Difference between revisions of "Proofs without words"

m (Miscellaneous: finish S^n = R^n)
m (Summations)
 
(36 intermediate revisions by 21 users not shown)
Line 6: Line 6:
 
http://mathoverflow.net/questions/8846/proofs-without-words
 
http://mathoverflow.net/questions/8846/proofs-without-words
 
http://gurmeet.net/computer-science/mathematical-recreations-proofs-without-words/
 
http://gurmeet.net/computer-science/mathematical-recreations-proofs-without-words/
 
sum  (2k-1)(-1)^{n-k} = n
 
  
 
hockey stick?
 
hockey stick?
 
+
Catalan numbers / reflection principle
Catalan numbers
+
geometric series: http://mahalanobis.twoday.net/stories/3472911/
 +
Fibonacci: sum_{i=1}^{2n-1} F_iF_{i+1} = F_{2n}^2
 +
sum of squares, http://legacy.lclark.edu/~mathsci/invitation.pdf
 +
weighted sum of triangle numbers: http://legacy.lclark.edu/~mathsci/weightedsum.pdf
 +
pentagonal/octagonal numbers: http://legacy.lclark.edu/~mathsci/octnos.pdf
  
 
geometry:
 
geometry:
 
Pythagorean Theorem (multiple ways - 5?)
 
Pythagorean Theorem (multiple ways - 5?)
 
+
smallest triangle in angle (Engel 315 5)
smallest triangle in angle
 
 
 
 
equilateral triangle in three lines
 
equilateral triangle in three lines
 
 
Fermat point problem
 
Fermat point problem
 
 
Fagnano problem
 
Fagnano problem
 
 
regular heptagon identity?
 
regular heptagon identity?
 
 
Law of Cosines
 
Law of Cosines
 
 
Law of Sines
 
Law of Sines
 
 
radical axes/homotheties?
 
radical axes/homotheties?
  
 +
Engel 313 E12, 316 6, 11, 323 4, 326/7
 +
arbelos/salinon: http://legacy.lclark.edu/~mathsci/arbelos.pdf, http://legacy.lclark.edu/~mathsci/salinon.pdf
 +
abc/4R: http://legacy.lclark.edu/~mathsci/euler.pdf
 +
area of triangle xy: http://legacy.lclark.edu/~mathsci/area-rt-tri.pdf
  
 
misc.  
 
misc.  
 
3D QM-GM
 
3D QM-GM
 
+
AM-GM from http://jwilson.coe.uga.edu/emt725/AMGM/AMGM.5.html
Jensen's inequality (draw curve, multiple points)
 
 
 
 
Minkowski's inequality
 
Minkowski's inequality
 
 
rt(a^2 - ab + b^2) rt(b^2 - bc + c^2) < rt(a^2 + ac + c^2)
 
rt(a^2 - ab + b^2) rt(b^2 - bc + c^2) < rt(a^2 + ac + c^2)
 
integration by parts
 
 
Homeomorphism between S^1 - (1,0) to R
 
 
cardinality of interval with R
 
cardinality of interval with R
 
 
domino tiling, per mathoverflow
 
domino tiling, per mathoverflow
 
+
hexagon tiling with cube (aops logo)
 
-->
 
-->
 
== Summations ==
 
== Summations ==
Line 54: Line 44:
 
int n = 6; pair shiftR = ((n+2),0); real r = 0.3;
 
int n = 6; pair shiftR = ((n+2),0); real r = 0.3;
 
pen colors(int i){ return rgb(i/n,0.4+i/(2n),1-i/n); } /* shading */
 
pen colors(int i){ return rgb(i/n,0.4+i/(2n),1-i/n); } /* shading */
void htick(pair A, pair B,pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); }
+
void htick(pair A, pair B, pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); }
  
