problem string | solution string | candidates list | tags list | metadata dict |
|---|---|---|---|---|
For the sequence \[S_n=\frac{1}{\sqrt{n^2+1^2}}+\frac{1}{\sqrt{n^2+2^2}}+\cdots+\frac{1}{\sqrt{n^2+n^2}},\]find the limit \[\lim_{n\to\infty}n\left(n\cdot\left(\log(1+\sqrt{2})-S_n\right)-\frac{1}{2\sqrt{2}(1+\sqrt{2})}\right).\] | An appropriate problem for the Euler-Maclaurin' formula implementation :) $$ S_n=\frac{1}{\sqrt{n^2+1^2}}+\frac{1}{\sqrt{n^2+2^2}}+\cdots+\frac{1}{\sqrt{n^2+n^2}}=\frac1n(S_0-1) $$ where, denoting $f(k)=\frac{1}{\sqrt{1+(\frac kn)^2}}$ $$ S_0=\sum_{k=0}^nf(k)\sim n\int_0^1\frac{dx}{\sqrt{1+x^2}}+\frac12\big(f(n)+f... | [
"Follows also from Abel's Partial Summation Lemma.",
"@above: I tried with partial summation but got stuck. Could you please explain in detail?"
] | [
"origin:aops",
"2023 Contests",
"2023 SEEMOUS"
] | {
"answer_score": 18,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 SEEMOUS/3028876.json"
} |
Prove that if $A{}$ is an $n\times n$ matrix with complex entries such that $A+A^*=A^2A^*$ then $A=A^*$ . (Here, we denote by $M^*$ the conjugate transpose $\overline{M}^t$ of the matrix $M{}$ ). | Let $spectrum(A)=(\lambda_i)$ and $j\leq n$ .
We may assume, up to an unitary change of basis, that $A$ is upper triangular with diagonal $[\cdots,\lambda_j]$ . $(A+A^*)[n,n]=(A^2A^*)[n,n]$ gives $\lambda_j+\bar{\lambda_j}=\lambda_j^2\bar{\lambda_j}$ .
Then, for every $j$ , $\lambda_j=0$ or $\pm\sqrt{2}$ ... | [
"Multiply the equation by $A^*$ from the left to get $AA^*+(A^*)^2=A^2(A^2)^*$ . Transpose the last equality and we obtain $AA^*+A^2=A^2(A^2)^*$ . Hence $(A^*)^2=A^2$ . Plug $A^2$ in the original to get $A+A^*=(A^*)^3$ , hence $A$ is a polynomial of $A^*$ , hence they commute, hence $A$ is normal. Chan... | [
"origin:aops",
"2023 Contests",
"2023 SEEMOUS"
] | {
"answer_score": 140,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 SEEMOUS/3028880.json"
} |
Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous, strictly decreasing function such that $f([0,1])\subseteq[0,1]$ .
[list=i]
[*]For all positive integers $n{}$ prove that there exists a unique $a_n\in(0,1)$ , solution of the equation $f(x)=x^n$ . Moreover, if $(a_n){}$ is the sequence defined as above, prove th... | For the first question a diffrent anwser is to define $ \displaystyle g_n\left ( x \right )=\left ( f\left ( x \right ) \right )^{\frac{1}{n}}$ prove that $\displaystyle g_n\left [ 0,1 \right ]\subseteq \left [ 0,1 \right ]$ and is strictly decreasing so by fixed point theorem there exist only one point such that ... | [
"<blockquote>just the main steps: too lazy to type ....\n*-> click to enlarge*</blockquote>\n\ncan someone write the above proof in LaTeX?"
] | [
"origin:aops",
"2023 Contests",
"2023 SEEMOUS"
] | {
"answer_score": 6,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 SEEMOUS/3028883.json"
} |
Given is a triangle $ABC$ with circumcenter $O$ and orthocenter $H$ . If $O_a, O_b, O_c$ denote the circumcenters of $\triangle AOH$ , $\triangle BOH$ , $\triangle COH$ , then prove that $AO_a, BO_b, CO_c$ are concurrent. | By Law of sines: $$ \frac{\sin\angle BAO_1}{BO_1} = \frac{\sin \angle ABO_1}{AO_1}, ~~ \frac{\sin\angle O_1AC}{CO_1} = \frac{\sin \angle ACO_1}{AO_1}. $$ We can easily see that $\angle BO_1C = 2\alpha$ , so $\angle O_1BC = \angle O_1CB = 90^\circ - \alpha$ , hence $$ \frac{\sin\angle BAO_1}{\sin \angle O_1AC} = \f... | [
"Sol:- Let the circumcenters of $AOH,BOH,COH$ be $O_A,O_B,O_C$ . The isogonals of $AO_A,BO_B,CO_C$ are the lines through $A,B,C$ perpendicular to $OH$ which concur at line at infinity. Since isogonal conjugate of line at infinity is circumcircle. $AO_A,BO_B,CO_C$ concur at $(ABC)$ .\n[center][img=5x5]ht... | [
"origin:aops",
"2023 Serbia National Math Olympiad",
"2023 Contests"
] | {
"answer_score": 16,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2023 Contests/2023 Serbia National Math Olympiad/3044622.json"
} |
Given a positive integer $n$ and a prime $q$ , prove that the number $n^q+(\frac{n-1}{2})^2$ can't be a power of $q$ . | Assume such $n,q$ existed. Obviously $n$ is odd. Let $n=2m+1$ , and so $(2m+1)^q+m^2=q^k$ for some $k \geq 1$ .
If $q=2,$ then $(2m+1)^2+m^2=2^k,$ and so $3m^2+4m+1=2^k$ , hence $3m^2+4m+1-2^k=0$ . Thus, the determinant must be a perfect square, i.e. $\Delta=4^2-3 \cdot 4 (1-2^k)=12 \cdot 2^k+4$ is a ... | [
"<blockquote>Given a positive integer $n$ and a prime $q$ , prove that the number $n^q+(\\frac{n-1}{2})^2$ can't be a power of $q$ .</blockquote>\n\nLet $n=2m+1$ then we have $(2m+1)^q+m^2=q^a$ by fermat theorym we have that $q|m+1$ .\nIf $q=2$ take a contradiction $mod4$ If $q=odd$ Let $m=q^x*y-1$ ... | [
"origin:aops",
"2023 Serbia National Math Olympiad",
"2023 Contests"
] | {
"answer_score": 70,
"boxed": false,
"end_of_proof": false,
"n_reply": 9,
"path": "Contest Collections/2023 Contests/2023 Serbia National Math Olympiad/3044624.json"
} |
Given is a triangle $ABC$ with incenter $I$ and circumcircle $\omega$ . The incircle is tangent to $BC$ at $D$ . The perpendicular at $I$ to $AI$ meets $AB, AC$ at $E, F$ and the circle $(AEF)$ meets $\omega$ and $AI$ at $G, H$ . The tangent at $G$ to $\omega$ meets $BC$ at $J$ and $AJ$ ... | Let $T$ be $A-$ mixtilinear touch point and $MT\cap BC=P$ . $IT\cap BC=L$ and let $N$ be the midpoint of arc $BAC$ . Let $Q$ be the midpoint of $BC$ . Let $S$ be $A-$ sharky devil.
----------------------------------------------------------------------------------------------------------------------------... | [
"What is $D$ ?\n\nEDIT: Ig it's the point of tangency of the $A-$ mixtilinear incircle and $\\omega$ ",
"<blockquote>Given is a triangle $ABC$ with incenter $I$ and circumcircle $\\omega$ . The incircle is tangent to $BC$ at $D$ . The perpendicular at $I$ to $AI$ meets $AB, AC$ at $E, F$ and t... | [
"origin:aops",
"2023 Serbia National Math Olympiad",
"2023 Contests"
] | {
"answer_score": 216,
"boxed": false,
"end_of_proof": true,
"n_reply": 9,
"path": "Contest Collections/2023 Contests/2023 Serbia National Math Olympiad/3044627.json"
} |
Given are positive integers $m, n$ and a sequence $a_1, a_2, \ldots, $ such that $a_i=a_{i-n}$ for all $i>n$ . For all $1 \leq j \leq n$ , let $l_j$ be the smallest positive integer such that $m \mid a_j+a_{j+1}+\ldots+a_{j+l_j-1}$ . Prove that $l_1+l_2+\ldots+l_n \leq mn$ . | The idea is that the partial sums have period $mn$ and that we can extend the definition of $l_j$ periodically.
-----**<span style="color:#f00">Lemma.</span>** Given are positive integers $m, n$ and a sequence $a_1, a_2, \ldots$ of integers in $[0,m-1]$ such that $a_i=a_{i-n}$ for all $i>n$ . For all $1 \... | [
"Induction on $n$ works.\nIf $n=1$ then the sequence $(a_i)$ is constant and $l_1 \\le m$ (because $a_1m =0\\mod m$ )\nSuppose the result is true for every $n$ -periodic sequence $(a_i)$ , note that this sequence is entirely defined by its first $n$ elements.\nNow we will show that there exists an inte... | [
"origin:aops",
"2023 Serbia National Math Olympiad",
"2023 Contests"
] | {
"answer_score": 218,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2023 Contests/2023 Serbia National Math Olympiad/3044632.json"
} |
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function which satisfies the following:
- $f(m)=m$ , for all $m\in\mathbb{Z}$ ;
- $f(\frac{a+b}{c+d})=\frac{f(\frac{a}{c})+f(\frac{b}{d})}{2}$ , for all $a, b, c, d\in\mathbb{Z}$ such that $|ad-bc|=1$ , $c>0$ and $d>0$ ;
- $f$ is monotonically increasing.
(a) ... | <blockquote>Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function which satisfies the following:
- $f(m)=m$ , for all $m\in\mathbb{Z}$ ;
- $f(\frac{a+b}{c+d})=\frac{f(\frac{a}{c})+f(\frac{b}{d})}{2}$ , for all $a, b, c, d\in\mathbb{Z}$ such that $|ad-bc|=1$ , $c>0$ and $d>0$ ;
- $f$ is monotonically incre... | [
"<blockquote>Let $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ be a function which satisfies the following:\n\n- $f(m)=m$ , for all $m\\in\\mathbb{Z}$ ;\n- $f(\\frac{a+b}{c+d})=\\frac{f(\\frac{a}{c})+f(\\frac{b}{d})}{2}$ , for all $a, b, c, d\\in\\mathbb{Z}$ such that $|ad-bc|=1$ , $c>0$ and $d>0$ ;\n- $f$ is... | [
"origin:aops",
"2023 Serbia National Math Olympiad",
"2023 Contests"
] | {
"answer_score": 174,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 Serbia National Math Olympiad/3062771.json"
} |
Given is a cube of side length $2021$ . In how many different ways is it possible to add somewhere on the boundary of this cube a $1\times 1\times 1$ cube in such a way that the new shape can be filled in with $1\times 1\times k$ shapes, for some natural number $k$ , $k\geq 2$ ? | This amazing problem comes from the Balkan MO 2022 Shortlist. Specifically, it is C5. This solution follows along the first official solution given for this wonderful problem.
Call a position of the added cubelet $k-$ good if the resulting solid can be covered with $k \times 1 \times 1$ cuboids . Partition the $20... | [
"For anyone wondering, this question is actually not that hard. \n\nFirstly, we can discard all composites because if they work for some position, so does one of its primes.\n\nThe roots of unity argument is just colouring in disguise, when we colour the cube mod $k$ such that any 1x1x $k$ takes each colour once... | [
"origin:aops",
"2023 Serbia National Math Olympiad",
"2023 Contests"
] | {
"answer_score": 290,
"boxed": false,
"end_of_proof": false,
"n_reply": 7,
"path": "Contest Collections/2023 Contests/2023 Serbia National Math Olympiad/3062774.json"
} |
The positive integers are partitioned into 2 sequences $a_1<a_2<\dots$ and $b_1<b_2<\dots$ such that $b_n=a_n+n$ for every positive integer $n$ .
Show that $a_n+b_n=a_{b_n}$ . | Here's my goofy solution from contest, there are other cooler solutions that I'll let someone else post.
<details><summary>Solution</summary>First notice the sequence is uniquely determined because $b_i>a_i$ and thus you can always find the next $a_i$ if you've determined all the previous $a_i$ and $b_i$ since... | [
"@gvole Can you please post the rest of the problems from Serbia TST (except for the ISL ones, if there are such)? Sorry for posting something not related to the problem.",
"I am checking if I am allowed to do so. \n\nThey haven't disclosed anything about the origin of the problems so far.",
"We can actually pr... | [
"origin:aops",
"2023 Contests",
"2023 Serbia Team Selection Test"
] | {
"answer_score": 58,
"boxed": false,
"end_of_proof": false,
"n_reply": 9,
"path": "Contest Collections/2023 Contests/2023 Serbia Team Selection Test/3076913.json"
} |
In a simple graph with 300 vertices no two vertices of the same degree are adjacent (boo hoo hoo).
What is the maximal possible number of edges in such a graph? | I claim that the answer is $\frac12 (299 + 298 \cdot 2 + \cdots 276 \cdot 24) = 42550$ .
Proof of Bound: Notice that we can have at most $k$ vertices with degree $300 - k$ , so the bound is obvious.
Construction: Label $k$ vertices with the label $k$ for $k$ between $1$ and $24$ . Then connect all pair... | [
"<blockquote>In a simple graph with 300 vertices no two vertices of the same degree are connected. \nWhat is the maximal possible number of edges in such a graph?</blockquote>\n\nI believe by connected you actually mean adjacent. \nBecause otherwise the answer is $0$ due to the following fact (which can be proved... | [
"origin:aops",
"2023 Contests",
"2023 Serbia Team Selection Test"
] | {
"answer_score": 16,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 Serbia Team Selection Test/3076921.json"
} |
A circle centered at $A$ intersects sides $AC$ and $AB$ of $\triangle ABC$ at $E$ and $F$ , and the circumcircle of $\triangle ABC$ at $X$ and $Y$ . Let $D$ be the point on $BC$ such that $AD$ , $BE$ , $CF$ concur. Let $P=XE\cap YF$ and $Q=XF\cap YE$ .
Prove that the foot of the perpendicula... | Let $K$ be the projection of $D$ on $EF$ and let $PQ$ intersect $EF$ at point $K'$ . We have the following two Claims:**Claim 1:** $K$ divides $FE$ at ratio $BF/CE$ .
