problem string | solution string | candidates list | tags list | metadata dict |
|---|---|---|---|---|
Let $p\geq 3$ be an odd positive integer. Show that $p$ is prime if and only if however we choose $(p+1)/2$ pairwise distinct positive integers, we can find two of them, $a$ and $b$ , such that $(a+b)/\gcd(a,b)\geq p.$ | I get the feel I have seen this somewhere before.
<details><summary>Solution</summary>First assume $p$ is prime and let the positive integers be $a_1, a_2, \cdots, a_{p+1/2}$ . Since the values $\frac{a+b}{\gcd(a, b)}$ remain unaffected by scaling, divide all the $a_i$ with $\gcd(a_1, a_2, \cdots, a_n)$ . Now s... | [
"[2007 China MO](https://artofproblemsolving.com/community/c6h379725p2100299)",
"when I see the problem,I notice that it is same as 2007 CMO P2.\n"
] | [
"origin:aops",
"2022 Contests",
"2022 Romania EGMO TST"
] | {
"answer_score": 56,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Romania EGMO TST/2781232.json"
} |
The incircle and the excircle of triangle $ABC$ touch the side $AC$ at points $P$ and $Q$ respectively. The lines $BP$ and $BQ$ meet the circumcircle of triangle $ABC$ for the second time at points $P'$ and $Q'$ respectively.
Prove that $$ PP' > QQ' $$ | I actually used power of point theorem and stewarts theorem to do this problem([Problem 6](https://drive.google.com/file/d/1PWgLmq5UlpxswUbVHiBdnoVo4Kh7-Qq7/view?usp=sharing))
Let, $\overline{AB}=c,\overline{BC}=a,\overline{CA}=b$ and $\text{Pow}_\omega(X)$ denote the power of $X$ with respect to a circle $\om... | [
"Well I did it basically with PoP and then Stewart to see why $BP<BQ$ .",
"Simple Problem - \n\n**Attachments:**\n\n[Sharygin_Geometry_Olympiad P6.pdf](https://cdn.artofproblemsolving.com/attachments/5/4/bfaa79180328e62ca4da287b94e841849f4907.pdf)",
"Well, to be honest, it should have been specified that $BA\... | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 1132,
"boxed": false,
"end_of_proof": true,
"n_reply": 7,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2794130.json"
} |
A square with center $F$ was constructed on the side $AC$ of triangle $ABC$ outside it. After this, everything was erased except $F$ and the midpoints $N,K$ of sides $BC,AB$ .
Restore the triangle. | Nice Problem! My first construction problem :)
First draw $\ell$ , the perpendicular from $F$ to $NK$ and then draw $\ell'$ a perpendicular to $\ell$ at $F$ . Then draw the angle bisectors at $F$ of $\angle(\ell,\ell')$ Now construct a parallelogram $NKFX$ where $X$ lies on $\ell'$ . Draw a circle c... | [
"Let $\\ell$ be the line through $F$ perpendicular to $KN$ , and let $M$ (the midpoint of $AC$ ) be a variable point on $\\ell$ . As $M$ moves along $\\ell$ , $A$ moves along the line $\\ell_A$ through $F$ with $\\measuredangle(\\ell, \\ell_A) = 45^\\circ$ , and $C$ moves along the line $\\ell... | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 148,
"boxed": false,
"end_of_proof": true,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2794132.json"
} |
Points $P,Q,R$ lie on the sides $AB,BC,CA$ of triangle $ABC$ in such a way that $AP=PR, CQ=QR$ . Let $H$ be the orthocenter of triangle $PQR$ , and $O$ be the circumcenter of triangle $ABC$ .
Prove that $$ OH||AC $$ . | We'll denote $\angle BAC=\alpha$ , $\angle ABC=\beta$ and $\angle BCA=\gamma$ . We have that: $$ \angle APR=180^{\circ}-\angle PAR-\angle PRA=180^{\circ}-2\angle PAR=180^{\circ}-2\alpha $$ $$ \angle CQR=180^{\circ}-\angle QCR-\angle QRC=180^{\circ}-2\angle QCR=180^{\circ}-2\gamma $$ Let $k_{1}$ be the circle... | [
"Vary $R$ on $AC$ , since $P,Q$ move linearly, $(APQ)$ goes through a fixed point. Taking $R = A, R = C$ , we see that this fixed point is $O$ . Observe that $\\angle PHQ = 180 - \\angle PRQ = \\angle ARP + \\angle CRQ = \\angle A + \\angle C = 180 - \\angle ABC$ so $(APQ)$ goes through $H$ too.\n\nT... | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 1118,
"boxed": false,
"end_of_proof": false,
"n_reply": 11,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2794133.json"
} |
The sides $AB, BC, CD$ and $DA$ of quadrilateral $ABCD$ touch a circle with center $I$ at points $K, L, M$ and $N$ respectively. Let $P$ be an arbitrary point of line $AI$ . Let $PK$ meet $BI$ at point $Q, QL$ meet $CI$ at point $R$ , and $RM$ meet $DI$ at point $S$ .
Prove that $P,N$ and... | Oops, angle chase works, but there are cases about the configuration (or you can be smart, unlike me, and use directed angles).
<details><summary>long angle chase</summary>There are differences in the construction and the angle chase for the cases if $P$ lies after point $I$ on $AI$ , that is, the points are $A,... | [
"Sketch: basically four time Menelaus' theorem using the incircle with equal tangents",
"Easily solvable using Moving points... Just Vary $P$ on line $AI$ and check it suffices it prove it for $3$ values of $P$ ..",
"pole and polar works too",
"One liner solution using symmetry and angle chase- $$ \\a... | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 1456,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2794137.json"
} |
Let $\omega_1$ be the circumcircle of triangle $ABC$ and $O$ be its circumcenter. A circle $\omega_2$ touches the sides $AB, AC$ , and touches the arc $BC$ of $\omega_1$ at point $K$ . Let $I$ be the incenter of $ABC$ .
Prove that the line $OI$ contains the symmedian of triangle $AIK$ . | We will prove that $OA$ and $OK$ are tangents to $\odot(AIK)$ from which the desired conclusion trivially follows.
Let $E$ and $F$ be the points such that $\omega_2$ is tangent to $AB$ and $AC$ at $E$ and $F$ respectively
It is well known that $BEIK$ and $CFIK$ are cyclic (Refer EGMO Lemma 4.40 ... | [
"Here's what I submitted in the contest..\n\n**Attachments:**\n\n[Sharygin_Geometry_Olympiad P-10.pdf](https://cdn.artofproblemsolving.com/attachments/9/2/6802db47d70123e613fb95c8e8e200e2509117.pdf)",
"Say $M$ is the midpoint of $\\widehat{BC}$ and $N$ is the midpoint of $\\widehat{BAC}$ .Clearly $\\{M,O,... | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 146,
"boxed": false,
"end_of_proof": false,
"n_reply": 8,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2794145.json"
} |
Let $ABC$ be a triangle with $\angle A=60^o$ and $T$ be a point such that $\angle ATB=\angle BTC=\angle ATC$ . A circle passing through $B,C$ and $T$ meets $AB$ and $AC$ for the second time at points $K$ and $L$ .Prove that the distances from $K$ and $L$ to $AT$ are equal. | Let $A_1$ be the point such that $A, A_1$ are on opposite sides of $BC$ and $\triangle A_1BC$ is equilateral. Let $\omega$ be the circumcircle of $\odot (A_1BC)$ . Then $T = A_1A \cap \omega \not \equiv A_1$ . Furthermore, let $P, Q$ be the feet of the perpendiculars from $K, L$ to line $AT$ .**Lemma:*... | [
"This was so cutee uwu\n<blockquote>Let $ABC$ be a triangle with $\\angle A=60^o$ and $T$ be a point such that $\\angle ATB=\\angle BTC=\\angle ATC$ . A circle passing through $B,C$ and $T$ meets $AB$ and $AC$ for the second time at points $K$ and $L$ .Prove that the distances from $K$ and $L$... | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 144,
"boxed": false,
"end_of_proof": false,
"n_reply": 7,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2794148.json"
} |
A line $l$ parallel to the side $BC$ of triangle $ABC$ touches its incircle and meets its circumcircle at points $D$ and $E$ . Let $I$ be the incenter of $ABC$ . Prove that $AI^2 = AD \cdot AE$ . | <blockquote><blockquote>Let $DE$ intersect $AC$ at $B'$ . By inversion around $A$ with radius $\sqrt{AD\cdot AE}$ , $B$ goes to $B'\Rightarrow AB\cdot AB' = AD \cdot AE$ Notice that triangles $ABI$ and $AIB'$ are similar which gives $AI^2=AB\cdot AB'=AD\cdot AE$ Q.E.D.</blockquote>
Why does $B$ go to ... | [
"Simply screams inversion\n<blockquote>A line $l$ parallel to the side $BC$ of triangle $ABC$ touches its incircle and meets its circumcircle at points $D$ and $E$ . Let $I$ be the incenter of $ABC$ . Prove that $AI^2 = AD \\cdot AE$ .</blockquote>\n\nDefine $M_A:= AI\\cap (ABC).$ We begin with the fo... | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 154,
"boxed": false,
"end_of_proof": false,
"n_reply": 13,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2794158.json"
} |
The products of the opposite sidelengths of a cyclic quadrilateral $ABCD$ are
equal. Let $B'$ be the reflection of $B$ about $AC$ . Prove that the circle passing through $A,B', D$ touches $AC$ | so this is mainly about completing well known configurations for instance consider Brazil 2011/5
<details><summary>solution</summary>[asy]
unitsize(1.2cm);
defaultpen(fontsize(10pt));
markscalefactor = 0.03;
pair X = (2.26, 5.36);
pair A = (0.06, -1.4);
pair B = (2.34, -3.88);
pair C = (9.78, -1.32);
pair D = (0.16, 3.... | [
"\n<blockquote>The products of the opposite sidelengths of a cyclic quadrilateral $ABCD$ are\nequal. Let $B'$ be the reflection of $B$ about $AC$ . Prove that the circle passing through $A,B', D$ touches $AC$ </blockquote>Since $$ AB\\times CD=BC\\times AD\\implies \\frac{AB}{BC}\\times \\frac{CD}{AD}=... | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 112,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2794161.json"
} |
Let $I$ be the incenter of triangle $ABC$ , and $K$ be the common point of $BC$ with the external bisector of angle $A$ . The line $KI$ meets the external bisectors of angles $B$ and $C$ at points $X$ and $Y$ . Prove that $\angle BAX = \angle CAY$ | Let $I$ be the incenter of $\triangle ABC$ , and $K$ be the common point of $BC$ with the external bisector of angle $A$ . The line $KI$ meets the external bisectors of angles $B$ and $C$ at points $X$ and $Y$ . Prove that $\angle BAX = \angle CAY$ . $\textbf{Solution}$ We'll use the Angle Bisector t... | [
"It is well known due to the Apollonius Circle theorem that, the vertices of triangle and feet of internal and external angle bisectors of the opposite angle form harmonic pair. Projecting through the excentre finishes the problem.",
"So cuttee uwu\n\n<blockquote>Let $I$ be the incenter of triangle $ABC$ , and... | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 1126,
"boxed": false,
"end_of_proof": false,
"n_reply": 12,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2794162.json"
} |
The circumcenter $O$ , the incenter $I$ , and the midpoint $M$ of a diagonal of a bicentral quadrilateral were marked. After this the quadrilateral was erased. Restore it.
| Does this work?
We divide the solution into two parts. One is the part where we examine the properties of $ABCD$ and the second one, where we present a construction. $\textbf{Observations:}$ Let $M$ be the midpoint of the diagonal $AC$ and $N$ be the midpoint of the diagonal $BD$ . Let the incircle of $ABCD... | [
"Let $ABCD$ be the quadrilateral, and suppose $M$ is the midpoint of $BD$ . Let $P$ be the intersection of the diagonals $AC,BD$ . It is known that $IO$ passes through $P$ . See [here](https://www.cut-the-knot.org/ctk/BicentricQuadri.shtml) for a reference. We can therefore locate $P$ by intersecting ... | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 1416,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2794163.json"
} |
An ellipse with focus $F$ is given. Two perpendicular lines passing through $F$ meet the ellipse at four points. The tangents to the ellipse at these points form a quadrilateral circumscribed around the ellipse. Prove that this quadrilateral is inscribed into a conic with focus $F$ | What a pretty problem! Took me a while to get through, but I enjoyed the process and it taught me a lot!