 
  /* triangle */
 
  /* triangle */
 
draw((-r,0)--(-r,-n+1)^^(r,-n+1)--(r,0),linetype("4 4"));
 
draw((-r,0)--(-r,-n+1)^^(r,-n+1)--(r,0),linetype("4 4"));
for(int i = 0; i < n; ++i)
+
for(int i = 0; i < n; ++i)  
 
  draw((-i,-i)--(i,-i));
 
  draw((-i,-i)--(i,-i));
 
for(int i = 0; i < n; ++i)
 
for(int i = 0; i < n; ++i)
Line 148: Line 138:
  
 
The sum of the first <math>n</math> positive integers is <math>{n+1 \choose 2}</math>.{{ref|1}}<br><br>
 
The sum of the first <math>n</math> positive integers is <math>{n+1 \choose 2}</math>.{{ref|1}}<br><br>
 +
</center>
 +
 +
<center><asy>unitsize(15); defaultpen(linewidth(0.7));
 +
int n = 5; pair shiftR = (2*n + 4, 1); real r = 0.35; pen sm = fontsize(10);
 +
pen colors(int i){ return rgb(i/n,0.4+i/(2n),1-i/n); } /* shading */
 +
void htick(pair A, pair B,pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); }
 +
 +
/* draw plus/minus circle at A */
 +
void plus  (pair A) { filldraw(CR(A,r),rgb(0.5,1,0.5)); MP("+",A,(0,0), sm); }
 +
void minus (pair A) { filldraw(CR(A,r),rgb(1,0.5,0.5)); MP("-",A,(0,0), sm); }
 +
 +
/* triangle 1 */
 +
path p1 = (-1,-1/2) -- (1,-1/2);
 +
draw((-n+1-r,-n+1-2*r)--(n-1+r,-n+1-2*r));
 +
for(int i = 0; i < n; ++i) {
 +
for(int j = 0; j < 2*i+1; ++j) {
 +
  if (i % 2 == 0)
 +
  plus((j-i,-i));
 +
  else
 +
  minus((j-i,-i));
 +
}
 +
if (i % 2 != 0)
 +
  p1 = (-i-1/2,-i-3/2)--(-i-1/2,-i+1/2)--p1--(i+1/2,-i+1/2)--(i+1/2,-i-3/2);
 +
plus((2*i - n + 1, -n + 1 - 4*r));
 +
}
 +
p1 = p1 -- cycle;
 +
draw(p1, rgb(0.85,0.85,0.85));
 +
 +
/* triangle 2 */
 +
n = n+1;
 +
path p2 = (-0.1,1/2) -- (0.1,1/2);
 +
draw(shiftR+ (-n+1-r,-n+1-2*r)-- shiftR+ (n-1+r,-n+1-2*r));
 +
for(int i = 0; i < n; ++i) {
 +
for(int j = 0; j < 2*i+1; ++j) {
 +
  if (i % 2 != 0)
 +
  plus(shiftR+ (j-i,-i));
 +
  else
 +
  minus(shiftR+ (j-i,-i));
 +
}
 +
if (i % 2 == 0)
 +
  p2 = (-i-1/2,-i-3/2)--(-i-1/2,-i+1/2)--p2--(i+1/2,-i+1/2)--(i+1/2,-i-3/2);
 +
plus(shiftR+ (2*i - n + 1, -n + 1 - 4*r));
 +
}
 +
p2 = p2 -- cycle;
 +
draw(shift(shiftR)*p2, rgb(0.85,0.85,0.85));
 +
htick(shiftR+(-n,r),shiftR+(-n,-n+1-r)); label("$n$",shiftR+(-n,(-n+1)/2),W,fontsize(10));
 +
</asy> <br>
 +
 +
The alternating sum of the first <math>n</math> odd natural numbers is <math>\sum_{k=1}^n (-1)^{n-k}(2k-1) = n</math>. ([http://www.math.hmc.edu/~benjamin/papers/OddPWW.pdf Source])<br><br>
 
</center>
 
</center>
  
Line 183: Line 222:
 
