https://artofproblemsolving.com/wiki/index.php?title=Special:NewPages&feed=atom&hideredirs=1&limit=20&offset=&namespace=0&username=&tagfilter=&size-mode=max&size=0AoPS Wiki - New pages [en]2024-03-29T15:56:30ZFrom AoPS WikiMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php/PaperMath%E2%80%99s_circlesPaperMath’s circles2024-03-27T21:19:55Z<p>Papermath: /* Proof */</p>
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<div>==PaperMath’s circles==<br />
This theorem states that for a <math>n</math> tangent externally tangent circles with equal radii in the shape of a <math>n</math>-gon, the radius of the circle that is externally tangent to all the other circles can be written as <math>\frac {r(1-\cos(\frac{90(n-2)}n)}{\cos(\frac{90(n-2)}n)}</math> and the radius of the circle that is internally tangent to all the other circles can be written as <math>\frac {r(1-\cos(\frac{90(n-2)}n)}{\cos(\frac{90(n-2)}n)}+2r</math> Where <math>r</math> is the radius of one of the congruent circles and where <math>n</math> is the number of tangent circles.<br />
<br />
Here is a diagram of what <math>n=5</math> would look like.<br />
<br />
<asy><br />
size(10cm); //Asymptote by PaperMath <br />
real s = 0.218;<br />
pair A, B, C, D, E;<br />
A = dir(90 + 0*72)*s/cos(36);<br />
B = dir(90 + 1*72)*s/cos(36);<br />
C = dir(90 + 2*72)*s/cos(36);<br />
D = dir(90 + 3*72)*s/cos(36);<br />
E = dir(90 + 4*72)*s/cos(36);<br />
draw(A--B--C--D--E--cycle);<br />
real r = 1; // Radius of the congruent circles is 1 unit<br />
draw(circle(A, r));<br />
draw(circle(B, r));<br />
draw(circle(C, r));<br />
draw(circle(D, r));<br />
draw(circle(E, r));<br />
pair P_center = (A + B + C + D + E) / 5;<br />
real R_central = 1/cos(pi/180*54) - 1; <br />
draw(circle(P_center, R_central));<br />
</asy><br />
<br />
Here is a diagram of what <math>n=8</math> would look like.<br />
<br />
<asy><br />
size(10cm); // Asymptote by PaperMath<br />
real s = 2.28;<br />
pair A, B, C, D, E, F, G, H;<br />
A = dir(90 + 0*45)*s/cos(22.5);<br />
B = dir(90 + 1*45)*s/cos(22.5);<br />
C = dir(90 + 2*45)*s/cos(22.5);<br />
D = dir(90 + 3*45)*s/cos(22.5);<br />
E = dir(90 + 4*45)*s/cos(22.5);<br />
F = dir(90 + 5*45)*s/cos(22.5);<br />
G = dir(90 + 6*45)*s/cos(22.5);<br />
H = dir(90 + 7*45)*s/cos(22.5);<br />
draw(A--B--C--D--E--F--G--H--cycle);<br />
real r = 1; // Radius of the congruent circles is 1 unit<br />
draw(circle(A, r));<br />
draw(circle(B, r));<br />
draw(circle(C, r));<br />
draw(circle(D, r));<br />
draw(circle(E, r));<br />
draw(circle(F, r));<br />
draw(circle(G, r));<br />
draw(circle(H, r));<br />
pair P_center = (A + B + C + D + E + F + G + H) / 8;<br />
real R_central = 1/cos(pi/180*67.5) - 1; // Updated radius of the central circle<br />
draw(circle(P_center, R_central));<br />
</asy><br />
<br />
==Proof==<br />
We can let <math>r</math> be the radius of one of the congruent circles, and let <math>x</math> be the radius of the externally tangent circle, which means the side length of the <math>n</math>-gon is <math>2r</math>. We can draw an apothem of the <math>n</math>-gon, which bisects the side length, forming a right triangle. The length of the base is half of <math>2r</math>, or <math>r</math>, and the hypotenuse is <math>x+r</math>. The angle adjacent to the base is half of an angle of a regular <math>n</math>-gon. We know the angle of a regular <math>n</math>-gon to be <math>\frac {180(n-2)}n</math>, so half of that would be <math>\frac {90(n-2)}n</math>. Let <math>a=\frac {90(n-2)}n</math> for simplicity. We now have <math>\cos a=\frac {adj}{hyp}</math>, or <math>\cos a = \frac {r}{x+r}</math>. Multiply both sides by <math>x+r</math> and we get <math>\cos a~x+\cos a~r=r</math>, and then a bit of manipulation later you get that <math>x=\frac {r(1-\cos a)}{\cos a}</math>, or when you plug in <math>a=\frac {90(n-2)}n</math>, you get <math>\frac {r(1-\cos(\frac{90(n-2)}n)}{\cos(\frac{90(n-2)}n)}</math>. Add <math>2r</math> to find the radius of the internally tangent circle to get <math>\frac {r(1-\cos)(\frac{90(n-2)}n)}{\cos(\frac{90(n-2)}n)}+2r</math>, and we are done.<br />
<br />
==Fun stuff==<br />
Let <math>r=1</math>, then <math>\lim_{n \to \infty} \frac {1-\cos(\frac{90(n-2)}n)}{\cos(\frac{90(n-2)}n)} = \infty</math>. How? Just plug in infinity to find out!<br />
<br />
==Notes==<br />
PaperMath’s circles was discovered by the aops user PaperMath, as the name implies.<br />
<br />
==See also==<br />
*[[PaperMath’s sum]]<br />
[[Category:Geometry]]<br />
[[Category:Theorems]]</div>Papermathhttps://artofproblemsolving.com/wiki/index.php/Fairfax_County_Math_LeagueFairfax County Math League2024-03-26T11:05:45Z<p>Hcps-sundarrap: Created page with "Hello"</p>
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<div>Hello</div>Hcps-sundarraphttps://artofproblemsolving.com/wiki/index.php/TheThe2024-03-25T00:53:30Z<p>Icyfire500: Created page with "The is a word."</p>
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<div>The is a word.</div>Icyfire500https://artofproblemsolving.com/wiki/index.php/Sharygin_Olympiads,_the_bestSharygin Olympiads, the best2024-03-24T05:37:17Z<p>Vvsss: /* 2024, Problem 16 */</p>
<hr />
<div><i><b>Igor Fedorovich Sharygin</b></i> (13/02/1937 - 12/03/2004, Moscow) - Soviet and Russian mathematician and teacher, specialist in elementary geometry, popularizer of science. He wrote many textbooks on geometry and created a number of beautiful problems. He headed the mathematics section of the Russian Soros Olympiads. After his death, Russia annually hosts the Geometry Olympiad for high school students. It consists of two rounds – correspondence and final. The correspondence round lasts 3 months.<br />
<br />
The best problems of these Olympiads will be published. The numbering contains the year of the Olympiad and the serial number of the problem. Solutions are often different from the original ones.<br />
<br />
==2024, Problem 23==<br />
[[File:2023 23 1.png|350px|right]]<br />
A point <math>P</math> moves along a circle <math>\Omega.</math> Let <math>A</math> and <math>B</math> be fixed points of <math>\Omega,</math> and <math>C</math> be an arbitrary point inside <math>\Omega.</math><br />
<br />
The common external tangents to the circumcircles of triangles <math>\triangle APC</math> and <math>\triangle BCP</math> meet at point <math>Q.</math><br />
<br />
Prove that all points <math>Q</math> lie on two fixed lines.<br />
<br />
<i><b>Solution</b></i><br />
<br />
Denote <math>A' = AC \cap \Omega, B' = BC \cap \Omega, \omega = \odot APC, \omega' = \odot BPC.</math><br />
<math>\theta = \odot ACB', \theta' = \odot BCA'.</math><br />
<br />
<math>O</math> is the circumcenter of <math>\triangle APC, O'</math> is the circumcenter of <math>\triangle BPC.</math><br />