*Proof:* Let $FE$ intersect $BC$ at point $S$ . Note that $(S,B,D,C)=-1$ in light of the complete quadrilateral $AFDE.BC$ , and s... | [
"Let $PQ \\cap EF = H$ , $XY \\cap EF = K$ . Since $((EF \\cap BC), D; B, C) = -1$ , all we need to prove is that $HD$ bisects $\\angle BHC$ , or $\\frac{FH}{HE}=\\frac{BF}{CE}$ .\nNote that $\\frac{FH}{HE}=\\frac{FK}{KE}$ due to polar/harmonic stuff. Denote $Pow(T, \\odot A) - Pow(T, \\odot (ABC))$ by ... | [
"origin:aops",
"2023 Contests",
"2023 Serbia Team Selection Test"
] | {
"answer_score": 178,
"boxed": false,
"end_of_proof": false,
"n_reply": 12,
"path": "Contest Collections/2023 Contests/2023 Serbia Team Selection Test/3076926.json"
} |
Let $p$ be a prime and $P\in \mathbb{R}[x]$ be a polynomial of degree less than $p-1$ such that $\lvert P(1)\rvert=\lvert P(2)\rvert=\ldots=\lvert P(p)\rvert$ . Prove that $P$ is constant. | <details><summary>Solution</summary>If $p=2$ , deg $P <1$ , so constant, done. Let $p >2$ . If $P(1)=\dots = P(p)=0$ , $P$ has atleast $p$ roots, which is more than its degree. So $P \equiv 0$ , done. Hence, WLOG scale $P$ to $\lvert P(1)\rvert=\lvert P(2)\rvert=\ldots=\lvert P(p)\rvert = 1$ and $P(p)=1$ .... | [
"Here's my solution from contest. I think this is the cleanest way to do it.\n\n<details><summary>Solution</summary>Apply a linear transform so that the two values of $P$ are 0 and 1, then take the $p-1$ -th finite difference to obtain:\\[\\sum(-1)^i\\binom{p-1}{i}=0\\]\nover all the $i$ where $P(i)=1$ . You ... | [
"origin:aops",
"2023 Contests",
"2023 Serbia Team Selection Test"
] | {
"answer_score": 60,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 Serbia Team Selection Test/3076929.json"
} |
For positive integers $a$ and $b$ , define \[a!_b=\prod_{1\le i\le a\atop i \equiv a \mod b} i\]
Let $p$ be a prime and $n>3$ a positive integer. Show that there exist at least 2 different positive integers $t$ such that $1<t<p^n$ and $t!_p\equiv 1\pmod {p^n}$ . | My solution from contest (the only one lol):
<details><summary>Solution</summary>If $p$ is odd, then $p^n-p+1$ and $p^n-p-1$ work.
The first one is just the product of all the residues which are $1 \pmod p$ , so it is equal to \[g^{p-1}g^{2(p-1)}\dots g^{(p^{n-1}-1)(p-1)}=g^{(p-1)p^{n-1}\frac{p^{n-1}-1}2}=1\] ... | [
"For $p=2$ it should be also possible to do it with the fact that $(-1)^a3^s$ for $a=0,1$ and $s=1,2,\\ldots,2^{n-2}$ cover all odd remainders $\\pmod {2^n}$ for $n\\geq 2$ . (The same also holds for $(-1)^a5^s$ .)"
] | [
"origin:aops",
"2023 Contests",
"2023 Serbia Team Selection Test"
] | {
"answer_score": 42,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Serbia Team Selection Test/3076934.json"
} |
There are $n^2$ segments in the plane (read walls), no two of which are parallel or intersecting.
Prove that there are at least $n$ points in the plane such that no two of them see each other (meaning there is a wall separating them). | **S**olved with **AdhityaMV, Siddharth03, starchan**
For each segment $S_1, S_2, \cdots, S_{n^2}$ , pick a point $P_i$ on it such that all of them have distinct $x$ coordinates. Reorder them by $x$ -coordinate, and call them $P_1, P_2, \cdots, P_{n^2}$ . Now to each $P_i$ , assign the slope of the line $S_i$ ... | [
"Can someone post the official solution?",
"I didn't come up with this solution, maybe someone smarter can shed some light on the intricacies of this problem.\n\nBasically, you want to introduce a coordinate system that isn't parallel to any of the segments (you want slopes to be defined). Also like make the segm... | [
"origin:aops",
"2023 Contests",
"2023 Serbia Team Selection Test"
] | {
"answer_score": 136,
"boxed": false,
"end_of_proof": true,
"n_reply": 5,
"path": "Contest Collections/2023 Contests/2023 Serbia Team Selection Test/3076938.json"
} |
Let $L$ be the midpoint of the minor arc $AC$ of the circumcircle of an acute-angled triangle $ABC$ . A point $P$ is the projection of $B$ to the tangent at $L$ to the circumcircle. Prove that $P$ , $L$ , and the midpoints of sides $AB$ , $BC$ are concyclic. | Let $O,N,H$ be the circumcenter, nine-point center, and orthocenter of $\triangle ABC$ , respectively.
Let $S,M$ be the midpoints of $AB,BC$ respectively.
We have $OT\perp BC$ and $TP\perp OT$ , so $TP\parallel BC\parallel SM$ .
As $N$ is the midpoint of $OH$ , and we have $HP\perp PL$ and $OT\perp PL$... | [
"I did some trig bash stuff to solve this :kekw:",
"Alternative solution: Let $M,N,K$ be the midpoints of $AB, BC, AC$ and $E$ the foot of the perpendicular from $B$ to $AC$ .\nSince $MN \\parallel LP \\parallel EK$ , the claim is equivalent to $MNLP$ being an isosceles trapezoid.\nBut since $EKLP$ ... | [
"origin:aops",
"2023 Contests",
"2023 Sharygin Geometry Olympiad"
] | {
"answer_score": 40,
"boxed": false,
"end_of_proof": false,
"n_reply": 8,
"path": "Contest Collections/2023 Contests/2023 Sharygin Geometry Olympiad/3025542.json"
} |
The diagonals of a rectangle $ABCD$ meet at point $E$ . A circle centered at $E$ lies inside the rectangle. Let $CF$ , $DG$ , $AH$ be the tangents to this circle from $C$ , $D$ , $A$ ; let $CF$ meet $DG$ at point $I$ , $EI$ meet $AD$ at point $J$ , and $AH$ meet $CF$ at point $L$ . Prove that... | <details><summary>Solution</summary>Since $E$ is the North Pole of $CID$ , we have that $CEID$ is cyclic. Now, also $AELJ$ is cyclic since $\angle LEA=90^\circ$ and
\[\angle LEJ=90^\circ-\angle JEA=90^\circ+\angle CEI=90^\circ-\angle IDC=\angle ADI=\angle LAJ.\]
Finally, from this and $\angle LEA=90^\circ$ w... | [
"One Pappus! $AJD$ and $ICL$ implies $LJ \\perp AD$ "
] | [
"origin:aops",
"2023 Contests",
"2023 Sharygin Geometry Olympiad"
] | {
"answer_score": 14,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Sharygin Geometry Olympiad/3025543.json"
} |
A circle touches the lateral sides of a trapezoid $ABCD$ at points $B$ and $C$ , and its center lies on $AD$ . Prove that the diameter of the circle is less than the medial line of the trapezoid. | <details><summary>Solution</summary>This can be directly computed: If $M$ is the midpoint of $BC$ , then we have two right triangles $ABE$ and $BEM$ with sides $AB=a, BE=b, AE=c=\sqrt{a^2+b^2}$ and $EM=h=\frac{ab}{c}$ so that $BM=\sqrt{b^2-h^2}=\frac{b^2}{c}$ .
So the diameter is $2b$ and the length of th... | [] | [
"origin:aops",
"2023 Contests",
"2023 Sharygin Geometry Olympiad"
] | {
"answer_score": 20,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Sharygin Geometry Olympiad/3025544.json"
} |
Points $D$ and $E$ lie on the lateral sides $AB$ and $BC$ respectively of an isosceles triangle $ABC$ in such a way that $\angle BED = 3\angle BDE$ . Let $D'$ be the reflection of $D$ about $AC$ . Prove that the line $D'E$ passes through the incenter of $ABC$ . | **General Problem**
Let $ABC$ be a triangle with incenter $I$ $D,E$ two points of $AB ,BC$ s.t. $DE\parallel AI $ $D'$ is the intersection of the parallel to $BC$ through $A$ and the parallel to $ BI $ through $ D$ .
Prove that $E,I,D'$ are collinear.
proof :
Let $d_1$ be the infinity point of... | [
"<details><summary>Solution</summary>If $D$ moves linearly on $AB$ , then also $D'$ and $E$ move linearly. So it suffices to consider the two extremal cases $D=A$ and $D=B$ .\nThe latter is trivial since then $E=B$ and $D'E$ is the angular bisector of $\\angle CBA$ , hence passing through $I$ .\nIf ... | [
"origin:aops",
"2023 Contests",
"2023 Sharygin Geometry Olympiad"
] | {
"answer_score": 44,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 Sharygin Geometry Olympiad/3025546.json"
} |
Let $ABCD$ be a cyclic quadrilateral. Points $E$ and $F$ lie on the sides $AD$ and $CD$ in such a way that $AE = BC$ and $AB = CF$ . Let $M$ be the midpoint of $EF$ . Prove that $\angle AMC = 90^{\circ}$ . | Parallel from $B$ intersects $(ABC)$ at $B'$ .
Notice that $B'CF$ and $B'AE$ are isoscele triangles. Let $P$ and $Q$ be the midpoints of $B'F$ and $B'E$ . $\triangle B'PC$ is similar to $\triangle AQB'$ and take $M'$ such that these triangles are similar to $\triangle AM'C$ There is a spiral simil... | [
"<blockquote>Let $ABCD$ be a cyclic quadrilateral. Points $E$ and $F$ lie on the sides $AD$ and $CD$ in such a way that $AE = BC$ and $AB = CF$ . Let $M$ be the midpoint of $EF$ . Prove that $\\angle AMC = 90^{\\circ}$ .</blockquote>\n\nLet point $N$ be the reflection of $A$ with respect to $M... | [
"origin:aops",
"2023 Contests",
"2023 Sharygin Geometry Olympiad"
] | {
"answer_score": 50,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 Sharygin Geometry Olympiad/3025549.json"
} |
Let $A_1, B_1, C_1$ be the feet of altitudes of an acute-angled triangle $ABC$ . The incircle of triangle $A_1B_1C_1$ touches $A_1B_1, A_1C_1, B_1C_1$ at points $C_2, B_2, A_2$ respectively. Prove that the lines $AA_2, BB_2, CC_2$ concur at a point lying on the Euler line of triangle $ABC$ . | By cevian nest, they are obviously concurrent now note that $ABC$ and $A_2B_2C_2$ are homothetic and the centre of homothety is $AA_2 \cap BB_2 \cap CC_2=X$ so $X$ , the circumcenter of $A_2B_2C_2 = H$ and the circumcenter of $ABC=O$ are collinear, done. $\blacksquare$ | [
" $B_2C_2\\perp HA_1\\Rightarrow BC \\parallel B_2C_2$ Similarly other sides are parallel too. $ABC$ and $A_2B_2C_2$ are homothetic and homothety center $J$ sends $O$ to $H$ Q.E.D.",
"Note that $H$ , the orthocenter of $\\triangle ABC$ , is also the incenter of $\\triangle A_1B_1C_1$ . Thus, $B_2C_2... | [
"origin:aops",
"2023 Contests",
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"answer_score": 114,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 Sharygin Geometry Olympiad/3025550.json"
} |
Let $A$ be a fixed point of a circle $\omega$ . Let $BC$ be an arbitrary chord of $\omega$ passing through a fixed point $P$ . Prove that the nine-points circles of triangles $ABC$ touch some fixed circle not depending on $BC$ . | Denote $\omega$ to be the unit circle. Let $a$ and $p$ be fixed.
Then $p=b+c-bc\overline{p}$ , i.e. $c=\dfrac{b-p}{b\overline{p}-1}$ .
Moreover, the midpoint of segment $AP$ is $m=\dfrac{a+p}{2}$ , and the $9-$ point center is $o_{N}=\dfrac{a+b+c}{2}$ .
After some manipulation, we obtain that $$ \left|... | [
"[very nice](https://artofproblemsolving.com/community/c6h107355p606889)"
] | [
"origin:aops",
"2023 Contests",
"2023 Sharygin Geometry Olympiad"
] | {
"answer_score": 32,
"boxed": false,
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"path": "Contest Collections/2023 Contests/2023 Sharygin Geometry Olympiad/3025551.json"
} |
A triangle $ABC$ $(a>b>c)$ is given. Its incenter $I$ and the touching points $K, N$ of the incircle with $BC$ and $AC$ respectively are marked. Construct a segment with length $a-c$ using only a ruler and drawing at most three lines. | <blockquote>A triangle $ABC$ $(a>b>c)$ is given. Its incenter $I$ and the touching points $K, N$ of the incircle with $BC$ and $AC$ respectively are marked. Construct a segment with length $a-c$ using only a ruler and drawing at most three lines.</blockquote>
Iran lemma prob | [] | [
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} |
It is known that the reflection of the orthocenter of a triangle $ABC$ about its circumcenter lies on $BC$ . Let $A_1$ be the foot of the altitude from $A$ . Prove that $A_1$ lies on the circle passing through the midpoints of the altitudes of $ABC$ . | lol headsolved
Dilate at $H$ , orthocentre, so the perpendicular bisector of $HA_1$ passes through $O$ . Extending $AA_1$ to the circumcircle at $D$ , we see that $A_1D=AH$ as $OA=OD,OH=OA_1$ . Yet well known $A_1D=A_1H$ , hence $H$ is midpoint of $AA_1$ . Further if $M$ is midpoint of $BC$ , easy to ... | [
"The only key observation is that $H$ (orthocentre) is the midpoint of altitude passing through $A$ .",
"One liner?\nConsider the anticomplementary triangle of $ABC$ since the orthocenter (since de-longchamps point is orthocenter of anticomplimentary triangle) lies on midline so same should be the case with ... | [
"origin:aops",
"2023 Contests",
"2023 Sharygin Geometry Olympiad"
] | {
"answer_score": 28,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 Sharygin Geometry Olympiad/3025553.json"
} |
Altitudes $BE$ and $CF$ of an acute-angled triangle $ABC$ meet at point $H$ . The perpendicular from $H$ to $EF$ meets the line $\ell$ passing through $A$ and parallel to $BC$ at point $P$ . The bisectors of two angles between $\ell$ and $HP$ meet $BC$ at points $S$ and $T$ . Prove that the c... | Suppose $M$ is the miquel point of the complete quadrilateral $EFBC$ . Now let $AP \cap (PMH)=K$ . I claim that $K$ is the point of tangency of $(PST)$ and $(ABC)$ .