Let the ellipse be $\Gamma$ . Let $X, Y, Z, W$ be the intersection of the perpendicular lines through $F$ with $\Gamma$ , in that order. Let $A = XX\cap WW, B = XX\cap YY, C = YY\cap ZZ, D = ZZ\cap WW$ . Let... | [
"I only proved that the diagonals of the bigger quadrilateral meet at F and they are perpendicular. After that ded..",
"After dual transformation with center $F$ ellipse becomes a circle, intersections of ellipse with perpendicular lines become sides of square, circumscribed around circle. Touch points of squar... | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 144,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2794169.json"
} |
Chords $A_1A_2, A_3A_4, A_5A_6$ of a circle $\Omega$ concur at point $O$ . Let $B_i$ be the second common point of $\Omega$ and the circle with diameter $OA_i$ . Prove that chords $B_1B_2, B_3B_4, B_5B_6$ concur. | Kinda strange that no one has posted a complex bash yet. Well, here it is:
We will use complex numbers. WLOG let $\Omega$ be the unit circle and let $a,b,c,x$ denote the complex numbers of $A_{1},A_{3},A_{5},O$ respectively and denote by $a_{2},a_{4},a_{6},b_{i}$ the complex numbers of $A_{2},A_{4},A_{6},B_{i... | [
"Fix a point $O$ inside the circle and say $A_1$ is a point on the circle, let $A_2 = A_1O \\cap \\Omega$ , define $B_1, B_2$ as in the problem, it suffices to show $B_1B_2$ goes through a fixed point as $A_1$ varies. Let $O'$ be the center of the circle and say $C_1, C_2$ are antipodes of $A_1, A_2... | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 212,
"boxed": false,
"end_of_proof": false,
"n_reply": 12,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2794175.json"
} |
Let $ABCD$ be a cyclic quadrilateral, $E = AC \cap BD$ , $F = AD \cap BC$ . The bisectors of angles $AFB$ and $AEB$ meet $CD$ at points $X, Y$ . Prove that $A, B, X, Y$ are concyclic.
| Let $EY\cap AB=Y_{1}$ , $FX\cap AB=X_{1}$ and $AB\cap CD=G$ (if $AB\parallel CD$ , then $ABCD$ is a trapezoid, so $X\equiv Y$ and therefore $A$ , $B$ , $X$ , $Y$ trivially lie on a circle, that is, the circumcircle of $\triangle ABX$ ). From the Miquel theorem for points $B\in \overline{AC}$ , $C\in \... | [
"Very Nice Problem -- I used POP + Menelaus + Ratio Bash..\n\n**Attachments:**\n\n[Sharygin Problem 16_compressed.pdf](https://cdn.artofproblemsolving.com/attachments/6/3/6542d6906a3d635b2abdd571c988975694d0b0.pdf)",
"Here is a sketch of my solution. Let $AX$ and $BY$ intersect $BD$ and $AC$ at $R$ and ... | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 220,
"boxed": false,
"end_of_proof": false,
"n_reply": 11,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2794178.json"
} |
Eight points in a general position are given in the plane. The areas of all $56$ triangles with vertices at these points are written in a row. Prove that it is possible to insert the symbols " $+$ " and " $-$ " between them in such a way
that the obtained sum is equal to zero. | Same as @above, but the quadruplets have to be $14$ , not $17$ ...
Let's label the points as $A_{1},A_{2},\ldots ,A_{8}$ . Consider four of those points, say, $A_{i_{1}}$ , $A_{i_{2}}$ , $A_{i_{3}}$ , $A_{i_{4}}$ where the indices are different numbers from $1$ to $8$ . We'll prove that we can put either a $+... | [
"Did I fake solve it? \n<blockquote>Eight points in a general position are given in the plane. The areas of all 56 triangles with vertices at these points are written in a row. Prove that it is possible to insert the symbols \"+\" and \"−\" between them in such a way\nthat the obtained sum is equal to zero.</blockq... | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 96,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2794188.json"
} |
Let $O$ , $I$ be the circumcenter and the incenter of $\triangle ABC$ ; $R$ , $r$ be the circumradius and the inradius; $D$ be the touching point of the incircle with $BC$ ; and $N$ be an arbitrary point of segment $ID$ . The perpendicular to $ID$ at $N$ meets the circumcircle of $ABC$ at points $X$ ... | The answer is $\boxed{Rr}$ .
We employ coordinates. Set the origin at $O$ and the $x$ -axis parallel to $BC$ . Clearly it's also parallel to $XY$ . Set $I=(u,v)$ , $X=(-a,h)$ , and $Y=(a,h)$ . Note that $a^2+h^2=R^2$ and $OI^2 = u^2+v^2$ .
[asy]
size(6cm);
pair A = dir(120), B = dir(210), C = dir(-30), I... | [
"Following lemma kills: Let centers of three non-concentric circles $C_1,C_2,C_3$ lie on $\\ell$ . Let $O_i$ be the center of $C_i$ . Let the radical axis of $C_2,C_3$ cut $\\ell$ at $A_1$ . Define $A_2,A_3$ , similarly. Then, $A_1A_2:A_2A_3=O_1O_2:O_2O_3$ .\n\nI have a proof which uses linearity of po... | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 1056,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2794242.json"
} |
Let a point $P$ lie inside a triangle $ABC$ . The rays starting at $P$ and crossing the sides $BC$ , $AC$ , $AB$ under the right angle meet the circumcircle of $ABC$ at $A_{1}$ , $B_{1}$ , $C_{1}$ respectively. It is known that lines $AA_{1}$ , $BB_{1}$ , $CC_{1}$ concur at point $Q$ . Prove that al... | We will show that desired point is circumcenter $O$ .
Define $A_2$ as intersection of $B_1C_1\cap BC$ and similarly define $B_2$ and $C_2$ .
Define $A_3$ as intersection of $BC_1\cap CB_1$ and similarly define $B_3$ and $C_3$ .
Since triangles $ABC$ and $A_1B_1C_1$ are perspective, $A_2B_2C_2$ are ... | [
"Throughout the solution we use some known properties of perspective-orthologic triangles. At first, line $PQ$ passes through the second center $R$ of orthology of $ABC,A_1B_1C_1.$ Define point $A_2$ as symmetric to $A_1$ over perpendicular bisector of $BC,$ and $B_2,C_2$ similarly. We see (arcs are o... | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 158,
"boxed": false,
"end_of_proof": true,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2794243.json"
} |
Let $O$ and $H$ be the circumcenter and the orthocenter respectively of triangle $ABC$ . Itis known that $BH$ is the bisector of angle $ABO$ . The line passing through $O$ and parallel to $AB$ meets $AC$ at $K$ . Prove that $AH = AK$ | I never imagined my geo skills could become so bad and rusty that even this problem troubled me :wallbash_red: . Anyways here's my solution.
<blockquote>
Let $O$ and $H$ be the circumcenter and the orthocenter respectively of triangle $ABC$ . It is known that $BH$ is the bisector of angle $ABO$ . The line pass... | [
"<details><summary>Hint</summary>Consider the point of intersection of $OK$ and $BH$ .</details>\n<details><summary>Solution</summary>Let $S$ be the intersection of $OK$ and $BH$ . The key observation is that $S$ lies on the circumcircle.\nIndeed, we have\n\\[\\angle BSO=\\angle HBA=\\angle OBS\\]\nby ass... | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 72,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2794244.json"
} |
Let $ABCD$ be a curcumscribed quadrilateral with incenter $I$ , and let $O_{1}, O_{2}$ be the circumcenters of triangles $AID$ and $CID$ . Prove that the circumcenter of triangle $O_{1}IO_{2}$ lies on the bisector of angle $ABC$ | Let $W,X,Y,Z$ be the tangency points. Invert with respect to the incircle. We obtain the following equivalent problem:
<blockquote>Let $I$ be the circumcenter of cyclic quadrilateral $WXYZ$ , and let $A,B,C,D$ be the midpoints of $XY,YZ,ZW,WX$ respectively. Let $O_1,O_2$ be the reflections of $I$ in lines... | [
"Here is a short solution without inversion: Equivalently, we need to prove that the tangent line to $(O_1IO_2)$ at $I$ is perpendicular to $BI$ , in other words $\\angle BIO_2=90^\\circ+\\angle IO_1O_2$ . But it is an easy angle chase to see that both sides are equal to $90^\\circ+\\frac{\\alpha}{2}$ .",
... | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 62,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2794246.json"
} |
A triangle $ABC$ is given. Let $C'$ and $C'_{a}$ be the touching points of sideline $AB$ with the incircle and with the excircle touching the side $BC$ . Points $C'_{b}$ , $C'_{c}$ , $A'$ , $A'_{a}$ , $A'_{b}$ , $A'_{c}$ , $B'$ , $B'_{a}$ , $B'_{b}$ , $B'_{c}$ are defined similarly. Consider the len... | We will first calculate the altitudes and then solve $a)$ and $b)$ .
<details><summary>long</summary>Denote $AB=c$ , $BC=a$ , $CA=b$ , $p=\frac{a+b+c}{2}$ , $\angle ABC=\beta$ , $\angle BCA=\gamma$ , $\angle CAB=\alpha$ . Let $H_{1}$ be the foot of the perpendicular from $A'$ to line $B'C'$ . Let $H_{2}... | [
"<details><summary>long</summary>The possible numbers of different lengths are 2, 3, 4, 5, and 6. Let $I,I_A,I_B,I_C$ be the incenter and three excenters of $ABC$ . **Claim.** $\\triangle BIC \\sim \\triangle B_a'A_a'C_a'$ , and symmetric variants. \n\n*Proof.* We have\n\\[\n\\angle A_a'C_a'B_a' = \\angle A_a'I... | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 1546,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2794253.json"
} |
Let $K$ , $L$ , $M$ , $N$ be the midpoints of sides $BC$ , $CD$ , $DA$ , $AB$ respectively of a convex quadrilateral $ABCD$ . The common points of segments $AK$ , $BL$ , $CM$ , $DN$ divide each of them into three parts. It is known that the ratio of the length of the medial part to the length of the who... | @above a much simpler counterexample exists, but ok. Here is my full solution, including the motivation behind how we come up with a counterexample:
The answer is $\textbf{No}$ . $\textbf{Solution:}$ I wanted to present my thought process and how I came up with the counterexample. In what's presented below I explai... | [] | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 1352,
"boxed": true,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2794309.json"
} |
Let $CD$ be an altitude of right-angled triangle $ABC$ with $\angle C = 90$ . Regular triangles $ AED$ and $CFD$ are such that $E$ lies on the same side from $AB$ as $C$ , and $F$ lies on the same side from $CD$ as $B$ . The line $EF$ meets $AC$ at $L$ . Prove that $FL = CL + LD$ | Pretty easy
Just notice that the spiral sim centred at $D$ takes $AE \longrightarrow BF$ which means that $D=\odot(LAE) \cap \odot(LBF)$ . Van schooten’s theorem finishes $\blacksquare$ | [
"We claim that $L$ lies on $(AED)$ and $(CDF)$ . Redefine $L$ to be the second intersection of $(AED)$ and $(CDF)$ . \n[asy]\nsize(8cm);\n\npair A = (-1, 0), B = (1, 0), C = dir(85), D = foot(C, A, B), E = rotate(60, A)*D, F = rotate(60, C)*D, L = extension(A, C, E, F);\n\ndraw(A--B--C--cycle, red); draw(... | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 108,
"boxed": false,
"end_of_proof": true,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2795458.json"
} |
Let $AA_1$ , $BB_1$ , $CC_1$ be the altitudes of acute angled triangle $ABC$ . $A_2$ be the touching point of the incircle of triangle $AB_1C_1$ with $B_1C_1$ , points $B_2$ , $C_2$ be defined similarly. Prove that the lines $A_1A_2$ , $B_1B_2$ , $C_1C_2$ concur. | By similarity of $ABC$ , $AC_1B_1$ , $BC_1A_1$ , $CA_1B_1$ , we have $$ \prod_{cyc}\frac{B_1A_2}{A_2C_1}= $$ $$ =\prod_{cyc}\frac{(B_1C_1+B_1A-C_1A)/2}{(B_1C_1+AC_1-AB_1)/2}= $$ $$ =\prod_{cyc}\frac{BC+AB-AC}{BC+AC-AB}=1 $$ So by Ceva's theorem the three lines $A_1A_2,B_1B_2,C_1C_2$ concur. | [
"When I first read the problem, I misread and thought we had to prove that $AA_2,BB_2,CC_2$ concur.\nWhich happens to be true as well, by a similar argument using Ceva (or trigonometric Ceva).",
"Note that $AA_2$ , $BB_2$ , and $CC_2$ concur at the isogonal conjugate of the Gergonne point of $ABC$ so we'r... | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 22,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2795459.json"
} |
Let the diagonals of cyclic quadrilateral $ABCD$ meet at point $P$ . The line passing through $P$ and perpendicular to $PD$ meets $AD$ at point $D_1$ , a point $A_1$ is defined similarly. Prove that the tangent at $P$ to the circumcircle of triangle $D_1PA_1$ is parallel to $BC$ . | Consider the negative inversion at $P$ that swaps $A,C$ and $B,D$ . The line $AD$ inverts to $\omega$ , the circumcircle of $\triangle BCP$ . Points $A_1$ and $D_1$ are sent to $A'$ , the second intersection of $A_1P$ with $\omega$ , and $D'$ , the second intersection of $D_1P$ with $\omega$ .
[as... | [
"Actually, a simple angle chase does the job as well:\nWe only need to prove that $\\angle CBD+\\angle DPA_1=\\angle PD_1A_1$ .\nBut\n\\[\\angle CBD+\\angle DPA_1=\\angle CAD+\\angle DPA_1=90^\\circ-\\angle AA_1P+\\angle DPA_1=90^\\circ-\\angle ADP=\\angle PD_1A_1.\\]\n",
"We extend $PD_1$ and $PA_1$ to meet... | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 62,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2795460.json"
} |
Let $OABCDEF$ be a hexagonal pyramid with base $ABCDEF$ circumscribed around a sphere $\omega$ . The plane passing through the touching points of $\omega$ with faces $OFA$ , $OAB$ and $ABCDEF$ meets $OA$ at point $A_1$ , points $B_1$ , $C_1$ , $D_1$ , $E_1$ and $F_1$ are defined similarly. Let $\... | A cone with vertex $O,$ circumscribed around $\omega,$ meets base by conic, inscribed in $ABCDEF;$ by the Brianchon's theorem $AD,BE,CF$ concur.