<br />
Let <math>K</math> and <math>L</math> be the midpoints of the arcs <math>\overset{\Large\frown}{CB'}</math> of <math>\theta.</math><br />
<br />
Let <math>K'</math> and <math>L'</math> be the midpoints of the arcs <math>\overset{\Large\frown}{CA'}</math> of <math>\theta'.</math><br />
<br />
These points not depends from position of point <math>P.</math><br />
<br />
Suppose, <math>P \in \overset{\Large\frown} {B'ABA'} (</math> see diagram).<br />
<cmath>\angle A'BC = 2 \alpha = \angle B'AC \implies \angle D'BC = \angle DAC = \alpha \implies \angle DOC = \angle D'O'C = 2 \alpha.</cmath><br />
<cmath>O'D' = O'C, OC = OD \implies \triangle OCD \sim \triangle O'D'C \implies OC||O'D'.</cmath><br />
Let <math>F= CD \cup OO' \implies \frac {FO}{FO'} = \frac {OC}{O'D'} \implies Q = F.</math><br />
<cmath>\angle LCB' = \alpha = \angle B'BL' \implies LC || L'B.</cmath><br />
Similarly, <math>AL || CL' \implies \triangle DLC \sim \triangle CL'D' \implies \frac {LC}{L'D'} = \frac {DC}{CD'} = \frac {OC}{O'D'}.</math><br />
<br />
Let <math>F' = LL' \cap DD' \implies \frac {F'C}{F'D'} = \frac {LC}{L'D'} = \frac {OC}{O'D'}= \frac {FC}{FD'} \implies F' = F.</math><br />
<br />
Therefore <math>Q \in LL'.</math> Similarly, if <math>P \in \overset{\Large\frown} {B'A'}</math> then <math>Q \in KK'.</math><br />
<br />
'''vladimir.shelomovskii@gmail.com, vvsss'''<br />
==2024, Problem 22==<br />
[[File:2023 22 2.png|350px|right]]<br />
[[File:2024 22.png|350px|right]]<br />
A segment <math>AB</math> is given. Let <math>C</math> be an arbitrary point of the perpendicular bisector to <math>AB, O</math> be the point on the circumcircle of <math>\triangle ABC</math> opposite to <math>C,</math> and an ellipse centered at <math>O</math> touche <math>AB, BC, CA.</math> <br />
<br />
Find the locus of touching points <math>P</math> of the ellipse with the line <math>BC.</math><br />
<br />
<i><b>Solution</b></i><br />
<br />
Denote <math>M</math> the midpoint <math>AB, D</math> the point on the line <math>CO, DO = MO, \alpha = \angle CBM, b = OM.</math><br />
<br />
<cmath>\angle CBO = 90^\circ \implies \angle COB = \alpha, MB = b \tan \alpha,</cmath><br />
<cmath>CB = \frac {b \sin \alpha}{\cos^2 \alpha}, CO = \frac {b} {\cos^2 \alpha}, CN = b \left (1 + \frac {1} {\cos^2 \alpha} \right ).</cmath><br />
In order to find the ordinate of point <math>P,</math> we perform an affine transformation (compression along axis <math>AB)</math> which will transform the ellipse <math>MPD</math> into a circle with diameter <math>MD.</math> The tangent of the <math>CP</math> maps into the tangent of the <math>CE, E = \odot CBO \cap \odot MD, PF \perp CO.</math><br />
<cmath>\angle OEF = \angle ECO \implies OF = OE \sin \angle OEF = OE \sin \angle ECO = b \cos^2 \alpha.</cmath><br />
<cmath>CP = \frac {CF}{\sin \alpha} = \frac {b}{\sin \alpha}\left ( \frac {1} {\cos^2 \alpha} - \cos^2 \alpha \right ) = b \sin \alpha \left ( \frac {1}{\cos^2 \alpha } + 1 \right).</cmath><br />
<cmath>\frac {CP}{CD} = \sin \alpha , \angle PCD = 90^\circ - \alpha \implies \angle CPD = 90^\circ.</cmath><br />
<cmath>BP = CP - CB = \sin \alpha </cmath><br />
Denote <math>Q = AB \cap DP \implies BQ = \frac {BP}{\cos \alpha} = b \tan \alpha = MB.</math><br />
<br />
So point <math>Q</math> is the fixed point (<math>P</math> not depends from angle <math>\alpha, \angle BPQ = 90^\circ ).</math><br />
<br />
Therefore point <math>P</math> lies on the circle with diameter <math>BQ</math> (except points <math>B</math> and <math>Q.)</math><br />
<br />
'''vladimir.shelomovskii@gmail.com, vvsss'''<br />
==2024, Problem 21==<br />
[[File:2024 21 0.png|350px|right]]<br />
[[File:2024 21 1.png|350px|right]]<br />
A chord <math>PQ</math> of the circumcircle of a triangle <math>ABC</math> meets the sides <math>BC, AC</math> at points <math>A', B',</math> respectively. The tangents to the circumcircle at <math>A</math> and <math>B</math> meet at point <math>X,</math> and the tangents at points <math>P</math> and <math>Q</math> meets at point <math>Y.</math> The line <math>XY</math> meets <math>AB</math> at point <math>C'.</math><br />
<br />
Prove that the lines <math>AA', BB',</math> and <math>CC'</math> concur.<br />
<br />
<i><b>Proof</b></i><br />
<br />
WLOG, <math>P \in \overset{\Large\frown} {AC}.</math> <br />
Denote <math>\Omega = \odot ABC, Z = BB' \cap AA', D = AQ \cap BP.</math><br />
<br />
Point <math>D</math> is inside <math>\Omega.</math><br />
<br />
We use Pascal’s theorem for quadrilateral <math>APQB</math> and get <math>D \in XY.</math><br />
<br />
We use projective transformation which maps <math>\Omega</math> to a circle and that maps the point <math>D</math> to its center.<br />
<br />
From this point we use the same letters for the results of mapping. Therefore the segments <math>AQ</math> and <math>BP</math> are the diameters of <math>\Omega, C'D \in XY || AP \implies C'</math> is the midpoint <math>AB.</math><br />
<br />
<math>AB|| PQ \implies AB || B'A' \implies C' \in CZ \implies</math><br />
<br />
preimage <math>Z</math> lies on preimage <math>CC'.\blacksquare</math><br />
<br />
'''vladimir.shelomovskii@gmail.com, vvsss'''<br />
==2024, Problem 19==<br />
[[File:2024 19 4.png|250px|right]]<br />
[[File:2024 19 2.png|250px|right]]<br />
[[File:2024 19 3.png|250px|right]]<br />
A triangle <math>ABC,</math> its circumcircle <math>\Omega</math>, and its incenter <math>I</math> are drawn on the plane.<br />
<br />
Construct the circumcenter <math>O</math> of <math>\triangle ABC</math> using only a ruler.<br />
<br />
<i><b>Solution</b></i><br />
<br />
We successively construct:<br />
<br />
- the midpoint <math>D = BI \cap \Omega</math> of the arc <math>AB,</math><br />
<br />
- the midpoint <math>E = CI \cap \Omega</math> of the arc <math>AC,</math><br />
<br />
- the polar <math>H'H''</math> of point <math>H \in DE,</math><br />
<br />
- the polar <math>G'G''</math> of point <math>G \in DE,</math><br />
<br />
- the polar <math>F = H'H'' \cap G'G''</math> of the line <math>DE,</math> <br />
<br />
- the tangent <math>FD || AC</math> to <math>\Omega,</math><br />
<br />
- the tangent <math>FE || AB</math> to <math>\Omega,</math><br />
<br />
- the trapezium <math>ACDF,</math><br />
<br />
- the point <math>K = AF \cap CD,</math><br />
<br />
- the point <math>L = AD \cap CF,</math><br />
<br />
- the midpoint <math>M = AC \cap KL</math> of the segment <math>AB,</math><br />
<br />
- the midpoint <math>M'</math> of the segment <math>AC,</math><br />
<br />
- the diameter <math>DM</math> of <math>\Omega,</math> <br />
<br />
- the diameter <math>EM'</math> of <math>\Omega,</math><br />
<br />
- the circumcenter <math>O = DM \cap EM'.</math><br />
<br />
