom{9}{1} \right ) + \left (\binom{12}{2}\binom{9}{1} + \binom{12}{2} \cdot \binom{9}{2} \right ) + \cdots
&= \sum_{i = 0}^{9} \binom{12}{i + 1}\binom{9}{i} + \sum_{i = 0}^{9} \binom{12}{i}\binom{9}{i} - 12
&= \sum_{i = 0}^{9} \binom{12}{11-i}\binom{9}{i} + \sum_{i = 0}^{9} \binom{12}{12-i}\binom{9}{i} - 12
&= \binom{21}{11} + \binom{21}{12} - 12 = \boxed{646634}
\end{align*} | [] | [
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Random sequences $a_1, a_2, . . .$ and $b_1, b_2, . . .$ are chosen so that every element in each sequence is chosen independently and uniformly from the set $\{0, 1, 2, 3, . . . , 100\}$ . Compute the expected value of the smallest nonnegative integer $s$ such that there exist positive integers $m$ and $n$ with $$ s =\sum^m_{i=1} a_i =\sum^n_{j=1}b_j . $$ | [] | [
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Consider permutations $(a_0, a_1, . . . , a_{2022})$ of $(0, 1, . . . , 2022)$ such that $\bullet$ $a_{2022} = 625$ , $\bullet$ for each $0 \le i \le 2022$ , $a_i \ge \frac{625i}{2022}$ , $\bullet$ for each $0 \le i \le 2022$ , $\{a_i, . . . , a_{2022}\}$ is a set of consecutive integers (in some order).
The number of such permutations can be written as $\frac{a!}{b!c!}$ for positive integers $a, b, c$ , where $b > c$ and $a$ is minimal. Compute $100a + 10b + c$ . | Typo: In the second condition, $a_i \le \frac{625i}{2022}$ should be $a_i \ge \frac{625i}{2022}$ . | [
"<details><summary>ans idk</summary>216695????????????????????????????????????????</details>"
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Let $S$ be a set of size $11$ . A random $12$ -tuple $(s_1, s_2, . . . , s_{12})$ of elements of $S$ is chosen uniformly at random. Moreover, let $\pi : S \to S$ be a permutation of $S$ chosen uniformly at random. The probability that $s_{i+1}\ne \pi (s_i)$ for all $1 \le i \le 12$ (where $s_{13} = s_1$ ) can be written as $\frac{a}{b}$ where $a$ and $b$ are relatively prime positive integers. Compute $a$ . | We claim that the probability is $\frac{10^{12}+4}{11^{12}}$ . We can count this as follows.
First, note that if it was not in a circle (without the condition at $i=12$ ), then we can fix the first element ( $a_i$ ), and then each element thereafter has a $\frac{10}{11}$ chance of not being the mapping of the previous element through $\pi$ . With this, we would get a "raw" probability of $\frac{10^{11}}{11^{11}}$ . However, we do have the condition holding at $i=12$ , so we have overcounted, since some of the cases we counted could have the sequence "loop around" so that the condition does not hold at $i=12$ .
To account for this, we can subtract such cases where $\pi(a_{12})=a_1$ . Theoretically, if the condition did not have to hold at $i=11$ , we could again similarly fix $a_{12}$ (which then fixes $a_1$ in turn, since $\pi(a_{12})=a_1$ , and the probability $a_1$ is what we want it to be is $\frac{1}{11}$ ), and each element thereafter would have a $\frac{10}{11}$ chance of not being the previous element's mapping through $\pi$ . This gives us the "raw" probability here of $\frac{10^{10}}{11^{10}}*\frac{1}{11}=\frac{10^{10}}{11^{11}}$ . Now, notice that we have again overcounted---we have a possibility of the sequence again "looping around" so that the condition does not hold at $i=11$ .