By properties of central projections $$ \overline{A_1D_1}\cap \overline{B_1E_1}=\overline{B_1E_1}\cap \overline{O(AD\cap BE)}=\overline{B_1E_1}\cap \overline{O(BE\c... | [] | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 14,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2795461.json"
} |
Let $BH$ be an altitude of right angled triangle $ABC$ ( $\angle B = 90^o$ ). An excircle of triangle $ABH$ opposite to $B$ touches $AB$ at point $A_1$ ; a point $C_1$ is defined similarly. Prove that $AC // A_1C_1$ . | Note that $A A_1 = s_1 - c$ $C C_1 = s_2 - a$ where $s_1, s_2$ are the semiperimeters of $A B H$ and $H B C$ respectively.
it suffices to prove that - $(B A)/(A A_1) = (B C)/(C C_1)$ which is trivial after substituting - $ A H = c^2/b $ $ C H = a^2/b $ | [
"Note that $BA_1 = p(\\triangle ABH)/2$ and $BC_1 = p(\\triangle ACH)/2$ , where $p$ is the perimeter function. Since these triangles are similar with the ratio being $BA/BC$ , we are done.",
"Trivial by Thales and a bit of length bash: $BA_1=\\frac{AB+BH+AH}{2}, BC_1=\\frac{BC+BH+CH}{2}$ , $BH=\\frac{ac}... | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 16,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2895851.json"
} |
Let circles $s_1$ and $s_2$ meet at points $A$ and $B$ . Consider all lines passing through $A$ and meeting the circles for the second time at points $P_1$ and $P_2$ respectively. Construct by a compass and a ruler a line such that $AP_1.AP_2$ is maximal. | A bit similar to the above one.
Let $\angle P_1BA= \varphi_1$ and $\angle P_2BA=\varphi_2$ . It's easy to see that $AP_1=2R_1sin \varphi_1, AP_2=2R_2sin \varphi_2$ and that $\varphi_1+\varphi_2=const$ , so we are to maximize $sin \varphi_1. sin \varphi_2=\frac{1}{2}(cos(\varphi_1-\varphi_2)-cos(\varphi_1+\varphi... | [
"Let the center of $s_i$ be $O_i$ . $\\angle O_1AP_1 + \\angle O_2AO_1 + \\angle P_2AO_2 = 0$ , so we need to maximize the product of cosines of two angles which sum to a constant angle. $\\cos x \\cos (A - x) = \\frac{\\cos A + \\cos (2x - A)}{2}$ , so the minimum is achieved when $x = \\frac{A}{2}$ . This c... | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 38,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2895853.json"
} |
A medial line parallel to the side $AC$ of triangle $ABC$ meets its circumcircle at points at $X$ and $Y$ . Let $I$ be the incenter of triangle $ABC$ and $D$ be the midpoint of arc $AC$ not containing $B$ .A point $L$ lie on segment $DI$ in such a way that $DL= BI/2$ . Prove that $\angle IXL = \an... | Let $E$ be the midpoint of $BI$ . Note that if $I_B$ is the $B-$ excenter of $ABC$ , then $L$ is the midpoint of $BI_B$ . Let $X'$ be the reflection of $X$ in $BI$ . Since $XY \parallel BC$ , $BD$ bisects $\angle XBY$ , so $B, X', Y$ are collinear. The problem reduces to showing that $BX \cdot BY ... | [
"Motivation can be found easily by taking inversion that swaps $X$ with $Y$ It swaps $A$ with midpoint of $BC$ , $I$ with $L$ etc.",
"Let $X'$ be reflection of $X$ w.r.t $BI$ . Now we need to prove $X'YLI$ is cyclic. Note that $XY \\parallel AC$ so $B,X',Y$ are collinear. Let $BY$ meet $AC$... | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 40,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2895854.json"
} |
Let $ABC$ be an isosceles triangle with $AB = AC$ , $P$ be the midpoint of the minor arc $AB$ of its circumcircle, and $Q$ be the midpoint of $AC$ . A circumcircle of triangle $APQ$ centered at $O$ meets $AB$ for the second
time at point $K$ . Prove that lines $PO$ and $KQ$ meet on the bisector of... | Let $R$ be the midpoint of minor arc $AC$ of $(ABC)$ . Denote by $J$ and $T$ intersection points of line $PO$ with $BR$ and $(ABC)$ , we need to prove that points $K,Q,J$ are collinear. Let $O'$ be the circumcenter of triangle $ABC$ . Note that $\angle PAQ=\angle PAC$ , so $\triangle POQ\sim\triang... | [
"Let the midpoint of minor arc $AC$ be $X$ and let $BX \\cap PO = Z$ . Since $$ \\angle QXZ = \\angle QXC - \\angle BXC = \\frac{AXC}{2} - (180 - \\angle ABC) = \\frac{3 \\angle ABC}{2} - 90 = \\angle OPQ = \\angle ZPQ $$ we have that $PQXZ$ is cyclic. This gives that $$ \\angle PQK = \\angle PAK = \\an... | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 180,
"boxed": false,
"end_of_proof": true,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2896795.json"
} |
Chords $AB$ and $CD$ of a circle $\omega$ meet at point $E$ in such a way that $AD = AE = EB$ . Let $F$ be a point of segment $CE$ such that $ED = CF$ . The bisector of angle $AFC$ meets an arc $DAC$ at point $P$ . Prove
that $A$ , $E$ , $F$ , and $P$ are concyclic. | First, since $\angle CBE = \angle CDA = \angle DEA = \angle CEB$ , we have that $CE = CB$ . Redefine $P$ as the point so $ABCP$ is an isosceles trapezoid. Also, $\angle PAD = \angle DAE + \angle PAE = 180 - 2 \angle ABC + \angle ABC = 180 - \angle ADC$ so $APCD$ is also an isosceles trapezoid. Redefine $F$ ... | [
"Cute, indeed. Here is my solution, which has a bit different finish. Redefine $P$ such that $ADCP$ is an isosceles trapezoid, and due to the equal segments, $AEFP$ also will be isosceles trapezoid. Hence we have to prove that $PF$ is bisector of $\\angle AFC$ . We have that $\\angle PFC = \\angle ADE= \\... | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 134,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2896803.json"
} |
Lateral sidelines $AB$ and $CD$ of a trapezoid $ABCD$ ( $AD >BC$ ) meet at point $P$ . Let $Q$ be a point of segment $AD$ such that $BQ = CQ$ . Prove that the line passing through the circumcenters of triangles $AQC$ and $BQD$ is perpendicular to $PQ$ . | A cute problem indeed. Here is a sketch of my solution at the contest:
The statement is equivalent to proving that $P$ lies on the radical axis of the circumcircles of triangles $AQC$ and $BQD$ . Let their second intersection be point $W$ . We will prove that $P, Q$ and $W$ are collinear. Easy anglechase giv... | [
"The formulation was kind of dumb imho. We just want to prove $PQ$ is radical axis of $(AQC)$ and $(BQD)$ . Let $(AQC)$ meet $PD$ at $D'$ and let $(BQD)$ meet $PA$ at $A'$ . Proving $A'D'DA$ is cyclic is sufficient by PoP. It is equivalent to $A'D'CB$ being cyclic by Reim's. Now, this is an easy... | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 66,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2896809.json"
} |
Let $H$ be the orthocenter of an acute-angled triangle $ABC$ . The circumcircle of triangle $AHC$ meets segments $AB$ and $BC$ at points $P$ and $Q$ . Lines $PQ$ and $AC$ meet at point $R$ . A point $K$ lies on the line $PH$ in such a way that $\angle KAC = 90^{\circ}$ . Prove that $KR$ is perpen... | Let $L$ be an intersection of $QH$ and the line through $C$ perpendicular to $AC$ .
And let $D$ be a midpoint of $AC$ , $ E = BH \cap AC, F=AH \cap BC, G = CH \cap AB, T = GF \cap AB, \alpha $ be a circumcircle of $\triangle AHC, \beta$ be a circumcircle of quadrilateral $BGHF$ , and $ O = \alpha \cap \... | [
"I can't believe that I couldn't prove for two hours that $KL, AC, PQ$ are concurrent (as I see now in the above post - by one line Desargue's theorem for triangles $CLQ$ and $PKA$ ), but I instead proved that $KL$ is perpendicular to $BM$ by trig-length bash much faster, and as I see the above solution, I... | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 136,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2896812.json"
} |
Several circles are drawn on the plane and all points of their intersection or touching are marked. Is it possible that each circle contains exactly five marked points and each point belongs to exactly five circles? | Here is an amusing fake disproof that the answer is No! :D (As far as I know, the answer is Yes but I have not looked for
a construction.)
Consider the bipartite graph $G$ with classes $C$ and $P$ in which there is an edge between a circle in $C$ and a point in $P$ if the point lies on the circle. The condit... | [
"Yes?????",
"Because your title literally states there are 5 points on each circle",
"<blockquote>Because your title literally states there are 5 points on each circle</blockquote>\n\nThat's the problem statement: we want to have **five points on each circle, five circles through each point**. It doesn't mean t... | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 38,
"boxed": false,
"end_of_proof": false,
"n_reply": 7,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2896814.json"
} |
Let $ABCD$ be a convex quadrilateral with $\angle B= \angle D$ . Prove that the midpoint of $BD$ lies on the common internal tangent to the incircles of triangles $ABC$ and $ACD$ . | The [official solution](https://geometry.ru/olimp/2022/final_sol_eng_2022.pdf) shows that (<details><summary>notations</summary>$N$ is midpoint of $BD$ ; $\omega_1,\omega_2$ incircles of $ABC$ , $ADC$ .</details>) $N$ lies outside both $\omega_1,\omega_2$ and the sum of lengths of tangents from $N$ to $\o... | [] | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 20,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2897117.json"
} |
Let $ABCD$ be a convex quadrilateral. The common external tangents to circles $(ABC)$ and $(ACD)$ meet at point $E$ , the common external tangents to circles $(ABD)$ and $(BCD)$ meet at point $F$ . Let $F$ lie on $AC$ , prove that $E$ lies on $BD$ . | Consider the inversion centered at $F$ with radius $r=FB=FD$ mapping $(ABD)$ to $(BCD)$ . So, the inversion fixes $B$ and $D$ while $A$ goes to $C$ under this inversion.
So, $A'B'=AB\cdot \frac{r^2}{FA\cdot FB}\implies BC=AB\cdot \frac{r^2}{FA\cdot FB}$ and $A'D'=AD\cdot \frac{r^2}{FA\cdot FD}\implies... | [
"Denote $O_1,O_2$ to be the orthocenter of $\\triangle ABC$ and $\\triangle ACD$ ,and then we can know that: $E$ lies on $BD \\Leftrightarrow \\angle O_1BD = \\angle O_2DB \\Leftrightarrow \\angle BDC - \\angle DBC = \\angle BAC = \\angle DAC \\cdots (*)$ Similarly we have: $F$ lies on $AC \\Leftrightarrow... | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 46,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2897120.json"
} |
$A_1A_2A_3A_4$ and $B_1B_2B_3B_4$ are two squares with their vertices arranged clockwise.The perpendicular bisector of segment $A_1B_1,A_2B_2,A_3B_3,A_4B_4$ and the perpendicular bisector of segment $A_2B_2,A_3B_3,A_4B_4,A_1B_1$ intersect at point $P,Q,R,S$ respectively.Show that: $PR\perp QS$ .
**Attachment... | [] | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2903134.json"
} | |
Let $ABCD$ be a convex quadrilateral with $\angle{BAD} = 2\angle{BCD}$ and $AB = AD$ . Let $P$ be a point such that $ABCP$ is a parallelogram. Prove that $CP = DP$ . | <details><summary>solution</summary>Construct the reflection of $A$ about $BD$ , $X$ and note that it is the circumcenter of $\triangle CBD$ . Thus if $Y$ was a point on the line through $C$ parallel to $BD$ and $H$ the foot from $C$ to $BD$ , we would have $$ \angle XCY = \angle HCY - \angle HCX = 9... | [
"Let E be such that $EBCD$ is a parallelogram. Then $A$ is the circumcircle of $EBD$ . We will use complex numbers. Let $A$ be the origin. We get that $C=b+d-e$ and $P=C+A-B=d-e$ . We want to prove that $\\mid P-C\\mid = \\mid P-D\\mid $ or $\\mid -c\\mid =\\mid -e\\mid $ which is clearly so.",
"Con... | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 46,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2933778.json"
} |
Let $ABCD$ be a right-angled trapezoid and $M$ be the midpoint of its greater lateral side $CD$ . Circumcircles $\omega_{1}$ and $\omega_{2}$ of triangles $BCM$ and $AMD$ meet for the second time at point $E$ . Let $ED$ meet $\omega_{1}$ at point $F$ , and $FB$ meet $AD$ at point $G$ . Prove tha... | Note that, $\angle BEM = 180 - \angle C = \angle D $ , which means that $E$ is a common point to cyclic quadrilaterals $EBCM$ and $AEMD$ and $E$ lies on line $\overline {AB}$ . $\angle EMC = 90^\circ$ and $\angle EFC = \angle EMC = 90^\circ$ as quad $EFMC$ is cyclic.