</asy><br>
 
</asy><br>
  
[[Nichomauss' Theorem]]: <math>n^3</math> can be written as the sum of <math>n</math> consecutive integers, and consequently that <math>1^3 + 2^3 + \cdots + n^3 = (1+2+\cdots + n)^2</math>. <br><br>
+
[[Nichomauss' Theorem]]: the sum of the first <math>n</math> cubes can be written as the square of the sum of the first <math>n</math> integers, a statement that can be written as <math>1^3+2^3+...+n^3=(1+2+...+n)^2</math>. <br><br>
 
Here, we use the same re-arrangement as the first proof on this page (the sum of first odd integers is a square). Here's another re-arrangement to see this:
 
Here, we use the same re-arrangement as the first proof on this page (the sum of first odd integers is a square). Here's another re-arrangement to see this:
 
<br><asy>defaultpen(linewidth(0.7)); unitsize(15); pen heavy = linewidth(2);  
 
<br><asy>defaultpen(linewidth(0.7)); unitsize(15); pen heavy = linewidth(2);  
Line 260: Line 299:
 
real h = 0.6;                            // scale factor of diagram
 
real h = 0.6;                            // scale factor of diagram
 
pair shiftR1 = (n*h+1,0),                // middle diagram shift offset
 
pair shiftR1 = (n*h+1,0),                // middle diagram shift offset
     shiftR2 = shiftR1 + (n*h+1,0);    // right diagram shift offset
+
     shiftR2 = shiftR1 + (n*h+1,-1);    // right diagram shift offset
  
 
int lvl(int i) { return ceil(((8*i+9)^.5-1)/2); } /* return level of square i */
 
int lvl(int i) { return ceil(((8*i+9)^.5-1)/2); } /* return level of square i */
Line 417: Line 456:
 
<center>[[#toc|Back to Top]]</center>
 
<center>[[#toc|Back to Top]]</center>
  
=== Geometric series ===
+
=== Geometric Series ===
 
<center><asy>defaultpen(linewidth(0.7)); unitsize(15);
 
<center><asy>defaultpen(linewidth(0.7)); unitsize(15);
 
int n = 10;  /* # of iterations */
 
int n = 10;  /* # of iterations */
Line 495: Line 534:
 
</asy><br>
 
</asy><br>
  
Another proof of the identity <math>\frac 14 + \frac {1}{4^2} + \frac {1}{4^3} + \cdots = \frac 13</math>. <br><br>
+
Another proof of the identity <math>\frac 14 + \frac {1}{4^2} + \frac {1}{4^3} + \frac {1}{4^4}+\cdots = \frac 13</math>. <br><br>
 
</center>
 
</center>
  
Line 550: Line 589:
 
}
 
}
 
</asy><br><br>
 
</asy><br><br>
The [[arithmetic-geometric series]] <math>\sum_{n=1}^{\infty} nr^n = \sum_{n=1}^{\infty} \sum_{i=n}^{\infty} r^i = \sum_{n=1}^{\infty} \frac{r^{-n}}{1-r} = \frac{r}{(1-r)^2}</math>, also known as Gabriel's staircase.{{ref|2}}<br><br></center>
+
The [[arithmetic-geometric series]] <math>\sum_{n=1}^{\infty} nr^n = \sum_{n=1}^{\infty} \sum_{i=n}^{\infty} r^i = \sum_{n=1}^{\infty} \frac{r^{n}}{1-r} = \frac{r}{(1-r)^2}</math>, also known as Gabriel's staircase.{{ref|2}}<br><br></center>
  
 
<center>[[#toc|Back to Top]]</center>
 
<center>[[#toc|Back to Top]]</center>
Line 808: Line 847:
 
In trapezoid <math>ABCD</math> with <math>\overline{AB} \parallel \overline{CD}</math>, then <math>\overline{AC} \perp \overline{BD} \Longleftrightarrow AC^2 + BD^2 = (AB + CD)^2</math>.<br><br></center>
 
In trapezoid <math>ABCD</math> with <math>\overline{AB} \parallel \overline{CD}</math>, then <math>\overline{AC} \perp \overline{BD} \Longleftrightarrow AC^2 + BD^2 = (AB + CD)^2</math>.<br><br></center>
  