'''vladimir.shelomovskii@gmail.com, vvsss'''<br />
==2024, Problem 18==<br />
[[File:2024 18 1.png|390px|right]]<br />
Let <math>AH, BH', CH''</math> be the altitudes of an acute-angled triangle <math>ABC, I_A</math> be its excenter corresponding to <math>A, I'_A</math> be the reflection of <math>I_A</math> about the line <math>AH.</math> Points <math>I'_B, I'_C</math> are defined similarly. Prove that the lines <math>HI'_A, H'I'_B, H''I'_C</math> concur.<br />
<br />
<i><b>Proof</b></i><br />
<br />
Denote <math>I</math> the incenter of <math>\triangle ABC.</math> Points <math>A, I, I_A</math> are collinear.<br />
We will prove that <math>I \in HI'_A.</math> <br />
Denote <math>D \in BC, ID \perp BC, D' \in I'_AI_A, ID' \perp BC, E = BC \cap AI_A,</math><br />
<math>F \in BC, I_AF \perp BC, AH = h_A, ID = r, I_AF = r_A, BC = a, s</math> - semiperimeter.<br />
<cmath>\frac {HD}{HF} = \frac {AI}{AI_A} = \frac {h_A - r}{h_A + r}.</cmath><br />
The area <math>[ABC] = r \cdot s = r_A (s - a) = \frac {a h_A}{2} \implies</math><br />
<cmath>\frac {1}{r} = \frac {1}{r_A} + \frac {2}{h_A} \implies \frac {h_A - r}{h_A+ r} = \frac {r_A}{r}.</cmath><br />
<cmath>\frac {I'_AD'}{HD} = \frac {HD + HF}{HD} = 1 + \frac {HF}{HD} = 1 + \frac {r_A}{r}= \frac {r_A + r}{r} \implies</cmath> <br />
Points <math>I, H, I'_a</math> are collinear, so the lines <math>HI'_A, H'I'_B, H''I'_C</math> concur at the point <math>I.</math><br />
<br />
'''vladimir.shelomovskii@gmail.com, vvsss'''<br />
==2024, Problem 16==<br />
[[File:2024 16 1.png|300px|right]]<br />
Let <math>AA', BB',</math> and <math>CC'</math> be the bisectors of a triangle <math>\triangle ABC.</math><br />
<br />
The segments <math>BB'</math> and <math>A'C'</math> meet at point <math>D.</math> Let <math>E</math> be the projection of <math>D</math> to <math>AC.</math><br />
<br />
Points <math>P</math> and <math>Q</math> on the sides <math>AB</math> and <math>BC,</math> respectively, are such that <math>EP = PD, EQ = QD.</math> <br />
<br />
Prove that <math>\angle PDB' = \angle EDQ.</math><br />
<br />
<i><b>Proof</b></i><br />
<br />
<math>\triangle PDQ = \triangle PEQ (DQ = EQ, DP = PF, PQ</math> is the common side) <math>\implies</math><br />
<br />
<math>PQ \perp DE, F = PQ \cap DE</math> is the midpoint <math>DE \implies</math><br />
<br />
<math>G = BB' \cap PQ</math> is the midpoint of <math>DB'.</math><br />
<cmath>\frac{BG}{BB'} =\frac {BQ}{BC} = \frac {BP}{BA} = \frac {BD}{BI}.</cmath><br />
(see [[Bisector | Division of bisector]] for details.)<br />
<br />
So <math>DQ || CC', PD || AA'.</math> Denote <math>\angle ACC' = \angle BCC' = \gamma, \angle A'AC = \alpha, B'BC = \beta.</math><br />
<cmath>\angle PDB' = \angle AIB' = \angle BB'C - \angle IAC = 180^\circ - \beta - 2 \gamma - \alpha = 90^\circ - \gamma.</cmath><br />
<cmath>\angle QDE = 90^\circ - \angle DQP = 90^\circ - \gamma = \angle PDB'.</cmath><br />
<br />
Another solution see [[Isogonal_conjugate | 2024_Sharygin_olimpiad_Problem_16]]<br />
<br />
'''vladimir.shelomovskii@gmail.com, vvsss'''</div>Vvssshttps://artofproblemsolving.com/wiki/index.php/Connecticut_MathcountsConnecticut Mathcounts2024-03-23T22:16:58Z<p>Gp102: /* CT State Team Members (National Written Place, CD place if known) */</p>
<hr />
<div>== National Team Awards and Placement ==<br />
* 2019 - 19th<br />
* 2018 - 20th<br />
* 2017 - 20th<br />
* 2012 - 15th<br />
<br />
== CT State Team Champions ==<br />
* 2024 - Timothy Edwards Middle School<br />
* 2023 - Greenwich Academy<br />
* 2022 - No Team Qualification for State<br />
* 2020 - East Lyme Middle School<br />
* 2019 - Brunswick Middle School<br />
* 2018 - Timothy Edwards Middle School<br />
* 2017 - Whisconier Middle School<br />
* 2015 - Dodd Middle School<br />
* 2014 - Eastern Middle School<br />
* 2013 - Eastern Middle School<br />
* 2012 - Eastern Middle School<br />
<br />
== CT State Team Members (National Written Place, CD place if known) ==<br />
* 2024 - Alexander Svoronos, Joseph Girotto, Girish Prasad, Arjun Leih, Coach: Beth Goldman<br />
<br />
* 2023 - Abby Kesmodel, Sohan Javeri (65th), Joseph Girotto, Girish Prasad, Coach: Will Roble<br />
<br />
* 2022 - Vikram Sarkar (53rd), Shaurya Ranjan, Sohan Javeri, Leon Jiang, Coach: Archi Rudra<br />
<br />
* 2021 - Kyle Zhang (10th), Leon Jiang, Vikram Sarkar, Enya Yu<br />
<br />
* 2020 - Mingwen Duan, Kyle Zhang, Andrew Tu, Jordan Lewkowitz, Coach: Shangquan Duan<br />
<br />
* 2019 - Adeethyia Shankar (55th), Mingwen Duan (28th), Ryan Yang, Alan Wag, Coach: Kevin Landesman<br />
<br />
* 2018 - Ryan Yang, Shangyu Xu, Mingwen Duan, Natalie Shell, Coach: William Denker<br />
<br />
* 2017 - Samuel Florin (54th), Luke Choi, Ryan Yang, Alex Devorsetz<br />
<br />
* 2015 - Prasik Mohanraj, Hizami Anuar, David Metrick, William Duan, Coach: Thomas Tuscano<br />
<br />
* 2014 - Mingfei Duan, Steven Ma, Koshik Maapatra, Shreyas Srinivasan, Coach: Sarah Hoenigmann<br />
<br />
* 2013 - Dale Yu, Mingfei Duan, Eliza Khokhar, Samual Bidwell<br />
<br />
* 2012 - Michael Kural (25th), Robert Kao, Jacob Klegar, Shangda Xu<br />
<br />
==National Countdown Round==<br />
* None Yet<br />
<br />
==Masters Round==<br />
<br />
* 1992 - Jenny Hoffman (2nd Place)<br />
<br />
==State Countdown Winners==<br />
* 2024 - Girish Prasad<br />
<br />
* 2023 - Girish Prasad<br />
<br />
* 2022 - Vikram Sarkar<br />
<br />
* 2020 - Mingwen Duan<br />
<br />
* 2019 - Jiarui Peng<br />
<br />
* 2018 - Aryan Kalia<br />
<br />
* 2017 - Luke Choi<br />
<br />
* 2014 - Shreyas Srinivasan<br />
<br />
* 2012 - Shangda Xu</div>Gp102https://artofproblemsolving.com/wiki/index.php/2025_USAJMO2025 USAJMO2024-03-22T19:50:34Z<p>Hungrycalculator: Created page with "oh no no no"</p>
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<div>oh no no no</div>Hungrycalculatorhttps://artofproblemsolving.com/wiki/index.php/2024_USAMO_Problems2024 USAMO Problems2024-03-21T02:33:55Z<p>Mathkiddie: </p>
<hr />
<div>==Day 1==<br />
===Problem 1===<br />
Find all integers <math>n \geq 3</math> such that the following property holds: if we list the divisors of <math>n !</math> in increasing order as <math>1=d_1<d_2<\cdots<d_k=n!</math>, then we have<br />
<br />
<cmath>d_2-d_1 \leq d_3-d_2 \leq \cdots \leq d_k-d_{k-1}.</cmath><br />
<br />
[[2024 USAMO Problems/Problem 1|Solution]]<br />
<br />
===Problem 2===<br />
Let <math>S_1, S_2, \ldots, S_{100}</math> be finite sets of integers whose intersection is not empty. For each non-empty <math>T \subseteq\left\{S_1, S_2, \ldots, S_{100}\right\}</math>, the size of the intersection of the sets in <math>T</math> is a multiple of the number of sets in <math>T</math>. What is the least possible number of elements that are in at least 50 sets?<br />
<br />