Thus, $\angle CFD = 90 ^\circ$ which... | [
"<details><summary>solution</summary>First of all note that $EM \\perp CD$ . We claim $DMGF$ concyclic, since $$ \\measuredangle MDG = \\measuredangle CDA =\\measuredangle DCB = \\measuredangle MCB = \\measuredangle MFB = \\measuredangle MFG $$ Clearly $MD = MF$ is equivalent to $\\angle DFC = 90$ , which ... | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 42,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2933779.json"
} |
A circle $\omega$ and a point $P$ not lying on it are given. Let $ABC$ be an arbitrary equilateral triangle inscribed into $\omega$ and $A', B', C'$ be the projections of $P$ to $BC, CA, AB$ . Find the locus of centroids of triangles $A' B'C'$ . | We claim that it is the midpoint of $OP$ where $O$ is the center of $\omega$ . For that, let's fix $ABC$ , and vary $P$ on a line perpendicular to $BC$ . Then the projections of $P$ onto $AB$ and $AC$ move with the same speed, with equal movements in the direction parallel to $BC$ . So the centroid of ... | [
"nice.\nLet $A = a, B = b, C = c$ be points on the unit circle in the complex plane and $P = p$ a point not lying on it(to make $A'B'C'$ a non-degenerated triangle). Denote by $G = g$ the centroid of $A'B'C'$ . Note that: $A' = \\frac{1}{2}(b+c+p-bc \\bar p), B' = \\frac{1}{2}(a+c+p-ac \\bar p), C' = \\fra... | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 40,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2933784.json"
} |
A line meets a segment $AB$ at point $C$ . Which is the maximal number of points $X$ of this line such that one of angles $AXC$ and $BXC$ is equlal to a half of the second one?
| [] | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2948415.json"
} | |
Let $ AB$ and $AC$ be the tangents from a point $A$ to a circle $ \Omega$ . Let $M$ be the midpoint of $BC$ and $P$ be an arbitrary point on this segment. A line $AP$ meets $ \Omega$ at points $D$ and $E$ . Prove that the common external tangents to circles $MDP$ and $MPE$ meet on the midline of ... | Note that $DP$ is symmedian in $DBC$ so $\angle PDC = \angle MDB$ .
Claim $: PMD$ and $PME$ are tangent to $CDB$ .
Proof $:$ Note that $\angle PMD = \angle MDB + \angle MBD = \angle EDC + \angle CBD = \angle EBD$ with same approach we have $\angle PME = \angle DBE$ Let $O$ be center of $DBC$ and $O... | [
"I know midline as midbase lol, wlog $BP \\ge BM$ , as the other case is similar.\nUsing the symedian at the problem we have by angle chase $$ \\angle DMP-\\angle DBC=\\angle BDM=\\angle CDP \\implies (PDM) \\; \\text{tangent to} \\; \\Omega $$ $$ \\angle EMP-\\angle EBC=\\angle BEM=\\angle CEP \\implies (PEM)... | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 70,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2948437.json"
} |
Let $O, I$ be the circumcenter and the incenter of triangle $ABC$ , $P$ be an arbitrary point on segment $OI$ , $P_A$ , $P_B$ , and $P_C$ be the second common points of lines $PA$ , $PB$ , and $PC$ with the circumcircle of triangle $ABC$ . Prove that the bisectors of angles... | Let $M_A,N_A$ be the midpoints of arc $BC,BAC$ ,and similarly denote $M_B,N_B,M_C,N_C$ .Now let $K = P_AN_A \cap IO$ ,so applying a [lemma](https://artofproblemsolving.com/community/u832107h2914887p26028863) for $(A)I(M_A)O(N_A)K(P_A)P(A)$ we will know that $N_BP_B,N_CP_C$ pass through $K$ ,so the three bise... | [
"Let $M_A,N_A$ be the midpoints of arc $BC$ and $BAC$ . Then applying pascal on $P_BN_BM_BP_CN_CM_C$ and $BM_BP_CCM_CP_B$ gives the desired concurrency."
] | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 18,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2948442.json"
} |
Several circles are drawn on the plane and all points of their meeting or touching are marked. May be that each circle contains exactly four marked points and exactly four marked points lie on each circle?
| [
"See here:\n[https://artofproblemsolving.com/community/c6h2896814p25792303](https://artofproblemsolving.com/community/c6h2896814p25792303)"
] | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2948444.json"
} | |
Let $ABCA'B'C'$ be a centrosymmetric octahedron (vertices $A$ and $A'$ , $B$ and $B'$ , $C$ and $C'$ are opposite) such that the sums of four planar angles equal $240^o$ for each vertex. The Torricelli points $T_1$ and $T_2$ of triangles $ABC$ and $A'BC$ are marked. Prove that the distance... | [] | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2948448.json"
} | |
Let $ABCD$ be a cyclic quadrilateral, $O$ be its circumcenter, $P$ be a common points of its diagonals, and $M , N$ be the midpoints of $AB$ and $CD$ respectively. A circle $OPM$ meets for the second time segments $AP$ and $BP$ at points $A_1$ and $B_1$ respectively and a circle $OPN$ meets... | [] | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2948452.json"
} | |
Ten points on a plane a such that any four of them lie on the boundary of some square. Is obligatory true that all ten points lie on the boundary of some square? | [] | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2948455.json"
} | |
An isosceles trapezoid $ABCD$ ( $AB = CD$ ) is given. A point $P$ on its circumcircle is such that segments $CP$ and $AD$ meet at point $Q$ . Let $L$ be tha midpoint of $ QD$ . Prove that the diagonal of the trapezoid is not greater than the sum of distances from the midpoints of the lateral sides to ana arb... | Let $M,N,K$ be midpoints of $AB,CD,CQ$ and $T$ be an arbitrary point on $PL$ . Let $N'$ be reflection of $N$ w.r.t $PL$ .
Claim $:PLKMA$ is cyclic.
Proof $:$ Note that $PQ.QC=QA.QD \implies PQ.QK=QA.QL$ so $PLKA$ is cyclic. Also $\angle LKM = \angle DNM = 180 - \angle A$ so $LKMA$ is cyclic.
Now... | [] | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 40,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2948457.json"
} |
Two circles meeting at points $A, B$ and a point $O$ lying outside them are given. Using a compass and a ruler construct a ray with origin $O$ meeting the first circle at point $C$ and the second one at point $D$ in such a way that the ratio $OC : OD$ be maximal.
| [] | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2948458.json"
} | |
An incircle of triangle $ABC$ touches $AB$ , $BC$ , $AC$ at points $C_1$ , $A_1$ , $ B_1$ respectively. Let $A'$ be the reflection of $A_1$ about $B_1C_1$ , point $C'$ is defined similarly. Lines $A'C_1$ and $C'A_1$ meet at point $D$ . Prove that $BD \parallel AC$ . | Pure angle chase ftw. $$ \angle A_1DC_1=180-\angle DA_1C_1-\angle DC_1A_1=180-\angle BAC-\angle BCA=\angle ABC \implies BDA_1C_1 \; \text{cyclic} $$ From here $\angle BAC=\angle DA_1C_1=180-\angle ABD$ so $BD \parallel AC$ thus we are done :D | [
"same solution as above, a quick motivation for it is noticing the spiral similarity at B that sends $A'C_1$ to $C'A_1$ "
] | [
"origin:aops",
"2022 Contests",
"2022 Sharygin Geometry Olympiad"
] | {
"answer_score": 8,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Sharygin Geometry Olympiad/2948460.json"
} |
Let $x,y,z$ be three positive integers with $\gcd(x,y,z)=1$ . If
\[x\mid yz(x+y+z),\]
\[y\mid xz(x+y+z),\]
\[z\mid xy(x+y+z),\]
and
\[x+y+z\mid xyz,\]
show that $xyz(x+y+z)$ is a perfect square.
*Proposed by usjl* | Let $p$ be an arbitrary prime divisor of $xyz$ , and let $v_p(x)$ be the $p$ -adic valuation of $x$ . Note that $p$ cannot divide all three of $x,y,z$ since $\gcd(x,y,z)=1$ . Now it suffices to show that $v_p(xyz(x+y+z))$ is even.**Case 1:** Suppose $p$ divides exactly one of $x,y,z$ .
WLOG let it be ... | [
"A proposer remark:\n\nIn fact this is a lemma of a much harder and much weirder problem. One version is the following:\n\nLet $A,B,C$ be three lattice points such that the incenter and three excenters of $\\triangle ABC$ are also lattice points. If the lengths of $AB,AC,BC$ are $x+y,x+z,y+z$ , respectively,... | [
"origin:aops",
"2022 Contests",
"2022 Taiwan Mathematics Olympiad"
] | {
"answer_score": 60,
"boxed": false,
"end_of_proof": false,
"n_reply": 10,
"path": "Contest Collections/2022 Contests/2022 Taiwan Mathematics Olympiad/2784234.json"
} |
Find all functions $f,g:\mathbb{R}^2\to\mathbb{R}$ satisfying that
\[|f(a,b)-f(c,d)|+|g(a,b)-g(c,d)|=|a-c|+|b-d|\]
for all real numbers $a,b,c,d$ .
*Proposed by usjl* | As the name suggests, we are equivalently finding isometries $F:\mathbb{R}^2\rightarrow \mathbb{R}^2$ with respect with the Manhattan metric (also called $L^1$ I think) where we define $d(A,B)=||\vec{AB}||_1=||\vec{B}-\vec{A}||_1=|x_B-x_A|+|y_B-y_A|$ , where $A=(x_A,y_A),B=(x_B,y_B)$ belong to $\mathbb{R}^2$ .
... | [
"[https://hackmd.io/@sine/TMO_22P3](https://hackmd.io/@sine/TMO_22P3)\nI am lazy to translate it to English :("
] | [
"origin:aops",
"2022 Contests",
"2022 Taiwan Mathematics Olympiad"
] | {
"answer_score": 100,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Taiwan Mathematics Olympiad/2784236.json"
} |
Two babies A and B are playing a game with $2022$ bottles of milk. Each bottle has a maximum capacity of $200$ ml, and initially each bottle holds $30$ ml of milk.
Starting from A, they take turns and do one of the following:
(1) Pick a bottle with at least $100$ ml of milk, and drink half of it.
(2) Pick two bott... | A proposer remark:
One other version of this is that we start with $1011$ bottles with $90$ ml of milk, and $1011$ bottles with $150$ ml of milk. The solution we have for this version is completely different from the one for the version in the contest.
One interesting question would thus be: determine the person ... | [
"I think we can make it easier if we simply consider that both A and B can ensure that there's only 120ml when some bottles have milk more than 100ml.This is because 2022 is an even number.Whenever 90ml appears,A or B can make it into 120ml for there is at least one bottle have 30ml.\n(Of course, in a limited time... | [
"origin:aops",
"2022 Contests",
"2022 Taiwan Mathematics Olympiad"
] | {
"answer_score": 8,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Taiwan Mathematics Olympiad/2784241.json"
} |
Let $J$ be the $A$ -excenter of an acute triangle $ABC$ . Let $X$ , $Y$ be two points on the circumcircle of the triangle $ACJ$ such that $\overline{JX} = \overline{JY} < \overline{JC}$ . Let $P$ be a point lies on $XY$ such that $PB$ is tangent to the circumcircle of the triangle $ABC$ . Let $Q$ be ... | <blockquote> $\textbf{Lemma}$ The Miquel point of $\mathcal{Q}\{PX,XP^*,P^*Y,YP\}$ is on $(ABC)$ i dont understand this. can u help me?</blockquote>
To prove this, a good insight is by perfect hexagon. A perfect hexagon $ABCDEF$ on complex plane is defined as $\dfrac{a-b}{b-c}\cdot\dfrac{c-d}{d-e}\dfrac{e-f}{f... | [
"This is my solution during the test. Let $N$ be the midpoint of arc $\\widehat{ABC}$ , $P^*$ be the isogonal conjugate of $P$ wrt $\\triangle ABC$ , $P'$ is on $\\overline{XY}$ and satisfying $\\overline{XP}=\\overline{YP'}$ . We have $\\overline {JX}=\\overline{JY}$ , which implies $AX, AY$ is i... | [
"origin:aops",
"2022 Contests",
"2022 Taiwan Mathematics Olympiad"
] | {
"answer_score": 132,
"boxed": false,
"end_of_proof": false,
"n_reply": 9,
"path": "Contest Collections/2022 Contests/2022 Taiwan Mathematics Olympiad/2784246.json"
} |
There are $2022$ black balls numbered $1$ to $2022$ and $2022$ white balls numbered $1$ to $2022$ as well. There are also $1011$ black boxes and white boxes each. In each box we put two balls that are the same color as the the box. Prove that no matter how the balls are distributed, we can always pick on... | Similar to IMO 2020 P3.**<span style="color:#00f">Lemma: </span>**If a graph's all degrees are $2$ , then this graph can be divided into several cycles such that each vertex belongs to exactly one of them.
