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2024-03-28T21:17:45Z
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https://artofproblemsolving.com/wiki/index.php/PaperMath%E2%80%99s_circles
PaperMath’s circles
2024-03-27T21:19:55Z
<p>Papermath: /* Proof */</p>
<hr />
<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>
Papermath
https://artofproblemsolving.com/wiki/index.php/Fairfax_County_Math_League
Fairfax County Math League
2024-03-26T11:05:45Z
<p>Hcps-sundarrap: Created page with "Hello"</p>
<hr />
<div>Hello</div>
Hcps-sundarrap
https://artofproblemsolving.com/wiki/index.php/The
The
2024-03-25T00:53:30Z
<p>Icyfire500: Created page with "The is a word."</p>
<hr />
<div>The is a word.</div>
Icyfire500
https://artofproblemsolving.com/wiki/index.php/Sharygin_Olympiads,_the_best
Sharygin Olympiads, the best
2024-03-24T05:37:17Z
<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>
Vvsss
https://artofproblemsolving.com/wiki/index.php/Connecticut_Mathcounts
Connecticut Mathcounts
2024-03-23T22:16:58Z
<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>
Gp102
https://artofproblemsolving.com/wiki/index.php/2025_USAJMO
2025 USAJMO
2024-03-22T19:50:34Z
<p>Hungrycalculator: Created page with "oh no no no"</p>
<hr />
<div>oh no no no</div>
Hungrycalculator
https://artofproblemsolving.com/wiki/index.php/2024_USAMO_Problems
2024 USAMO Problems
2024-03-21T02:33:55Z
<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-tsuruko
https://artofproblemsolving.com/wiki/index.php/2024_USAMO_Problems/Problem_1
2024 USAMO Problems/Problem 1
2024-03-21T02:33:01Z
<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-tsuruko
https://artofproblemsolving.com/wiki/index.php/2024_USAMO_Problems/Problem_2
2024 USAMO Problems/Problem 2
2024-03-21T02:32:39Z
<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-tsuruko
https://artofproblemsolving.com/wiki/index.php/2024_USAMO_Problems/Problem_3
2024 USAMO Problems/Problem 3
2024-03-21T02:32:24Z
<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-tsuruko
https://artofproblemsolving.com/wiki/index.php/2024_USAMO_Problems/Problem_4
2024 USAMO Problems/Problem 4
2024-03-21T02:32:06Z
<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-tsuruko
https://artofproblemsolving.com/wiki/index.php/2024_USAMO_Problems/Problem_5
2024 USAMO Problems/Problem 5
2024-03-21T02:30:58Z
<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-tsuruko
https://artofproblemsolving.com/wiki/index.php/2024_USAMO_Problems/Problem_6
2024 USAMO Problems/Problem 6
2024-03-21T02:28:32Z
<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-tsuruko
https://artofproblemsolving.com/wiki/index.php/2024_USAMO
2024 USAMO
2024-03-21T02:19:51Z
<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 agarwal
https://artofproblemsolving.com/wiki/index.php/2024_USAJMO_Problems/Problem_5
2024 USAJMO Problems/Problem 5
2024-03-20T23:59:24Z
<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>
Flyingpig7
https://artofproblemsolving.com/wiki/index.php/2024_USAJMO_Problems/Problem_4
2024 USAJMO Problems/Problem 4
2024-03-20T19:33:32Z
<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>
Logicus14
https://artofproblemsolving.com/wiki/index.php/2024_USAJMO_Problems/Problem_3
2024 USAJMO Problems/Problem 3
2024-03-20T01:43:54Z
<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>
Ryanjwang
https://artofproblemsolving.com/wiki/index.php/2024_USAJMO_Problems/Problem_2
2024 USAJMO Problems/Problem 2
2024-03-20T01:37:37Z
<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>
Ryanjwang
https://artofproblemsolving.com/wiki/index.php/2024_USAJMO_Problems/Problem_1
2024 USAJMO Problems/Problem 1
2024-03-20T01:36:05Z
<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>
Ryanjwang
https://artofproblemsolving.com/wiki/index.php/2024_USAJMO
2024 USAJMO
2024-03-20T01:31:01Z
<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>
Ryanjwang
https://artofproblemsolving.com/wiki/index.php/Classroom_hacks
Classroom hacks
2024-03-20T00:01:45Z
<p>Lostinbali: Created blank page</p>
<hr />
<div></div>
Lostinbali
https://artofproblemsolving.com/wiki/index.php/Gg_boiiis
Gg boiiis
2024-03-18T18:18:39Z
<p>Ambertiger79: Ambertiger79 created the page Gg boiiis using a non-default content model "plain text": idk</p>
<hr />
<div></div>
Ambertiger79
https://artofproblemsolving.com/wiki/index.php/Challenge_Math_Online_Program
Challenge Math Online Program
2024-03-15T16:30:34Z
<p>Liqiank: </p>
<hr />
<div>Developed by Noetic Learning, [https://www.noetic-learning.com/gifted/index.jsp Challenge Math Online] aims to sharpen young students' mathematical problem-solving and logical reasoning abilities. Featuring non-routine questions, the program presents challenges akin to those found in prominent math competitions, including the esteemed [[Noetic Learning Math Contest]]. This innovative approach not only prepares students for competitive success but also deeply enriches their understanding and application of mathematical concepts<br />
<br />