[[2024 USAMO Problems/Problem 2|Solution]]<br />
<br />
===Problem 3===<br />
Let <math>m</math> be a positive integer. A triangulation of a polygon is <math>m</math>-balanced if its triangles can be colored with <math>m</math> colors in such a way that the sum of the areas of all triangles of the same color is the same for each of the <math>m</math> colors. Find all positive integers <math>n</math> for which there exists an <math>m</math>-balanced triangulation of a regular <math>n</math>-gon.<br />
<br />
Note: A triangulation of a convex polygon <math>\mathcal{P}</math> with <math>n \geq 3</math> sides is any partitioning of <math>\mathcal{P}</math> into <math>n-2</math> triangles by <math>n-3</math> diagonals of <math>\mathcal{P}</math> that do not intersect in the polygon's interior.<br />
<br />
[[2024 USAMO Problems/Problem 3|Solution]]<br />
<br />
==Day 2==<br />
===Problem 4===<br />
Let <math>m</math> and <math>n</math> be positive integers. A circular necklace contains <math>m n</math> beads, each either red or blue. It turned out that no matter how the necklace was cut into <math>m</math> blocks of <math>n</math> consecutive beads, each block had a distinct number of red beads. Determine, with proof, all possible values of the ordered pair <math>(m, n)</math>.<br />
<br />
[[2024 USAMO Problems/Problem 4|Solution]]<br />
<br />
===Problem 5===<br />
Point <math>D</math> is selected inside acute triangle <math>A B C</math> so that <math>\angle D A C=</math> <math>\angle A C B</math> and <math>\angle B D C=90^{\circ}+\angle B A C</math>. Point <math>E</math> is chosen on ray <math>B D</math> so that <math>A E=E C</math>. Let <math>M</math> be the midpoint of <math>B C</math>.<br />
Show that line <math>A B</math> is tangent to the circumcircle of triangle <math>B E M</math>.<br />
<br />
[[2024 USAMO Problems/Problem 5|Solution]]<br />
<br />
===Problem 6===<br />
Let <math>n>2</math> be an integer and let <math>\ell \in\{1,2, \ldots, n\}</math>. A collection <math>A_1, \ldots, A_k</math> of (not necessarily distinct) subsets of <math>\{1,2, \ldots, n\}</math> is called <math>\ell</math>-large if <math>\left|A_i\right| \geq \ell</math> for all <math>1 \leq i \leq k</math>. Find, in terms of <math>n</math> and <math>\ell</math>, the largest real number <math>c</math> such that the inequality<br />
<cmath>\sum_{i=1}^k \sum_{j=1}^k x_i x_j \frac{\left|A_i \cap A_j\right|^2}{\left|A_i\right| \cdot\left|A_j\right|} \geq c\left(\sum_{i=1}^k x_i\right)^2</cmath><br />
holds for all positive integers <math>k</math>, all nonnegative real numbers <math>x_1, \ldots, x_k</math>, and all <math>\ell</math>-large collections <math>A_1, \ldots, A_k</math> of subsets of <math>\{1,2, \ldots, n\}</math>.<br />
<br />
Note: For a finite set <math>S,|S|</math> denotes the number of elements in <math>S</math>.<br />
<br />
[[2024 USAMO Problems/Problem 6|Solution]]<br />
<br />
==See Also==<br />
{{USAMO newbox|year=2024|before=[[2023 USAMO Problems]]|after=[[2025 USAMO Problems]]}}<br />
{{MAA Notice}}</div>Anyu-tsurukohttps://artofproblemsolving.com/wiki/index.php/2024_USAMO_Problems/Problem_12024 USAMO Problems/Problem 12024-03-21T02:33:01Z<p>Anyu-tsuruko: Created page with "Find all integers <math>n \geq 3</math> such that the following property holds: if we list the divisors of <math>n !</math> in increasing order as <math>1=d_1<d_2<\cdots<d_k=n..."</p>
<hr />
<div>Find all integers <math>n \geq 3</math> such that the following property holds: if we list the divisors of <math>n !</math> in increasing order as <math>1=d_1<d_2<\cdots<d_k=n!</math>, then we have<br />
<cmath><br />
d_2-d_1 \leq d_3-d_2 \leq \cdots \leq d_k-d_{k-1} .<br />
</cmath></div>Anyu-tsurukohttps://artofproblemsolving.com/wiki/index.php/2024_USAMO_Problems/Problem_22024 USAMO Problems/Problem 22024-03-21T02:32:39Z<p>Anyu-tsuruko: Created page with "Let <math>S_1, S_2, \ldots, S_{100}</math> be finite sets of integers whose intersection is not empty. For each non-empty <math>T \subseteq\left\{S_1, S_2, \ldots, S_{100}\rig..."</p>
<hr />
<div>Let <math>S_1, S_2, \ldots, S_{100}</math> be finite sets of integers whose intersection is not empty. For each non-empty <math>T \subseteq\left\{S_1, S_2, \ldots, S_{100}\right\}</math>, the size of the intersection of the sets in <math>T</math> is a multiple of the number of sets in <math>T</math>. What is the least possible number of elements that are in at least 50 sets?</div>Anyu-tsurukohttps://artofproblemsolving.com/wiki/index.php/2024_USAMO_Problems/Problem_32024 USAMO Problems/Problem 32024-03-21T02:32:24Z<p>Anyu-tsuruko: Created page with "Let <math>m</math> be a positive integer. A triangulation of a polygon is <math>m</math>-balanced if its triangles can be colored with <math>m</math> colors in such a way that..."</p>
<hr />
<div>Let <math>m</math> be a positive integer. A triangulation of a polygon is <math>m</math>-balanced if its triangles can be colored with <math>m</math> colors in such a way that the sum of the areas of all triangles of the same color is the same for each of the <math>m</math> colors. Find all positive integers <math>n</math> for which there exists an <math>m</math>-balanced triangulation of a regular <math>n</math>-gon.<br />
Note: A triangulation of a convex polygon <math>\mathcal{P}</math> with <math>n \geq 3</math> sides is any partitioning of <math>\mathcal{P}</math> into <math>n-2</math> triangles by <math>n-3</math> diagonals of <math>\mathcal{P}</math> that do not intersect in the polygon's interior.</div>Anyu-tsurukohttps://artofproblemsolving.com/wiki/index.php/2024_USAMO_Problems/Problem_42024 USAMO Problems/Problem 42024-03-21T02:32:06Z<p>Anyu-tsuruko: Created page with "Let <math>m</math> and <math>n</math> be positive integers. A circular necklace contains <math>m n</math> beads, each either red or blue. It turned out that no matter how the..."</p>
<hr />
<div>Let <math>m</math> and <math>n</math> be positive integers. A circular necklace contains <math>m n</math> beads, each either red or blue. It turned out that no matter how the necklace was cut into <math>m</math> blocks of <math>n</math> consecutive beads, each block had a distinct number of red beads. Determine, with proof, all possible values of the ordered pair <math>(m, n)</math>.</div>Anyu-tsurukohttps://artofproblemsolving.com/wiki/index.php/2024_USAMO_Problems/Problem_52024 USAMO Problems/Problem 52024-03-21T02:30:58Z<p>Beef7: </p>
<hr />
<div>{{duplicate|[[2024 USAMO Problems/Problem 5|2024 USAMO/5]] and [[2024 USAJMO Problems/Problem 6|2024 USAJMO/6]]}}<br />
__TOC__<br />
<br />
== Problem ==<br />