Again, to account for this, we can add back the cases where $\pi(a_{11})=a_{12}$ and $\pi(a_{12})=a_1$ . Again, here we can also "ignore" the $i=10$ condition, and we would fix $a_{11}$ (in turn fixing $a_{12}$ and $a_1$ , occuring as we want with probability $\frac{1}{11^2}$ ) and the rest follow as we want each with probability $\frac{10}{11}$ . Multiplying these together as we want, we end up with a "raw probability" here of $\frac{10^{9}}{11^{9}}*\frac{1}{11^2}=\frac{10^{9}}{11^{11}}$ .
We can continue this pattern of overcounting and adding/subtracting back, until we end up with an answer of
\[\frac{10^{11}-10^{10}+10^9-10^8+\dots-10^2+10-1}{11^{11}},\]
before we add back our final necessary case, which is if the permutation is such that $\pi(a_i)=a_{i+1}$ holds for all $i$ . We want to know the probability of which this will happen.
First, notice that if this holds, then if we fix what $a_1$ is, the entirety of the $12$ -tuple must be fixed. Additionally, in order for the permutation mappings to loop around in a cycle, we must also have that the "cycle length" of the cycle $a_1$ belongs to in the permutation $\pi$ must be a divisor of $6$ . Essentially, we want to figure out the probability of $a_1$ being in a "cycle" of length $1$ , $2$ , $3$ , $4$ , or $6$ in the permutation $\pi$ , times $\frac{1}{11^{11}}$ to account for the probability of the other $11$ elements of the $12$ -tuple aligning with the values we want. Now, to solve this, I make the following claim.
<span style="color:#960000">***</span>**<span style="color:#960000">Claim.</span>** The probability of $a_1$ being in a cycle of length $k$ (for any $1\leq k\leq 11$ ) in the permutation $\pi$ is exactly $\frac{1}{11}$ .
*Proof.*
If we want $a_1$ to be in a cycle of length $k$ , then we need $k-1$ more elements to be in the cycle, and we need to choose which maps to which in the situation. First, to choose which elements map in which order in the cycle, we have a $10*9*\dots*(12-k)$ ways to choose them. Then, notice that $a_1$ maps to the first element with probability $\frac{1}{11}$ , the first element maps to the second element with probability $\frac{1}{10}$ , and so on (in particular, note that the very last probability accounts for the mapping from the $k$ -th element back to $a_1$ to complete the cycle, which has a probability of $\frac{1}{12-k}$ ). Multiplying all of these out, we get a total probability of $a_1$ being in a cycle of length exactly $k$ in $\pi$ is
\[\frac{10*9*\dots*(12-k)}{11*10*9*\dots*(12-k)}=\frac{1}{11},\]
as we wished to prove.
<span style="color:#960000">***</span>
This means that the probability of $a_1$ being in a "cycle" of length $1$ , $2$ , $3$ , $4$ , or $6$ in the permutation $\pi$ , times $\frac{1}{11^{11}}$ is equal to exactly
\[5*\frac{1}{11}*\frac{1}{11^{11}}=\frac{5}{11^{12}},\]
which we can add back to our original summation. Using geometric series summation, we end up with
\[\frac{10^{11}-10^{10}+10^9-10^8+\dots-10^2+10-1}{11^{11}}+\frac{5}{11^{12}}=\frac{10^{12}-1}{11^{12}}+\frac{5}{11^{12}},\]
or a final probability of $\frac{10^{12}+4}{11^{12}}$ , which we claimed in the first place. This means that $a=10^{12}+4$ (I'm too lazy to type out the whole number and count the number of $0$ 's), finishing the problem.