Let $B_1,\dots,B_{1011}$ and $W_1,\dots,W_{1011}$ be black and white boxes. Draw an edge between $B_i$ and... | [
"Write $N$ for $2022$ .\n\nWe make a graph with the numbers $1, 2, \\dots, N$ as vertices. Whenever two black balls numbered $a, b$ are put in the same black box, we add a black edge between the vertices $a$ and $b$ ; similarly add white edges corresponding to white boxes.\n\nIn the resulting graph, each ... | [
"origin:aops",
"2022 Contests",
"2022 Taiwan Mathematics Olympiad"
] | {
"answer_score": 130,
"boxed": false,
"end_of_proof": true,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Taiwan Mathematics Olympiad/2949651.json"
} |
Given an acute angle triangle $ABC$ with circumcircle $\Gamma$ and circumcenter $O$ . A point $P$ is taken on the line $BC$ but not on $[BC]$ . Let $K$ be the reflection of the second intersection of the line $AP$ and $\Gamma$ with respect to $OP$ . If $M$ is the intersection of the lines $AK$ and ... | I'm not as smart as #above so i found a different solution from others
Let $\angle BAC=\alpha , \angle ABC=\beta , \angle ACB=\gamma , AP \cap \Gamma={K'}$ <span style="color:#00f">Claim1.</span> $K \in \Gamma$
<details><summary>Proof</summary>Since $K$ -is the reflection of $K'$ over $OP$ we know that: $PK'... | [
"<details><summary>Solution</summary>We have $$ \\angle OLM=\\angle OKM=\\angle OAM\\Rightarrow OALM\\text{ is cyclic.} $$ Then, $$ \\angle LPM=\\angle LMO-\\angle PLM=\\angle ALM-\\angle LAO= $$ $$ \\angle ALM-\\angle ALO=\\angle OLM\\Rightarrow |OL|^2=|OM|\\cdot |OP| $$ Hence, $$ \\angle MBO=\\angle OPB... | [
"origin:aops",
"2022 Contests",
"2022 Turkey EGMO TST"
] | {
"answer_score": 282,
"boxed": false,
"end_of_proof": true,
"n_reply": 5,
"path": "Contest Collections/2022 Contests/2022 Turkey EGMO TST/2802959.json"
} |
We are given some three element subsets of $\{1,2, \dots ,n\}$ for which any two of them have at most one common element. We call a subset of $\{1,2, \dots ,n\}$ *nice* if it doesn't include any of the given subsets. If no matter how the three element subsets are selected in the beginning, we can add one more eleme... | Assume otherwise, then there is a violating subset for each potential added element. This amounts to at least $n-29$ violating subsets. But each subset includes two elements in the 29 element nice subset, so there are at most ${29 \choose 2}$ of these. Therefore $n \leq {30 \choose 2}$ .
Claim: There is such an e... | [
"<blockquote>Assume otherwise, then there is a violating subset for each potential added element. This amounts to at least $n-29$ violating subsets. But each subset includes two elements in the 29 element nice subset, so there are at most ${29 \\choose 2}$ of these. Therefore $n \\leq {30 \\choose 2}$ .\n\nCla... | [
"origin:aops",
"2022 Contests",
"2022 Turkey EGMO TST"
] | {
"answer_score": 18,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Turkey EGMO TST/2802960.json"
} |
On a table there are $100$ red and $k$ white buckets for which all of them are initially empty. In each move, a red and a white bucket is selected and an equal amount of water is added to both of them. After some number of moves, there is no empty bucket and for every pair of buckets that are selected together at l... | Generalize to $a, b$ red and white buckets.
If the subgraph of $K_{a, b}$ of buckets selected together is connected, then $a = b$ since the total amount of water on each side is equal.
If it isn't connected, simply split into connected components, so the answer is $k = 100$ . Construction is trivial. | [] | [
"origin:aops",
"2022 Contests",
"2022 Turkey EGMO TST"
] | {
"answer_score": 8,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Turkey EGMO TST/2802961.json"
} |
We are given three points $A,B,C$ on a semicircle. The tangent lines at $A$ and $B$ to the semicircle meet the extension of the diameter at points $M,N$ respectively. The line passing through $A$ that is perpendicular to the diameter meets $NC$ at $R$ , and the line passing through $B$ that is perpendicu... | Nice ratio chasing exercise.
Let $Z$ be the intersection of $MN$ and the tangent to the semicircle from $C$ . Let the feet of perpendiculars from $A,B,C$ to $MN$ be $U,V,W$ respectively. From Menelaus it is enough to prove that $\frac{ZM}{ZN}=\frac{SM}{SC}\cdot \frac{RC}{RN}$ . From Thales we know that $\... | [
"Invert wrt the semicircle helps well\nPole and polar helps too.\nBy the way, nice problem!",
"Can you give me a soultion\n",
"<blockquote>Invert wrt the semicircle helps well\nPole and polar helps too.\nBy the way, nice problem!</blockquote>\n\nCan you give me a solution"
] | [
"origin:aops",
"2022 Contests",
"2022 Turkey EGMO TST"
] | {
"answer_score": 32,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Turkey EGMO TST/2802963.json"
} |
Find all pairs of integers $(a,b)$ satisfying the equation $a^7(a-1)=19b(19b+2)$ . | $x(x+2)$ reminds of $(x+1)^2 - 1$ so we have $a^8 - a^7 + 1 = (19b+1)^2$ . from here we need to rewrite $a^8 - a^7 + 1 = xy$ such that $\gcd{(x,y)} = 1$ . we know $a^8 - a^7 + 1 = (a^2-a+1)(a^6 - a^4 - a^3 + a + 1)$ and also with Euclidean algorithm we have $\gcd{(a^2-a+1,a^6 - a^4 - a^3 + a + 1)} = {1,19}$ ... | [
"We have $$ (19b+1)^2=a^8-a^7+1=(a^2-a+1)(a^6-a^4-a^3+a+1) $$ Also $gcd(a^2-a+1,a^6-a^4-a^3+a+1)=1,\\; 19$ . But, $19$ doesn’t divide $(19b+1)^2$ . Hence, the gcd should be $1$ . This means both of the numbers are perfect squares.\nThen, $(2a-1)^2+3$ is a perfect square. Hence, $a=0,1$ and for these valu... | [
"origin:aops",
"2022 Contests",
"2022 Turkey EGMO TST"
] | {
"answer_score": 32,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Turkey EGMO TST/2802965.json"
} |
Let $x,y,z$ be positive real numbers satisfying the equations $$ xyz=1\text{ and }\frac yz(y-x^2)+\frac zx(z-y^2)+\frac xy(x-z^2)=0 $$ What is the minimum value of the ratio of the sum of the largest and smallest numbers among $x,y,z$ to the median of them. | See that $$ \left(\frac{x^2}y-1\right)\left(\frac{y^2}z-1\right)\left(\frac{z^2}x-1\right)=0 $$ WLOG $y^2=z$ . Then, $1=xyz=xy^3$ . Hence; $(x,y,z)=\left(t^3, \frac 1t, \frac 1{t^2}\right)$ for a positive real number $t$ .
Regardless of the value of $t$ , the median of these numbers is $\frac 1t$ . Hence, we ... | [] | [
"origin:aops",
"2022 Contests",
"2022 Turkey EGMO TST"
] | {
"answer_score": 24,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Turkey EGMO TST/2802966.json"
} |
Find all pairs of prime numbers $(p,q)$ for which
\[2^p = 2^{q-2} + q!.\] | Clearly the case when $p<q$ yields a contradiction
If $p=q$ The equation is transformed into $2^p=2^{p-2}+p!\Longleftrightarrow2^p-2^{p-2}=p!\Longrightarrow3\cdot2^{p-2}=p!$ However notice that $3\cdot2^{p-2}\equiv3,5\text{ or }6\pmod 7$ while $p!\equiv1\pmod 7$ for $p=5$ and $p!\equiv0\pmod 7$ for $p>5$ .... | [
"<details><summary>Hint</summary>mod 7 and $v_2$</details>\n<details><summary>Solution</summary>It is easy to see that $(p, q)=(3, 3)$ and $(p, q)=(7, 5)$ are solutions. Now assume that $q \\ge 7$ and $p\\ge 13$ . $\\text{mod 7}$ gives us $p \\equiv q-2 \\pmod{3}$ . Since both $p$ and $q$ are differe... | [
"origin:aops",
"2022 Contests",
"2022 Turkey Team Selection Test"
] | {
"answer_score": 1184,
"boxed": false,
"end_of_proof": false,
"n_reply": 7,
"path": "Contest Collections/2022 Contests/2022 Turkey Team Selection Test/2800667.json"
} |
Find all functions $f: \mathbb{Q^+} \rightarrow \mathbb{Q}$ satisfying $f(x)+f(y)= \left(f(x+y)+\frac{1}{x+y} \right) (1-xy+f(xy))$ for all $x, y \in \mathbb{Q^+}$ . | <details><summary>It's bash time</summary>Let $P(x,y)$ denote the assertion. $P(1,1)$ gives us $2f(1)=\left(f(2)+\frac 12\right)f(1)$ .
<span style="color:#f00">**i)**</span> $f(1)=0$ . $P\left(x, \frac 1x\right)$ gives us $f(x)+f\left(\frac 1x\right)=0$ ......(1) $P(x,1)$ gives us $f(x)=\left(f(x+1)+\frac 1{x... | [
"This problem is not completely bashy actually and Q+ can be changed to R+. Here is another solution for $f(1)=0$ (sorry for not writing completely, I do not remember the equations and I am lazy to rediscover them):\n\n1_ $P(x,1)$ gives the relation between $f(x)$ and $f(x+1)$ .\n\n2_ $P(x,1-x)$ gives the r... | [
"origin:aops",
"2022 Contests",
"2022 Turkey Team Selection Test"
] | {
"answer_score": 114,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Turkey Team Selection Test/2800669.json"
} |
In a triangle $ABC$ , the incircle centered at $I$ is tangent to the sides $BC, AC$ and $AB$ at $D, E$ and $F$ , respectively. Let $X, Y$ and $Z$ be the feet of the perpendiculars drawn from $A, B$ and $C$ to a line $\ell$ passing through $I$ . Prove that $DX, EY$ and $FZ$ are concurrent. | <blockquote>In a triangle $ABC$ , the incircle centered at $I$ is tangent to the sides $BC, AC$ and $AB$ at $D, E$ and $F$ , respectively. Let $X, Y$ and $Z$ be the feet of the perpendiculars drawn from $A, B$ and $C$ to a line $\ell$ passing through $I$ . Prove that $DX, EY$ and $FZ$ are concu... | [
"<details><summary>Trigonometric Solution</summary>See that we have $FXIE$ is cyclic.\nNow, let’s restate this problem.\n\nGiven a triangle $DEF$ with circumcenter $I$ . A line $\\ell$ passing through $I$ intersects $(FIE), (FID)$ and $(DIE)$ at $X, Y$ and $Z$ , respectively. Prove that $DX, EY$ a... | [
"origin:aops",
"2022 Contests",
"2022 Turkey Team Selection Test"
] | {
"answer_score": 336,
"boxed": false,
"end_of_proof": false,
"n_reply": 9,
"path": "Contest Collections/2022 Contests/2022 Turkey Team Selection Test/2800673.json"
} |
We have three circles $w_1$ , $w_2$ and $\Gamma$ at the same side of line $l$ such that $w_1$ and $w_2$ are tangent to $l$ at $K$ and $L$ and to $\Gamma$ at $M$ and $N$ , respectively. We know that $w_1$ and $w_2$ do not intersect and they are not in the same size. A circle passing through $K... | Because of my amazing geometry skills, I couldn't see the direct angle chasing. Anyway, by Monge Theorem, we know that the center of homotety sending $\omega_1$ to $\omega_2$ is the intersection of $KL$ and $MN$ . Take the second intersections $X$ , $Y$ of $MN$ and $\omega_1$ , $\omega_2$ , angle chasing ... | [] | [
"origin:aops",
"2022 Contests",
"2022 Turkey Team Selection Test"
] | {
"answer_score": 28,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Turkey Team Selection Test/2800674.json"
} |
On a circle, 2022 points are chosen such that distance between two adjacent points is always the same. There are $k$ arcs, each having endpoints on chosen points, with different lengths. Arcs do not contain each other. What is the maximum possible number of $k$ ? | <span style="color:#134F5C">The answer is $k=1011$ .</span>
Let's number the points with $1, 2 \dots ,2022$ .