[https://www.noetic-learning.com/gifted/index.jsp Challenge Math Online] is primarily designed for students gifted in math who are seeking additional challenges and support. The program:<br />
<br />
* Provides challenges beyond the regular school curriculum,<br />
* Strengthens creative problem solving, and logical reasoning skills,<br />
* Further develops gifted students' intellect in math, and<br />
* Helps students excel in national math competitions.<br />
<br />
== Problem-Solving Strategies Covered ==<br />
* Model with Diagrams<br />
* Draw a Picture<br />
* Make an Organized List<br />
* Find a Pattern <br />
* Use logical Reasoning<br />
* Work Backwards <br />
* Guess and Check | Solve a Simpler Problem<br />
<br />
== Topics Covered ==<br />
* Numbers and Operations <br />
* Geometry and Measurement <br />
* Probability and Statistics <br />
* Data Analysis <br />
* Algebraic Thinking</div>
Liqiank
https://artofproblemsolving.com/wiki/index.php/Noetic_Learning_Math_Contest
Noetic Learning Math Contest
2024-03-15T14:23:33Z
<p>Liqiank: /* Resources */</p>
<hr />
<div>== Overview ==<br />
The [http://www.noetic-learning.com/mathcontest/ Noetic Learning Math Contest] is a popular national biannual problem-solving competition for elementary and middle school students. The contest aims to encourage students' interest in mathematics and develop problem-solving skills. It is open to students in grades 2 through 8. It is held twice a year, in the fall and spring. The competition consists of a 45-minute timed test, comprising 20 math problems. It currently offers both paper/pencil and online version.<br />
<br />
== Awards and Recognitions ==<br />
Students who participate in the Noetic Learning Math Contest can earn the following awards and recognition, based on their performance:<br />
<br />
*Team Winner Medal: The top scorer on the team. <br />
*National Honor Roll Medal: Awarded to the top 10% of participants in each grade category. <br />
*National Honorable Mention Ribbon: Awarded to students who score in the top 50% of participants in their grade category.<br />
<br />
School teams can earn the following award:<br />
<br />
*Team Achievement Plaque: The top 10% of teams in each grade.<br />
<br />
== Resources ==<br />
* [http://www.noetic-learning.com/mathcontest Contest Home Page]<br />
* [http://www.noetic-learning.com/mathcontest/pasttests.jsp Noetic Learning Math Contest Past Tests]<br />
* [http://www.noetic-learning.com/gifted Challenge Math Online]<br />
* [https://www.noetic-learning.com/pow.jsp Problem of the Week]<br />
* [https://www.noetic-learning.com/videos.jsp Video Library - Problem Solving Strategies]<br />
<br />
== See Also ==<br />
* [https://www.youtube.com/@NoeticLearning Noetic Learning Youtube Channel]<br />
<br />
[[Category:Mathematics competitions]]<br />
[[Category:Introductory mathematics competitions]]</div>
Liqiank
https://artofproblemsolving.com/wiki/index.php/Noetic_Learning_Math_Contests
Noetic Learning Math Contests
2024-03-15T14:11:00Z
<p>Liqiank: /* Noetic Learning Math Contest Overview */</p>
<hr />
<div>== Noetic Learning Math Contest Overview ==<br />
<br />
<br />
The [http://www.noetic-learning.com/mathcontest/ Noetic Learning Math Contest] is a popular national biannual problem-solving competition for elementary and middle school students. The contest aims to encourage students' interest in mathematics and develop problem-solving skills. It is open to students in grades 2 through 8. It is held twice a year, in the fall and spring. The competition consists of a 45-minute timed test, comprising 20 math problems.<br />
<br />
== Awards and recognition ==<br />
<br />
Students who participate in the Noetic Learning Math Contest can earn the following awards and recognition, based on their performance:<br />
<br />
*Team Winner: The top scorer on the team. <br />
*National Honor Roll: Awarded to the top 10% of participants in each grade category. <br />
*Honorable Mention: Awarded to students who score in the top 50% of participants in their grade category.<br />
<br />
School teams can earn the following award:<br />
<br />
*Team Achievement: The top 10% of teams in each grade.<br />
<br />
== Resources ==<br />
* [http://www.noetic-learning.com/mathcontest Noetic Learning Math Contest Home Page]<br />
* [http://www.noetic-learning.com/gifted Noetic Learning Challenge Math Online]</div>
Liqiank
https://artofproblemsolving.com/wiki/index.php/Point_redefinition
Point redefinition
2024-03-15T01:01:25Z
<p>Marcus zhang: Created blank page</p>
<hr />
<div></div>
Marcus zhang
https://artofproblemsolving.com/wiki/index.php/Alex
Alex
2024-03-14T00:22:19Z
<p>Arjunkatta: Created page with "Alex is a hammerhead shark who has four arms he neat and organized. later in the b5 books he becomes a director he is a well-known character"</p>
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<div>Alex is a hammerhead shark who has four arms he neat and organized. later in the b5 books he becomes a director he is a well-known character</div>
Arjunkatta
https://artofproblemsolving.com/wiki/index.php/Grog
Grog
2024-03-14T00:17:58Z
<p>Arjunkatta: </p>
<hr />