 +
<center><asy>
 +
// feel free to change these four points!
 +
pair A = (0,0), B = (3, -2), C = (5,1), D = (1,3);
 +
 +
//
 +
// Rest of code
 +
//
 +
size(200); defaultpen(linewidth(0.9)); pen lightgreen = rgb(0.6,1,0.6), lightred = rgb(1,0.6,0.6), smdash = linewidth(0.7)+linetype("2 2");
 +
 +
pair E = IP(A--C,B--D), AB = (A+B)/2, BC = (B+C)/2, CD = (C+D)/2, DA = (D+A)/2, ABE = 2*AB-E, BCE = 2*BC-E, CDE = 2*CD-E, DAE = 2*DA-E;
 +
path midpts = AB--BC--CD--DA--cycle;
 +
filldraw(shift(-E)*scale(2)*midpts,lightgreen); filldraw(midpts,lightred);
 +
draw(A--B--C--D--cycle); draw(A--C); draw(B--D); draw((A+ABE)/2--(C+BCE)/2,smdash); draw((B+ABE)/2--(D+DAE)/2,smdash); draw((B+BCE)/2--(D+CDE)/2,smdash); draw((A+DAE)/2--(C+CDE)/2,smdash);
 +
 +
dot(A); dot(B); dot(C); dot(D); label("$A$",A,W);label("$B$",B,S);label("$C$",C,E);label("$D$",D,N);
 +
</asy><br>
 +
 +
[[Varignon's theorem]]: the area of the outer parallelogram is twice the area of the quadrilateral and four times the area of the midpoint parallelogram, so the midpoint parallelogram of a (convex) quadrilateral has area <math>1/2</math> of the quadrilateral.
 +
</center>
 +
 +
Proof for Volume of a Cone: http://www.mathematische-basteleien.de/wuerfel16.gif
 
<center>[[#toc|Back to Top]]</center>
 
<center>[[#toc|Back to Top]]</center>
  
Line 827: Line 887:
 
draw((0,1)--(pi/2,1),linewidth(1));
 
draw((0,1)--(pi/2,1),linewidth(1));
 
</asy>
 
</asy>
<math>\int_0^{\pi/2} \sin^2 x = \int_0^{\pi/2} \cos^2 x = \frac {\pi}{4}</math> from <math>{sin2x+cos2x=1sinx=cos(π/2x)</math>. ([http://www.kkuniyuk.com/M15005Proof.pdf Source])
+
<math>\int_0^{\pi/2} \sin^2 x \, dx = \int_0^{\pi/2} \cos^2 x \, dx = \frac {\pi}{4}</math> from <math>{sin2x+cos2x=1sinx=cos(π/2x)</math>. ([http://www.kkuniyuk.com/M15005Proof.pdf Source])
 
<br><br></center>
 
<br><br></center>
  
 +
<center><asy>
 +
import graph; size(170); defaultpen(linewidth(0.7)); Label k; k.p=fontsize(8);
 +
 +
real xmax = 1.35, xmin = -0.35, ymax = 1.35, ymin = -0.35, lblpt = 1/2 + 0.15, alpha = 2, epsilon = 0.015;
 +
 +
real f(real x) { return x^alpha; }
 +
 +
xaxis(xmin,xmax,Ticks(k, 0.25, NoZero),Arrows(6)); yaxis(ymin,ymax,Ticks(k, 0.25, NoZero),Arrows(6));
 +
filldraw(graph(f,0,1)--(1,0)--(0,0)--cycle,gray(0.7),linewidth(1));
 +
draw(graph(f,1,1.1),linewidth(0.7),EndArrow(4));
 +
draw((lblpt-epsilon,f(lblpt))--(0+epsilon,f(lblpt)),Arrows(4)); draw((lblpt,f(lblpt)-epsilon)--(lblpt,0+epsilon),Arrows(4));
 +
label("$x = y^{1/\alpha}$",(lblpt/2,f(lblpt)),N,fontsize(10));
 +
label("$y = x^{\alpha}$",(lblpt,f(lblpt)/2),E,fontsize(10));
 +
draw((0,1)--(1,1),linewidth(1));
 +
</asy>
 +
<math>\alpha > 0 \Longrightarrow \int_0^{1} \left(x^\alpha + x^{1/\alpha}\right) \, dx = 1.</math> ([http://www.math.ubc.ca/~morey/talk/proofwowords.html Source])
 +
<br><br></center>
  