<span style="color:#960000">***</span>
In general, the **<span style="color:#960000">Claim</span>**'s value of $11$ can be extended to any integer $n$ with a similar argument. Additionally, for this problem, using a similar approach, we can find that if we instead have a $n$ -tuple and a permutation of $k$ -elements (for example, in our problem, $n$ was $12$ and $k$ was $11$ ), if $m$ is the number of factors of $n$ less than or equal to $k$ , then the probability that the condition described in the original problem (extended to match the numbers here) is satisfied is then exactly
\[p=\frac{(k-1)^n+(m-1)}{k^n}.\]
In particular, one should notice that $m$ is always at least $1$ , since $1\leq k$ and $1\mid n$ always. This means that the probability is always at least $\left(\frac{k-1}{k}\right)^n$ , which I thought was cool. | [
"<details><summary>Answer</summary>1000000000004</details>\n<details><summary>Solution</summary>Let's count the number $N$ of ordered pairs $(T, \\pi)$ , where $T$ is the 12-tuple, satisfying the given condition.\n\nWe will use inclusion-exclusion on the 12 constraints of the form $s_{i+1} \\neq \\pi(s_i)$ . ... | [
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"boxed": false,
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"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2804653.json"
} |
Find all natural numbers $n$ for which there is a permutation $\sigma$ of $\{1,2,\ldots, n\}$ that satisfies:
\[
\sum_{i=1}^n \sigma(i)(-2)^{i-1}=0
\] | Awesome problem, the hardest and nicest problem of the test, very similar to <details><summary>past IMO problem</summary>IMO 2012, P6</details> in terms of the ideas involved. Solved with $\textbf{Aryan-23}$ . $\textbf{Lemma 1:}$ $n \equiv 1 \pmod{3}$ do not work. $\textbf{Proof}$
Assume for the sake of contradiction that some $n \equiv 1 \pmod{3}$ works.
Look at the equation $\pmod{3}$ , then $$ 0 = \sum_{i=1}^{n} \sigma(i) \cdot (-2)^{i-1} = \sum_{i=1}^{n} i = \frac{n(n+1)}{2} \equiv 1 \pmod{3} $$ which is a contradiction. $\blacksquare$
We present the same construction as in #4 with some motivation afterwards. $\textbf{Lemma 2}$ $n \equiv 0,2 \pmod{3}$ work. $\textbf{Proof}$
We induct with base case permutations $(2,1), (2,3,1)$ for $n = 2,3$ , respectively.
Assume that $(\sigma(1), \cdots, \sigma(n))$ work for some $n \in \mathbb{N}$ , $n \equiv 0,2 \pmod{3}$ , then take $\sigma_1(1) = 2, \sigma_1(2) = 3, \sigma(m+3) = 1$ and $\sigma_1(i) = \sigma(i-2)+3$ for $i \in \{3, \cdots, n+2\}$ which can be verified to work for $n+3$ as $(\sigma_1(1), \cdots, \sigma_1(n+3))$ . $\blacksquare$
Summing up our conclusions from $\textbf{Lemma 1}$ and $\textbf{Lemma 2}$ , we can conclude that only $n \equiv 0,2 \pmod{3}$ work.
<details><summary>Terribly Written Motivation</summary>I actually started the problem myself before Aryan joined me and was able to conclude using the same global strategy as IMO 2012, P6 by taking $\pmod{3}$ which is motivated by the fact that $(-2)^{i-1} \equiv 1 \pmod{3}$ giving us that $n \equiv 0,2 \pmod{3}$ , this leaves us to construct, and there should intuitively be some construction; after finding the construction for $n =2,3,5,6$ , we imposed the conditions that $\sigma(1) = 2, \sigma(2) = 3$ and guessed that $i$ gets inserted into the $i-2-$ th place for each new thing and when there is a jump of two we insert them in decreasing order consecutively. This gives the constructions $(2,3,5,6,8,7,4,1)$ and $(2,3,5,6,8,9,7,4,1)$ for $8,9$ , respectively, the rest of the problem is to just formalize the inductive construction which turns out to contain $0,2 \pmod{3}$ in ascending and $1 \pmod{3}$ in descending order.</details> | [
"It's easy to get $n \\equiv 0,2 \\pmod 3$ and then you die for the rest of the contest. ",
"NVT's problem!!\n\nThis was easily the best problem on the test according to me. \n\nSoln. Looking mod 3, we get that $3|n^2+n$ and thus, $n\\not\\equiv 1 \\pmod{3}$ . This is kinda a thing that's easy to miss and th... | [
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"path": "Contest Collections/2022 Contests/2022 India National Olympiad/2795409.json"
} |
Let $D$ be an interior point on the side $BC$ of an acute-angled triangle $ABC$ . Let the circumcircle of triangle $ADB$ intersect $AC$ again at $E(\ne A)$ and the circumcircle of triangle $ADC$ intersect $AB$ again at $F(\ne A)$ . Let $AD$ , $BE$ , and $CF$ intersect the circumcircle of triangle $ABC$ again at $D_1(\ne A)$ , $E_1(\ne B)$ and $F_1(\ne C)$ , respectively. Let $I$ and $I_1$ be the incentres of triangles $DEF$ and $D_1E_1F_1$ , respectively. Prove that $E,F, I, I_1$ are concyclic. | Guys I found a solution heavily using the properties of miquel point !