<div>He is the main character across the beast academy guidebooks he loves to draw and is a gigantic purple beast in most endings he asks controversial questions that are unsolved.</div>
Arjunkatta
https://artofproblemsolving.com/wiki/index.php/2025_AMC_8_Problems/Problem_7
2025 AMC 8 Problems/Problem 7
2024-03-13T05:10:49Z
<p>Riemann1826: Created page with "The 2025 AMC 8 is held. Please do not post false problems."</p>
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<div>The 2025 AMC 8 is held. Please do not post false problems.</div>
Riemann1826
https://artofproblemsolving.com/wiki/index.php/2025_AMC_8_Problems/Problem_6
2025 AMC 8 Problems/Problem 6
2024-03-13T05:09:58Z
<p>Riemann1826: </p>
<hr />
<div>The 2025 AMC 8 is held. Please do not post false problems.</div>
Riemann1826
https://artofproblemsolving.com/wiki/index.php/Emojis
Emojis
2024-03-11T21:25:25Z
<p>Yuanqitan: I need help ot get emojis.</p>
<hr />
<div>Where do you find the place where it has all the emojis?</div>
Yuanqitan
https://artofproblemsolving.com/wiki/index.php/Rules_of_exponents
Rules of exponents
2024-03-11T20:39:52Z
<p>Akmolloy: Blanked the page</p>
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Akmolloy
https://artofproblemsolving.com/wiki/index.php/B
B
2024-03-10T17:17:52Z
<p>Whatsoever281: Created page with "https://artofproblemsolving.com/wiki/index.php/A"</p>
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<div>https://artofproblemsolving.com/wiki/index.php/A</div>
Whatsoever281
https://artofproblemsolving.com/wiki/index.php/A
A
2024-03-08T01:48:06Z
<p>Whatsoever281: </p>
<hr />
<div>LEAVE THIS PAGE<br />
<br />
maybe<br />
Go to kill.com<br />
https://artofproblemsolving.com/wiki/index.php?title=Main_Page<br />
<cmath>\sqrt{2}=1.4142135623730950488016887242096980785696718753769480731766797379907324784621<br />
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85007603610091594656706768836055717400767569050961367194013249356052401859991050<br />
62108163597726431380605467010293569971042425105781749531057255934984451126922780<br />
34491350663756874776028316282960553242242695753452902883876844642917328277088831<br />
80870253398523381227499908123718925407264753678503048215918018861671089728692292<br />
01197599880703818543332536460211082299279293072871780799888099176741774108983060<br />
80032631181642798823117154363869661702999934161614878686018045505553986913115186<br />
01038637532500455818604480407502411951843056745336836136745973744239885532851793<br />
08960373898915173195874134428817842125021916951875593444387396189314549999906107<br />
58704909026088351763622474975785885836803745793115733980209998662218694992259591<br />
32764236194105921003280261498745665996888740679561673918595728886424734635858868<br />
64496822386006983352642799056283165613913942557649062065186021647263033362975075</cmath></div>
Whatsoever281
https://artofproblemsolving.com/wiki/index.php/University_of_Maryland_High_School_Mathematics_Competition
University of Maryland High School Mathematics Competition
2024-03-06T15:17:57Z
<p>Clarkculus: changed difficulty</p>
<hr />
<div>The '''University of Maryland High School Mathematics Competition''' (UMD Math Competition) is a mathematics competition for high school students in Maryland and D.C. offered by the University of Maryland since 1997. It is split into an AMC-style 25 question multiple choice exam in October and a 5 question proof exam for the top 15% of scorers in the first part in November. Part I is worth 6 points per question, with 2 points per question left unanswered. Part II is worth 30 points per question with potential partial credit. The students with the 50 highest combined scores are awarded Honorable Mentions, and the top 3 overall and county winners are announced.<br />
<br />
{{Contest Info|name=UMD Math Competition Part I|region=Chesapeake|type=Multiple Choice|difficulty=1-3|breakdown=1-1.5 (Problems 1-10)<br>1.5-2.5 (Problems 11-20)<br>2.5-3 (Problems 21-25)}}<br />
{{Contest Info|name=UMD Math Competition Part II|region=Chesapeake|type=Free Response|difficulty=1-5|breakdown=1-2 (Problems 1/2)<br>2-3.5 (Problems 3/4)<br>5 (Problem 5)}}<br />
<br />
== How to Register ==<br />
Ask your math teacher to register your school with the UMD.<br />
<br />
== Resources ==<br />
* [https://mathcomp.math.umd.edu/ UMD Homepage]<br />
<br />
== See also ==<br />
* [[Maryland mathematics competitions]]</div>
Clarkculus
https://artofproblemsolving.com/wiki/index.php/Beestar_math_competition
Beestar math competition
2024-03-02T15:45:05Z
<p>Bsming: Created page with "Are you familiar with the Beestar Math Competition? It's an online math contest held twice a year, in spring and fall, geared towards sparking interest in math among elementar..."</p>
<hr />