Point <math>D</math> is selected inside acute triangle <math>ABC</math> so that <math>\angle DAC=\angle ACB</math> and <math>\angle BDC=90^\circ+\angle BAC</math>. Point <math>E</math> is chosen on ray <math>BD</math> so that <math>AE=EC</math>. Let <math>M</math> be the midpoint of <math>BC</math>. Show that line <math>AB</math> is tangent to the circumcircle of triangle <math>BEM</math>.<br />
<br />
== Solution 1 ==<br />
<br />
<br />
==See Also==<br />
{{USAMO newbox|year=2024|num-b=4|num-a=6}}<br />
{{MAA Notice}}</div>Anyu-tsurukohttps://artofproblemsolving.com/wiki/index.php/2024_USAMO_Problems/Problem_62024 USAMO Problems/Problem 62024-03-21T02:28:32Z<p>Anyu-tsuruko: Created page with "Let <math>n>2</math> be an integer and let <math>\ell \in\{1,2, \ldots, n\}</math>. A collection <math>A_1, \ldots, A_k</math> of (not necessarily distinct) subsets of <math>\..."</p>
<hr />
<div>Let <math>n>2</math> be an integer and let <math>\ell \in\{1,2, \ldots, n\}</math>. A collection <math>A_1, \ldots, A_k</math> of (not necessarily distinct) subsets of <math>\{1,2, \ldots, n\}</math> is called <math>\ell</math>-large if <math>\left|A_i\right| \geq \ell</math> for all <math>1 \leq i \leq k</math>. Find, in terms of <math>n</math> and <math>\ell</math>, the largest real number <math>c</math> such that the inequality<br />
<cmath><br />
\sum_{i=1}^k \sum_{j=1}^k x_i x_j \frac{\left|A_i \cap A_j\right|^2}{\left|A_i\right| \cdot\left|A_j\right|} \geq c\left(\sum_{i=1}^k x_i\right)^2<br />
</cmath><br />
holds for all positive integers <math>k</math>, all nonnegative real numbers <math>x_1, \ldots, x_k</math>, and all <math>\ell</math>-large collections <math>A_1, \ldots, A_k</math> of subsets of <math>\{1,2, \ldots, n\}</math>.<br />
Note: For a finite set <math>S,|S|</math> denotes the number of elements in <math>S</math>.</div>Anyu-tsurukohttps://artofproblemsolving.com/wiki/index.php/2024_USAMO2024 USAMO2024-03-21T02:19:51Z<p>Ayush agarwal: Created page with "The 53rd '''USAMO''' was held on March 19 and March 20, 2024. The first link will contain the full set of test problems. The rest will contain each individual problem and its..."</p>
<hr />
<div>The 53rd '''USAMO''' was held on March 19 and March 20, 2024. The first link will contain the full set of test problems. The rest will contain each individual problem and its solution.<br />
<br />
[[2024 USAMO Problems]]<br />
* [[2024 USAMO Problems/Problem 1]]<br />
* [[2024 USAMO Problems/Problem 2]]<br />
* [[2024 USAMO Problems/Problem 3]]<br />
* [[2024 USAMO Problems/Problem 4]]<br />
* [[2024 USAMO Problems/Problem 5]]<br />
* [[2024 USAMO Problems/Problem 6]]<br />
<br />
== See Also ==<br />
* [[Mathematics competitions]]<br />
* [[Mathematics competition resources]]<br />
* [[Math books]]<br />
* [[USAMO]]<br />
<br />
{{USAMO newbox|year= 2024 |before=[[2023 USAMO]]|after=[[2025 USAMO]]}}<br />
{{MAA Notice}}</div>Ayush agarwalhttps://artofproblemsolving.com/wiki/index.php/2024_USAJMO_Problems/Problem_52024 USAJMO Problems/Problem 52024-03-20T23:59:24Z<p>Virjoy2001: /* Solution 1 */</p>
<hr />
<div>__TOC__<br />
<br />
== Problem ==<br />
Find all functions <math>f:\mathbb{R}\rightarrow\mathbb{R}</math> that satisfy<br />
<cmath>f(x^2-y)+2yf(x)=f(f(x))+f(y)</cmath><br />
for all <math>x,y\in\mathbb{R}</math>.<br />
<br />
== Solution 1 ==<br />
<br />
I will denote the original equation <math>f(x^2-y)+2yf(x)=f(f(x))+f(y)</math> as OE.<br />
<br />
I claim that the only solutions are <math>f(x) = -x^2, f(x) = 0,</math> and <math>f(x) = x^2.</math><br />
<br />
Lemma 1: <math>f(0) = 0.</math><br />
<br />
Proof of Lemma 1:<br />
<br />
We prove this by contradiction. Assume <math>f(0) = k \neq 0.</math><br />
<br />
By letting <math>x=y=0</math> in the OE, we have <cmath>f(0) = f^2(0) + f(0) \Longrightarrow f^2(0) = 0 \Longrightarrow f(k) = 0.</cmath><br />
<br />
If we let <math>x = 0</math> and <math>y = k^2</math> in the OE, we have<br />
<cmath>f(-k^2) + 2k^2f(0) = f^2(0) + f(k^2) \Longrightarrow f(-k^2) + 2k^3 = f(k^2)</cmath> and if we let <math>x = k</math> and <math>y = k^2</math> in the OE, we get<br />
<br />
<cmath>f(0) + 2k^2f(k) = f^2(k) + f(k^2) \Longrightarrow k = k + f(k^2) \Longrightarrow f(k^2) = 0 \Longrightarrow f(-k^2) = 2k^3. </cmath><br />
<br />
However, upon substituting <math>x = k</math> and <math>y = -k^2</math> in the OE, this implies<br />
<br />
<cmath>f(0) -2k^2f(k) = f^2(k) + f(-k^2) \Longrightarrow k = k + 2k^3 \Longrightarrow 2k^3 = 0.</cmath><br />
<br />
This means <math>k = 0,</math> but we assumed <math>k \neq 0,</math> contradiction, which proves the Lemma.<br />
<br />
Substitute <math>y = 0</math> in the OE to obtain<br />
<cmath>f(x^2) = f^2(x) + f(0) = f^2(x)</cmath><br />
and let <math>y = x^2</math> in the OE to get<br />
<cmath>f(0) + 2x^2 f(x) = f^2(x) + f(x^2) = 2f(x^2) = 2x^2 f(x) \Longrightarrow \dfrac{f(x)}{x^2} = \dfrac{f(x^2)}{x^4} \Longrightarrow f(x) \propto x^2.</cmath><br />
<br />
Thus we can write <math>f(x) = kx^2</math> for some <math>k.</math> By <math>f(x^2) = f^2(x),</math> we have <cmath>kx^4 = k^3x^4,</cmath> so <math>k = -1, 0, 1,</math> yielding the solutions <cmath>f(x) = -x^2, f(x) = 0, f(x) = x^2. \blacksquare</cmath><br />
<br />
- [https://artofproblemsolving.com/wiki/index.php/User:Spectraldragon8 spectraldragon8]<br />
<br />
==See Also==<br />
{{USAJMO newbox|year=2024|num-b=4|num-a=6}}<br />
{{MAA Notice}}</div>Flyingpig7https://artofproblemsolving.com/wiki/index.php/2024_USAJMO_Problems/Problem_42024 USAJMO Problems/Problem 42024-03-20T19:33:32Z<p>Eevee9406: finish solution 1</p>
<hr />
<div>__TOC__<br />
<br />
== Problem ==<br />
Let <math>n \ge 3</math> be an integer. Rowan and Colin play a game on an <math>n \times n</math> grid of squares, where each square is colored either red or blue. Rowan is allowed to permute the rows of the grid, and Colin is allowed to permute the columns of the grid. A grid coloring is <math>orderly</math> if:<br />
<br />
*no matter how Rowan permutes the rows of the coloring, Colin can then permute the columns to restore the original grid coloring; and<br />
<br />
*no matter how Colin permutes the column of the coloring, Rowan can then permute the rows to restore the original grid coloring; <br />
<br />
In terms of <math>n</math>, how many orderly colorings are there?<br />
<br />
== Solution 1 ==<br />
We focus on the leftmost column for simplicity. Let <math>m</math> be the number of red squares in this column. We then have five cases:<br />
<br />
<br />
1. <math>m=1</math><br />
<br />
When Rowan permutes the rows of the coloring, we consider only the first column, which by the above contains <math>m=1</math> red colors, so there are <math>{n \choose 1}=n</math> ways to permute the first column’s rows. Thus every other column will have to contain one different permutation of the first column; otherwise, there will be at least one permutation of which there is no corresponding column.<br />