[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(9cm); ... | [
"Let $PA$ recuts $(ABC)$ at $D$ , $AH,AO$ recuts $(ABC)$ at $H',O'$ then $ADO'H'$ rectangle $DH'$ and $AO$ are symmetric wrt the perpendicular bisector of $BC$ ; \nlet $DH'$ hits $BC$ at $K$ ; \n since $H,H'$ are symmetric wrt $BC$ then $DK$ and $KH$ are symmetric wrt ... | [
"origin:aops",
"2023 Contests",
"2023 Sharygin Geometry Olympiad"
] | {
"answer_score": 288,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 Sharygin Geometry Olympiad/3025554.json"
} |
Let $H$ be the orthocenter of an acute-angled triangle $ABC$ ; $E$ , $F$ be points on $AB, AC$ respectively, such that $AEHF$ is a parallelogram; $X, Y$ be the common points of the line $EF$ and the circumcircle $\omega$ of triangle $ABC$ ; $Z$ be the point of $\omega$ opposite to $A$ . Prove that... | What the heck is written about Sharky Devil in #13? I don't agree man... Anyways, here's a solution with some trigonometry and little harmonics.
*Solution:* Let $M$ be the midpoint of $\overline{EF}$ and define $D \coloneqq ZH \cap \odot(ABC) \ne Z$ . Denote $\odot(ABC)$ by $\omega$ . Let $\ell$ be the line... | [
"Let D be the A-foot on BC and O be the center of $\\omega$ .\nClearly $AFH \\sim BHC$ . Let $M$ be the midpoint of AH. Then from the similarity we can say $M$ and $N$ correspond each other in this similar figures(the parallelograms). Thus $\\angle FEA=\\angle CHZ$ . Thus a rotation of $ABC$ by angle of ... | [
"origin:aops",
"2023 Contests",
"2023 Sharygin Geometry Olympiad"
] | {
"answer_score": 186,
"boxed": false,
"end_of_proof": true,
"n_reply": 11,
"path": "Contest Collections/2023 Contests/2023 Sharygin Geometry Olympiad/3025555.json"
} |
Let $ABC$ be a triangle with obtuse angle $B$ , and $P, Q$ lie on $AC$ in such a way that $AP = PB, BQ = QC$ . The circle $BPQ$ meets the sides $AB$ and $BC$ at points $N$ and $M$ respectively. $\qquad\textbf{(a)}$ (grades 8-9) Prove that the distances from the common point $R$ of $PM$ and $NQ$ ... | Sketch of part b: you can define $A'$ , the reflection of the orthocenter among $AC$ in $\triangle ABC$ . Then $A',O,S$ turn out to be collinear by Tales and similarity. Let $L$ be the center of $(PQRO)$ , then you can show that $LR \parallel A'O$ by the similarity of $PQR$ , $CBA$ and angle chasing which... | [
"In b, there is a short solution by writing trig ceva in BQP with trig ratio lemma to find QS/PS and writing trig ratio in OQP to find QS’/PS’. They turn out to be equal.(of course you have to calculate the angles). Any synthetic solution?",
"Another (similar to the above one) approach for b) is to notice that by... | [
"origin:aops",
"2023 Contests",
"2023 Sharygin Geometry Olympiad"
] | {
"answer_score": 18,
"boxed": false,
"end_of_proof": false,
"n_reply": 7,
"path": "Contest Collections/2023 Contests/2023 Sharygin Geometry Olympiad/3025556.json"
} |
The base $AD$ of a trapezoid $ABCD$ is twice greater than the base $BC$ , and the angle $C$ equals one and a half of the angle $A$ . The diagonal $AC$ divides angle $C$ into two angles. Which of them is greater? | Why so complicated? Let $AB \cap CD=P$ , just let $\angle BCA=\alpha$ , now sine law in $\bigtriangleup PDA$ gives $\frac{AD}{CD}=\frac{2\sin \frac{\alpha}{2}}{\sin \alpha}$ , so $CD=AD\cos \frac{\alpha}{2}$ so $CD<AD$ , done. $\blacksquare$ | [
"It just suffices to compare $AD$ and $CD$ . ;)",
"This problem was such a nice trig bash, the first 3 lines of my solution were:\n<blockquote>\nWe first prove the following: \n\nLemma: For any real $\\theta \\in (\\pi/3]$ , it holds that \\[4\\cos 2\\theta + \\frac{1}{\\cos^2 \\theta} - 1 < 4 \\cos^2 \\theta... | [
"origin:aops",
"2023 Contests",
"2023 Sharygin Geometry Olympiad"
] | {
"answer_score": 114,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 Sharygin Geometry Olympiad/3025557.json"
} |
Suppose that a closed oriented polygonal line $\mathcal{L}$ in the plane does not pass through a point $O$ , and is symmetric with respect to $O$ . Prove that the winding number of $\mathcal{L}$ around $O$ is odd.
The winding number of $\mathcal{L}$ around $O$ is defined to be the following sum of the orie... | Don't boo spam the brilliant emote! Stop this guy from brillianting! Complex Analysis FTW.
Anyways, let $\mathcal{\chi}(\mathcal L(t))$ denote the winding number of the polygonal line $\mathcal L(t): [-1,1]\rightarrow \mathbb C$ WRT $O$ . Deform $\mathcal L$ in a sufficiently small neighborhood around the verti... | [
"Woah winding numbers! As luck would have it, Bubu is the only one who has posted a solution.",
"It suffices to show that any odd map $S^1 \\to S^1$ has odd degree. It turns out [this is true in general](https://planetmath.org/proofofborsukulamtheorem) for any $S^n$ (but I don't think this link proves it for ... | [
"origin:aops",
"2023 Contests",
"2023 Sharygin Geometry Olympiad"
] | {
"answer_score": 130,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 Sharygin Geometry Olympiad/3025558.json"
} |
Let $ABCD$ be a convex quadrilateral. Points $X$ and $Y$ lie on the extensions beyond $D$ of the sides $CD$ and $AD$ respectively in such a way that $DX = AB$ and $DY = BC$ . Similarly points $Z$ and $T$ lie on the extensions beyond $B$ of the sides $CB$ and $AB$ respectively in such a way that... | [
"15 and 16 easiest probs on the test",
"imagine not bary\nEdit: I got a 7 for my bary bash!!!! :first:"
] | [
"origin:aops",
"2023 Contests",
"2023 Sharygin Geometry Olympiad"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
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"path": "Contest Collections/2023 Contests/2023 Sharygin Geometry Olympiad/3025559.json"
} | |
Let $AH_A$ and $BH_B$ be the altitudes of a triangle $ABC$ . The line $H_AH_B$ meets the circumcircle of $ABC$ at points $P$ and $Q$ . Let $A'$ be the reflection of $A$ about $BC$ , and $B'$ be the reflection of $B$ about $CA$ . Prove that $A',B', P,Q$ are concyclic. | Let $M$ and $N$ be the reflections of $H$ over $H_B$ and $H_A$ respectively.
Using Orthocentre-reflection and PoP, $A'H_A\cdot HH_A=AH_A\cdot H_AN=QH_A\cdot H_AP\implies A'\in (PQH)$ $B'H_B\cdot HH_B=BH_B\cdot H_BM=QH_B\cdot H_BP\implies B'\in (PQH)$ Therefore, $A', B', P, Q$ are concyclic. | [
"My solution…\n\n**Attachments:**\n\n[Sharygin_P16.pdf](https://cdn.artofproblemsolving.com/attachments/1/2/efb4f5fbec9195f73dc435db3cda5f50618ec3.pdf)"
] | [
"origin:aops",
"2023 Contests",
"2023 Sharygin Geometry Olympiad"
] | {
"answer_score": 16,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Sharygin Geometry Olympiad/3025560.json"
} |
A common external tangent to circles $\omega_1$ and $\omega_2$ touches them at points $T_1, T_2$ respectively. Let $A$ be an arbitrary point on the extension of $T_1T_2$ beyond $T_1$ , and $B$ be a point on the extension of $T_1T_2$ beyond $T_2$ such that $AT_1 = BT_2$ . The tangents from $A$ to $\... | Here is a not entirely synthetic solution using some (not too much) trig bash:
Let $O_1$ and $O_2$ be the centers of $\omega_1$ and $\omega_2$ respectively and let $M$ be the midpoint of $T_1T_2$ . Then $O_1T_1T_2O_2$ is clearly a right-angled trapezoid. Also let $O_1T_1=r_1$ and $O_2T_2=r_2$ . Let $W... | [
"@above $A, B$ need to be sufficiently far away for it to work, otherwise it just becomes a Gergonne cevian, interestingly. And yeah, I don't see a way to do this without using Cartesian Coordinates.",
"In $\\triangle ABC$ , draw the incircle $\\omega$ . Let it hit $\\overline{AB}$ at $D$ , and let the opp... | [
"origin:aops",
"2023 Contests",
"2023 Sharygin Geometry Olympiad"
] | {
"answer_score": 1150,
"boxed": true,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 Sharygin Geometry Olympiad/3025563.json"
} |
Restore a bicentral quadrilateral $ABCD$ if the midpoints of the arcs $AB,BC,CD$ of its circumcircle are given. | <blockquote>consider the labellings as in the diagram below . We are given $Q,R,S$ and we need to restore $ABCD$ . It is well known that $UW \perp VX \iff ABCD$ is bicentric . $T$ is midpoint of arc $AD$ . Note that $QRST$ and $VWXU$ are homothetic so $RT \perp SQ$ so we construct $T$ from here . Now ... | [
"Okay, look I might get cancelled for this solution, but it's all fair game. No one said you can't bash lengths using constructions. \n\n**Attachments:**\n\n[Sharygin-Problem-18.pdf](https://cdn.artofproblemsolving.com/attachments/7/a/53e74f202b29d627019e3e75b0671b7ea9193f.pdf)"
] | [
"origin:aops",
"2023 Contests",
"2023 Sharygin Geometry Olympiad"
] | {
"answer_score": 38,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Sharygin Geometry Olympiad/3025564.json"
} |
A cyclic quadrilateral $ABCD$ is given. An arbitrary circle passing through $C$ and $D$ meets $AC,BC$ at points $X,Y$ respectively. Find the locus of common points of circles $CAY$ and $CBX$ . | Let $O$ and $O_1$ be the circumcenters of $ABC$ and $CXY$ , $l$ be the perpendicular bisector of $CD$ .
Perpendicular from $O_1$ to $AC$ and perpendicular from $O$ to $BC$ intersect at $P$ . Similarly define $Q$ .
Notice $P$ and $Q$ are circumcenters of $CBX$ and $CAY$
So the other inter... | [
"<details><summary>Huh.</summary>The locus is line $CU$ , where $U$ is the point on the circumcircle of quadrilateral $ABCD$ such that $(AB;DU) = -1$ . Proof is to just invert at $C$ and use Menelaus or whatever floats your boat.</details>",
"Invert about $C$ with arbitrary radius.\nLet $P'$ denote th... | [
"origin:aops",
"2023 Contests",
"2023 Sharygin Geometry Olympiad"
] | {
"answer_score": 158,
"boxed": false,
"end_of_proof": false,
"n_reply": 9,
"path": "Contest Collections/2023 Contests/2023 Sharygin Geometry Olympiad/3025565.json"
} |
Let a point $D$ lie on the median $AM$ of a triangle $ABC$ . The tangents to the circumcircle of triangle $BDC$ at points $B$ and $C$ meet at point $K$ . Prove that $DD'$ is parallel to $AK$ , where $D'$ is isogonally conjugated to $D$ with respect to $ABC$ . | <blockquote>Consider the following lemma:
<span style="color:#f00">**Lemma:**</span> Let unique $b, y, i, j \in \mathbb P^1$ be fixed and $a \in \mathbb P^1$ be variable. There is a unique involution on $\mathbb P^1$ swapping $i \leftrightarrow j$ and $a \leftrightarrow b$ . Suppose, this involution swaps $x... | [
"Consider the following lemma:\n\n<span style=\"color:#f00\">**Lemma:**</span> Let unique $b, y, i, j \\in \\mathbb P^1$ be fixed and $a \\in \\mathbb P^1$ be variable. There is a unique involution on $\\mathbb P^1$ swapping $i \\leftrightarrow j$ and $a \\leftrightarrow b$ . Suppose, this involution swaps... | [
"origin:aops",
"2023 Contests",
"2023 Sharygin Geometry Olympiad"
] | {
"answer_score": 216,
"boxed": false,
"end_of_proof": false,
"n_reply": 9,
"path": "Contest Collections/2023 Contests/2023 Sharygin Geometry Olympiad/3025566.json"
} |
Let $ABCD$ be a cyclic quadrilateral; $M_{ac}$ be the midpoint of $AC$ ; $H_d,H_b$ be the orthocenters of $\triangle ABC,\triangle ADC$ respectively; $P_d,P_b$ be the projections of $H_d$ and $H_b$ to $BM_{ac}$ and $DM_{ac}$ respectively. Define similarly $P_a,P_c$ for the diagonal $BD$ . Prove th... | Here is a solution avoiding Humpty points, though in its main idea it feels the same as the above:
Our main goal will be to prove that lines $AC, BD, P_aP_c$ and $P_bP_d$ have a common point. To prove this, we will use the following lemmas:
Lemma 1. $H_a, H_c, B$ and $D$ are concyclic. (and similarly so are ... | [
"Walkthrough: 1) Pa,Pb,Pc,Pd are the Humpty points so $H_aP_aBH_cP_cD$ and $H_bP_bAH_dP_dC$ are concyclic. 2)Also $M_{ac}P_d.M_{ac}B=M_{ac}D.M_{ac}P_b=M_{ac}C^2$ .\n3)This means $P_d,B,P_b,D$ are concyclic. Now applying Radical Center on $P_dBP_bD$ , $ABCD$ , $H_bP_bAH_dP_dC$ we get $AC,BD,P_bP_d$ concur... | [
"origin:aops",
"2023 Contests",
"2023 Sharygin Geometry Olympiad"
] | {
"answer_score": 112,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 Sharygin Geometry Olympiad/3025567.json"
} |
Let $ABC$ be a scalene triangle, $M$ be the midpoint of $BC,P$ be the common point of $AM$ and the incircle of $ABC$ closest to $A$ , and $Q$ be the common point of the ray $AM$ and the excircle farthest from $A$ . The tangent to the incircle at $P$ meets $BC$ at point $X$ , and the tangent to the ... | A nice problem indeed, here is my solution:
Let $I_1$ be the center of the incircle of $ABC$ and let $I_2$ be the center of the $A-$ excircle of $ABC$ . Let $R$ be the intersection of $PI_1$ and $QI_2$ .