<div>Are you familiar with the Beestar Math Competition? It's an online math contest held twice a year, in spring and fall, geared towards sparking interest in math among elementary and middle school students while enhancing their mathematical skills. Since 2014, the competition has drawn participants from across the country, spanning seven grade levels from 2nd to 8th grade, with tailored questions for each level. Unlike other math contests, Beestar focuses on fundamental concepts and skills taught in school curriculums, serving as a comprehensive assessment of students' grasp of grade-appropriate knowledge. It's designed to nurture interest, fuel enthusiasm, and ultimately elevate mathematical proficiency for the majority of students.<br />
<br />
Results and rankings are announced on the website, with awards including gold, silver, and bronze medals, as well as encouragement awards, fostering a sense of achievement among participants. The fall competition for 2024 is scheduled to run online from March 18th to March 21st. Parents are encouraged to register their children, with registration fees set at <math>10 for grades 2-5 and </math>12 for grades 6-8.<br />
<br />
The competition comprises two rounds, totaling eight sections. Sample questions are provided to familiarize participants with the format and content. Questions for grades 2-5 can be found [link]. The competition follows the format of the AMC (American Mathematics Competition), featuring DASH and SPARK rounds, assessing students' ability to solve problems quickly and accurately, as well as their capacity for in-depth and comprehensive problem analysis. It lays the groundwork and offers ample preparation for future participation in AMC and MathCounts competitions.</div>
Bsming
https://artofproblemsolving.com/wiki/index.php/MehtA%2B_Ethical_Hacking_Camp
MehtA+ Ethical Hacking Camp
2024-03-01T07:34:41Z
<p>BPM14: Created page with "MehtA+ Intro to Ethical Hacking Camp is a virtual course for 7th-12th graders (ages 12-18) located anywhere in the world. Did you know that cybercrime inflicted damages global..."</p>
<hr />
<div>MehtA+ Intro to Ethical Hacking Camp is a virtual course for 7th-12th graders (ages 12-18) located anywhere in the world. Did you know that cybercrime inflicted damages globally totalling $6 trillion USD, just in 2021? This camp is for anyone interested in learning how to defend oneself against these threats and identify and fix vulnerabilities preemptively.<br />
<br />
===Camp Description=== <br />
<br />
In this 1 week hands-on self-paced course, MehtA+ instructors will teach Grades 7-12 students the basics of penetration testing (ethical hacking), networking, security and cybersecurity best practices.<br />
<br />
Students will learn how to ethically hack using Browser Exploitation Framework (BeEF) on Kali Linux to evaluate system security. They will specifically learn the various ways in which malicious actors steal credentials and data, and how to protect oneself against it.<br />
<br />
Students will complete fourteen labs and write a penetration test report for their final project. Upon completion, students will receive a certificate. No prior experience is required. Windows or Mac laptop/computer with around ~50 GB available storage necessary.<br />
<br />
===More Information + FAQs===<br />
<br />
For more information and to sign up: https://mehtaplus.com/courses/ai-in-visual-arts-camp/</div>
BPM14
https://artofproblemsolving.com/wiki/index.php/MehtA%2B_Competitive_Programming_Bootcamp
MehtA+ Competitive Programming Bootcamp
2024-03-01T07:06:14Z
<p>BPM14: </p>
<hr />
<div>MehtA+ Competitive Programming Bootcamp is held virtually, every summer for 7th-12th graders (ages 12-18) the basics of data structures and algorithms in Python. Many entry-level computer scientist positions offer a six-figure salary. In order to land such coveted jobs, it is necessary to pass several difficult coding interviews. This camp is designed to help students prepare for a variety of topics at the USACO Bronze, Silver and Gold level as well as coding interviews for internships & jobs at the collegiate level. At the end of the camp, students will participate in a MehtA+ coding competition.<br />
<br />
===Locations and Dates===<br />
<br />
The camp will be offered during the weekdays from June 24 – July 5, 2024 from 9 – 12 am CST. <br />
<br />
===Camp Description=== <br />
<br />
The instructors are MIT, Stanford, Harvard and UIUC computer science alumni and students who have done well at the USACO Gold Level and who have passed coding interviews at the topmost software engineering companies (Meta, Google, Amazon & Microsoft, etc.)<br />
<br />
Topics covered include, but not limited to are data structures and control structures, algorithmic complexity, recursion, sorting and searching algorithms, hashing, dynamic programming, graph theory & object oriented programming.<br />
<br />
Students who successfully complete the camp will receive a certificate. <br />
<br />
Prerequisites: Basic familiarity with Python is required.<br />
<br />
To check if you have all the prerequisites necessary for the camp, go to W3schools. If you understand the content between Python Syntax and Python Dictionaries (see left sidebar on W3schools website), then you are a good fit for the camp. <br />
<br />