<br />
<br />
Furthermore, each permutation will be different, so each row will contain one and only one red square, which also fulfills the case of if Colin permutes the coloring first. Thus there are <math>n\cdot (n-1)\cdot(n-2)\cdot\cdot\cdot2\cdot1=n!</math> different colorings for this case (the same as choosing squares such as no square is in the same row or column as any other square).<br />
<br />
<br />
<br />
2. <math>m=n-1</math><br />
<br />
This is essentially the same as case 1 except for the coloring; now there is one blue square and the rest are red squares. Thus there are also <math>n!</math> different colorings for this case.<br />
<br />
<br />
<br />
3. <math>m=0</math><br />
<br />
Since we have an entirely blue column, we are unable to have a column with <math>1</math> red square only as doing so would leave one permutation that is not covered by at least one column (that space is being taken for the blank column). We are also unable to have a completely blue column as doing so would allow for Colin to shift the columns and in doing so fail for Rowan to shift back the columns. We also cannot have a column with any other number of red squares other than <math>0</math> as will be shown below, so there is <math>1</math> case here in which the entire coloring is red.<br />
<br />
<br />
<br />
4. <math>m=n</math><br />
<br />
This is the same is an entire blue column, and, similar to above, we have <math>1</math> coloring.<br />
<br />
<br />
<br />
5. <math>1<m<n-1</math><br />
<br />
This is the final case and is equivalent to permuting for <math>{n \choose m}</math> different ways. We must prove that this is greater than <math>n</math> to show that the columns are not able to contain every possible permutation of this column for all values of <math>n</math> such that <math>n>3</math> (when <math>n=3</math>, there is no such positive integer <math>m</math> that satisfies the conditions). Note that if we have any column with a different number of red squares, it is an unattainable column and is thus not optimal.<br />
<br />
<br />
Lemma: Given that <math>m</math> and <math>n</math> are positive integers such that <math>1<m<n-1</math> and <math>n>3</math>, it is true for all <math>m</math> and <math>n</math> that <math>{n \choose m}>n</math>.<br />
<br />
Proof: Assume that <math>m<\frac{n-1}{2}</math>.<br />
<br />
<math>\Leftrightarrow</math> <math>m+1<n-m</math><br />
<br />
<math>\Leftrightarrow</math> <math>(m+1)!(n-m-1)!<m!(n-m)!</math><br />
<br />
<math>\Leftrightarrow</math> <math>\frac{n!}{m!(n-m)!}<\frac{n!}{(m+1)!(n-m-1)!}</math><br />
<br />
<math>\Leftrightarrow</math> <math>{n \choose m}<{n \choose m+1}</math><br />
<br />
Similarly, we can prove that <math>{n \choose m}>{n \choose m+1}</math> for <math>m>\frac{n-1}{2}</math>.<br />
<br />
Now we split our proof into two cases.<br />
<br />
Case 1: <math>n</math> is even.<br />
<br />
The largest integer less than <math>\frac{n-1}{2}</math> is <math>\frac{n}{2}-1</math>, so we know that:<br />
<br />
<math>{n \choose \frac{n}{2}}>{n \choose \frac{n}{2}-1}>\cdot\cdot\cdot>{n \choose 2}</math><br />
<br />
by induction. On the other hand, the smallest integer greater than <math>\frac{n-1}{2}</math> is <math>\frac{n}{2}</math>, so we know that:<br />
<br />
<math>{n \choose \frac{n}{2}}>{n \choose \frac{n}{2}+1}>\cdot\cdot\cdot>{n \choose n-2}</math><br />
<br />
also by induction. Thus out of the given range for <math>m</math> we know that <math>{n \choose 2}</math> and <math>{n \choose n-2}</math> are the minimum values, and all that is left is to prove that they are both greater than <math>n</math>. Furthermore, since <math>{n \choose 2}={n \choose n-2}</math>, we only have to prove that <math>{n \choose 2}>n</math>.<br />
<br />
We start with the given: <math>n>3</math><br />
<br />
<math>\Leftrightarrow</math> <math>\frac{n-1}{2}>1</math><br />
<br />
<math>\Leftrightarrow</math> <math>\frac{n(n-1)}{2}>n</math><br />
<br />
<math>\Leftrightarrow</math> <math>\frac{n!}{2!(n-2)!}>n</math><br />
<br />
<math>\Leftrightarrow</math> <math>{n \choose 2}>n</math><br />
<br />
Thus we have proven the inequality for all even <math>n</math>.<br />
<br />
Case 2: <math>n</math> is odd.<br />
<br />
The greatest integer less than <math>\frac{n-1}{2}</math> is <math>\frac{n-3}{2}</math>, so we know that:<br />
<br />
<math>{n \choose \frac{n-1}{2}}>{n \choose \frac{n-3}{2}}>\cdot\cdot\cdot>{n \choose 2}</math><br />
<br />
by induction. On the other hand, the smallest integer greater than <math>\frac{n-1}{2}</math> is <math>\frac{n+1}{2}</math>, so we know that:<br />
<br />
<math>{n \choose \frac{n+1}{2}}>{n \choose \frac{n+3}{2}}>\cdot\cdot\cdot>{n \choose n-2}</math><br />
<br />
also by induction. Since <math>{n \choose \frac{n+1}{2}}={n \choose \frac{n-1}{2}}</math>, we know that once again, <math>{n \choose n-2}={n \choose 2}</math> is the minimum of the given range for <math>m</math>, and the same proof applies. Thus, the inequality holds true for odd and in turn all positive integers <math>n>3</math>.<br />
<br />
<br />
As a result, due to our lemma, there are always more permutations of the columns than the number of columns itself, so there will always exist a permutation of the column such that there are no corresponding original columns of which to match with. Thus there are no solutions for this case.<br />
<br />
<br />
In conclusion, there are a total of <math>2\cdot n!+2</math> different colorings for which the above apply.<br />
<br />
<br />
~eevee9406<br />
<br />
==See Also==<br />
{{USAJMO newbox|year=2024|num-b=3|num-a=5}}<br />
{{MAA Notice}}</div>Logicus14https://artofproblemsolving.com/wiki/index.php/2024_USAJMO_Problems/Problem_32024 USAJMO Problems/Problem 32024-03-20T01:43:54Z<p>Bronzetruck2016: /* Solution 1 */</p>
<hr />
<div>__TOC__<br />
<br />
== Problem ==<br />
Let <math>a(n)</math> be the sequence defined by <math>a(1)=2</math> and <math>a(n+1)=(a(n))^{n+1}-1</math> for each integer <math>n\geq1</math>. Suppose that <math>p>2</math> is prime and <math>k</math> is a positive integer. Prove that some term of the sequence <math>a(n)</math> is divisible by <math>p^k</math>.<br />
<br />
== Solution 1 ==<br />
<br />
Lemma <math>1</math>:<br />
<br />
Given a prime <math>p</math>, a positive integer <math>k</math>, and an even <math>m</math> such that <math>p^k|a(m)</math>, we must have that <math>p^{k+1}|a(m+2)</math>.<br />
<br />
<br />
Proof of Lemma <math>1</math>:<br />
<br />
<math>a(m+1)\equiv (a(m))^{m+1}-1\equiv -1 \mod p^{k+1}</math><br />
<br />
Then, <math>a(m+2)\equiv (a(m+1))^{m+2}-1\equiv (-1)^{m+2}-1\equiv 0 \mod p^{k+1}</math><br />
<br />
<br />