 
<center><asy>unitsize(15); defaultpen(linewidth(0.7));
 
<center><asy>unitsize(15); defaultpen(linewidth(0.7));
Line 945: Line 1,022:
 
pen gridpen = dotted;
 
pen gridpen = dotted;
 
int gridmin = -2, gridmax = 2, gridlines = 6;
 
int gridmin = -2, gridmax = 2, gridlines = 6;
pair projections[] = {(0.5,-1.2),(-0.5,-1.5),(-1.5,-0.5),(1.2,-0.25)};
+
pair projections[] = {(1,-1),(1,-2),(2,-2),(2,-1)};
 +
triple IPs[] = new triple[4];
 +
draw((1,-1,-1)--(1,-2,-1)--(2,-2,-1)--(2,-1,-1)--cycle);
  
 
for(int i = 0; i < gridlines; ++i) {
 
for(int i = 0; i < gridlines; ++i) {
Line 964: Line 1,043:
 
  dot(IP, yellow);
 
  dot(IP, yellow);
 
  draw(circle(IP, 0.06, (0,1,0)));
 
  draw(circle(IP, 0.06, (0,1,0)));
 +
IPs[i] = IP;
 
  dot(OP, red);
 
  dot(OP, red);
 
  draw(circle(OP, 0.06, (0,1,0)));
 
  draw(circle(OP, 0.06, (0,1,0)));
 +
if(i != 0)
 +
  draw(IPs[i] -- IPs[i-1], dotted);
 
}
 
}
 +
draw(IPs[0]--IPs[3], dotted);
  
 
dot((0,0,1),green);
 
dot((0,0,1),green);
Line 974: Line 1,057:
 
<br><br></center>
 
<br><br></center>
  
 +
Sum of arctangents formula:
 +
 +
<asy>
 +
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
 +
import graph; usepackage("amsmath"); size(13cm);
 +
real labelscalefactor = 0.5; /* changes label-to-point distance */
 +
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
 +
pen dotstyle = black; /* point style */
 +
real xmin = -0.4717093412177357, xmax = 7.405441345585962, ymin = -1.1854534297865673, ymax = 7.342957746870971;  /* image dimensions */
 +
pen uququq = rgb(0.25098039215686274,0.25098039215686274,0.25098039215686274); pen aqaqaq = rgb(0.6274509803921569,0.6274509803921569,0.6274509803921569); pen qqwuqq = rgb(0.,0.39215686274509803,0.); pen cqcqcq = rgb(0.7529411764705882,0.7529411764705882,0.7529411764705882);
  