Choose $1$ length arc from point $1$ , $2$ length arc from point $2$ and so on to $1011$ length arc from point $1011$ . This gives the desired $k=1011$ , since every problem condition is satisfie... | [
"Sketch of the solution (will add details later)**<span style=\"color:#f00\">Answer:</span>** $1011$ **<span style=\"color:#f00\">Proof of the bound:</span>**\nLet the selected arcs be $a_1,a_2,...,a_k$ First, observe that by decreasing the length of all arcs clockwise by $1$ , we can assume the length of the sm... | [
"origin:aops",
"2022 Contests",
"2022 Turkey Team Selection Test"
] | {
"answer_score": 146,
"boxed": false,
"end_of_proof": true,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Turkey Team Selection Test/2800675.json"
} |
For a polynomial $P(x)$ with integer coefficients and a prime $p$ , if there is no $n \in \mathbb{Z}$ such that $p|P(n)$ , we say that polynomial $P$ *excludes* $p$ . Is there a polynomial with integer coefficients such that having degree of 5, excluding exactly one prime and not having a rational root? | Here's a similar construction:
Let $P(x)=((2x)^2+3)((2x)^3-3)$ . We claim this polynomial works. Firstly, note that $P$ excludes $2$ , as all of its outputs are odds. Now, let $p>2$ be a prime. We distinguish three cases.**Case 1:** $p=3$ . Then, we may take $x=3$ .**Case 2:** $p \equiv 1 \pmod 3$ . Then, we ... | [
"<details><summary>Solution</summary>Unfortunately and surprisingly the answer is yes.\n\nThe polynomial $P(x)=((3x)^2+3x+1)((3x)^3+2)$ works. \n\nPrime numbers in the form $3k+1$ divides $(3x)^2+3x+1$ and prime numbers in the form $3k+2$ divides $(3x)^3+2$ . Clearly, $3$ doesn’t divide this polynomial ... | [
"origin:aops",
"2022 Contests",
"2022 Turkey Team Selection Test"
] | {
"answer_score": 42,
"boxed": false,
"end_of_proof": false,
"n_reply": 9,
"path": "Contest Collections/2022 Contests/2022 Turkey Team Selection Test/2800676.json"
} |
What is the minimum value of the expression $$ xy+yz+zx+\frac 1x+\frac 2y+\frac 5z $$ where $x, y, z$ are positive real numbers? | Let $x, y, z$ are positive real numbers. Prove that $$ xy+yz+zx+\frac 1x+\frac 2y+\frac 5z\geq 3\sqrt[3]{36} $$ <details><summary>Good.</summary><blockquote><details><summary>Answer</summary>$3\sqrt[3]{36}$</details>
<details><summary>Equality case</summary>$(x,y,z)=\left(\frac{\sqrt[3]{6}}3, \frac{\sqrt[3]{6}}2, ... | [
"<details><summary>Answer</summary>$3\\sqrt[3]{36}$</details>\n<details><summary>Equality case</summary>$(x,y,z)=\\left(\\frac{\\sqrt[3]{6}}3, \\frac{\\sqrt[3]{6}}2, \\sqrt[3]{6}\\right)$</details>\n<details><summary>Proof</summary>By AM-GM inequality, $$ xy+\\frac{1}{3x}+\\frac{1}{2y}\\ge \\frac{\\sqrt[3]{36}}2 ... | [
"origin:aops",
"2022 Contests",
"2022 Turkey Team Selection Test"
] | {
"answer_score": 64,
"boxed": false,
"end_of_proof": false,
"n_reply": 7,
"path": "Contest Collections/2022 Contests/2022 Turkey Team Selection Test/2800677.json"
} |
$ABC$ triangle with $|AB|<|BC|<|CA|$ has the incenter $I$ . The orthocenters of triangles $IBC, IAC$ and $IAB$ are $H_A, H_A$ and $H_A$ . $H_BH_C$ intersect $BC$ at $K_A$ and perpendicular line from $I$ to $H_BH_B$ intersect $BC$ at $L_A$ . $K_B, L_B, K_C, L_C$ are defined similarly. Prove th... | Let $D,E,F$ be the tangency points of incircle. Let $D',E',F'$ be their antipodes and $AD'\cap (I)=D_0$ etc. We have $(H_A,I;H_AI\cap EF,D)=-1=(E_0,E';F,D)\overset{E}{=}(EE_0\cap ID,I;EF\cap ID,D)$ hence $E,E_0,H_A$ are collinear so $H_B,H_C$ lie on $DD_0$ . Thus, $K_A=D$ and $L_A$ is the midpoint of ... | [
"How many problems are in a Turkey TST?",
"There are 3 days each containing exactly 3 problems which gives a total of 9 problems.",
"**Lemma 1** $H_{B},D,H_{C}$ are collinear.**Proof 1** Since $BH_{C}$ and $CH_{B}$ are parallel and $\\frac{BD}{DC}=\\frac{BF}{CE}=\\frac{BH_{C}}{CH_{B}}$ $H_{B},D,H_{C}$ ... | [
"origin:aops",
"2022 Contests",
"2022 Turkey Team Selection Test"
] | {
"answer_score": 122,
"boxed": false,
"end_of_proof": true,
"n_reply": 7,
"path": "Contest Collections/2022 Contests/2022 Turkey Team Selection Test/2800678.json"
} |
In every acyclic graph with 2022 vertices we can choose $k$ of the vertices such that every chosen vertex has at most 2 edges to chosen vertices. Find the maximum possible value of $k$ . | induct to winduct
Let the graph be $T$ and call a vertex subset *good* if the induced subgraph formed by it has maximum degree $2$ . Replace $2022$ with $n$ . Then the answer is $\lceil \tfrac{3}{4}n \rceil$ .
<span style="color:#00f">**Construction:**</span> Let $n=4a+b$ with $0 \leq b<4$ . Then construct... | [
"Take the star graph so we have $k \\le 3$ . To show we can always pick $3$ , choose a leaf, its neighbour and an arbitrary other vertex, so nothing can be adjacent to all three of these, done.",
"Every chosen vertex must have at most 2 chosen neighbors, not every vertex.",
"I guess the answer is 1517 ,is it?... | [
"origin:aops",
"2022 Contests",
"2022 Turkey Team Selection Test"
] | {
"answer_score": 232,
"boxed": false,
"end_of_proof": true,
"n_reply": 8,
"path": "Contest Collections/2022 Contests/2022 Turkey Team Selection Test/2800680.json"
} |
Find all pairs of primes $(p, q)$ for which $p-q$ and $pq-q$ are both perfect squares. | Does this work?
We claim the answer is $\boxed{(3,2)}$ .
Case 1: $q=2$ .
Then let $p-2=a^2$ and $2p-2=2(p-1)=4b^2$ .
So we have \[p=(2p-2)-(p-2)=4b^2-a^2=(2b-a)(2b+a)\]
Thus, $2b-a=1$ and $2b+a=p$ .
This implies $4b=p+1\implies b=\frac{p+1}{4}$ .
Then $a=\frac{p-1}{2}$ .
Now \[p-2=\frac{(p-1)^2}{... | [
"Ans (3,2). Subtract and factor by difference of squares. Then p must go into one of the two. Make cases and then do bounding.",
"(3,2) was the only answer",
"Also JMO 2022/5\nFor some reason this took me like 2 hours in contest and I got it with 20 minutes left. Unlike j2 I managed to finish the writeup in tim... | [
"origin:aops",
"2022 USAJMO",
"2022 Contests"
] | {
"answer_score": 1152,
"boxed": false,
"end_of_proof": true,
"n_reply": 148,
"path": "Contest Collections/2022 Contests/2022 USAJMO/2808839.json"
} |
Let $ABCD$ be a rhombus, and let $K$ and $L$ be points such that $K$ lies inside the rhombus, $L$ lies outside the rhombus, and $KA = KB = LC = LD$ . Prove that there exist points $X$ and $Y$ on lines $AC$ and $BD$ such that $KXLY$ is also a rhombus.
*Proposed by Ankan Bhattacharya* | We present **two** solutions.
<span style="color:#f00">**Solution 1:**</span> *Cartesian Coordinates.*
Let $O=AC \cap BD$ be the origin, and consequently $A=(-a,0),C=(a,0),B=(0,b), D=(0,-b).$ [asy]
import olympiad;
unitsize(50);
defaultpen(fontsize(8.5pt)); dotfactor*=1;
pair A,B,C,D,O,K,L,X,Y,M,R,S,T,U;
A = (-1,... | [
"Quite easy for a J4. Once you add the midpoints of $AB$ and $CD$ and realize there's a translation taking $ABK$ to $DCL$ , you can use parallelograms and angle chase to finish.**One Thing:** I don't know if you can automatically assume the triangles are directly similar (based on the inside/outside conditio... | [
"origin:aops",
"2022 USAJMO",
"2022 Contests"
] | {
"answer_score": 368,
"boxed": false,
"end_of_proof": false,
"n_reply": 58,
"path": "Contest Collections/2022 Contests/2022 USAJMO/2808841.json"
} |
For which positive integers $m$ does there exist an infinite arithmetic sequence of integers $a_1, a_2, . . .$ and an infinite geometric sequence of integers $g_1, g_2, . . .$ satisfying the following properties?
- $a_n - g_n$ is divisible by $m$ for all integers $n \ge 1$ ;
- $a_2 - a_1$ is not divisible ... | The answer is $\boxed{m\text{ is not squarefree }}$ .
Proof of necessity: Suppose $m$ was squarefree. Let the common difference be $d$ . We will show that $d^2\mid m$ .
Consider three terms $a-d,a,a+d$ . By the first condition, we must have $(a-d)(a+d)\equiv a^2\pmod m$ . So \[a^2-d^2\equiv a^2\pmod m\implie... | [
"The solution set is all $m$ such that there exist a prime p with $v_p(m) \\geq 2$ .\n\n First we show that all such $m$ works. Assume $m = p^e \\cdot q$ , where $p$ is prime and $p\\not | \\ q$ . We can then construct $a_n = 1+(n-1)p^{e-1}q$ and $g_n =(p^{e-1}q+1)^{n-1}$ . It is obvious that the seco... | [
"origin:aops",
"2022 USAJMO",
"2022 Contests"
] | {
"answer_score": 1124,
"boxed": false,
"end_of_proof": false,
"n_reply": 38,
"path": "Contest Collections/2022 Contests/2022 USAJMO/2808845.json"
} |
Let $a$ and $b$ be positive integers. The cells of an $(a+b+1)\times (a+b+1)$ grid are colored amber and bronze such that there are at least $a^2+ab-b$ amber cells and at least $b^2+ab-a$ bronze cells. Prove that it is possible to choose $a$ amber cells and $b$ bronze cells such that no two of the $a+b$ ... | Define a set $S$ of cells in the $(a + b + 1) \times (a + b + 1)$ grid to be "non-conflicting" if it satisfies the following properties:
1. Each column has exactly 1 cell in $S$ 2. Each row has exactly 1 cell in $S$ These conditions imply that $|S| = a+b+1$ .**Claim**: there exists a non-conflicting set $A$ wi... | [
"Define permutations $\\pi \\in [a+b+1]$ on the grid in an obvious way. We want a permutation which has either $a$ A's and $b+1$ B's or $a+1$ A's and $b$ B's. Define the score of a permutation $s_A(\\pi)$ to be the number of A's in it. We want $s_A(\\pi) \\in \\{a,a+1\\}$ for some $\\pi$ . \n\n<span ... | [
"origin:aops",
"2022 USAJMO",
"2022 Contests"
] | {
"answer_score": 230,
"boxed": false,
"end_of_proof": true,
"n_reply": 52,
"path": "Contest Collections/2022 Contests/2022 USAJMO/2808846.json"
} |
Let $a_0, b_0, c_0$ be complex numbers, and define \begin{align*}a_{n+1} &= a_n^2 + 2b_nc_n b_{n+1} &= b_n^2 + 2c_na_n c_{n+1} &= c_n^2 + 2a_nb_n\end{align*}for all nonnegative integers $n.$
Suppose that $\max{\{|a_n|, |b_n|, |c_n|\}} \leq 2022$ for all $n.$ Prove that $$ |a_0|^2 + |b_0|^2 + |c_0|^2 \leq... | Let $S_n=|a_n|+|b_n|+|c_n|$ for all $n$ **Key claim:** $S_{n+1} \geq S_n^2$ for all $n$ .
*Proof:* Equivalently, we want to prove (let $a_n=x$ , $b_n=y$ and $c_n=z$ ) that $$ |x^2+2yz|^2+|y^2+2zx|^2+|z^2+2xy|^2 \geq (|x|^2+|y|^2+|z|^2) $$ However, $|x^2+2yz|^2+|y^2+2zx|^2+|z^2+2xy|^2-(|x|^2+|y|^2+|z|^2)^2=... | [
"<details><summary>One-liner</summary>$|a_{n+1}|^2+|b_{n+1}|^2+|c_{n+1}|^2=(|a_n|^2+|b_n|^2+|c_n|^2)^2+2|a_n\\overline{b_n}+b_n\\overline{c_n}+c_n\\overline{a_n}|^2$</details>",
"this was a joke problem 6, what happened to maa this year.",
"ok dotted got the same thing as me\nthankfully no fakesolve :D",
"@2a... | [
"origin:aops",
"2022 USAJMO",
"2022 Contests"
] | {
"answer_score": 140,
"boxed": false,
"end_of_proof": true,
"n_reply": 27,
"path": "Contest Collections/2022 Contests/2022 USAJMO/2808878.json"
} |
Let $b\geq2$ and $w\geq2$ be fixed integers, and $n=b+w$ . Given are $2b$ identical black rods and $2w$ identical white rods, each of side length 1.
We assemble a regular $2n-$ gon using these rods so that parallel sides are the same color. Then, a convex $2b$ -gon $B$ is formed by translating the black r... | umm overexplained and nonrigorous solution below...
Change white and black to red and blue, rename $b$ and $w$ as $r$ and $b.$ We claim if we swap the colors of two adjacent sides, along with their opposite counterparts, the difference of areas will remain the same. Notice this is sufficient to solve the proble... | [
"complex bash lol\n\n**Attachments:**\n\n[2022-03-22 18-03-compressed.pdf](https://cdn.artofproblemsolving.com/attachments/4/b/3a219ea90dcbda012f6878c17eea47996df8e9.pdf)",
"Trivial by vectors.",
"<details><summary>Solution</summary>Define $\\theta = \\frac{2\\pi}{n}$ radians. Fix a side $s$ of the $2n$ -g... | [
"origin:aops",
"2022 USAJMO",
"2022 Contests"
] | {
"answer_score": 232,
"boxed": false,
"end_of_proof": false,
"n_reply": 40,
"path": "Contest Collections/2022 Contests/2022 USAJMO/2808883.json"
} |
There are $2022$ users on a social network called Mathbook, and some of them are Mathbook-friends. (On Mathbook, friendship is always mutual and permanent.)