Therefore, by induction, if there exists an even integer <math>m</math> such that <math>p|a(m)</math>, then for all integers <math>k</math>, <math>p^k|a(m+2k-2)</math>, so we are done if there exists an even <math>m</math> such that <math>p|a(m)</math>.<br />
<br />
<br />
Now, consider the case where there is some prime <math>p>2</math> such that there are no even integers <math>m</math> such that <math>p|a(m)</math>.<br />
<br />
Lemma <math>2</math>:<br />
<br />
In this case, we must have that <math>p|a(m)</math> if <math>m\equiv -1 \mod p-1</math> for all integers <math>m</math>.<br />
<br />
Proof of Lemma <math>2</math>:<br />
<br />
Suppose for the sake of contradiction that there exists some <math>m</math> such that <math>m\equiv -1\mod p-1</math> and <math>p</math> does not divide <math>a(m)</math>. Then, we have <math>a(m+1)\equiv (a(m))^{m+1}-1\equiv 1-1\equiv 0\mod p</math>, by Fermat's Little Theorem. Since for all <math>p>2</math>, <math>p-1</math> is even, then <math>m+1</math> would be even. However this results in a contradiction.<br />
<br />
Then, we get that if <math>m\equiv -1\mod p-1</math>, then <math>0\equiv a(m)\equiv (a(m-1))^m-1\mod p\implies (a(m-1))^{m+1}\equiv 1\equiv a(m-1)\mod p</math>.<br />
<br />
Then, by LTE, <math>v_p(a(m))=v_p((a(m-1))^m-1)=v_p(a(m-1)-1)+v_p(m)>v_p(m)</math>. Since <math>\gcd(p-1,p)=1</math>, then <math>\gcd(p-1,p^k)=1</math> for all positive integers <math>k</math>, so then by Chines Remainder Theorem there exists integers <math>m</math> such that <math>m\equiv -1\mod p-1</math> and <math>m\equiv 0\mod p^k</math>, so we are done <math>\square</math><br />
<br />
<br />
Remark: I think this is a very cool NT problem.<br />
<br />
<br />
-bronzetruck2016<br />
<br />
==See Also==<br />
{{USAJMO newbox|year=2024|num-b=2|num-a=4}}<br />
{{MAA Notice}}</div>Ryanjwanghttps://artofproblemsolving.com/wiki/index.php/2024_USAJMO_Problems/Problem_22024 USAJMO Problems/Problem 22024-03-20T01:37:37Z<p>Ejbohsolon: /* Problem */</p>
<hr />
<div>__TOC__<br />
<br />
=== Problem ===<br />
Let <math>m</math> and <math>n</math> be positive integers. Let <math>S</math> be the set of integer points <math>(x,y)</math> with <math>1\leq x\leq2m</math> and <math>1\leq y\leq2n</math>. A configuration of <math>mn</math> rectangles is called ''happy'' if each point in <math>S</math> is a vertex of exactly one rectangle, and all rectangles have sides parallel to the coordinate axes. Prove that the number of happy configurations is odd.<br />
<br />
<br />
<br />
=== Solution 1 ===<br />
<br />
==See Also==<br />
{{USAJMO newbox|year=2024|num-b=1|num-a=3}}<br />
{{MAA Notice}}</div>Ryanjwanghttps://artofproblemsolving.com/wiki/index.php/2024_USAJMO_Problems/Problem_12024 USAJMO Problems/Problem 12024-03-20T01:36:05Z<p>Oinava: /* Solution 3 */</p>
<hr />
<div>__TOC__<br />
<br />
==Problem==<br />
<br />
Let <math>ABCD</math> be a cyclic quadrilateral with <math>AB = 7</math> and <math>CD = 8</math>. Points <math>P</math> and <math>Q</math> are selected on segment <math>AB</math> such that <math>AP = BQ = 3</math>. Points <math>R</math> and <math>S</math> are selected on segment <math>CD</math> such that <math>CR = DS = 2</math>. Prove that <math>PQRS</math> is a cyclic quadrilateral.<br />
<br />
==Solution 1==<br />
<br />
First, let <math>E</math> and <math>F</math> be the midpoints of <math>AB</math> and <math>CD</math>, respectively. It is clear that <math>AE=BE=3.5</math>, <math>PE=QE=0.5</math>, <math>DF=CF=4</math>, and <math>SF=RF=2</math>. Also, let <math>O</math> be the circumcenter of <math>ABCD</math>. <br />
<br />
<asy> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br />
import graph; size(12cm); <br />
real labelscalefactor = 0.5; /* changes label-to-point distance */<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br />
pen dotstyle = black; /* point style */ <br />
real xmin = -12.19, xmax = 24.94, ymin = -15.45, ymax = 6.11; /* image dimensions */<br />
pen wrwrwr = rgb(0.38,0.38,0.38); <br />
/* draw figures */<br />
draw(circle((2.92,-3.28), 5.90), linewidth(2) + wrwrwr); <br />
draw((-2.52,-1.01)--(3.46,2.59), linewidth(2) + wrwrwr); <br />
draw((7.59,-6.88)--(-0.29,-8.22), linewidth(2) + wrwrwr); <br />
draw((3.46,2.59)--(7.59,-6.88), linewidth(2) + wrwrwr); <br />
draw((-0.29,-8.22)--(-2.52,-1.01), linewidth(2) + wrwrwr); <br />
draw((0.03,0.52)--(1.67,-7.89), linewidth(2) + wrwrwr); <br />
draw((5.61,-7.22)--(0.89,1.04), linewidth(2) + wrwrwr); <br />
/* dots and labels */<br />
dot((2.92,-3.28),dotstyle); <br />
label("$O$", (2.43,-3.56), NE * labelscalefactor); <br />
dot((-2.52,-1.01),dotstyle); <br />
label("$A$", (-2.91,-0.91), NE * labelscalefactor); <br />
dot((3.46,2.59),linewidth(4pt) + dotstyle); <br />
label("$B$", (3.49,2.78), NE * labelscalefactor); <br />
dot((7.59,-6.88),dotstyle); <br />
label("$C$", (7.82,-7.24), NE * labelscalefactor); <br />
dot((-0.29,-8.22),linewidth(4pt) + dotstyle); <br />
label("$D$", (-0.53,-8.62), NE * labelscalefactor); <br />
dot((0.03,0.52),linewidth(4pt) + dotstyle); <br />
label("$P$", (-0.13,0.67), NE * labelscalefactor); <br />
dot((0.89,1.04),linewidth(4pt) + dotstyle); <br />
label("$Q$", (0.62,1.16), NE * labelscalefactor); <br />
dot((5.61,-7.22),linewidth(4pt) + dotstyle); <br />
label("$R$", (5.70,-7.05), NE * labelscalefactor); <br />
dot((1.67,-7.89),linewidth(4pt) + dotstyle); <br />
label("$S$", (1.75,-7.73), NE * labelscalefactor); <br />
dot((0.46,0.78),linewidth(4pt) + dotstyle); <br />
label("$E$", (0.26,0.93), NE * labelscalefactor); <br />
dot((3.64,-7.55),linewidth(4pt) + dotstyle); <br />
label("$F$", (3.73,-7.39), NE * labelscalefactor); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br />
/* end of picture */</asy><br />
<br />
By properties of cyclic quadrilaterals, we know that the circumcenter of a cyclic quadrilateral is the intersection of its sides' perpendicular bisectors. This implies that <math>OE\perp AB</math> and <math>OF\perp CD</math>. Since <math>E</math> and <math>F</math> are also bisectors of <math>PQ</math> and <math>RS</math>, respectively, if <math>PQRS</math> is indeed a cyclic quadrilateral, then its circumcenter is also at <math>O</math>. Thus, it suffices to show that <math>OP=OQ=OR=OS</math>. <br />
<br />
Notice that <math>PE=QE</math>, <math>EO=EO</math>, and <math>\angle QEO=\angle PEO=90^\circ</math>. By SAS congruency, <math>\Delta QOE\cong\Delta POE\implies QO=PO</math>. Similarly, we find that <math>\Delta SOF\cong\Delta ROF</math> and <math>OS=OR</math>. We now need only to show that these two pairs are equal to each other. <br />
<br />
Draw the segments connecting <math>O</math> to <math>B</math>, <math>Q</math>, <math>C</math>, and <math>R</math>. <br />
<br />
<asy> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br />