<blockquote>Let $D$ be an interior point on the side $BC$ of an acute-angled triangle $ABC$ . Let the circumcircle of triangle $ADB$ intersect $AC$ again at $E(\ne A)$ and the circumcircle of triangle $ADC$ intersect $AB$ again at $F(\ne A)$ . Let $AD$ , $BE$ , and $CF$ intersect the circumcircle of triangle $ABC$ again at $D_1(\ne A)$ , $E_1(\ne B)$ and $F_1(\ne C)$ , respectively. Let $I$ and $I_1$ be the incentres of triangles $DEF$ and $D_1E_1F_1$ , respectively. Prove that $E,F, I, I_1$ are concyclic.</blockquote>
First note that $I_1 \equiv BE \cap CF$ by a simple angle chase.
Now Consider the complete quadrilateral, $BFAECI_1$ clearly $D(\odot(\triangle ABE)\cap \odot(\triangle ACF))$ is its miquel point.
Since $D \in BC$ we get that quadrilateral $AEI_1F$ must be cyclic.Now its well known that if $O$ is the centre of $\odot(AEI_1F)$ , $D \in \odot(\triangle OEF)$ (Check E.G.M.O proposition $10.14$ ) and $OD$ bisects $\angle EDF$ (E.G.M.O proposition $10.15$ ).
So by the well known Fact 5/Incentre-Excentre Lemma we get that $I \equiv OD \cap \odot(AEI_1F)$ .The End $\blacksquare$ | [
"<details><summary>Thanks INMO :)</summary>\n\n- Angle chase to show $BE_1$ and $CF_1$ are angle bisectors in $D_1E_1F_1$ .\n- Chase more to show $\\angle EDF = \\angle E_1D_1F_1$ (infact the corresponding lines are parallel).\n- Apply $\\angle BIC = 90 + \\angle A/2$ in $DEF$ , $D_1E_1F_1$ to finish.\n... | [
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"boxed": false,
"end_of_proof": true,
"n_reply": 40,
"path": "Contest Collections/2022 Contests/2022 India National Olympiad/2795412.json"
} |
For a positive integer $N$ , let $T(N)$ denote the number of arrangements of the integers $1, 2, \cdots N$ into a sequence $a_1, a_2, \cdots a_N$ such that $a_i > a_{2i}$ for all $i$ , $1 \le i < 2i \le N$ and $a_i > a_{2i+1}$ for all $i$ , $1 \le i < 2i+1 \le N$ . For example, $T(3)$ is $2$ , since the possible arrangements are $321$ and $312$ (a) Find $T(7)$ (b) If $K$ is the largest non-negative integer so that $2^K$ divides $T(2^n - 1)$ , show that $K = 2^n - n - 1$ .
(c) Find the largest non-negative integer $K$ so that $2^K$ divides $T(2^n + 1)$ | For the sake of completeness, here's a full solution.
Consider a graph with vertices as $1,2,\cdots,N$ and draw an arrow $i \rightarrow j$ if $j = 2i$ or $j = 2i+1$ . For $N = 2^n - 1$ , this actually is a binary tree with $n-1$ levels with all numbers in the range $[2^k, 2^{k+1})$ on the $k$ th level.