If you have knowledge of any other object-oriented programming language, please review the concepts on W3schools for Python before attending this camp.<br />
<br />
===Application=== <br />
<br />
Application here: https://mehtaplustutoring.com/ai-in-visual-arts-application/<br />
<br />
Deadline: <br />
<br />
The deadline to register is Wednesday, June 19, 2024 at 11:59 pm CST.<br />
<br />
===Daily Schedule===<br />
<br />
The camp will be offered during the weekdays from June 24 – July 5, 2024 from 9 – 12 am CST. <br />
<br />
Interested students from Grades 7-12 from anywhere around the world are eligible to attend the camp. <br />
<br />
For class activities, students will be divided based on their prior experience and familiarity with concepts, which will be determined based on the results of a short survey/quiz. <br />
<br />
===Syllabus===<br />
<br />
The material covered in this camp is asked during entry-level software engineer coding interviews. It is also asked during several middle school and high school programming competitions. So why is it important to participate in programming competitions? Learn the answer to this question here: https://mehtaplus.medium.com/the-importance-of-pursuing-computer-science-competitions-6bbe4729a8bf!<br />
<br />
Good coding practices will be taught and enforced in the camp.<br />
<br />
The following topics in Python will be covered:<br />
<br />
Data Structures (Arrays, Strings, Stacks, Queues, Trees), Conditional statements, Loops, Functions, Time Complexity, Recursion, Sorting, Searching (1-D & 2-D), Hashing, Dynamic Programming problems, Graph Theory (Types of graphs) and Graph Traversal Algorithms (BFS, DFS, Dijkstra, Bellman-Ford), Object Oriented Programming Concepts, <br />
<br />
The camp will involve lectures as well as group activities, homework competitions in order to prepare students adequately for programming competitions.<br />
<br />
===More Information + FAQs===<br />
<br />
For more information: https://mehtaplus.com/courses/competitive-program/</div>
BPM14
https://artofproblemsolving.com/wiki/index.php/19329_contest_math:_intro_to_geometry
19329 contest math: intro to geometry
2024-02-29T03:48:56Z
<p>Nutellabrot: Created page with "hello"</p>
<hr />
<div>hello</div>
Nutellabrot
https://artofproblemsolving.com/wiki/index.php/The_WTGF
The WTGF
2024-02-28T19:03:17Z
<p>Themathkidthatlikesaops: Created page with "The WiseTiger Gaming Forum is a forum founded by WiseTigerJ1. The co-creator was WiseTigerJ2."</p>
<hr />
<div>The WiseTiger Gaming Forum is a forum founded by WiseTigerJ1. The co-creator was WiseTigerJ2.</div>
Themathkidthatlikesaops
https://artofproblemsolving.com/wiki/index.php/Full_of_Baloney
Full of Baloney
2024-02-28T19:00:20Z
<p>Themathkidthatlikesaops: </p>
<hr />
<div>Full of Baloney © is a brand new forum founded by [[User:VeritasTian|VeritasTian]]. The welcome was created with the help of [[User:sximoz|sximoz]]. It was created on February 27, 4:34 PM. A lot of people started posting on this forum the second day.</div>
Themathkidthatlikesaops
https://artofproblemsolving.com/wiki/index.php/Getting_Started_with_JavaScript_Programming
Getting Started with JavaScript Programming
2024-02-27T02:03:39Z
<p>Kawhi2: Created page with "rect(200, 200, 200, 200);"</p>
<hr />
<div>rect(200, 200, 200, 200);</div>
Kawhi2
https://artofproblemsolving.com/wiki/index.php/2024_USAJMO_Problems
2024 USAJMO Problems
2024-02-25T00:50:14Z
<p>Oinava: /* Problem 4 */</p>
<hr />
<div>__TOC__<br />
<br />
==Day 1==<br />
===Problem 1===<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 />
[[2024 USAJMO Problems/Problem 1|Solution]]<br />
===Problem 2===<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 />
[[2024 USAJMO Problems/Problem 2|Solution]]<br />
===Problem 3===<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 />
[[2024 USAJMO Problems/Problem 3|Solution]]<br />
==Day 2==<br />
===Problem 4===<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 columns 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 />
[[2024 USAJMO Problems/Problem 4|Solution]]<br />
<br />
===Problem 5===<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 />
[[2024 USAJMO Problems/Problem 5|Solution]]<br />
===Problem 6===<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 />
[[2024 USAJMO Problems/Problem 6|Solution]]<br />
== See also ==<br />
<br />
{{USAJMO box|year=2024|before=[[2023 USAJMO Problems]]|after=[[2025 USAJMO Problems]]}}<br />
{{MAA Notice}}</div>
Mathkiddie
https://artofproblemsolving.com/wiki/index.php/The_Clan_Gathering
The Clan Gathering
2024-02-23T22:59:25Z
<p>Lostinbali: Created page with "The Clan Gathering"</p>
<hr />
<div>The Clan Gathering</div>
Lostinbali
https://artofproblemsolving.com/wiki/index.php/The_clan_gathering
The clan gathering
2024-02-23T22:59:06Z
<p>Lostinbali: Blanked the page</p>
<hr />
<div></div>
Lostinbali
https://artofproblemsolving.com/wiki/index.php/2025_AMC_8_Problems/Problem_10
2025 AMC 8 Problems/Problem 10
2024-02-23T13:39:54Z
<p>Theumbrellaacademy: 2025 AMC 8 Question 10</p>
<hr />