import graph; size(12cm); <br />
real labelscalefactor = 0.5; /* changes label-to-point distance */<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br />
pen dotstyle = black; /* point style */ <br />
real xmin = -12.19, xmax = 24.94, ymin = -15.45, ymax = 6.11; /* image dimensions */<br />
pen wrwrwr = rgb(0.38,0.38,0.38); <br />
/* draw figures */<br />
draw(circle((2.92,-3.28), 5.90), linewidth(2) + wrwrwr); <br />
draw((-2.52,-1.01)--(3.46,2.59), linewidth(2) + wrwrwr); <br />
draw((7.59,-6.88)--(-0.29,-8.22), linewidth(2) + wrwrwr); <br />
draw((3.46,2.59)--(7.59,-6.88), linewidth(2) + wrwrwr); <br />
draw((-0.29,-8.22)--(-2.52,-1.01), linewidth(2) + wrwrwr); <br />
draw((0.03,0.52)--(1.67,-7.89), linewidth(2) + wrwrwr); <br />
draw((5.61,-7.22)--(0.89,1.04), linewidth(2) + wrwrwr); <br />
draw((0.46,0.78)--(2.92,-3.28), linewidth(2) + wrwrwr); <br />
draw((2.92,-3.28)--(3.64,-7.55), linewidth(2) + wrwrwr); <br />
draw((2.92,-3.28)--(7.59,-6.88), linewidth(2) + wrwrwr); <br />
draw((5.61,-7.22)--(2.92,-3.28), linewidth(2) + wrwrwr); <br />
draw((2.92,-3.28)--(3.46,2.59), linewidth(2) + wrwrwr); <br />
draw((2.92,-3.28)--(0.89,1.04), linewidth(2) + wrwrwr); <br />
/* dots and labels */<br />
dot((2.92,-3.28),dotstyle); <br />
label("$O$", (2.43,-3.56), NE * labelscalefactor); <br />
dot((-2.52,-1.01),dotstyle); <br />
label("$A$", (-2.91,-0.91), NE * labelscalefactor); <br />
dot((3.46,2.59),linewidth(1pt) + dotstyle); <br />
label("$B$", (3.49,2.78), NE * labelscalefactor); <br />
dot((7.59,-6.88),dotstyle); <br />
label("$C$", (7.82,-7.24), NE * labelscalefactor); <br />
dot((-0.29,-8.22),linewidth(1pt) + dotstyle); <br />
label("$D$", (-0.53,-8.62), NE * labelscalefactor); <br />
dot((0.03,0.52),linewidth(1pt) + dotstyle); <br />
label("$P$", (-0.13,0.67), NE * labelscalefactor); <br />
dot((0.89,1.04),linewidth(1pt) + dotstyle); <br />
label("$Q$", (0.62,1.16), NE * labelscalefactor); <br />
dot((5.61,-7.22),linewidth(1pt) + dotstyle); <br />
label("$R$", (5.70,-7.05), NE * labelscalefactor); <br />
dot((1.67,-7.89),linewidth(1pt) + dotstyle); <br />
label("$S$", (1.75,-7.73), NE * labelscalefactor); <br />
dot((0.46,0.78),linewidth(1pt) + dotstyle); <br />
label("$E$", (0.26,0.93), NE * labelscalefactor); <br />
dot((3.64,-7.55),linewidth(1pt) + dotstyle); <br />
label("$F$", (3.73,-7.39), NE * labelscalefactor); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br />
/* end of picture */</asy><br />
<br />
Also, let <math>r</math> be the circumradius of <math>ABCD</math>. This means that <math>AO=BO=CO=DO=r</math>. Recall that <math>\angle BEO=90^\circ</math> and <math>\angle CFO=90^\circ</math>. Notice the several right triangles in our figure. <br />
<br />
Let us apply Pythagorean Theorem on <math>\Delta BEO</math>. We can see that <math>EO^2+EB^2=BO^2\implies EO^2+3.5^2=r^2\implies EO=\sqrt{r^2-12.25}.</math> <br />
<br />
Let us again apply Pythagorean Theorem on <math>\Delta QEO</math>. We can see that <math>QE^2+EO^2=QO^2\implies0.5^2+r^2-12.25=QO^2\implies QO=\sqrt{r^2-12}.</math> <br />
<br />
Let us apply Pythagorean Theorem on <math>\Delta CFO</math>. We get <math>CF^2+OF^2=OC^2\implies4^2+OF^2=r^2\implies OF=\sqrt{r^2-16}</math>. <br />
<br />
We finally apply Pythagorean Theorem on <math>\Delta RFO</math>. This becomes <math>OF^2+FR^2=OR^2\implies r^2-16+2^2=OR^2\implies OR=\sqrt{r^2-12}</math>. <br />
<br />
This is the same expression as we got for <math>QO</math>. Thus, <math>OQ=OR</math>, and recalling that <math>OQ=OP</math> and <math>OR=OS</math>, we have shown that <math>OP=OQ=OR=OS</math>. We are done. QED<br />
<br />
~Technodoggo<br />
<br />
==Solution 2==<br />
<br />
We can consider two cases: <math>AB \parallel CD</math> or <math>AB \nparallel CD.</math> The first case is trivial, as <math>PQ \parallel RS</math> and we are done due to symmetry. For the second case, WLOG, assume that <math>A</math> and <math>C</math> are located on <math>XB</math> and <math>XD</math> respectively. Extend <math>AB</math> and <math>CD</math> to a point <math>X,</math> and by Power of a Point, we have <cmath>XA\cdot XB = XC \cdot XD,</cmath> which may be written as <cmath>XA \cdot (XA+7) = XC \cdot (XC+8),</cmath> or <cmath>XA^2 + 7XA = XC^2 + 8XC.</cmath> We can translate this to <cmath>XA^2 + 7XA +12 = XC^2 + 8XC +12,</cmath> so <cmath>XP\cdot XQ = (XA+3)(XA+4)=(XC+2)(XC+6)= XR\cdot XS,</cmath> and therefore by the Converse of Power of a Point <math>PQRS</math> is cyclic, and we are done.<br />
<br />
- [https://artofproblemsolving.com/wiki/index.php/User:Spectraldragon8 spectraldragon8]<br />
<br />
==Solution 3==<br />
<br />
All 4 corners of <math>PQRS</math> have equal power of a point (<math>12</math>) with respect to the circle <math>(ABCD)</math>, with center <math>O</math>.<br />
<br />
Draw diameters (of length <math>AQ</math>) of circle <math>(ABCD)</math> through <math>Q</math> and <math>S</math>, with length <math>A</math>. Let <math>q</math> be the distance from <math>Q</math> to the circle along a diameter, and likewise <math>s</math> be distance from <math>S</math> to the circle.<br />
<br />
Then <math>q(AQ-q) = s(AQ-s) = 12</math> and <math>q,s < AQ/2</math> (radius). Therefore, <math>q=s</math> and <math>AQ/2 -q = AQ/2 -s</math>. But <math>AQ-q=OQ</math>, <math>AQ-s=OS</math>, <math>OQ = OP</math> and <math>OS = OR</math> by symmetry around the perpendicular bisectors of <math>PQ</math> and <math>RS</math>, so <math>P,Q,R,S</math> are all equidistant from <math>O</math>, forming a circumcircle around <math>PQRS</math>.<br />
<br />
-BraveCobra22aops and oinava<br />
<br />
==See Also==<br />
{{USAJMO newbox|year=2024|before=First Question|num-a=2}}<br />
{{MAA Notice}}</div>Ryanjwanghttps://artofproblemsolving.com/wiki/index.php/2024_USAJMO2024 USAJMO2024-03-20T01:31:01Z<p>Erinb28lms: </p>
<hr />
<div>The 15th USAJMO was held on March 19th and 20th, 2024. The first link will contain the full set of test problems. The rest will contain each individual problem and its solution.<br />
<br />
[[2024 USAJMO Problems]]<br />
* [[2024 USAJMO Problems/Problem 1]]<br />
* [[2024 USAJMO Problems/Problem 2]]<br />
* [[2024 USAJMO Problems/Problem 3]]<br />
* [[2024 USAJMO Problems/Problem 4]]<br />
* [[2024 USAJMO Problems/Problem 5]]<br />
* [[2024 USAJMO Problems/Problem 6]]<br />
<br />
== See Also ==<br />
* [[Mathematics competitions]]<br />
* [[Mathematics competition resources]]<br />
* [[Math books]]<br />
* [[USAJMO]]<br />
<br />
{{USAJMO newbox|year= 2024 |before=[[2023 USAJMO]]|after=[[2025 USAJMO]]}}</div>Ryanjwanghttps://artofproblemsolving.com/wiki/index.php/Classroom_hacksClassroom hacks2024-03-20T00:01:45Z<p>Lostinbali: Created blank page</p>
<hr />
<div></div>Lostinbali