 +
draw((0.,0.)--(0.,1.)--(1.,1.)--cycle);
 +
draw((1.,1.)--(0.,3.)--(0.,1.)--cycle, uququq);
 +
draw((0.,3.)--(6.,6.)--(1.,1.)--cycle, aqaqaq);
 +
draw(arc((1.,1.),0.3101240427875472,180.,225.)--(1.,1.)--cycle, qqwuqq);
 +
draw(arc((1.,1.),0.3101240427875472,116.56505117707799,180.)--(1.,1.)--cycle, qqwuqq);
 +
draw(arc((1.,1.),0.3101240427875472,45.,116.56505117707799)--(1.,1.)--cycle, qqwuqq);
 +
/* draw grid of horizontal/vertical lines */
 +
pen gridstyle = linewidth(0.7) + cqcqcq; real gridx = 1., gridy = 1.; /* grid intervals */
 +
for(real i = ceil(xmin/gridx)*gridx; i <= floor(xmax/gridx)*gridx; i += gridx)
 +
draw((i,ymin)--(i,ymax), gridstyle);
 +
for(real i = ceil(ymin/gridy)*gridy; i <= floor(ymax/gridy)*gridy; i += gridy)
 +
draw((xmin,i)--(xmax,i), gridstyle);
 +
/* end grid */
 +
 +
/* draw figures */
 +
draw((0.,0.)--(0.,1.));
 +
draw((0.,1.)--(1.,1.));
 +
draw((1.,1.)--(0.,0.));
 +
draw((1.,1.)--(0.,3.), uququq);
 +
draw((0.,3.)--(0.,1.), uququq);
 +
draw((0.,1.)--(1.,1.), uququq);
 +
draw((0.,3.)--(6.,6.), aqaqaq);
 +
draw((6.,6.)--(1.,1.), aqaqaq);
 +
draw((1.,1.)--(0.,3.), aqaqaq);
 +
label("$\arctan 1 + \arctan 2 + \arctan 3 = \pi$",(1.544096936901321,0.5925910821953678),SE*labelscalefactor,fontsize(10));
 +
/* dots and labels */
 +
dot((0.,0.),linewidth(3.pt) + dotstyle);
 +
dot((0.,1.),dotstyle);
 +
dot((1.,1.),dotstyle);
 +
dot((0.,3.),dotstyle);
 +
dot((6.,6.),dotstyle);
 +
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
 +
/* end of picture */
 +
</asy>
  
 
<center>[[#toc|Back to Top]]</center>
 
<center>[[#toc|Back to Top]]</center>

Latest revision as of 15:07, 16 September 2024

The following demonstrate proofs of various identities and theorems using pictures, inspired from this gallery.

Summations

[asy]unitsize(15); defaultpen(linewidth(0.7)); int n = 6; pair shiftR = ((n+2),0); real r = 0.3; pen colors(int i){ return rgb(i/n,0.4+i/(2n),1-i/n); } /* shading */ void htick(pair A, pair B, pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); }   /* triangle */ draw((-r,0)--(-r,-n+1)^^(r,-n+1)--(r,0),linetype("4 4")); for(int i = 0; i < n; ++i)   draw((-i,-i)--(i,-i)); for(int i = 0; i < n; ++i)  for(int j = 0; j < 2*i+1; ++j)   filldraw(CR((j-i,-i),r),colors(i));   /* square */ draw(r*expi(pi/4)+shiftR--(n-1,-n+1)+r*expi(pi/4)+shiftR^^r*expi(5*pi/4)+shiftR--r*expi(5*pi/4)+(n-1,-n+1)+shiftR,linetype("4 4")); for(int i = 0; i < n; ++i)  draw(shiftR+(0,-i)--shiftR+(i,-i)--shiftR+(i,0)); for(int i = 0; i < n; ++i)  for(int j = 0; j < n; ++j)   filldraw(CR((j,-i)+shiftR,r),colors((i>j)?i:j));  htick(shiftR+(-1,r),shiftR+(-1,-n+1-r)); label("$n$",shiftR+(-1,(-n+1)/2),W,fontsize(10)); [/asy]

The sum of the first $n$ odd natural numbers is $n^2$.