<div>There are fifty \$1, \$2, and \$5 notes. There are two more \$1 notes than \$2 notes. Given that the total value of the notes is \$116, find the number of \$5 notes.<br />
<br />
<math>\textbf{(A) } 9\qquad\textbf{(B) } 10\qquad\textbf{(C) } 11\qquad\textbf{(D) } 12\qquad\textbf{(E) } 13</math></div>
Theumbrellaacademy
https://artofproblemsolving.com/wiki/index.php/2025_AMC_8_Problems/Problem_11
2025 AMC 8 Problems/Problem 11
2024-02-23T11:06:13Z
<p>Theumbrellaacademy: 2025 AMC 8 Problem 11</p>
<hr />
<div>Find the smallest positive integer <math>k</math> such that <math>(2^{91}+k)</math> is divisible by <math>127</math>.<br />
<br />
<math>\textbf{(A)}\ 122 \qquad \textbf{(B)}\ 123 \qquad \textbf{(C)}\ 124 \qquad \textbf{(D)}\ 125 \qquad \textbf{(E)}\ 126</math></div>
Theumbrellaacademy
https://artofproblemsolving.com/wiki/index.php/Polya%E2%80%99s_method_for_extremums
Polya’s method for extremums
2024-02-22T15:24:36Z
<p>Vvsss: /* Clever point elephant in hollow tetrahedron */</p>
<hr />
<div>==Polya’s method==<br />
People involved in spirituality believe that we are created and placed in the best of all possible worlds.<br />
<br />
Therefore, the laws governing our world are based on the principles of optimality.<br />
<br />
Natural scientists have found a number of empirical laws and methods that describe nature and therefore optimal laws. <br />
<br />
Pólya proposed a method in which the researcher replaces the geometric problem of finding the optimum with a mechanical or optical model and applies empirical dependencies to this model.<br />
<br />
<br />
==The segment of the shortest length==<br />
[[File:Shortest segment.png|400px|right]]<br />
The segment <math>AB</math> has the ends on the sides of a right angle and contains a point <math>C(x_C, y_C).</math> Find the shortest length of such a segment.<br />
<br />
<i><b>Solution</b></i> <br />
<br />
Let's imagine that <math>AB</math> is a spring rod that cannot bend, but tends to shorten its length. <br />
<br />
The rod is fixed at point <math>C</math> on a hinge without friction. The hinge allows the rod to rotate and slide.<br />
<br />
The ends of the rod can slide without friction along the grooves - the sides of the corner. <br />
<br />
Let the rod be balanced, and the force pulling it together is equal to <math>T.</math> The grooves can create a force only along the normal, so they act on the rod with forces <br />
<cmath>F_B = \frac {T}{\sin \alpha}, F_A = \frac {T}{\cos \alpha}.</cmath> <br />
For the rod to be balanced, it is required that the moments of the forces be equal relative to point <math>C.</math> The moments of forces are:<br />
<cmath>F_A \cdot AC \cdot \sin \alpha = F_B \cdot BC \cdot \cos \alpha \implies \frac {BC}{AC} = \tan^2 \alpha,</cmath><br />
<cmath>\frac {y_C}{x_C} = \frac {BC \cdot \sin \alpha}{AC \cdot \cos \alpha} = \tan^3 \alpha \implies</cmath><br />
<cmath>AB = \left ( x_C^ {\frac {2}{3}} + y_C^ {\frac {2}{3}}\right ) ^{\frac {3}{2}}.</cmath><br />
'''vladimir.shelomovskii@gmail.com, vvsss'''<br />
<br />
==The circle inside a fixed plane angle==<br />
[[File:Circle and triangle.png|400px|right]]<br />
Let the plane angle with vertex <math>A</math> and the circle inside the angle be given. A straight line is drawn through point M of this circle, tangent to the circle and intersecting the sides of the angle at points <math>B</math> and <math>C.</math> Find the condition under which the area of <math>\triangle ABC</math> is the smallest.<br />
<br />
<i><b>Solution</b></i> <br />
<br />
Let us imagine that <math>AB</math> is a rod that cannot bend, and whose ends slide freely along the sides of the angle, and at point <math>M</math> it rests on a convex curve. <br />
<br />
Let <math>\triangle ABC</math> be covered with a soap film, which, as usual, tends to reduce its area proportional to the energy. <br />
Let ABC be a section of a triangular prism, two faces of which AB and AC are fixed, and the third (BC) is a piston whose width can be changed so that it fits hermetically to the faces AB and AC and can slide along them. The piston rests on a cylinder whose cross-section is a circle with center O.<br />
<br />
Air has been removed from the prism. External air presses on the piston and it comes to an equilibrium position. In this case, the energy of the system will be minimal if the volume of the prism becomes minimal.<br />
In the equilibrium position, the moments of forces applied to the piston <math>BC</math> on the segments <math>BM</math> and <math>MC</math> are equal. <br />
These moments are proportional to the square of the length of the segment, that is, the equilibrium condition is the equality <math>BM = MC</math> or <math>M</math> is the midpoint <math>BC</math> (the center of mass of a homogeneous segment <math>BC.</math>)<br />
<br />
'''vladimir.shelomovskii@gmail.com, vvsss'''<br />
==The sphere inside a fixed trihedral angle==<br />
There is a sphere inside a fixed trihedral angle with vertex <math>A.</math> A plane is drawn through point <math>M</math> of this sphere, tangent to the sphere and intersecting the edges of the angle at points <math>B, C,</math> and <math>D.</math><br />