Starting now, Mathbook will only allow a new friendship to be formed between two users if they have *at least two* friends in common. What is the minimum numbe... | <details><summary>old</summary>My original sol was incomprehensible so I wrote a new comprehensible (hopefully) sol.
Generalize the problem for $n$ users. Let users be vertices of a graph $G$ . Two vertices are connected if and only if the users the vertices correspond to are friends. The operation turns a $C_4$ ... | [
"If this is correct, then on my first official USAMO I solved problem 6. WOW \n\nThe answer is $f(4k-2)=6k-5, f(4k-1)=6k-3, f(4k)=6k-2$ and $f(4k+1)=6k$ . Note the operation basically turns a $C_4$ into a $K_4$ .\n\n\nWe consider an alternate process that correspond to cliques of (at least 4) vertices merging... | [
"origin:aops",
"2022 Contests",
"2022 USAMO"
] | {
"answer_score": 1448,
"boxed": false,
"end_of_proof": false,
"n_reply": 15,
"path": "Contest Collections/2022 Contests/2022 USAMO/2808836.json"
} |
A function $f: \mathbb{R}\to \mathbb{R}$ is *essentially increasing* if $f(s)\leq f(t)$ holds whenever $s\leq t$ are real numbers such that $f(s)\neq 0$ and $f(t)\neq 0$ .
Find the smallest integer $k$ such that for any 2022 real numbers $x_1,x_2,\ldots , x_{2022},$ there exist $k$ essentially increasin... | The answer is $\boxed{11}=\log_2(2022+1)$ .**Bound.** Let $x_n=2023-n$ for all $n\in [2022]$ . For each $n\in[2022]$ , let $I_n\subseteq [k]$ be the set of all $i$ such that $f_i(n)$ is nonzero. If there exist integers $1\leq n_1<n_2\leq 2022$ such that $I_{n_1}=I_{n_2}$ , then
\[
x_{n_1}=f_1(n_1)+\cdots+... | [
"The answer is $11 = \\lceil \\log_2(2022) \\rceil$ . **<span style=\"color:#00f\">Proof of bound:</span>** Take a strictly decreasing sequence $(x_1,\\ldots,x_{2022})$ which never hits $0$ . Suppose there existed working functions $f_1,\\ldots,f_{10}$ . Define $v(n) = (f_1(n),\\ldots,f_{10}(n))$ . Simply con... | [
"origin:aops",
"2022 Contests",
"2022 USAMO"
] | {
"answer_score": 1202,
"boxed": true,
"end_of_proof": true,
"n_reply": 26,
"path": "Contest Collections/2022 Contests/2022 USAMO/2808837.json"
} |
Let $\mathbb{R}_{>0}$ be the set of all positive real numbers. Find all functions $f:\mathbb{R}_{>0} \to \mathbb{R}_{>0}$ such that for all $x,y\in \mathbb{R}_{>0}$ we have
\[f(x) = f(f(f(x)) + y) + f(xf(y)) f(x+y).\] | Let $\mathbb{R}_{>0}$ be the set of all positive real numbers. Find all functions $f:\mathbb{R}_{>0} \rightarrow \mathbb{R}_{>0}$ such that for all $x, y \in \mathbb{R}_{>0}$ we have $$ f(x) = f (f(f(x)) + y) + f(xf(y))f(x + y). $$ .
Solved with hint from rama1728 (the solution set)
Let $P(x,y)$ denote the ... | [
"We have $f(x)>f(f^2(x)+y)$ . We'll derive a LOT of information from this alone. Equivalently, this says:\n\\[ C>f^2(x) \\implies f(C)<f(x) \\quad \\forall C,x. \\qquad (\\spadesuit)\\]\nIf $x>f^2(x)$ , then $C=x$ in $(\\spadesuit)$ gives $f(x)<f(x)$ , contradiction. Hence $\\boxed{f^2(x) \\ge x}$ . So $f... | [
"origin:aops",
"2022 Contests",
"2022 USAMO"
] | {
"answer_score": 1236,
"boxed": false,
"end_of_proof": true,
"n_reply": 44,
"path": "Contest Collections/2022 Contests/2022 USAMO/2808852.json"
} |
Let $n\ge 2$ be an integer. Thibaud the Tiger lays $n$ $2\times 2$ overlapping squares out on a table, such that the centers of the squares are equally spaced along the line $y=x$ from $(0,0)$ to $(1,1)$ (including the two endpoints). For example, for $n=4$ the resulting figure is shown below, and it cove... | Note that the figure consists of $n-1$ L-shapes with area $\frac{4n-5}{(n-1)^2}$ and one square with area $4$ . Hence, we require $\frac{4n-5}{n-1} + 4\ge \sqrt{63}$ $\iff (8-\sqrt{63})n\ge 9-\sqrt{63}$ $\iff n\ge 9+\sqrt{63}$ so the desired minimum is $\lceil 9+\sqrt{63}\rceil = \boxed{17}$ . | [] | [
"origin:aops",
"2022 Contests",
"2022 Utah Mathematical Olympiad"
] | {
"answer_score": 1014,
"boxed": true,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Utah Mathematical Olympiad/2810504.json"
} |
Let $x$ and $y$ be relatively prime integers. Show that $x^2+xy+y^2$ and $x^2+3xy+y^2$ are relatively prime. | Let $p$ be a prime dividing $x^2+xy+y^2$ and $x^2+3xy+y^2$ .
Notice that $(x,y) = (O,E), (E,O), $ or $(O,O)$ where $O$ denotes an odd number and $E$ an even. This is because if $x$ and $y$ were both even, then $2$ divides both of them, which isn't ok.
In any of those $3$ cases, $x^2+xy+y^2$ i... | [
"Suppose FTSOC that there exists a prime $p$ dividing both $x^2 + xy + y^2$ and $x^2 + 3xy + y^2$ . Then $p$ divides both $x^2 + 3xy + y^2 - (x^2 + xy + y^2) = 2xy$ and $x^2 + xy + y^2 + x^2 + 3xy + y^2 = 2(x+y)^2$ .\n\nCase 1: $p$ is odd. Then $p | xy$ , so WLOG let $p | x$ . Then $p | (x+y) - x = y$ ,... | [
"origin:aops",
"2022 Contests",
"2022 Utah Mathematical Olympiad"
] | {
"answer_score": 54,
"boxed": false,
"end_of_proof": false,
"n_reply": 12,
"path": "Contest Collections/2022 Contests/2022 Utah Mathematical Olympiad/2810505.json"
} |
Find all sequences $a_1, a_2, a_3, \dots$ of real numbers such that for all positive integers $m,n\ge 1$ , we have
\begin{align*}
a_{m+n} &= a_m+a_n - mn \text{ and}
a_{mn} &= m^2a_n + n^2a_m + 2a_ma_n.
\end{align*} | <blockquote>Find all sequences $a_1, a_2, a_3, \dots$ of real numbers such that for all positive integers $m,n\ge 1$ , we have
\begin{align*}
a_{m+n} &= a_m+a_n - mn \text{ and}
a_{mn} &= m^2a_n + n^2a_m + 2a_ma_n.
\end{align*}</blockquote>
We have $a_1=2a_1+2a_1^2\Rightarrow a_1\in\left\{0,-\frac{1}{2}\right\}$... | [] | [
"origin:aops",
"2022 Contests",
"2022 Utah Mathematical Olympiad"
] | {
"answer_score": 16,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Utah Mathematical Olympiad/2810508.json"
} |
Alpha and Beta are playing a game on a $10\times 100$ grid of squares. At each turn, they can fold the grid along any of the interior horizontal or vertical gridlines, which creates a smaller (folded) grid of squares (on the first move, they can choose one of $9$ horizontal or $99$ vertical gridlines). The person... | [] | [
"origin:aops",
"2022 Contests",
"2022 Utah Mathematical Olympiad"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/2022 Contests/2022 Utah Mathematical Olympiad/2810510.json"
} | |
$2022$ lily pads are arranged in a circle. Each lily pad starts with height $1$ . A frog starts on one of the lily pads, and jumps around clockwise as follows: if the frog is on a lily bad of height $k$ , the lily pad grows by $1$ (becoming $k+1$ ), and then the frog jumps $k$ lily pads clockwise (i.e. jumping... | [] | [
"origin:aops",
"2022 Contests",
"2022 Utah Mathematical Olympiad"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/2022 Contests/2022 Utah Mathematical Olympiad/2810513.json"
} | |
An $m \times n$ grid of squares (with $m$ rows and $n$ columns) has some of its squares colored blue. The grid is called *fish-friendly* if a fish can swim from the left edge of the grid to the right edge of the grid only moving through blue squares. In other words, there is a sequence of blue squares, each horiz... | [] | [
"origin:aops",
"2022 Contests",
"2022 Utah Mathematical Olympiad"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/2022 Contests/2022 Utah Mathematical Olympiad/2810514.json"
} | |
Let $a$ be a non-negative real number and a sequence $(u_n)$ defined as: $u_1=6,u_{n+1} = \frac{2n+a}{n} + \sqrt{\frac{n+a}{n}u_n+4}, \forall n \ge 1$ a) With $a=0$ , prove that there exist a finite limit of $(u_n)$ and find that limit
b) With $a \ge 0$ , prove that there exist a finite limit of $(u_n)$ | For part a) we easy have $u_n>5\quad\forall n\ge 1$ and $$ |u_{n+1}-5|=\frac{|u_n-5|}{\sqrt{u_n+4}+3}<\frac{1}{5}|u_n-5|<\cdots<\left(\frac{1}{5}\right)^n|u_1-5|\implies \lim u_n=5. $$ For part b) we use the well-known lemma
If $z_n$ is positive sequence and $z_{n+1}\le cz_n+q_n$ with $\lim q_n=0$ and $c\in(... | [
"DNCT1. You can prove that lemma: \nIf $z_n$ is positive sequence and $z_{n+1}\\le cz_n+q_n$ with $\\lim q_n=0$ and $c\\in(0,1)$ then we have $\\lim z_n=0$ ."
] | [
"origin:aops",
"2022 Contests",
"2022 Vietnam National Olympiad"
] | {
"answer_score": 58,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Vietnam National Olympiad/2794224.json"
} |
Find all function $f:\mathbb R^+ \rightarrow \mathbb R^+$ such that:
\[f\left(\frac{f(x)}{x}+y\right)=1+f(y), \quad \forall x,y \in \mathbb R^+.\] | <blockquote>Find all function $f:\mathbb R^+ \rightarrow \mathbb R^+$ such that: $f(\frac{f(x)}{x}+y)=1+f(y), \forall x,y \in \mathbb R^+$ </blockquote>
Let $P(x,y)$ be the assertion $f(\frac{f(x)}x+y)=1+f(y)$ If $f(x)$ is injective : $P(x,y)$ and $P(1,y)$ imply $f(\frac{f(x)}x+y)=f(f(1)+y)$ and so, sinc... | [
"<blockquote><blockquote>Find all function $f:\\mathbb R^+ \\rightarrow \\mathbb R^+$ such that: $f(\\frac{f(x)}{x}+y)=1+f(y), \\forall x,y \\in \\mathbb R^+$ </blockquote>\nLet $P(x,y)$ be the assertion $f(\\frac{f(x)}x+y)=1+f(y)$ If $f(x)$ is injective : $P(x,y)$ and $P(1,y)$ imply $f(\\frac{f(x)}x+... | [
"origin:aops",
"2022 Contests",
"2022 Vietnam National Olympiad"
] | {
"answer_score": 1066,
"boxed": false,
"end_of_proof": false,
"n_reply": 13,
"path": "Contest Collections/2022 Contests/2022 Vietnam National Olympiad/2794225.json"
} |
Let $ABC$ be a triangle. Point $E,F$ moves on the opposite ray of $BA,CA$ such that $BF=CE$ . Let $M,N$ be the midpoint of $BE,CF$ . $BF$ cuts $CE$ at $D$ a) Suppost that $I$ is the circumcenter of $(DBE)$ and $J$ is the circumcenter of $(DCF)$ , Prove that $MN \parallel IJ$ b) Let $K$ be the m... | a) Let $MI$ intersects $NJ$ at $P$ . Note that $MI$ and $NJ$ are perpendicular bisectors of $BE$ and $CF$ , we have $PB=PE$ and $PC=PF$ . Combining with $BF=CE$ we get $PEC \cong PBF$ . Hence, $PEB \sim PCF$ and $(I,J),(M,N)$ are corresponding points. So $\tfrac{PI}{PM}=\tfrac{PJ}{PN}$ and there... | [
"The idea of this problem is complete quadrilateral\na) Use similar triangle and a bit of angle chasing\nb) note that $HK$ is the Steiner line of quadrilateral $BCFE$ ",
"a) Let $S$ be second intersection of $(DBE)$ and $(DCF)$ . We have $\\angle{SFD} = \\angle{SCD}$ and $\\angle{SED} = \\angle{SBD}$ .... | [
"origin:aops",
"2022 Contests",
"2022 Vietnam National Olympiad"
] | {
"answer_score": 80,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Vietnam National Olympiad/2794227.json"
} |
For every pair of positive integers $(n,m)$ with $n<m$ , denote $s(n,m)$ be the number of positive integers such that the number is in the range $[n,m]$ and the number is coprime with $m$ . Find all positive integers $m\ge 2$ such that $m$ satisfy these condition:
i) $\frac{s(n,m)}{m-n} \ge \frac{s(1,m)}{m... | Pardon me if I'm extremely tired, but here's what I think is a solution.