Lemma 1. $RP=RQ$ .
Proof.
Let $P'$ be the second intersection of $AM$ and the $A-$ excircle ... | [
"Very beautiful problem.\nThis Lemma kills it.\nhttps://tieba.baidu.com/p/7951343410#/",
"@above indeed. Rather i feel the problem is misplaced as far as difficulty is concerned.\nLet AQ intersect the excircle again at R. We consider the homothety centered at A and mapping the incircle to the excircle as $\\math... | [
"origin:aops",
"2023 Contests",
"2023 Sharygin Geometry Olympiad"
] | {
"answer_score": 202,
"boxed": false,
"end_of_proof": false,
"n_reply": 12,
"path": "Contest Collections/2023 Contests/2023 Sharygin Geometry Olympiad/3025569.json"
} |
An ellipse $\Gamma_1$ with foci at the midpoints of sides $AB$ and $AC$ of a triangle $ABC$ passes through $A$ , and an ellipse $\Gamma_2$ with foci at the midpoints of $AC$ and $BC$ passes through $C$ . Prove that the common points of these ellipses and the orthocenter of triangle $ABC$ are collinear... | Denote by $l_1,l_2$ the directrix of the eclipses. We have (let $X,Y$ be the two intersections) $\frac{XM_b}{d_{x-l_1}}=e_1,\frac{XM_b}{d_{x-l_2}}=e_2$ ,so $\frac{d_{x-l_1}}{d_{x-l_2}}=\frac{e_2}{e_1}$ ,the same as $Y$ ,so it suffices to verify $H$ for this, which is trivial. | [
"[link to this problem](https://artofproblemsolving.com/community/q1h2968421p26593343) quite weird seeing this problem posted 3 days before the start of the test.",
"[ another relevant thread ]( https://artofproblemsolving.com/community/c6h2686348)",
"This is a well-known problem.",
"Well, even worse right [h... | [
"origin:aops",
"2023 Contests",
"2023 Sharygin Geometry Olympiad"
] | {
"answer_score": 12,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2023 Contests/2023 Sharygin Geometry Olympiad/3025570.json"
} |
A tetrahedron $ABCD$ is give. A line $\ell$ meets the planes $ABC,BCD,CDA,DAB$ at points $D_0,A_0,B_0,C_0$ respectively. Let $P$ be an arbitrary point not lying on $\ell$ and the planes of the faces, and $A_1,B_1,C_1,D_1$ be the second common points of lines $PA_0,PB_0,PC_0,PD_0$ with the spheres $PBCD... | I am very happy that this interesting 3D geometry problem of mine was selected by the Organizing Committee. The official solution is published [here](https://geometry.ru/olimp/2023/zaoch_sol_eng_2023.pdf). I am sending the solution I submitted as follows:
Let $\mathcal{S}$ be the circumsphere of the tetrahedron $AB... | [
"Sketch:\nDraw the circumshpere of ABCD and plane PA0B0C0D0. Let l cross shpere ABCD at X and Y. Use monge theorum and we can get A0 B0 C0 D0 P X Y are concyclic."
] | [
"origin:aops",
"2023 Contests",
"2023 Sharygin Geometry Olympiad"
] | {
"answer_score": 114,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Sharygin Geometry Olympiad/3025572.json"
} |
The ratio of the median $AM$ of a triangle $ABC$ to the side $BC$ equals $\sqrt{3}:2$ . The points on the sides of $ABC$ dividing these side into $3$ equal parts are marked. Prove that some $4$ of these $6$ points are concyclic. | Another quick complex solution could be done by setting $M=0,B=1,C=-1,A=a$ s.t $a.\overline{a}=3$ .
By section formula the trisection points are $\frac{2a+1}{3},\frac{a+2}{3},\frac{1}{3},\frac{-1}{3},\frac{2a-1}{3},\frac{a-2}{3}$ .
Now it could be easily proved that, $$ \frac{\frac{a+2}{3}-\frac{-1}{3}}{\frac{a+2}... | [
"I failed the finals but still am uploading my complex bash solution for this problem since I am shameless. Took me forever for me to make the asy diag cuz of me being so useless in life. :wheelchair:\n\n[asy]\n pair M = (0,0);\n pair B = (-1/sqrt(3),0);\n pair C = (1/sqrt(3),0);\n pair ... | [
"origin:aops",
"2023 Contests",
"2023 Sharygin Geometry Olympiad"
] | {
"answer_score": 110,
"boxed": false,
"end_of_proof": true,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 Sharygin Geometry Olympiad/3124982.json"
} |
Can a regular triangle be placed inside a regular hexagon in such a way that all vertices of the triangle were seen from each vertex of the hexagon? (Point $A$ is seen from $B$ , if the segment $AB$ dots not contain internal points of the triangle.) | <details><summary>Method with Cartesian coordinate system</summary>Assume it is possible. By rotation and homothety, we can put the vertices of the equilateral triangle to $A~(1,0)$ , $B~(-1,0)$ and $C~\left(0,\sqrt3\right)$ . Then the vertices of the regular hexagon $V_1V_2V_3V_4V_5V_6$ (counterclockwise orienta... | [
"cute problem\n<details><summary>solution</summary>The answer is no. Suppose otherwise and let $PQR$ be the regular triangle inscribed in the regular hexagon $A_1A_2A_3A_4A_5A_6$ . Extend lines $PQ, QR, RP$ infinitely and observe that the region from where all three vertices can be seen is precisley the union ... | [
"origin:aops",
"2023 Contests",
"2023 Sharygin Geometry Olympiad"
] | {
"answer_score": 138,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Sharygin Geometry Olympiad/3124983.json"
} |
Points $A_1$ , $A_2$ , $B_1$ , $B_2$ lie on the circumcircle of a triangle $ABC$ in such a way that $A_1B_1 \parallel AB$ , $A_1A_2 \parallel BC$ , $B_1B_2 \parallel AC$ . The line $AA_2$ and $CA_1$ meet at point $A'$ , and the lines $BB_2$ and $CB_1$ meet at point $B'$ . Prove that all lines $A'B'... | Sorry to correct but only these collinearities don't imply all of them lie on a single line. However, there is an easy fix.
By angle chasing $A'A_1A_2$ and $B'BC$ are homothetic because their corresponding sides are parallel. So we can say that $LA'B'$ are collinear too. Similiarly, $P$ lies on this line. Comb... | [
"Hint: use similars triangles and Menelaus theorem\nThis point, in this $A'B'$ concur., lies on line $AB$ ",
"Also, it is possible to solve it by moving points",
"Wooohooo!! Comeback after a looong time with the 670th post!(and did a geo problem after more than half a year) \nLet $B_2C \\cap B_1A \\equiv L,... | [
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The incircle $\omega$ of a triangle $ABC$ centered at $I$ touches $BC$ at point $D$ . Let $P$ be the projection of the orthocenter of $ABC$ to the median from $A$ . Prove that the circle $AIP$ and $\omega$ cut off equal chords on $AD$ . | Let $N$ and $M$ be the midpoints of $AD$ and $BC$ , the incircle touches $AC$ and $AB$ at $E$ and $F$ , $L=EF\cap BC$ .
We will prove that $N$ is on the radical axis of $\omega$ and $(AIP)$ which will finish the problem.
Claim 1. $N$ , $I$ , and $M$ are colinear.
Proof. Let $DI$ intersect ... | [
"here's a somewhat motivatable solution described to me by **mueller.25**\n\n<details><summary>solution</summary>First, a lemma:\n\nLemma: Let $ABC$ be a triangle and $X$ a point in its interior. Let $AX$ meet $BC$ at point $K$ . Then for any circle $\\omega$ through $A$ and $X$ the quantity \\[\\tex... | [
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} |
A point $D$ lie on the lateral side $BC$ of an isosceles triangle $ABC$ . The ray $AD$ meets the line passing through $B$ and parallel to the base $AC$ at point $E$ . Prove that the tangent to the circumcircle of triangle $ABD$ at $B$ bisects $EC$ . | <blockquote>A point $D$ lie on the lateral side $BC$ of an isosceles triangle $ABC$ . The ray $AD$ meets the line passing through $B$ and parallel to the base $AC$ at point $E$ . Prove that the tangent to the circumcircle of triangle $ABD$ at $B$ bisects $EC$ .</blockquote>
Well I did solve this that ... | [
"Let $BB$ and $EC$ meet at $M$ . Let parallel through $B$ to $EC$ meets $(B D A)$ at point $K$ and $AC$ at point $T$ . And parallel through $B$ to $AC$ meets $(B D A)$ at point $N$ . Then we have:\n $\\angle BCA=\\angle EBC=\\angle DKN$ and $\\angle DNK=\\angle CBT$ which means that $DKN... | [
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Let $ABC$ be acute-angled triangle with circumcircle $\Gamma$ . Points $H$ and $M$ are the orthocenter and the midpoint of $BC$ respectively. The line $HM$ meets the circumcircle $\omega$ of triangle $BHC$ at point $N\not= H$ . Point $P$ lies on the arc $BC$ of $\omega$ not containing $H$ in suc... | Solved with $\textbf{egxa}$ Let $HM$ intersects arc $BC$ of $\Gamma$ not containing $A$ at the point $D$ , and arc $BC$ of $\Gamma$ containing $A$ at the point $E$ . Let intersection of $HQ$ and $\Gamma$ be $R \neq Q$ . Consider homothety with center $M$ and ratio $-1$ . Then $H,B,C$ will be... | [
"Hint: Let ray $HM \\cap (ABC) = A'$ \nThen $A'$ lies on $(PQN)$ and $A'$ is center of $(AB'C')$ .\nIn some moment I remembered idea in problem 10 from Correspondence round and after this I solved this problem...",
"Nice and not easy problem, i will just post what i needed to solve. Let HM intersects (A... | [
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Let $H$ be the orthocenter of triangle $\mathrm T$ . The sidelines of triangle $\mathrm T_1$ pass through the midpoints of $\mathrm T$ and are perpendicular to the corresponding bisectors of $\mathrm T$ . The vertices of triangle $\mathrm T_2$ bisect the bisectors of $\mathrm T$ . Prove that the lines joinin... | Now I will write more detailes.
Let $AL_a=l_a, BL_b=l_b, CL_c=l_c$ be a angles bisectors, $AM_a=m_a,BM_b=m_b,CM_c=m_c$ be a midpoints of sides of $T=ABC$ , $I$ be its incenter. Also let $X_a,X_b,X_c$ be a vertices of $T_2$ and $P_a,P_b,P_c$ be a vertices of $T_1$ . If we will prove that $HX_a^2-HX_c^2=P_... | [
"Not very hard by Carno, medians and bisectors formulas\nMore details, maybe, I will add later",
"Also, at the some moment very good will be using this lemma:\nIn triangle $ABC$ $A_1, C_1$ lies on sides $BC, AB$ , Then $H$ lies on radial axe circles with diameters $AA_1,CC_1$ ."
] | [
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Let $ABC$ be a triangle with $\angle A = 120^\circ$ , $I$ be the incenter, and $M$ be the midpoint of $BC$ . The line passing through $M$ and parallel to $AI$ meets the circle with diameter $BC$ at points $E$ and $F$ ( $A$ and $E$ lie on the same semiplane with respect to $BC$ ). The line passing ... | Yayyy this was what gave me the H.M.,insane day 2 I will never forget this day in my life.
Here's an overview of the solution I found at the contest which uses five phantom points :lol:
<details><summary>Click to expand</summary>Consider $P'$ and $Q'$ as reflection of $C$ and $B$ over $BI$ , $CI$ respectivel... | [
"Very nice problem! \nHere is my solution, but some detailes I didn't wrote\nThe key of solving this problem is understand that points $B$ and $Q$ are symmetric onto line $CI$ and analogy with point $C$ . To prove it we can say that $C_1=CI \\cap (BEC), B_1 = BI \\cap (BEC)$ , after that let $Q'$ be a poi... | [
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} |
Let $ABC$ be an isosceles obtuse-angled triangle, and $D$ be a point on its base $AB$ such that $AD$ equals to the circumradius of triangle $BCD$ . Find the value of $\angle ACD$ . | Let $O$ be the center of $(BCD)$ and let $(BCD)$ intersect $AC$ at $T\neq C$ . Then by angle chasing $AD=DT$ and $DO=OT$ so triangle $DTO$ is equilateral. Now $\angle ACD=\angle TCD=\frac{1}{2}\angle TOD =30^{\circ}$ . | [
"which side is isos",
"Note that since $BC = AC$ , we have: $$ \\frac{BC}{\\sin \\angle BDC} = \\frac{AC}{\\sin \\angle ADC} $$ However this implies that (by the sine rule): $$ \\frac{AD}{\\sin \\angle ACD} = 2 \\cdot AD $$ which further implies that $\\angle ACD = 30^\\circ$ ."
] | [
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The bisectors of angles $A$ , $B$ , and $C$ of triangle $ABC$ meet for the second time its circumcircle at points $A_1$ , $B_1$ , $C_1$ respectively. Let $A_2$ , $B_2$ , $C_2$ be the midpoints of segments $AA_1$ , $BB_1$ , $CC_1$ respectively. Prove that the triangles $A_1B_1C_1$ and $A_2B_2C_2$ a... | Here’s a simple synthetic one.
Let $O,I$ be the circumcenter, incenter, respectively. Note that $OA_2\perp AA_1$ , so $\angle OA_2I=90^{\circ}$ . Thus, $A_2$ lies on the circle with diameter $OI$ . Similarly, $B_2,C_2$ also lie on that circle. Hence, $A_2,B_2,C_2,I$ are concyclic. We finish by angle chasin... | [
"Straightforward with complex bash :D ",
"<details><summary>complex bash!</summary>Let the circumircle of ABC be the unit circle and the complex coordinates of $A, B, C$ be $a^2, b^2, c^2$ respectively. Then $a_1= -bc, b_1= -ca, c_1= -ab$ , and $a_2= \\frac{a^2-bc}{2}, b_2= \\frac{b^2-ca}{2}, c_2= \\frac{c^... | [
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} |
The altitudes of a parallelogram are greater than $1$ . Does this yield that the unit square may be covered by this parallelogram? | [] | [
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Let $ABC$ be an acute-angled triangle, $O$ be its circumcenter, $BM$ be a median, and $BH$ be an altitude. Circles $AOB$ and $BHC$ meet for the second time at point $E$ , and circles $AHB$ and $BOC$ meet at point $F$ . Prove that $ME = MF$ . | Official Solution.