<br />
Find the condition under which the volume of the pyramid <math>ABCD</math> is the smallest.<br />
<br />
<i><b>Solution</b></i> <br />
<br />
Let us imagine the plane <math>BCD</math> in the form of the piston covering the evacuated volume <math>ABCD.</math> There is no friction of the piston against the walls.<br />
<br />
In the equilibrium position, the volume of the evacuated part of the system is minimal. The equilibrium condition is the equality of the moments of pressure forces, which is equivalent to the equality of the moments of gravity for a homogeneous plate <math>BCD.</math> So the point <math>M</math> of the contact with the sphere (or any convex solid) must be at the center of mass of the polygon <math>BCD.</math><br />
<br />
'''vladimir.shelomovskii@gmail.com, vvsss'''<br />
<br />
==The smallest inscribed equilateral triangle==<br />
[[File:Rotation triangle.png|400px|right]]<br />
A right triangle with sides <math>a</math> and <math>b</math> be given. Find the area of the smallest regular triangle that can be inscribed in it. All vertices of the required triangle must be located on different sides of this triangle.<br />
<br />
<i><b>Solution</b></i> <br />
<br />
Let's imagine the desired triangle as a cross-section of a drill drilling the plane of a given triangle. The axis of rotation does not have to be in the center of the desired triangle.<br />
<br />
Let's turn the drill at a small angle clockwise so that the points of the drill touching the sides of this triangle move along the sides.<br />
<br />
The required triangle is minimal, which means that all its vertices cannot lie on its sides. One or more will go inside. When turning counterclockwise the situation is similar.<br />
<br />
This means that there is an instantaneous axis of rotation of the drill for such turns. Since the required regular triangle is pedal for a point on the instantaneous axis of rotation, this point is the first Apollonius point of given triangle.<br />
<br />
Using the properties of this point, we find that the required area is<br />
<cmath>\frac {\sqrt{3}}{4} \frac { a^2 b^2}{a^2 + b^2 + \sqrt{3}ab}.</cmath><br />
'''vladimir.shelomovskii@gmail.com, vvsss'''<br />
== Curve of the smallest length==<br />
[[File:Shortest line in triangle B.png|280px|right]]<br />
[[File:Shortest line in triangle A.png|320px|right]]<br />
Find the length of the curve of the smallest length dividing a given equilateral triangle with side <math>a</math> into two equal parts.<br />
<br />
<i><b>Solution</b></i> <br />
<br />
Note that the shortest (continuous) curve has ends <math>D</math> and <math>E</math> on two sides of the triangle, which we will call blue and red. Let us denote the common vertex of these two sides as <math>B.</math><br />
<br />
Let us consider a set of six identical regular triangles in each of which the desired (identical!) curve is drawn.<br />
<br />
We place the first of them in an arbitrary way in a certain plane.<br />
<br />
We apply the second triangle to the first in the same plane so that the two blue sides and common vertices <math>B</math> of these triangles coincide. In this case, the ends <math>D</math> of the curve on the blue line will coincide, that is, the set of these two curves is also a continuous curve.<br />
<br />
We will apply the third triangle to the second so that the red sides, vertices <math>B</math> and points <math>E</math> are coincide. We will similarly apply the fourth, fifth and sixth triangles.<br />
<br />
As a result, we obtain a regular hexagon formed by sides opposite to vertex <math>B,</math> inside which a continuous curve limits an area whose area is equal to half the area of the hexagon. It is equal to the area of the three given triangles.<br />
<br />
It is clear that this is a circle, its area (and radius) are known. The required length of the curve is a sixth of the length of its circumference.<br />
<cmath>\pi r^2 = \frac {3 \sqrt {3}}{4} a^2 \implies l = \frac{\pi r}{3} = \frac {a \sqrt{\pi}}{2 \sqrt[4]{3}}</cmath><br />
<br />
'''vladimir.shelomovskii@gmail.com, vvsss'''<br />
== Triangle of the biggest area==<br />
[[File:Max area triangle.png|300px|right]]<br />
A fixed point <math>H</math> in the plane <math>ABC</math> be given. The distances from <math>H</math> to the vertices of triangle <math>ABC</math> are <math>a,b,c.</math> Find the condition that the area of triangle ABC is maximum.<br />
<br />
<i><b>Solution</b></i> <br />
<br />
Let's imagine that at points <math>A, B, C</math> there are rods fixed on hinges that cannot bend or move out of a predetermined plane, but with minimal force they lengthen or shorten. The ends of the rods are connected in pairs in the form of a triangle <math>DEF</math> and slide freely together by changing the length of the rods.<br />
<br />
Let there be a soap film located in the <math>DEF</math> plane outside the <math>DEF</math> triangle, which, as usual, tends to reduce its area proportional to the energy.<br />
<br />