Let $p>2$ be the smallest prime dividing $m$ (clearly $2^2\nmid 2022^n+1$ ). Thus, we get that
\[\frac{s(1,m)-(p-1)}{m-p}=\frac{s(p,m)}{m-p}\geq\frac{s(1,m)}m\]
Upon rearranging, we get
\[m(p-1)\geq p\varphi(m)=ps(1,m)\geq m(p-1)\]
Thus, all... | [
"bump......",
"I believe that we need $m^2|2022^m+1$ rather than $m|2022^m+1$ .\n\nEdit: I never claim that it’s not possible to fix.",
"You're right, but isn't that easily fixable? The only change is that there are finitely many $k$ (again by LTE). "
] | [
"origin:aops",
"2022 Contests",
"2022 Vietnam National Olympiad"
] | {
"answer_score": 24,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Vietnam National Olympiad/2794232.json"
} |
Consider 2 non-constant polynomials $P(x),Q(x)$ , with nonnegative coefficients. The coefficients of $P(x)$ is not larger than $2021$ and $Q(x)$ has at least one coefficient larger than $2021$ . Assume that $P(2022)=Q(2022)$ and $P(x),Q(x)$ has a root $\frac p q \ne 0 (p,q\in \mathbb Z,(p,q)=1)$ . Prove th... | Because $Q(2022)=P(2022)$ , let $Q(x)-P(x)=(x-2022)(b_m.x^m+b_{m-1}x^{m-1}+...+b_0)=b_m x^{m+1}+(-2022b_m+b_{m-1})x^m+...+(-2022b_1+b0)x+(-2022b_0)$ The coffients of $P(x)$ is not larger than $2021$ and $Q(x)$ has at least one coffient larger than $2021$ , we get $Q(x)-P(x)$ is not constant (which means $m ... | [
"Well, @above, you proved that $b_i<1$ for all $i$ , but that doesn't give $b_i \\leq 0$ for all $i$ since the coefficients are not integers. Correct me if I am missing something, but either the statement is missing the condition for integer coefficients, or the above solution doesn't work for non-integer co... | [
"origin:aops",
"2022 Contests",
"2022 Vietnam National Olympiad"
] | {
"answer_score": 86,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Vietnam National Olympiad/2794824.json"
} |
We are given 4 similar dices. Denote $x_i (1\le x_i \le 6)$ be the number of dots on a face appearing on the $i$ -th dice $1\le i \le 4$ a) Find the numbers of $(x_1,x_2,x_3,x_4)$ b) Find the probability that there is a number $x_j$ such that $x_j$ is equal to the sum of the other 3 numbers
c) Find the probabi... | <blockquote>We are given 4 similar dices. Denote $x_i (1\le x_i \le 6)$ be the number of dots on a face appearing on the $i$ -th dice $1\le i \le 4$ a) Find the numbers of $(x_1,x_2,x_3,x_4)$ b) Find the probability that there is a number $x_j$ such that $x_j$ is equal to the sum of the other 3 numbers
c) Find... | [
"<details><summary>Comment</summary>I don't know why this year the combinatoric problem is easy.\nAnd also, there is only one combinatoric problem. Sad :([/hide=Comment]</details>",
"<blockquote><details><summary>Comment</summary>I don't know why this year the combinatoric problem is easy.\nAnd also, there is onl... | [
"origin:aops",
"2022 Contests",
"2022 Vietnam National Olympiad"
] | {
"answer_score": 16,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Vietnam National Olympiad/2794825.json"
} |
Let $ABC$ be an acute triangle, $B,C$ fixed, $A$ moves on the big arc $BC$ of $(ABC)$ . Let $O$ be the circumcenter of $(ABC)$ $(B,O,C$ are not collinear, $AB \ne AC)$ , $(I)$ is the incircle of triangle $ABC$ . $(I)$ tangents to $BC$ at $D$ . Let $I_a$ be the $A$ -excenter of triangle $AB... | a) Relabel $E$ be $S$ and $F$ be $R$ . Let $E, F$ be contact points of $(I)$ with $CA, AB;$ $B_e$ be Bevan point of $\triangle ABC$ . We have $\dfrac{\overline{LS}}{\overline{LR}} = \dfrac{\overline{LD}}{\overline{LI_a}} = \dfrac{\overline{LI}}{\overline{LB_e}}$ , so $R$ $\in$ $(B_e, B_eI_a)$ . He... | [
"a)If $S,R,N,T,A'$ are $A-$ sharkydevil, midpoints of arcs $BC,BAC$ , $A-$ mixtilinear touchpoint and antipode of $A$ on $(ABC)$ , then Pascal at $NRSA'AT$ gives that $MD,AT,IO$ concur. Since $A,E,T$ are collinear, $AE,IO,DR$ concur. If $I'$ is the reflection of $I$ over $A$ , since $DE\\paral... | [
"origin:aops",
"2022 Contests",
"2022 Vietnam National Olympiad"
] | {
"answer_score": 240,
"boxed": false,
"end_of_proof": true,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Vietnam National Olympiad/2794829.json"
} |
Let $T_n$ be the number of non-congruent triangles with positive area and integer side lengths summing to $n$ . Prove that $T_{2022}=T_{2019}$ .
*Proposed by Constantinos Papachristoforou* | Let $S_n$ denote the number of non-congruent scalene triangles with positive integer sides .
<span style="color:#f00">**Claim:**</span> For $n\ge 6$ we have $S_n=T_{n-6}$ *Proof.* Let $a<b<c$ are a length of a scalene triangle with $a+b+c=n$ .Then $a-1\le b-2 \le c-3$ are side-lengths of a triangle whose per... | [
" $T_n$ is the cardinality of the set $S_n = \\{(a, b, c) \\in \\Bbb Z_{> 0}: a + b + c = n, a \\leq b \\leq c, a + b > c\\}$ .\n\nWe define a map $f$ from $S_n$ to $S_{n + 3}$ sending $(a, b, c)$ to $(a + 1, b + 1, c + 1)$ .\nIt is clear that the image of $f$ lies in the set $R_{n + 3}$ , where we de... | [
"origin:aops",
"2022 Contests",
"ICMC 5"
] | {
"answer_score": 156,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/ICMC 5/2731127.json"
} |
Find all integers $n$ for which there exists a table with $n$ rows, $2022$ columns, and integer entries, such that subtracting any two rows entry-wise leaves every remainder modulo $2022$ .
*Proposed by Tony Wang* | The answer is $n \leq 2$ .
Suppose $(a_i), (b_i), (c_i)$ are three rows. We write $x = \sum (c_i - a_i)$ , $y = \sum (a_i - b_i)$ , $z = \sum (b_i - c_i)$ .
Write $m = 1011$ .
By assumption, $x \equiv y \equiv z \equiv 0 + 1 + \cdots + (2m - 1) \equiv m \pmod{2m}$ .
But $x + y + z = 0$ , contradiction. | [] | [
"origin:aops",
"2022 Contests",
"ICMC 5"
] | {
"answer_score": 16,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/ICMC 5/2731131.json"
} |
Let $\mathcal M$ be the set of $n\times n$ matrices with integer entries. Find all $A\in\mathcal M$ such that $\det(A+B)+\det(B)$ is even for all $B\in\mathcal M$ .
*Proposed by Ethan Tan* | We can generalize as follows. $\textbf{Proposition.}$ Let $p$ be an integer $\geq 2$ and $A\in M_n(\mathbb{Z})$ be s.t., for every $X\in M_n(\mathbb{Z})$ , $\det(A+X)-\det(X)=0 \mod p$ .
Then $A=0\mod p$ . $\textbf{Proof.}$ There are $U,V\in M_n(\mathbb{Z})$ with $\det=1$ s.t. $UAV=D=diag(d_i)$ is diag... | [
"<details><summary>Sketch</summary>Since determinants are polynomials of its variables it suffices to consider matrices with entries $0,1$ only, and the only polynomial that works is the one with all $0$ s which clearly works.\n\n Suppose $A$ has a $1$ somewhere. Then we can perform row/col operations while ... | [
"origin:aops",
"2022 Contests",
"ICMC 5"
] | {
"answer_score": 58,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2022 Contests/ICMC 5/2731135.json"
} |
Let $p$ be a prime number. Find all subsets $S\subseteq\mathbb Z/p\mathbb Z$ such that
1. if $a,b\in S$ , then $ab\in S$ , and
2. there exists an $r\in S$ such that for all $a\in S$ , we have $r-a\in S\cup\{0\}$ .
*Proposed by Harun Khan* | The answer is all the $S$ such that $S \setminus\{0\}$ is either a subset of $\{1\}$ or a subgroup of $(\Bbb Z/p\Bbb Z)^\times$ containing $-1$ .
-----
We first note that $S = \emptyset, \{0\}, \{1\}, \{0, 1\}$ are all possible solutions (for $S = \{0, 1\}$ , we take $r = 1$ ). In the following, we exclude... | [] | [
"origin:aops",
"2022 Contests",
"ICMC 5"
] | {
"answer_score": 88,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/ICMC 5/2731136.json"
} |
A *tanned vector* is a nonzero vector in $\mathbb R^3$ with integer entries. Prove that any tanned vector of length at most $2021$ is perpendicular to a tanned vector of length at most $100$ .
*Proposed by Ethan Tan* | <details><summary>Sketch</summary>I saw a similar one from an ancient China source before [here](https://artofproblemsolving.com/community/c6h286948). Will be skipping the algebra; Consider all the vectors with distance at most $50$ from the origin, show that the dot product with the original vector is bounded by som... | [] | [
"origin:aops",
"2022 Contests",
"ICMC 5"
] | {
"answer_score": 2,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/ICMC 5/2731139.json"
} |
Is it possible to cover a circle of area $1$ with finitely many equilateral triangles whose areas sum to $1.01$ , all pointing in the same direction?
*Proposed by Ethan Tan* | <blockquote>I agree that it is possible, but I am unfortunately not too convinced by the above argument, specifically the very first paragraph</blockquote>
Suppose the original circle $C$ has center $O$ and radius $r$ . We draw a larger circle $C'$ with center $O$ and of radius $r + \delta$ , where $\delta ... | [
"Answer: it is possible.\n\nFirst tile the plane with equilateral triangles [in the usual way](https://en.wikipedia.org/wiki/Euclidean_tilings_by_convex_regular_polygons#/media/File:1-uniform_n11.svg). Make the tiling fine enough so that all the equilateral triangles that have nonempty intersection with the circle ... | [
"origin:aops",
"2022 Contests",
"ICMC 5"
] | {
"answer_score": 42,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/ICMC 5/2731141.json"
} |
Let $S$ be a set of $2022$ lines in the plane, no two parallel, no three concurrent. $S$ divides the plane into finite regions and infinite regions. Is it possible for all the finite regions to have integer area?
*Proposed by Tony Wang* | <details><summary>Solution</summary>The answer is yes.
It suffices to show that we can make all finite regions have *rational* area, since they scaling by the HCF of all the denominators of areas will give us integer areas.
Note that if $m_1 \neq m_2$ as well as $c_1$ and $c_2$ are all rationals, then so is the... | [
"Also Bulgaria EGMO TST 2024 Problem 3.",
"Let each line pass through two lattice points; then the intersection of any two lines has rational coordinates, so by Shoelace the area of each finite region is rational, so by dilating everything with a suitable scale factor we're done."
] | [
"origin:aops",
"2022 Contests",
"ICMC 5"
] | {
"answer_score": 18,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/ICMC 5/2792537.json"
} |
Evaluate
\[\frac{1/2}{1+\sqrt2}+\frac{1/4}{1+\sqrt[4]2}+\frac{1/8}{1+\sqrt[8]2}+\frac{1/16}{1+\sqrt[16]2}+\cdots\]
*Proposed by Ethan Tan* | We can write the sum as a telescoping series: $$ \sum_{n=1}^{\infty} \frac{1/2^{n}}{1+2^{1/2^{n}}}=\sum_{n=1}^{\infty} \frac{1/2^{n}}{2^{1/2^{n}}-1}-\frac{1/2^{n-1}}{2^{1/2^{n-1}}-1}=-1+\underbrace{\lim\limits_{n \rightarrow \infty}\frac{1/2^{n}}{2^{1/2^n}-1}}_{(\ast)} $$ Letting $x=2^{-n}$ and using L'Hopital's ru... | [] | [
"origin:aops",
"2022 Contests",
"ICMC 5"
] | {
"answer_score": 1014,
"boxed": true,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/ICMC 5/2792544.json"
} |
A set of points has *point symmetry* if a reflection in some point maps the set to itself. Let $\cal P$ be a solid convex polyhedron whose orthogonal projections onto any plane have point symmetry. Prove that $\cal P$ has point symmetry.
*Proposed by Ethan Tan* | [
"We may assume that \\(\\mathcal P\\) refers to the boundary of the solid. Consider two non-parallel planes \\(\\pi_1\\) and \\(\\pi_2\\), project \\(\\mathcal P\\) onto them to get the convex polygons \\(\\mathcal P_1\\) and \\(\\mathcal P_2\\), find their respective symmetry centers \\(O_1\\) and \\(O_2\\). Let t... | [
"origin:aops",
"2022 Contests",
"ICMC 5"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/ICMC 5/2792546.json"
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.