Let the extension of $B H$ meet the circumcircle at point $D$ . Prove that $E$ lies also on circles $D C O$ and $A D H$ . In fact let $E^{\prime}$ be the second common point of circles $A B O$ and $D C O$ . Then $\angle B E^{\prime} C=2 \pi-\angle B E^{\prime} O-$ $\angle C E^{\prime... | [
"This problem is easy, making an $\\sqrt{ac}$ - inversion , and radical center of 2 tangent circles and a secant that contains F, we obtain that B, F, E'(inverse of E), are collinear. I will post the solution in a few days. :D :P",
"<blockquote>This problem is easy, making an $\\sqrt{ac}$ - inversion , and radica... | [
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} |
The median $CM$ and the altitude $AH$ of an acute-angled triangle $ABC$ meet at point $O$ . A point $D$ lies outside the triangle in such a way that $AOCD$ is a parallelogram. Find the length of $BD$ , if $MO= a$ , $OC = b$ . | [
"Let the reflection of $O$ across $M$ be $P$ . $AO=BP=CD$ implies $BD=CP=b+2a$ "
] | [
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For which $n$ the plane may be paved by congruent figures bounded by $n$ arcs of circles? | [] | [
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The bisector of angle $A$ of triangle $ABC$ meet its circumcircle $\omega$ at point $W$ . The circle $s$ with diameter $AH$ ( $H$ is the orthocenter of $ABC$ ) meets $\omega$ for the second time at point $P$ . Restore the triangle $ABC$ if the points $A$ , $P$ , $W$ are given. | <blockquote>The bisector of angle $A$ of triangle $ABC$ meet its circumcircle $\omega$ at point $W$ . The circle $s$ with diameter $AH$ ( $H$ is the orthocenter of $ABC$ ) meets $\omega$ for the second time at point $P$ . Restore the triangle $ABC$ if the points $A$ , $P$ , $W$ are given.</blockqu... | [
"Assuming u mean, u can re-store $\\triangle ABC$ using only ruler and compass, then yes.\nClearly its known that we can draw perpendicular bisectors so we can draw the circumcenter of $\\triangle ABC$ , call it $O$ , now we draw $AO \\cap \\omega=A'$ , also using compass we can also draw the reflection of a p... | [
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Two circles $\omega_1$ and $\omega_2$ meeting at point $A$ and a line $a$ are given. Let $BC$ be an arbitrary chord of $\omega_2$ parallel to $a$ , and $E$ , $F$ be the second common points of $AB$ and $AC$ respectively with $\omega_1$ . Find the locus of common points of lines $BC$ and $EF$ . | Let $K$ be the second intersection of $\omega_1$ and $\omega_2$ . Let $KC$ meet $\omega_2$ at $T$ . Let the line parallel to $BC$ through $A$ meet $\omega_2$ at $L$ . Let $KL$ meet $EF$ at $D$ . Applying pascal on $EFALKT$ implies $D$ and $C$ and intersection of $AL,TE$ are collinear. Note... | [
"Let $D$ be the other intersection of the two circles. Now, $D$ is clearly the miquel point of $BCFE$ . Thus, if $X=BC\\cap EF$ then $X\\in (DBE)$ . Thus, $\\angle (DX, a)=\\angle DXB=\\angle DEB=\\angle DEA$ . Thus, $X$ lies on the line $\\ell$ through $D$ such that $\\angle (\\ell, a)$ is the sam... | [
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Let $M$ be the midpoint of cathetus $AB$ of triangle $ABC$ with right angle $A$ . Point $D$ lies on the median $AN$ of triangle $AMC$ in such a way that the angles $ACD$ and $BCM$ are equal. Prove that the angle $DBC$ is also equal to these angles. | Let $K$ be point such a $AKBC$ is parallelogram. So, $C, M, K$ are collinear. Since $\angle A = \pi/2$ , we have $\angle DAC = \angle ACN = \angle BKC$ . Also $\angle ACD=\angle BCM =\angle AKC$ .
We have $\Delta KBC \sim \Delta ADC$ . So $CB/CD=CK/CA \Rightarrow \Delta CBD \sim CKA \Rightarrow \angle CBD =... | [
"Let $K$ be the midpoint of $BC$ , then $MK//AC$ then we have $\\angle NAC=\\angle NCA=\\angle NMK$ and $\\angle KCM=\\angle DCA$ , which gives us that the triangles $ACD$ and $MCK$ are similar. Then $DC/CK=AC/MC$ , then $DC/(2CK)=(AC/2)/MC$ , then $DC/BC=MK/MC$ (Because $MK$ is midline) $(1)$ . ... | [
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"path": "Contest Collections/2023 Contests/2023 Sharygin Geometry Olympiad/3125013.json"
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The Euler line of a scalene triangle touches its incircle. Prove that this triangle is obtuse-angled. | <details><summary>There were mistakes in the previous post, now it is corrected.</summary>Let $s$ , $R$ and $r$ be the semi-perimeter, circumradius and inradius of $\triangle A{}BC$ . Let ${}I$ , ${}O$ , $G$ be the incentre, circumcentre and centroid of $\triangle A{}BC$ .
Then $OG$ is the Euler line and
\... | [] | [
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Let $\omega$ be the circumcircle of triangle $ABC$ , $O$ be its center, $A'$ be the point of $\omega$ opposite to $A$ , and $D$ be a point on a minor arc $BC$ of $\omega$ . A point $D'$ is the reflection of $D$ about $BC$ . The line $A'D'$ meets for the second time at point $E$ . The perpendicul... | Let $H$ be the orthocenter, $AH$ intersects $(ABC)$ again in $H'$ and $DH'$ intersects $D'H$ on $BC$ at $X$ .
We will prove that $H$ is the isogonal conjugate of $O$ in quad $FGBC$ , if so, then $\angle FOG=180^{\circ}-\angle BOC=180^{\circ}-2\angle BAC$ . To do that, we'll show that $X$ lies o... | [
" $\\angle AED'=\\angle AEA'=90^{\\circ}$ , so the midpoint of $AD'$ is the foot from $O$ onto the perpendicular bisector of $D'E$ , or the foot from $O$ to $\\overline{FG}$ is the midpoint of $AD'$ . Since $D'$ lies on the reflection of $D$ across $\\overline{BC}$ , the midpoint of $AD'$ is on the... | [
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"path": "Contest Collections/2023 Contests/2023 Sharygin Geometry Olympiad/3125018.json"
} |
Let $ABC$ be a Poncelet triangle, $A_1$ is the reflection of $A$ about the incenter $I$ , $A_2$ is isogonally conjugated to $A_1$ with respect to $ABC$ . Find the locus of points $A_2$ . | <details><summary>Solution of 10.4</summary>We claim that the desired locus is the radical axis of $I$ and the circumcircle. Let $M=AI \cap (ABC)$ and $I_a$ be the $A$ -excenter. We will show that $A_2I^2=A_2A \cdot A_2M$ , which will finish the problem. As $BI$ bisects $\angle A_2BA_1$ and $\angle IBI_a=9... | [
"Umm not sure if this works so tell me if I'm wrong\nLet $I_a$ be $A-$ excenter in $ABC$ . One can easily observe that $(I,I_a ; A1,A2) =-1$ . Now consider an inversion with center $I$ and radius $-2\\sqrt{pow(I,(ABC))}$ which swaps $A,I_a$ and it sends $A_1$ to the further intersection of $IA_1$ wit... | [
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"path": "Contest Collections/2023 Contests/2023 Sharygin Geometry Olympiad/3125019.json"
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The incircle of a triangle $ABC$ touches $BC$ at point $D$ . Let $M$ be the midpoint of arc $\widehat{BAC}$ of the circumcircle, and $P$ , $Q$ be the projections of $M$ to the external bisectors of angles $B$ and $C$ respectively. Prove that the line $PQ$ bisects $AD$ . | <details><summary>Solution of 10.5</summary>Add the excenters $I_b, I_c$ and notice that $P, Q$ are midpoints of $BI_c, CI_b$ . We shall show that $PQ$ is the perpendicular bisector of $AD$ , so we can rephrase the problem to the following orthocenter configuration:
Let $ABC$ be a triangle with orthic trian... | [
"Sketch: Draw $M_B, M_C$ sre the midpoints of minor arcs $AC, AB$ in $(BAC)$ respecitivily, let $B', C'$ the B, C antipodes in $(ABC)$ , since its known that $MM_BIM_C$ is a paralelogram (use I-E lemma to prove this) we have $M, M_C, P$ colinear and $M, M_B, Q$ colinear, let $AB' \\cap MM_C=K$ and ... | [
"origin:aops",
"2023 Contests",
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"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 Sharygin Geometry Olympiad/3125021.json"
} |
Let $E$ be the projection of the vertex $C$ of a rectangle $ABCD$ to the diagonal $BD$ . Prove that the common external tangents to the circles $AEB$ and $AED$ meet on the circle $AEC$ . | If $S$ is the intersection of the common external tangents, then $S$ is the exsimilicenter of $(AEB), (AED)$ . Let $O_B, O_D$ be their respective centers; then by spiral similarity, $\triangle{AO_BO_D} \sim \triangle{ABD}$ , so $(AES)$ is the Circle of Apollonius containing all points with distances to $O_B,... | [
"<details><summary>Solution of 10.6</summary>Let $T$ be the exsimilicenter of $(ABE)$ and $(ADE)$ and let $TA \\cap (ABE)=X, TE \\cap (ADE)=Y$ . Obviously $EX \\parallel AY$ due to the homothety. By angle chasing, we have that $\\angle ATE=\\angle AXE-\\angle XET=\\angle ABE-\\angle AYE=\\angle ABE-\\angl... | [
"origin:aops",
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"path": "Contest Collections/2023 Contests/2023 Sharygin Geometry Olympiad/3125022.json"
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There are $43$ points in the space: $3$ yellow and $40$ red. Any four of them are not coplanar. May the number of triangles with red vertices hooked with the triangle with yellow vertices be equal to $2023$ ? Yellow triangle is hooked with the red one if the boundary of the red triangle meet the part of the plan... | I can't find problems with this but it doesn't seem right.
The answer is no. Label the red vertices $1,2,\dots ,40$ and for each $i$ let $d_i$ be the number of red points $j$ for which $ij$ passes through the yellow triangle. $\sum d_i$ is even because it counts each pair of vertices whose segment passes t... | [
"I have solved this problem during the final round and I came up with a solution that I think is a little bit more elegant than the official solution using bipartite graphs and complex calculations. \n<details><summary>solution</summary>The answer is no. Let the yellow triangle be $T$ and its plane be $\\alpha$ ... | [
"origin:aops",
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"path": "Contest Collections/2023 Contests/2023 Sharygin Geometry Olympiad/3125023.json"
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A triangle $ABC$ is given. Let $\omega_1$ , $\omega_2$ , $\omega_3$ , $\omega_4$ be circles centered at points $X$ , $Y$ , $Z$ , $T$ respectively such that each of lines $BC$ , $CA$ , $AB$ cuts off on them four equal chords. Prove that the centroid of $ABC$ divides the segment joining $X$ and the ra... | Very Amazing Problem, **S**olved with **rjp08****<span style="color:#00f">Claim:</span>** $\odot(ABC)$ is Nine point circle of $\triangle ZTY$ **<span style="color:#f00">SubClaim</span>** : If $M$ is midpoint of $ZT$ . Then $M$ lie on $\cdot(ABC)$ .
Let $M_a,,Z',T'$ be feet of perpendicular from $M,Z,T$ t... | [
"Truly marvelous problem. Congratulations to the proposer!**Main claim:** $X, Y, Z, T$ form an orthocenter pair and their NPC is $\\odot ABC$ .\nPick any two of $X, Y, Z, T$ , wlog $X, Y$ . Consider the midpoint of $XY$ :\nBy the equal chords condition, its three pedals on $AB, BC, CA$ are on the radical ax... | [
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"path": "Contest Collections/2023 Contests/2023 Sharygin Geometry Olympiad/3125025.json"
} |
In a scalene triangle $ABC$ with centroid $G$ and circumcircle $\omega$ centred at $O$ , the extension of $AG$ meets $\omega$ at $M$ ; lines $AB$ and $CM$ intersect at $P$ ; and lines $AC$ and $BM$ intersect at $Q$ . Suppose the circumcentre $S$ of the triangle $APQ$ lies on $\omega$ and $A... | Beautiful problem! $\angle ACS = 90^{\circ}$ and because $AS = SQ$ , then $AC = CQ$ . Let $X = AM \cap BC$ and $Y = BG \cap AC$ . Using Menelaus' Theorem on $\triangle BCQ$ with the transversal $AM$ we find that $BM : MQ = 1 : 2$ . We apply Menelaus' again on $\triangle BYQ$ with the transversal $AM$ to... | [
"Well, let $X, Y$ be the intersections of $AM$ with $BC, PQ$ . Since $AS$ is diameter, we have $AB \\perp BS$ and $AC \\perp CS$ .\nThis implies that $AB = BP, AC = CQ$ . Thus we get $XM := d, MY = 2d, AX = 3d$ by looking at $\\triangle{APQ}$ .\n\nSince $G$ is centroid we also get $AG = 2GX = 2d$ ... | [
"origin:aops",
"2023 Contests",
"2023 Sinapore MO Open"
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"answer_score": 126,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2023 Contests/2023 Sinapore MO Open/3102141.json"
} |
A grid of cells is tiled with dominoes such that every cell is covered by exactly one domino. A subset $S$ of dominoes is chosen. Is it true that at least one of the following 2 statements is false?
(1) There are $2022$ more horizontal dominoes than vertical dominoes in $S$ .
(2) The cells covered by the dominoes ... | Solution with just 1 coloring:
Let $(i,j)$ be colored $0$ if $i$ and $j$ are even, $1$ if $i$ is odd and $j$ is even, $3$ if $i$ is even and $j$ is odd, and $2$ if both $i$ and $j$ are odd.