The forces applied to the rod <math>DE</math> are directed perpendicular to <math>DE,</math> therefore <math>BH \perp DE.</math><br />
<br />
In the equilibrium position, the moments of forces applied to the rod <math>DE</math> on segments <math>BD</math> and <math>BE</math> are equal, which means that В is the midpoint of <math>DE.</math><br />
<br />
This means that the area <math>[DEF]</math> is maximum if <math>H</math> is the center of the circumcircle <math>\odot DEF.</math><br />
<br />
The area <math>[DEF]</math> is <math>4</math> times larger than the area <math>[ABC],</math> respectively, the area <math>[ABC]</math> is maximum if <math>H</math> is the orthocenter of <math>\triangle ABC.</math><br />
<br />
'''vladimir.shelomovskii@gmail.com, vvsss'''<br />
<br />
== Clever point elephant in hollow tetrahedron==<br />
[[File:Tetrachedron in cylinder 1.png|300px|right]]<br />
[[File:Tetrachedron in cylinder 2.png|300px|right]]<br />
[[File:Tetrachedron in cyl.png|300px|right]]<br />
[[File:Tetrachedron in cylinder 3.png|300px|right]]<br />
Let the hollow regular unit tetrahedron <math>ABCD</math> be given. A clever point elephant wants to fly along the shortest route that has points belonging to each face of the tetrahedron.<br />
<br />
Find the length of this path.<br />
<br />
<i><b>Solution</b></i> <br />
<br />
Denote <math>AB = a.</math><br />
<br />
Let <math>AM'</math> and <math>DM</math> be the medians of the faces <math>ABC</math> and <math>ABD. AM = \frac {\sqrt {3} a}{2}.</math><br />
<br />
Let <math>\frac {AE}{EM'} = \frac{DE_1}{E_1M} = \frac {3}{2}.</math><br />
<br />
It is known that the common perpendicular to these medians is <math>EE_1, EE_1 = \frac {a}{\sqrt{10}}.</math><br />
<br />
1. Note that <math>CE^2 = CM'^2 + M'E^2 = AE^2 + EE_1^2 = \frac {37 a^2}{100}.</math><br />
<br />
Let <math>E_0</math> be symmetrical to <math>E_1</math> with respect to <math>E.</math><br />
<br />
Then <math>AE_1 = AE_0 \implies CE_0 \perp EE_1, CE_0 = AE = DE_1.</math><br />
<br />
We similarly prove that <math>EE_1 = EE_0 = E_1E_2, BE_2 = AE.</math><br />
<br />
This means that all the vertices of the tetrahedron are located on circles of equal radius with centers on straight line <math>EE_1,</math> and the distances between the centers of the circles are equal.<br />
<br />
Consequently, the tetrahedron is inscribed in a cylinder <math>CYL</math> of radius <math>AE</math> with axis <math>E.</math><br />
<br />
2. Let <math>\varphi</math> be the angle between <math>\vec {AE}</math> and <math>\vec {DE_1}.</math><br />
<br />
<cmath>AD^2 = EE_1^2 + (AE^2 + DE_1^2 - 2 AE \cdot DE_1 \cdot\cos \varphi \implies \cos \varphi = -\frac{2}{3}</cmath> <cmath> \implies \cos 2\varphi = -\frac{1}{9}.</cmath> <br />
Let <math>I</math> be the center of the <math>ABD</math> face, <math>IE_1 = \frac{a}{10 \sqrt{3}} = - CE_0 \cos 2\varphi.</math><br />
<br />
Let point <math>F</math> be symmetrical to <math>C</math> with respect to <math>I, E_3</math> is symmetrical to <math>E_1</math> with respect to <math>E_2.</math><br />
<br />
<math>BF = a, IE_1 \perp EE_1 \implies F</math> is located in the plane containing <math>E_3</math> perpendicular to the straight line <math>EE_1 \implies FE_3 = \frac {IE_1}{\cos 2\varphi} = CE_0</math> that is, this point lies on <math>CYL</math> and the tetrahedron <math>ABFD = ABCD.</math><br />
<br />
Similarly, we will construct an arbitrarily long chain of equal tetrahedra inscribed in a cylinder whose median faces contain a sequence of points <math>E_i.</math><br />
<br />
It follows that the shortest path of the elephant cannot be shorter than <cmath>E_0E_3 = 4 EE_1 = \frac {4a}{\sqrt{10}}.</cmath><br />
There are some such ways inside the tetrahedron. One of them is shown in the diagram.<br />
<br />
'''vladimir.shelomovskii@gmail.com, vvsss'''</div>
Vvsss
https://artofproblemsolving.com/wiki/index.php/Planar_figures
Planar figures
2024-02-22T03:32:02Z
<p>Altonz118: /* Planar Figures */</p>
<hr />
<div>==Planar Figures==<br />
<br />
A plane figure is a flat shape. It does not have depth or thickness. Plane figures only have two dimensions (length and width). Thus, they are also known as <math>2D</math> figures or two-dimensional shapes. Some common examples of geometric plane figures are squares, rectangles, circles, triangles, etc.<br />
<br />
- iamcalifornia'sresident</div>
Altonz118
https://artofproblemsolving.com/wiki/index.php/SPINLAGI_:_Jackpot_Besar_Menunggumu_di_situs_Spinlagi_Situs_Slot
SPINLAGI : Jackpot Besar Menunggumu di situs Spinlagi Situs Slot
2024-02-19T20:33:42Z
<p>Marianasinta: awdawd</p>
<hr />
<div>adwd</div>
Marianasinta
https://artofproblemsolving.com/wiki/index.php/Slant_height
Slant height
2024-02-19T19:45:43Z
<p>Char0221: </p>
<hr />
<div><h3>Definition</h3><br />
The '''slant height''' of a cone is defined as the length of any segment connecting the vertex to the circumference of the base. In [[right cone]]s the slant height can be found via the [[Pythagorean theorem]].<br />
<br />
{{stub}}<br />
[[Category:Definition]]<br />
[[Category:Geometry]]</div>
Char0221