Clearly, we must have $a_1$ to be greater than any of the others and so $a_1 = 2^n - 1$ . Note that to arrange the $a_i$ , only the relative order matters, and not the fact that the numbers are $1,2,\cdots, N$ . So we choose $2^{n-1} - 1$ numbers for each subtree, which can be done in $\binom{2^n - 2}{2^{n-1} - 1}$ ways. Now, each the numbers of each subtree can be arranged in $T(2^{n-1} - 1)$ ways, so we have the recurrence $$ \boxed{T(2^n - 1) = \binom{2^n - 2}{2^{n-1} - 1}T(2^{n-1} - 1)^2} $$ Let $v(N) = v_2(T(N))$ , so we have $v(2^n - 1) = 2v(2^{n-1} -1) + v_2 \left(\binom{2^n - 2}{2^{n-1} - 1} \right)$ But we have $v_2(N!) = N - s_2(N)$ so the second term is just $2s_2(2^{n-1} - 1) - s_2(2^n - 2) = 2(n-1) - (n-1) = n-1$ , so we have $v(2^n - 1) = 2v(2^{n-1} - 1) + n-1 = 2(2^{n-1} - (n-1) - 1) + n-2 = 2^n - n - 1$ , by induction (base case being $n = 2$ ), so part (b) done.
Now consider $N = 2^n +1$ , this is the same as the previous case except we have two edges extra on one subtree, but the subtrees have sizes $2^{n-1} - 1$ and $2^{n-1} + 1$ so recursing again seems like it would work. In this case, once again, $a_1 = N$ and we choose numbers in $\binom{2^n}{2^{n-1} - 1}$ ways and arrange them in $T(2^{n-1}-1)$ and $T(2^{n-1}+1)$ ways respectively, so we have $$ \boxed{T(2^n + 1) = \binom{2^n}{2^{n-1}-1} T(2^{n-1} -1)T(2^{n-1}+1)} $$ We have $v_2$ of the binomial in this case is $s_2(2^{n-1}+1) + s(2^{n-1}-1) - s(2^n) = 2 + n-1 - 1 = n$ and using the previous result we had, we have $v(2^n + 1) = n + (2^{n-1} - (n-1) - 1) + T(2^{n-1} + 1) = 2^{n-1} + T(2^{n-1} + 1)$ , I claim $T(2^n +1) = 2^n - 1$ and this just follows by induction with base case $n = 1$ and using the previous line., that finishes part (c).
Finally for part (a), we have $T(7) = \binom{6}{3} T(3)^2 = 20(2)^2 = 80$ , done. $\blacksquare$ | [
"Legendre Formula , and Counting works... I am not sure that my solution works , but I got a general formula for $\\nu _{2} ( T(2k+1))$ .. ( and this was the only problem I could solve in Test :stink:",
" $T(7)=80$ . Could not formally write the complete proof of second part but it required to be deal with pow... | [
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"answer_score": 1178,
"boxed": false,
"end_of_proof": false,
"n_reply": 21,
"path": "Contest Collections/2022 Contests/2022 India National Olympiad/2795422.json"
} |
Given a monic quadratic polynomial $Q(x)$ , define \[ Q_n (x) = \underbrace{Q(Q(\cdots(Q(x))\cdots))}_{\text{compose $n$ times}} \]
for every natural number $n$ . Let $a_n$ be the minimum value of the polynomial $Q_n(x)$ for every natural number $n$ . It is known that $a_n > 0$ for every natural number $n$ and there exists some natural number $k$ such that $a_k \neq a_{k+1}$ .
(a) Prove that $a_n < a_{n+1}$ for every natural number $n$ .
(b) Is it possible to satisfy $a_n < 2021$ for every natural number $n$ ?
*Proposed by Fajar Yuliawan* | <details><summary>Official Solution (by Fajar Yuliawan)</summary><details><summary>(a)</summary>Let $Q(a) = a_1$ , then $Q(x) = (x - a)^2 + a_1$ .