Assume for contradiction that both (1) and (2) can be true. Each horizontal domino covers squares such... | [
"Routine coloring problem :)\nAssume that it is possible for $S$ have $2022$ more horizontal dominoes than vertical dominoes, and the cells covered by the dominoes in $S$ can be fully tiled only with L-tetrominoes.\nLet the number of vertical dominoes be $n$ . Thus there are $2022+n$ horizontal dominoes an... | [
"origin:aops",
"2023 Contests",
"2023 Sinapore MO Open"
] | {
"answer_score": 40,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 Sinapore MO Open/3102143.json"
} |
Find all functions $f: \mathbb{Z} \to \mathbb{Z}$ , such that $$ f(x+y)((f(x) - f(y))^2+f(xy))=f(x^3)+f(y^3) $$ for all integers $x, y$ . | There are $4$ solutions work, which is $f\equiv x$ , $f=\begin{cases} \ 1\ \text{if} \ x\equiv 1\pmod 2 \ 2 \ \text{if} \ x\equiv 0\pmod {2} \end{cases}, \ f(x)\equiv 2, \ f(x)\equiv 0.$ Let $P(x,y)$ denote the origianl assertion : $f(x+y)((f(x) - f(y))^2+f(xy))=f(x^3)+f(y^3), \quad\forall x,y\in\mathbb{Z}$ .... | [
"Quite tedious :( \n\nAs always, let $P(x, y)$ denote the original assertion when $x, y \\in \\mathbb{Z}$ . Also note that throughout the entire solution, when we use modulo some number $z$ , we're also checking that $z \\neq 0$ . $P(x, x): f(2x)f(x^2) = 2f(x^3)$ meaning that $$ f(2)f(1) = 2f(1), \\,\\, f(... | [
"origin:aops",
"2023 Contests",
"2023 Sinapore MO Open"
] | {
"answer_score": 1164,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2023 Contests/2023 Sinapore MO Open/3102144.json"
} |
Let $n \geq 2$ be a positive integer. For a positive integer $a$ , let $Q_a(x)=x^n+ax$ . Let $p$ be a prime and let $S_a=\{b | 0 \leq b \leq p-1, \exists c \in \mathbb {Z}, Q_a(c) \equiv b \pmod p \}$ . Show that $\frac{1}{p-1}\sum_{a=1}^{p-1}|S_a|$ is an integer. | [
"[gghx is sad right now](https://artofproblemsolving.com/community/c6h2876088p25555422)",
"smh why repeat last year's qn at least very little people memorised the solution"
] | [
"origin:aops",
"2023 Contests",
"2023 Sinapore MO Open"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Sinapore MO Open/3102145.json"
} | |
Determine all real numbers $x$ between $0$ and $180$ such that it is possible to partition an equilateral triangle into finitely many triangles, each of which has an angle of $x^{o}$ .
| [
"If you're wondering how hard this is, a Thai IMO team member who is good at combi couldn't solve this in 3 hours",
"WHY IS MOVED HERE IT NEEDS TO BE MOVED BACK TO COLLEGE MATH",
"i'm actually interested to know the solution for this, anyone has it ...?",
"<blockquote>If you're wondering how hard this is, a T... | [
"origin:aops",
"2023 Contests",
"2023 Sinapore MO Open"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 Sinapore MO Open/3102149.json"
} | |
A $3\times 3\times 3$ cube is made of 27 unit cube pieces. Each piece contains a lamp, which can be on or off. Every time a piece is pressed (the center piece cannot be pressed), the state of that piece and the pieces that share a face with it changes. Initially all lamps are off. Determine which of the following sta... | [
"<details><summary>Solution</summary>(2) is possible by pressing all faces and edge cubes exactly once. \n(1) and (3) are not possible as the parity of the central one is always the same as of the sum of the six faces.</details>"
] | [
"origin:aops",
"2023 Contests",
"2023 Spain Mathematical Olympiad"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Spain Mathematical Olympiad/3029549.json"
} | |
Let $ABC$ be an acute scalene triangle with incenter $I$ and orthocenter $H$ . Let $M$ be the midpoint of $AB$ . On the line $AH$ we consider points $D$ and $E$ , such that the line $MD$ is parallel to $CI$ and $ME$ is perpendicular to $CI$ . Prove that $AE=DH$ . | Let $N$ be the midpoint of $AH$ . It suffices to show that $EN = ND$ . Let $X = MD \cap BC$ , and notice that $\angle HDX = \angle MDE = 90^\circ - \angle C/2$ . Now notice that by Thales $MN \parallel BH \iff \angle MND = \angle C$ . Since $MD \parallel CI$ , then $ME \perp MD$ and by angle chasing the resul... | [
"Let $K$ be a midpoint of $AH$ . It is enough to prove that $MK=KD$ . But this is evident, since $\\angle KDM=\\angle KMD=90^\\circ-\\frac{1}{2} \\angle ACB$ .",
"Alternatively, you can let $N$ be the midpoint of $ED$ and by angle chasing we can deduce $\\angle MNA = 180^{\\circ} - \\angle C = \\angle B... | [
"origin:aops",
"2023 Contests",
"2023 Spain Mathematical Olympiad"
] | {
"answer_score": 118,
"boxed": false,
"end_of_proof": true,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 Spain Mathematical Olympiad/3029551.json"
} |
Find all quadruples $(a,b,c,d)$ of positive integers satisfying that $a^2+b^2=c^2+d^2$ and such that $ac+bd$ divides $a^2+b^2$ . | The answer is $$ \boxed{(a,b,c,d)=(1,k,1,k)\quad k\in \mathbb{N}^+} $$ | [
"Probably something with $(a^2+b^2)(c^2+d^2) = (ac+bd)^2 - (ad-bc)^2$ Thus we also need $ac+bd|ad-bc$ then maybe something with $|ac+bd| \\le |ad - bc|$ ",
"Note that $(ac+bd)^2 + (ad-bc)^2 = (a^2+b^2)^2$ \nSo $\\exists r,s$ with $gcd(r,s)=1$ so that $ac+bd = g(r^2-s^2)$ ; $ad-bc = 2grs $ ; $a^2 +b^2... | [
"origin:aops",
"2023 Contests",
"2023 Spain Mathematical Olympiad"
] | {
"answer_score": 1004,
"boxed": false,
"end_of_proof": false,
"n_reply": 7,
"path": "Contest Collections/2023 Contests/2023 Spain Mathematical Olympiad/3029555.json"
} |
Let $x_1\leq x_2\leq x_3\leq x_4$ be real numbers. Prove that there exist polynomials of degree two $P(x)$ and $Q(x)$ with real coefficients such that $x_1$ , $x_2$ , $x_3$ and $x_4$ are the roots of $P(Q(x))$ if and only if $x_1+x_4=x_2+x_3$ . | I claim that, for $x_1+x_4=a=x_2+x_3$ and any real $b$ , $Q(x)=x^2-ax+b$ and $P(x)=(x-Q(x_1))(x-Q(x_2))$ satisfy the condition.
<details><summary>Proof</summary>( $\implies$ ) Note that if $x_1,x_2,x_3,x_4$ are the roots of $P(Q(x))$ then $Q(x_1),Q(x_2),Q(x_3),Q(x_4)$ are the roots of $P(x)$ ; however, ... | [
"If $a_1, a_2$ are the roots of $P$ , then the $x$ -s are the roots of $Q(x) - a_i$ and the conditions follows from Vieta's sum of roots formula. For construction take $Q(x) = x^2 - sx + p$ and $a_1 = 0$ , $a_2 = p-q$ where $s$ is the common sum, $p = x_1x_2$ and $q = x_3x_4$ ."
] | [
"origin:aops",
"2023 Contests",
"2023 Spain Mathematical Olympiad"
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"answer_score": 50,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Spain Mathematical Olympiad/3030715.json"
} |
We have a row of 203 cells. Initially the leftmost cell contains 203 tokens, and the rest are empty. On each move we can do one of the following:
1)Take one token, and move it to an adjacent cell (left or right).
2)Take exactly 20 tokens from the same cell, and move them all to an adjacent cell (all left or all right).... | <blockquote>
This is not correct, because the blocks of $20$ tokens do not need to be disjoint. In fact, it is possible to reach the final position in $2024$ moves without any token moving to the left nine times. For simplicity, let us do this for $40$ tokens. Your bound would imply that $240$ moves are necessa... | [
"Bump....",
"I will post my solution later.",
"We're glad you informed us :)",
"I posted that so that I don't forget xD",
"bumpity bump<blockquote>Bump....</blockquote>\n\n",
"Here's the long awaited solution.\n\nFind the minimum number of moves that could have taken place between any two adjacent cells b... | [
"origin:aops",
"2023 Contests",
"2023 Spain Mathematical Olympiad"
] | {
"answer_score": 108,
"boxed": false,
"end_of_proof": false,
"n_reply": 13,
"path": "Contest Collections/2023 Contests/2023 Spain Mathematical Olympiad/3030718.json"
} |
In an acute scalene triangle $ABC$ with incenter $I$ , the line $AI$ intersects the circumcircle again at $D$ , and let $J$ be a point such that $D$ is the midpoint of $IJ$ . Consider points $E$ and $F$ on line $BC$ such that $IE$ and $JF$ are perpendicular to $AI$ . Consider points $G$ on $AE$... | This is PEAK geo!
Let $X = AI \cap BC$ .
<span style="color:#f00">**Claim 1:**</span> $H, G \in (ABC)$ *Proof:* By similar triangles it's easy to see that $FH \cdot FA = FJ^2$ . Hence, we would like to show that $FH \cdot FA = FJ^2 = FC \cdot FB$ which by PoP would be sufficient. Indeed, by Fact 5 $BICJ$ is cyc... | [
" $G$ is just the point on $(ABC)$ such that $AI \\perp IG$ , and similarly, $H$ is just the point on $(ABC)$ such that $AJ \\perp JH$ (here $J$ is clearly the $A-excenter$ The proof of this is to consider phantom points, let $G$ be the point on $(ABC)$ such that $AI \\perp IG$ , and let $AG \\ca... | [
"origin:aops",
"2023 Contests",
"2023 Spain Mathematical Olympiad"
] | {
"answer_score": 154,
"boxed": false,
"end_of_proof": true,
"n_reply": 6,
"path": "Contest Collections/2023 Contests/2023 Spain Mathematical Olympiad/3030722.json"
} |
Let $ABC$ be an acute triangle with incenter $I$ . On its circumcircle, let $M_A$ , $M_B$ and $M_C$ be the midpoints of minor arcs $BC, CA$ and $AB$ , respectively. Prove that the reflection $M_A$ over the line $IM_B$ lies on the circumcircle of the triangle $IM_BM_C$ . | Rephrase with $ABC$ as $M_AM_BM_C$ .
<blockquote>Given triangle $ABC$ with orthocenter $H$ and orthic triangle $DEF$ , show that $A'$ , the reflection of $A$ about $F$ , lies on $(BHC)$ .</blockquote>Proof: Obvious since $\angle BA'C=\angle BAC=\angle FHB$ . | [
"**Outline from Contest:** $M_AM_C$ is perpendicular bisector of $BI$ and angle chase - nothing substantive. **Remark:**\nProblems are ordered by difficulty (mod 4), only $5$ of the problems were released due to confidentiality. "
] | [
"origin:aops",
"2023 Contests",
"2023 Switzerland - Final Round"
] | {
"answer_score": 20,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Switzerland - Final Round/3030776.json"
} |
The wizard Albus and Brian are playing a game on a square of side length $2n+1$ meters surrounded by lava. In the centre of the square there sits a toad. In a turn, a wizard chooses a direction parallel to a side of the square and enchants the toad. This will cause the toad to jump $d$ meters in the chosen directio... | **Outline from Contest:**
Brian mirrors Albus's moves, then take a screenshot of the square before each of Albus's moves and notice that the toad forms a path of unit distance steps parallel to the grid-lines; then it will take at least $n$ for the toad to reach the edge of the board in this case, in particular, Bria... | [] | [
"origin:aops",
"2023 Contests",
"2023 Switzerland - Final Round"
] | {
"answer_score": 10,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Switzerland - Final Round/3030777.json"
} |
Let $x,y$ and $a_0, a_1, a_2, \cdots $ be integers satisfying $a_0 = a_1 = 0$ , and $$ a_{n+2} = xa_{n+1}+ya_n+1 $$ for all integers $n \geq 0$ . Let $p$ be any prime number. Show that $\gcd(a_p,a_{p+1})$ is either equal to $1$ or greater than $\sqrt{p}$ . | **Outline from Contest:**
Very standard idea, you can do similar things for Pisano period for Fibonacci, let $q$ be a prime.
If $q$ is a divisor of $y$ then $q \nmid gcd(a_n,a_{n+1})$ for $n \geq 1$ is easy to see.
If $gcd(q,y) = 1$ , then take by pigeonhole some pair of pairs $(a_i,a_{i+1}) = (a_j,a_{j+1... | [
"We uploaded our solution [https://calimath.org/pdf/SwissMO2023-3.pdf](https://calimath.org/pdf/SwissMO2023-3.pdf) on youtube [https://youtu.be/AzB3Q0aSshU](https://youtu.be/AzB3Q0aSshU)."