[asy] defaultpen(linewidth(0.7)); unitsize(15); int n = 6; pair shiftR = ((n+2),0); real r = 0.3; pen colors(int i){ return rgb(0.4+i/(2n),i/n,1-i/n); } /* shading */ void htick(pair A, pair B,pair ticklength = (0.15,0)){  draw(A--B);  draw(A-ticklength--A+ticklength);  draw(B-ticklength--B+ticklength); }   /* triangle */ draw((0.5,0)--(n-0.5,-n+1),linetype("4 4")); for(int i = 0; i < n; ++i)  draw((0,-i)--(i,-i)); for(int i = 0; i < n; ++i)  for(int j = 0; j <= i; ++j)   filldraw(CR((j,-i),r),colors(i));    /* arc arrow */ draw( arc((n,-n+1)/2, (1.5,-1.5), (n-1.5,-1.5), CW) ); fill((n-1.5,-1.5) -- (n-1.5,-1.5)+r*expi(5.2*pi/6) -- (n-1.5,-1.5)+r*expi(3.3*pi/6) -- cycle); /* manual arrowhead? avoid resizing */   /* square */ draw(shiftR+(0.5,0)--shiftR+(n-0.5,-n+1),linetype("4 4")); for(int i = 0; i < n; ++i)  draw(shiftR+(0,-i)--shiftR+(i,-i)^^shiftR+(n,-n+1)-(0,-i)--shiftR+(n,-n+1)-(i,-i)); for(int i = 0; i < n; ++i)  for(int j = 0; j < n+1; ++j)   filldraw(CR((j,-i)+shiftR,r),colors((j <= i) ? i : n-1-i));   /* labeling and ticks */ htick(shiftR+(-1,r),shiftR+(-1,-n+1-r)); label("$n$",shiftR+(-1,(-n+1)/2),W,fontsize(10)); htick(shiftR+(-r,-n),shiftR+(n+r-1,-n),(0,0.15)); label("$n$",shiftR+((n-1)/2,-n),S,fontsize(10)); htick(shiftR+(n-r,-n),shiftR+(n+r,-n),(0,0.15)); label("$1$",shiftR+(n,-n),S,fontsize(10)); [/asy]

The sum of the first $n$ positive integers is $n(n+1)/2$.

[asy]unitsize(15); defaultpen(linewidth(0.7)); int n = 6; real r = 0.35, h = 3/4; /* radius size and horizontal spacing */ pair shiftR = (h*(n+1)+r, 0);  pen colors(int i){  /* shading */  if(i == n) return red;   return rgb(5/n,0.4+5/(2n),1-5/n);  }  void htick(pair A, pair B, pair ticklength = (0.15,0)){   draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength);  } void makeshiftarrow(pair A, real dir, real arrowlength = r){  /* Arrow option resizes */   fill(A--A+arrowlength*expi(dir+pi/8)--A+arrowlength*expi(dir-pi/8)--cycle); } pair getCenter(int i, int j){ return ((2*j-i)*h,-i);}   /* triangle */ for(int i = 0; i < n+1; ++i){  draw((-i*h,-i)--(i*h,-i));   /* horizontal lining */  for(int j = 0; j <= i; ++j)   filldraw(circle(getCenter(i,j),r), colors(i)); }   /* fill in circle in row 4, column 3 */ filldraw(circle(getCenter(3,2),r),blue); draw(getCenter(n,2)-- getCenter(3,2)-- getCenter(n,n+2-3));  makeshiftarrow(getCenter(n,2),pi/4,0.5); makeshiftarrow(getCenter(n,n+2-3),3*pi/4,0.5);   htick(shiftR+(-1,r),shiftR+(-1,-n+1-r)); label("$n$",shiftR+(-1,(-n+1)/2),E,fontsize(10)); [/asy]

The sum of the first $n$ positive integers is ${n+1 \choose 2}$.[1]