Claim 1: $a_1 > a$ Proof. Suppose (FTSOC) $a \geq a_1$ . Notice that $Q_1(a) = a_1$ . Assume that $Q_n(b_n) = a_1$ where $b_n \geq a$ for some $n \geq 1$ . Let $b_{n+1} = a + \sqrt{b_n - a_1} \geq a$ , hence $Q(b_{n+1}) = b_n$ and
\[ Q_{n+1} (b_{n+1}) = Q_n (Q(b_{n+1})) = Q_n(b_n) = a_1. \]
Since $Q_n(x) \geq a_1$ for every natural number $n$ , we obtain that $a_n = a_1$ for all $n$ , a contradiction!
Claim 2: $a_n = Q_n(a)$ Proof. For $n=1$ it's true since $a_1 = Q_1(a)$ . Assume that $a_n = Q_n(a)$ for some $n \geq 1$ . Thus, $Q_n(x) \geq Q_n(a) \geq a_1 > a$ , so
\[ Q_{n+1} (x) = Q(Q_n(x)) = (Q_n(x) - a)^2 + a_1 \geq (Q_n(a) - a)^2 + a_1 = Q_{n+1} (a). \] So $a_{n+1} = Q_{n+1} (a)$ . Claim 2 is proven by induction.
Claim 3: $a_{n+1} > a_n$ for every natural number $n$ .
Proof. Since $a_1 > a$ , then
\[ a_2 = Q_2(a) = Q(a_1) = (a_1 - a)^2 + a_1 > a_1 > a. \]
Suppose it holds that $a_{n+1} > a_n > a$ for some natural number $n$ , then
\[ a_{n+2} = Q(a_{n+1}) = (a_{n+1} - a)^2 + a_1 > (a_n - a)^2 + a_1 = a_{n+1} > a. \]
So claim 3 is proven by induction. $\blacksquare$</details>
<details><summary>(b)</summary>It is possible. Take $Q_1(x) = x^2 + \frac{1}{4}$ . Notice that $a_1 = \frac{1}{4}$ and
\[ a_{n+1} = Q_{n+1} (0) = Q(Q_n(0)) = Q(a_n) = a_n^2 + \frac{1}{4}. \]
Claim. $a_n < \frac{1}{2}$ for every $n \geq 1$ .
Proof. It obviously holds for $n = 1$ . If $a_n < \frac{1}{2}$ , then
\[ a_{n+1} = a_n^2 + \frac{1}{4} < \frac{1}{4} + \frac{1}{4} = \frac{1}{2}. \]
So the claim is proven by induction. $\blacksquare$</details></details>
Translated by yours truly, from Indonesian. :) | [
"Nice one. Who is the author?",
"We do this more generally for a strictly convex twice differentiable function $f$ with a single minimum in $x_0$ .\nWe classify $3$ possible behaviours for $a_n$ . $I$ ) $f(x)>x$ $\\forall x\\in\\mathbb{R}$ . Therefore we have $t=inf(im(f(x)-x))=min(im(f(x)-x))>0$ , or in... | [
"origin:aops",
"2022 Contests",
"2022 Indonesia TST"
] | {
"answer_score": 168,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Indonesia TST/2735502.json"
} |
Five numbers are chosen from $\{1, 2, \ldots, n\}$ . Determine the largest $n$ such that we can always pick some of the 5 chosen numbers so that they can be made into two groups whose numbers have the same sum (a group may contain only one number). | I think,answer is a $$ n
\leq15 $$ {1,2,4,8,16} for $$ n\geq 16 $$ | [
"<blockquote>I think,answer is a15 ${1,2,4,8,16}$ for $n>=16$ </blockquote>\n\nWrong:7,8,9,11,14",
"I think answer is 12\n6,9,11,12,13 for n >=13",
"Bump this ",
"Bump bump bump",
"bump bump\n",
"Soooo, this is actually a pretty nice problem. \nFirst we show that n>=13 is impossible. \nWe can easily che... | [
"origin:aops",
"2022 Contests",
"2022 Indonesia TST"
] | {
"answer_score": 8,
"boxed": false,
"end_of_proof": false,
"n_reply": 7,
"path": "Contest Collections/2022 Contests/2022 Indonesia TST/2735505.json"
} |
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