] | [
"origin:aops",
"2023 Contests",
"2023 Switzerland - Final Round"
] | {
"answer_score": 58,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Switzerland - Final Round/3030778.json"
} |
Determine the smallest possible value of the expression $$ \frac{ab+1}{a+b}+\frac{bc+1}{b+c}+\frac{ca+1}{c+a} $$ where $a,b,c \in \mathbb{R}$ satisfy $a+b+c = -1$ and $abc \leqslant -3$ | **Outline from Contest:**
This solution is quite creative, you do not need to be this creative. We replace $1$ by $(a+b+c)^2$ and let $S$ be the sum. We claim that $S = 3$ is the minimum which is achieved when $(a,b,c)$ is a permutation of $(-3,1,1)$ . It is then possible to see that if we define $x = a+b, ... | [
"Let $a,b,c$ be reals such that $a+b+c =1$ and $abc\\geq 3.$ Prove that $$ \\frac{ab+1}{a+b}+\\frac{bc+1}{b+c}+\\frac{ca+1}{c+a}\\leq -3 $$ $$ \\frac{ab+c}{a+b}+\\frac{bc+a}{b+c}+\\frac{ca+b}{c+a}\\leq -6 $$ \n\n<details><summary>Problem 502</summary>Let $a$ , $b$ and $c$ be non-negative real n... | [
"origin:aops",
"2023 Contests",
"2023 Switzerland - Final Round"
] | {
"answer_score": 46,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2023 Contests/2023 Switzerland - Final Round/3030779.json"
} |
Let $D$ be the set of real numbers excluding $-1$ . Find all functions $f: D \to D$ such that for all $x,y \in D$ satisfying $x \neq 0$ and $y \neq -x$ , the equality $$ (f(f(x))+y)f \left(\frac{y}{x} \right)+f(f(y))=x $$ holds. | <blockquote>Let $D$ be the set of real numbers excluding $-1$ . Find all functions $f: D \to D$ such that for all $x,y \in D$ satisfying $x \neq 0$ and $y \neq -x$ , the equality $$ (f(f(x))+y)f \left(\frac{y}{x} \right)+f(f(y))=x $$ holds.</blockquote> $\color{blue}\boxed{\textbf{Answer:}f\equiv \frac{1-... | [
"**Outline from Contest:**\nTake $y = x$ , $y=0$ and $x=1$ and conclude that $f(x) = \\frac{2}{x+1}-1$ - boring problem. **Remark:**\nProblems are ordered by difficulty (mod 4), only $5$ of the problems were released due to confidentiality. \n\n@below\nI meant do these three things separately but yes rules... | [
"origin:aops",
"2023 Contests",
"2023 Switzerland - Final Round"
] | {
"answer_score": 1244,
"boxed": true,
"end_of_proof": true,
"n_reply": 5,
"path": "Contest Collections/2023 Contests/2023 Switzerland - Final Round/3030781.json"
} |
Find all positive integers $n$ satisfying the following conditions simultaneously:
(a) the number of positive divisors of $n$ is not a multiple of $8$ ;
(b) for all integers $x$ , we have
\[x^n \equiv x \mod n.\]
*Proposed by usjl* | The answers are $1$ and all primes.
The verification is easy, so we prove that these are the only solutions.
We begin by showing that $n$ is squarefree. Let $p$ be an arbitrary prime divisor of $n$ . Substituting $x=p$ gives us $p^n \equiv p \pmod{n}$ , so we have $\gcd(p^n, n)=\gcd(p, n)$ . It follows tha... | [
"Remark: As the title suggests, this is can be solved by a really well-known statement, basically saying that the Carmichael number is square-free has at least three distinct prime factors.",
"Made a small mistake in the test, might cost me TST qualification\n\n<details><summary>Solution</summary>It's easy to see... | [
"origin:aops",
"2023 Contests",
"2023 Taiwan Mathematics Olympiad"
] | {
"answer_score": 140,
"boxed": false,
"end_of_proof": false,
"n_reply": 8,
"path": "Contest Collections/2023 Contests/2023 Taiwan Mathematics Olympiad/3011005.json"
} |
Let $O$ be the center of circle $\Gamma$ , and $A$ , $B$ be two points on $\Gamma$ so that $O, A$ and $B$ are not collinear. Let $M$ be the midpoint of $AB$ . Let $P$ and $Q$ be points on $OA$ and $OB$ , respectively, so that $P \neq A$ and $P, M, Q$ are collinear. Let $X$ be the intersectio... | <details><summary>complex bash</summary>Identify $\Gamma$ with the unit circle, and assume $X$ lies on $\Gamma$ , so that $$ |a|=|b|=|x|=1 $$ $$ o = 0 $$ $$ m = \frac{a+b}2 $$ Now $P$ lies on line $OA$ and also on the line through $X$ parallel to $AB$ , so $$ p = \frac{a(-a)\left(x+\frac{ab}x\righ... | [
"Note that $Y$ lies on $\\Gamma$ if and only if $$ AO - X = OX - B + 90^\\circ \\iff 3X = 2A + B + 90^\\circ. $$ Let $P'$ be the reflection point of $P$ with respect to $OM$ . Then $$ X(O, B; P', Q) = (O, B; P', Q) = M(O, B; P', P) = -1, $$ i.e., $XP'$ is one of the angle bisectors of $\\angle O... | [
"origin:aops",
"2023 Contests",
"2023 Taiwan Mathematics Olympiad"
] | {
"answer_score": 220,
"boxed": false,
"end_of_proof": false,
"n_reply": 7,
"path": "Contest Collections/2023 Contests/2023 Taiwan Mathematics Olympiad/3011008.json"
} |
Let $n$ and $k$ be positive integers. Let $A$ be a set of $2n$ distinct points on the Euclidean plane such that no three points in $A$ are collinear. Some pairs of points in $A$ are linked with a segment so that there are $n^2 + k$ distinct segments on the plane. Prove that there exists at least $\frac{4... | Consider a graph $G=(V, E)$ with $V=A$ and $E$ containing the $n^2+k$ segments.
Let $u, v$ be two adjacent vertices, then the number of triangles (suppose is $f(u, v)$ ) that contains $u, v$ is the number of common neighbors of $u, v$ . $\Rightarrow f(u, v)\geq(\deg(u)-1)+(\deg(v)-1)-(2n-2)=\deg(u)+\deg(v... | [
"This is pretty standard. See here for an even stronger bound:\n[https://math.stackexchange.com/questions/3575265/lower-bound-for-number-of-triangles-in-simple-graph-very-hard-exercise?fbclid=IwAR3VbBPeeqv5J1a-Dk8QzMQiGRAoxMQLCNVSjlYabhxx205WuIlmZKcrIdc](https://math.stackexchange.com/questions/3575265/lower-bound-... | [
"origin:aops",
"2023 Contests",
"2023 Taiwan Mathematics Olympiad"
] | {
"answer_score": 32,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Taiwan Mathematics Olympiad/3011009.json"
} |
Let $m$ be a positive integer, and real numbers $a_1, a_2,\ldots , a_m$ satisfy
\[\frac{1}{m}\sum_{i=1}^{m}a_i = 1,\]
\[\frac{1}{m}\sum_{i=1}^{m}a_i ^2= 11,\]
\[\frac{1}{m}\sum_{i=1}^{m}a_i ^3= 1,\]
\[\frac{1}{m}\sum_{i=1}^{m}a_i ^4= 131.\]
Prove that $m$ is a multiple of $7$ .
*Proposed by usjl*
| <blockquote>Consider $b_i=a_i+0.5$ , we have $$ \sum b_i=\frac{3m}{2},\quad\sum b_i^2=\frac{49m}{4},\quad\sum b_i^3=\frac{147m}{8},\quad\sum b_i^4=\frac{2401m}{16} $$ by simple calculation. Note that $$ m\left(\sum b_i^4\right)=\left(\sum b_i^2\right)^2. $$ By Cauchy’s inequality, $b_i^2$ is constant in $i$ ,... | [
"A remark: The original statement asks to find all positive integers $m$ so that there are real numbers $a_1,\\ldots, a_m$ satisfying the equations. ",
"Consider $b_i=a_i+0.5$ , we have $$ \\sum b_i=\\frac{3m}{2},\\quad\\sum b_i^2=\\frac{49m}{4},\\quad\\sum b_i^3=\\frac{147m}{8},\\quad\\sum b_i^4=\\frac{240... | [
"origin:aops",
"2023 Contests",
"2023 Taiwan Mathematics Olympiad"
] | {
"answer_score": 44,
"boxed": false,
"end_of_proof": false,
"n_reply": 11,
"path": "Contest Collections/2023 Contests/2023 Taiwan Mathematics Olympiad/3011010.json"
} |
Let $n$ and $m$ be positive integers. The daycare nanny uses $n \times m$ square floor mats to construct an $n \times m$ rectangular area, with a baby on each of the mats. Each baby initially faces toward one side of the rectangle. When the nanny claps, all babies crawl one mat forward in the direction it is fa... | <details><summary>Solution</summary>All possible values of $n$ and $m$ are $(n, m)=(2k, 2l)$ for some $k, l\in\mathbb N$ .
Construction: color the rectangle floor mats like this: $$ \begin{array}{|c|c|c|c|c|c|c|}\hline
R&D&R&D&\cdots&R&D\hline
U&L&U&L&\cdots&U&L\hline
R&D&R&D&\cdots&R&D\hline
U&L&U&L&\cdots&U&... | [
"<blockquote>Let $n$ and $m$ be positive integers. The daycare nanny uses $1\\times 1$ square floor mats to construct an $n \\times m$ rectangular area, with a baby on each of the mats. Each baby initially faces toward one side of the rectangle. When the nanny claps, all babies crawl one mat forward in the ... | [
"origin:aops",
"2023 Contests",
"2023 Taiwan Mathematics Olympiad"
] | {
"answer_score": 52,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 Taiwan Mathematics Olympiad/3021144.json"
} |
Let $f:\mathbb{N}\to\mathbb{R}_{>0}$ be a given increasing function that takes positive values. For any pair $(m,n)$ of positive integers, we call it *disobedient* if $f(mn)\neq f(m)f(n)$ . For any positive integer $m$ , we call it *ultra-disobedient* if for any nonnegative integer $N$ , there are always infinit... | Let's prove by contradiction, and suppose that there exists some disobedient pair, but there doesn't exist some ultra-disobedient positive integer.
We can see that $f(1)=1$ , otherwise $1$ is ultra-disobedient.
If $m$ is not ultra-disobedient, then $\exists N$ s.t. $\not\exists n$ s.t. $(m, n), (m, n+1), \d... | [
"<blockquote>Let $f:\\mathbb{N}\\to\\mathbb{R}_{>0}$ be a given increasing function that takes positive values. For any pair $(m,n)$ of positive integers, we call it *disobedient* if $f(mn)\\neq f(m)f(n)$ . For any positive integer $m$ , we call it *ultra-disobedient* if for any nonnegative integer $N$ , the... | [
"origin:aops",
"2023 Contests",
"2023 Taiwan TST Round 1"
] | {
"answer_score": 94,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 Taiwan TST Round 1/3032689.json"
} |
Let $ABC$ be a triangle. Let $ABC_1, BCA_1, CAB_1$ be three equilateral triangles that do not overlap with $ABC$ .
Let $P$ be the intersection of the circumcircles of triangle $ABC_1$ and $CAB_1$ .
Let $Q$ be the point on the circumcircle of triangle $CAB_1$ so that $PQ$ is parallel to $BA_1$ . Let $R... | Let the second intersection of $\odot (ABC_1)$ and $BC$ be $X$ , the second intersection of $\odot(CAB_1)$ and $BC$ be $Y$ .
Then it is easy to check $AX \parallel CA_1 \parallel PR$ and $AY\parallel BA_1 \parallel PQ$ by angle chasing. This shows that $\triangle AXY$ is an equilateral triangle.
Therefo... | [
"Suppose the barycentric coordinates of points $P, Q, R$ are $(p_a, p_b, p_c), (q_a, q_b, q_c), (r_a, r_b, r_c)$ , respectively. It suffices to prove that $\\frac{1}{3}(p_a +q_a +r_a)=\\frac{1}{3}$ , which is equivalent as the following equation: $$ | \\triangle PBC| + | \\triangle QBC| + | \\triangle RBC| = |... | [
"origin:aops",
"2023 Contests",
"2023 Taiwan TST Round 1"
] | {
"answer_score": 172,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 Taiwan TST Round 1/3032690.json"
} |
Given some monic polynomials $P_1, \ldots, P_n$ with real coefficients, for any real number $y$ , let $S_y$ be the set of real number $x$ such that $y = P_i(x)$ for some $i = 1, 2, ..., n$ . If the sets $S_{y_1}, S_{y_2}$ have the same size for any two real numbers $y_1, y_2$ , show that $P_1, \ldots, P_n... | <details><summary>Solution</summary>W.l.o.g. no two of the $P_i$ are equal. Then there are only finitely many $x$ with $P_{i_1}(x)=P_{i_2}(x)$ for some $i_1 \ne i_2$ .
In particular, for all but finitely many $y$ , the size of $S_y$ is just the sum of the sizes of the $P_i^{-1}(y)$ .
Hence, if $n_1$ of the... | [
"Really similar solution to @above\n\n<details><summary>Solution</summary>Claim 1: all polynomials have odd degree.\nProof: Let there be some even degree polynomials.Let the largest extremal value of all $P_i$ be $T$ , and the smallest be $T'$ . Now take $y_1=T+1$ and $y_2=T'-1$ to find that $|S_{y_1}|>|S_... | [
"origin:aops",
"2023 Contests",
"2023 Taiwan TST Round 1"
] | {
"answer_score": 64,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Taiwan TST Round 1/3033414.json"
} |
There are $n$ cities on each side of Hung river, with two-way ferry routes between some pairs of cities across the river. A city is “convenient” if and only if the city has ferry routes to all cities on the other side. The river is “clear” if we can find $n$ different routes so that the end points of all these rout... | The answer is $\boxed{n-1}$ .Just use **Hall theorem**.
It's too easy for a TST exam. | [
"<details><summary>Solution</summary>Construct a graph whose vertices are cities and edges are ferry routes.\n\nThe answer is $n-1$ .\n\nLet $A, B$ be the set of the cities on the left side and the right side of Hung river, respectively.\nLet $C, D$ be the set of convenient cities in $A, B$ , respectively.\nL... | [
"origin:aops",
"2023 Contests",
"2023 Taiwan TST Round 1"
] | {
"answer_score": 1002,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 Taiwan TST Round 1/3033415.json"
} |
Let $\Omega$ be the circumcircle of an isosceles trapezoid $ABCD$ , in which $AD$ is parallel to $BC$ . Let $X$ be the reflection point of $D$ with respect to $BC$ . Point $Q$ is on the arc $BC$ of $\Omega$ that does not contain $A$ . Let $P$ be the intersection of $DQ$ and $BC$ . A point $E$ s... | Let $\measuredangle$ denote directed angles modulo $180^\circ$ .
After some angle chasing, we can reduce the problem to the following.
<blockquote>In $\triangle QBC$ , let $P$ be a point on $\overline{BC}$ , and let the isogonal of $\overline{QP}$ in $\angle BQC$ intersect the circumcircle $\Omega$ of $Q... | [
"Let point $R$ on $\\odot\\Omega$ satisfy $AQ=AR$ . Easy to see that $DR\\parallel PX\\parallel EQ$ .\nLet’s be on the coordinate plane that the unit circle is $\\odot\\Omega$ and y-axis is parallel to $EQ$ .\nLet $A(\\cos\\alpha,\\sin\\alpha)$ and $Q(\\cos(\\alpha+2\\beta),\\sin(\\alpha+2\\beta))$ .\nT... | [
"origin:aops",
"2023 Contests",
"2023 Taiwan TST Round 1"
] | {
"answer_score": 94,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 Taiwan TST Round 1/3033428.json"
} |
Let $\mathbb{Q}_{>1}$ be the set of rational numbers greater than $1$ . Let $f:\mathbb{Q}_{>1}\to \mathbb{Z}$ be a function that satisfies
\[f(q)=\begin{cases}
q-3&\textup{ if }q\textup{ is an integer,}
\lceil q\rceil-3+f\left(\frac{1}{\lceil q\rceil-q}\right)&\textup{ otherwise.}
\end{cases}\]
Show that for any ... | Repeatedly use the fact $f( \frac{ak-r}{k} )=a-3+ f(\frac{k}{r})$ and do induction on $x$ which we can assume $a=\frac{x}{x-y}$ and $b=\frac{x}{y}$ . | [
"But how was the problem proposed?",
"I will say more about the motivation later :3",
"Sorry for the long wait! Section 3.9 in [https://www.overleaf.com/read/bfbzksthmydz](https://www.overleaf.com/read/bfbzksthmydz) explains where it comes from: the tl;dr is that it comes from properties of Hirzebruch-Jung cont... | [
"origin:aops",
"2023 Contests",
"2023 Taiwan TST Round 1"
] | {
"answer_score": 8,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 Taiwan TST Round 1/3034051.json"
} |
For every positive integer $M \geq 2$ , find the smallest real number $C_M$ such that for any integers $a_1, a_2,\ldots , a_{2023}$ , there always exist some integer $1 \leq k < M$ such that
\[\left\{\frac{ka_1}{M}\right\}+\left\{\frac{ka_2}{M}\right\}+\cdots+\left\{\frac{ka_{2023}}{M}\right\}\leq C_M.\]
Here,... | Hopefully this is at least close:
Answer: $1011 + 1/M$ Construction: $(x, x, .... M-x, M-x, .. M-x)$ where $(M, x) = 1$ and there are 1011 of x.
Bound:
Define the sum when we plug in $x$ to be $F(x)$ . It's easy to see that $E[F(x)] \leq 1011 + \frac{1}{2}$ . Define P = $a_1 + a_2 + ... + a_{2023}$ and $(M,... | [
"Cleaner way to get bound:\nDefine $F(x)$ as previously. So $F(x) + F(M-x) \\leq 2023$ for all $x$ .\nDefine $P = a_1 + a_2 + ... a_{2023}$ and $N = (M, P)$ . \nCase 1. N = 1. Take x such that fractional part of $F(x)$ is $\\frac{1}{M}$ then $F(x) + F(M-x) \\geq (1012 + \\frac{1}{M}) + (1011 + \\frac{M... | [
"origin:aops",
"2023 Contests",
"2023 Taiwan TST Round 1"
] | {
"answer_score": 42,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 Taiwan TST Round 1/3034666.json"
} |
Let $k$ be a positive integer, and set $n=2^k$ , $N=\{1, 2, \cdots, n\}$ . For any bijective function $f:N\rightarrow N$ , if a set $A\subset N$ contains an element $a\in A$ such that $\{a, f(a), f(f(a)), \cdots\} = A$ , then we call $A$ as a cycle of $f$ . Prove that: among all bijective functions $f:N\r... | We can prove a much better bound then $2k-1$ .
Costruct the function in the following way:
Choose $a$ at random.
Choose $f(a)$ .
Set $b=f(a)$ and choose $f(b)$ .
And so on.
When we have $f(x)=a$ we have formed a new cicle.
Now choose randomly a different element and repeat this.
At every step we create a new ... | [
"The cases $k=1$ and $k=2$ can be checked readily, we focus only on $k\\ge 3$ .\n\nConsider a random function. Let $X$ be the number of cycles it contains, and let $X_i$ be the number of cycles it contains of length $i$ . Note that we have\n\\[ \\mathbb E[X_i] = \\binom{n}{i} \\frac {(i-1)! (n-i)!}{n!}=\\... | [
"origin:aops",
"2023 Contests",
"2023 Taiwan TST Round 1"
] | {
"answer_score": 80,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 Taiwan TST Round 1/3034699.json"
} |
Find all $f:\mathbb{N}\to\mathbb{N}$ satisfying that for all $m,n\in\mathbb{N}$ , the nonnegative integer $|f(m+n)-f(m)|$ is a divisor of $f(n)$ .
*Proposed by usjl* | Solved with **dragoon**.
The answer is $\boxed{f(x) = cx}$ for any positive integer $c$ . These work. Now we prove they are the only solutions.
Let $P(m,n)$ denote the assertion that \[ f(m+n) - f(m) \mid f(n)\] $P(m,1): f(m+1) - f(m) \mid f(1)$ . $P(1,m): f(m+1) - f(1) \mid f(m)$ .
For $a>b$ , $P(b, a-... | [
"This was proposed to 2022 IMO but sadly did not make to ISL. I am a bit disappointed after seeing how whack the N section of 2022 ISL is :/",
"Since $0$ is not a divisor of any positive integer we get that $f(m+n)\\neq f(m)$ so $f$ is injective.\n\nNow, for $m=1$ : $f(n+1)-f(1)|f(n)$ $n=1$ : $f(m+1)-f(... | [
"origin:aops",
"2023 Contests",
"2023 Taiwan TST Round 1"
] | {
"answer_score": 1198,
"boxed": true,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2023 Contests/2023 Taiwan TST Round 1/3109051.json"
} |
Let $f_n$ be a polynomial with real coefficients for all $n \in \mathbb{Z}$ . Suppose that
\[f_n(k) = f_{n+k}(k) \quad n, k \in \mathbb{Z}.\]
(a) Does $f_n = f_m$ necessarily hold for all $m,n \in \mathbb{Z}$ ?
(b) If furthermore $f_n$ is a polynomial with integer coefficients for all $n \in\mathbb{Z}$ , does... | <details><summary>Solution for (a)</summary>The answer is **No**.**<span style="color:#f00">Lemma 1.</span>** Let $n$ be a positive integer. Let $x_1,\ldots,x_n$ be pairwise distinct nonnegative real numbers and let $y_1,\ldots,y_n$ be real numbers. Then there exists a nonconstant *even* polynomial $f(x)\in\math... | [
"Do you mean that $\\{f_n(x)\\}_{n\\in\\mathbb Z}$ is a sequence of polynomials in $\\mathbb R[x]$ ?",
"Take $f_0(x)=0$ ... $f_{m} (x) =\\prod\\limits_{k=0}^{|m|} (x^2-k^2)$ ",
"Above: are you sure? $f_1(2)=12$ whereas $f_3(2)=0$ .",
"<blockquote>Do you mean that $\\{f_n(x)\\}_{n\\in\\mathbb Z}$ is ... | [
"origin:aops",
"2023 Contests",
"2023 Taiwan TST Round 2"
] | {
"answer_score": 282,
"boxed": false,
"end_of_proof": false,
"n_reply": 10,
"path": "Contest Collections/2023 Contests/2023 Taiwan TST Round 2/3046235.json"
} |
Find all functions $f : \mathbb{R} \to \mathbb{R}$ , such that $$ f\left(xy+f(y)\right)f(x)=x^2f(y)+f(xy) $$ for all $x,y \in \mathbb{R}$ *Proposed by chengbilly* | <blockquote>Find all functions $f : \mathbb{R} \to \mathbb{R}$ , such that $$ f\left(xy+f(y)\right)f(x)=x^2f(y)+f(xy) $$ for all $x,y \in \mathbb{R}$ *Proposed by chengbilly*</blockquote>
Let $P(x,y)$ be the assertion of the problem. One answer is $\boxed{f(x)=0\; \forall x\in \mathbb R}$ . So Let's assume that... | [
"<details><summary>Solution</summary>Let $P(x,y)$ be the assertion. $P(x,0) \\implies f(f(0))f(x)=x^2f(0)+f(0)$ , if $f(0)\\neq 0$ then we have $f(x)=cx^2+d$ for some $c,d\\in \\mathbb{R}$ , plugging back gives $f\\equiv 0$ , contradicting $f(0)\\neq 0$ , thus $f(0)=0$ .\n\nIf $f(t)=0$ for some $t\\neq... | [
"origin:aops",
"2023 Contests",
"2023 Taiwan TST Round 2"
] | {
"answer_score": 1134,
"boxed": true,
"end_of_proof": false,
"n_reply": 10,
"path": "Contest Collections/2023 Contests/2023 Taiwan TST Round 2/3046244.json"
} |
Is there a scalene triangle $ABC$ similar to triangle $IHO$ , where $I$ , $H$ , and $O$ are the incenter, orthocenter, and circumcenter, respectively, of triangle $ABC$ ?
*Proposed by Li4 and usjl.* | <blockquote>No
<details><summary>A cute diagram</summary>[asy]
import graph; size(25.cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-10.,xmax=15.,ymin=-6.,ymax=7.;
pair A=(-2.,4.), B=(-4.,-4.), C=(6.,-4.), I=(-0.53374862387472,-1.2936327068545446), H=(-2.,-2.), O=(1.,-... | [
"Lol I really think this problem is funny. Give it a try!",
"I guess the triangle doesn't exist but I don't have a proof. :(",
"<blockquote>I guess the triangle doesn't exist but I don't have a proof. :(</blockquote>\n\nIs this problem a counterexample to LoloChen's theorem?",
"No\n<details><summary>A cute di... | [
"origin:aops",
"2023 Contests",
"2023 Taiwan TST Round 2"
] | {
"answer_score": 90,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2023 Contests/2023 Taiwan TST Round 2/3046871.json"
} |
For each positive integer $k$ greater than $1$ , find the largest real number $t$ such that the following hold:
Given $n$ distinct points $a^{(1)}=(a^{(1)}_1,\ldots, a^{(1)}_k)$ , $\ldots$ , $a^{(n)}=(a^{(n)}_1,\ldots, a^{(n)}_k)$ in $\mathbb{R}^k$ , we define the score of the tuple $a^{(i)}$ as
\[\prod_{... | The reason I name this "Joints type inequality" is that as observed by YaWNeeT, the problem can actually be solved by modifying the proof of the following joints theorem: Corollary 6.8 in [https://arxiv.org/abs/2307.15380](https://arxiv.org/abs/2307.15380).
If you care enough and think about how this theorem relates t... | [] | [
"origin:aops",
"2023 Contests",
"2023 Taiwan TST Round 2"
] | {
"answer_score": 12,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Taiwan TST Round 2/3046874.json"
} |
Find all polynomials $P$ with real coefficients satisfying that there exist infinitely many pairs $(m, n)$ of coprime positives integer such that $P(\frac{m}{n})=\frac{1}{n}$ .
*Proposed by usjl* | If $P(x)$ is constant, then we must have $P(x)\equiv \frac 1k$ for some positive integer $k$ .
Assume that $deg(P)\ge 1$ . Clearly, there are infinitely many rational numbers $r$ such that $P(r)\in\mathbb{Q}$ . Then, by the Lagrange Interpolation formula, we know that $P$ has rational coefficients. Let $P(... | [
"<blockquote>If $P(x)$ is constant, then we must have $P(x)\\equiv \\frac 1k$ for some positive integer $k$ .\n\nAssume that $deg(P)\\ge 1$ . Clearly, there are infinitely many rational numbers $r$ such that $P(r)\\in\\mathbb{Q}$ . Then, by the Lagrange Interpolation formula, we know that $P$ has rationa... | [
"origin:aops",
"2023 Contests",
"2023 Taiwan TST Round 2"
] | {
"answer_score": 198,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2023 Contests/2023 Taiwan TST Round 2/3046876.json"
} |
Let $\Omega$ be the circumcircle of an acute triangle $ABC$ . Points $D$ , $E$ , $F$ are the midpoints of the inferior arcs $BC$ , $CA$ , $AB$ , respectively, on $\Omega$ . Let $G$ be the antipode of $D$ in $\Omega$ . Let $X$ be the intersection of lines $GE$ and $AB$ , while $Y$ the intersection... | Since $DA\perp BC,EB\perp AC,FC\perp AB$ we can rephrase the problem like below.**<span style="color:#0f0">New Problem Statement: </span>** $ABC$ is a triangle with altitudes $AD,BE,CF$ with $D,E,F\in (ABC)$ . $H$ is the orthocenter of $ABC$ . $A'$ is the antipode of $A$ and $A'B,A'C$ intersect $DE,DF$ ... | [
"Let $K$ and $L$ be the midpoints of the major arcs $AB$ and $AC$ . Simple angle-chasing gives us $GE//FC//LD$ and $GF//EB//KD$ . Let $P$ and $Q$ be the feet of the altitudes drawn from $D$ to $FC$ and $EB$ . Let $M,N,U,V$ be the midpoints of $[FY], [FC], [EX], [EB].$ We need to show that $\\... | [
"origin:aops",
"2023 Contests",
"2023 Taiwan TST Round 2"
] | {
"answer_score": 154,
"boxed": false,
"end_of_proof": true,
"n_reply": 6,
"path": "Contest Collections/2023 Contests/2023 Taiwan TST Round 2/3047540.json"
} |
Integers $n$ and $k$ satisfy $n > 2023k^3$ . Kingdom Kitty has $n$ cities, with at most one road between each pair of cities. It is known that the total number of roads in the kingdom is at least $2n^{3/2}$ . Prove that we can choose $3k + 1$ cities such that the total number of roads with both ends being a c... | **s**olved with Rushil, Adhitya, Ananda, Kanav, Siddharth, Malay
<details><summary>solution</summary>Lemma: If I have a graph $G$ on $m$ vertices with more than $m^{3/2}$ edges, then I can find a square in $G$ .
Proof: For every vertex $u$ I first mark with $u$ all pairs of vertices $(v, w)$ for $v \ne w... | [
"We present a sketch; the approximations here can easily be made more rigorous.\n\nObviously, the average degree is $4\\sqrt n$ . The expected number of $K_{1,2}$ s in the graph is\n\\[ \\sum_v \\binom{d(v)}2 \\ge \\frac{n(4\\sqrt n)^2}{2}=8n^2\\] by Jensen.\n\nIf $k=1$ , then choose two vertices at random. The ... | [
"origin:aops",
"2023 Contests",
"2023 Taiwan TST Round 2"
] | {
"answer_score": 168,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 Taiwan TST Round 2/3047605.json"
} |
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