[asy]unitsize(15); defaultpen(linewidth(0.7)); int n = 5; pair shiftR = (2*n + 4, 1); real r = 0.35; pen sm = fontsize(10); pen colors(int i){ return rgb(i/n,0.4+i/(2n),1-i/n); } /* shading */ void htick(pair A, pair B,pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); }  /* draw plus/minus circle at A */ void plus  (pair A) { filldraw(CR(A,r),rgb(0.5,1,0.5)); MP("+",A,(0,0), sm); } void minus (pair A) { filldraw(CR(A,r),rgb(1,0.5,0.5)); MP("-",A,(0,0), sm); }   /* triangle 1 */ path p1 = (-1,-1/2) -- (1,-1/2); draw((-n+1-r,-n+1-2*r)--(n-1+r,-n+1-2*r)); for(int i = 0; i < n; ++i) {  for(int j = 0; j < 2*i+1; ++j) {   if (i % 2 == 0)     plus((j-i,-i));   else     minus((j-i,-i));  }  if (i % 2 != 0)    p1 = (-i-1/2,-i-3/2)--(-i-1/2,-i+1/2)--p1--(i+1/2,-i+1/2)--(i+1/2,-i-3/2);  plus((2*i - n + 1, -n + 1 - 4*r)); } p1 = p1 -- cycle; draw(p1, rgb(0.85,0.85,0.85));   /* triangle 2 */ n = n+1; path p2 = (-0.1,1/2) -- (0.1,1/2); draw(shiftR+ (-n+1-r,-n+1-2*r)-- shiftR+ (n-1+r,-n+1-2*r)); for(int i = 0; i < n; ++i) {  for(int j = 0; j < 2*i+1; ++j) {   if (i % 2 != 0)     plus(shiftR+ (j-i,-i));   else     minus(shiftR+ (j-i,-i));  }  if (i % 2 == 0)    p2 = (-i-1/2,-i-3/2)--(-i-1/2,-i+1/2)--p2--(i+1/2,-i+1/2)--(i+1/2,-i-3/2);  plus(shiftR+ (2*i - n + 1, -n + 1 - 4*r)); } p2 = p2 -- cycle; draw(shift(shiftR)*p2, rgb(0.85,0.85,0.85)); htick(shiftR+(-n,r),shiftR+(-n,-n+1-r)); label("$n$",shiftR+(-n,(-n+1)/2),W,fontsize(10)); [/asy]

The alternating sum of the first $n$ odd natural numbers is $\sum_{k=1}^n (-1)^{n-k}(2k-1) = n$. (Source)

[asy]defaultpen(linewidth(0.7)); unitsize(15); pen heavy = linewidth(2);  int n2 = 4, n = floor(n2*(n2+1)/2); real h = 0.6; pair shiftR1 = (n*h+1,0), shiftR2 = shiftR1 + (n*h+1,0); /* global configurable variables */  int lvl(int i){ return ceil(((8*i+9)^.5-1)/2); } pen colors(int i){ return rgb(0.5-lvl(i)/5,0.3+lvl(i)/7,1-lvl(i)/6); } /* shading */  void htick(pair A, pair B,pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); }    /* gradient triangle */ for(int i = 0; i < n; ++i){  for(int j = 0; j < 2*i+1; ++j){   filldraw(shift(shiftR1)*xscale(h)*yscale(h)*shift((j-i,-i))*unitsquare,colors(i));    if(j % lvl(i) == 0 && j != lvl(i)^2)    draw(shift(shiftR1)*xscale(h)*yscale(h)*shift((j-i,-i))*((0,0)--(0,1)--(1,1)), heavy);   if(j == 2*i)  /* right border */    draw(shift(shiftR1)*xscale(h)*yscale(h)*shift((j-i,-i))*((1,0)--(1,1)--(0,1)), heavy);  } }   for(int i = 0; i < n2; ++i)   draw(shift(shiftR1)*xscale(h)*yscale(h)*shift((-i*(i+1)/2,-i*(i+1)/2))*((0,1)--(2*i*(i+1)/2+1,1)), heavy); draw(shift(shiftR1)*xscale(h)*yscale(h)*shift((-n2*(n2+1)/2,-n2*(n2+1)/2))*((1,1)--(2*n2*(n2+1)/2,1)), heavy);   /* gradient square */ for(int i = 0; i < n; ++i)  for(int j = 0; j < n; ++j)   filldraw(shift(shiftR2)*xscale(h)*yscale(h)*shift((j,-i))*unitsquare,colors((i>j)?i:j));   /* n nxn squares */ for(int i = 0; i < n2; ++i){  filldraw(xscale(h)*yscale(h)*shift((-i,-(i+1)*(i+2)/2+1))*xscale(i+1)*yscale(i+1)*unitsquare, colors(floor(i*(i+1)/2)), heavy);   } [/asy]

Nichomauss' Theorem: the sum of the first $n$ cubes can be written as the square of the sum of the first $n$ integers, a statement that can be written as $1^3+2^3+...+n^3=(1+2+...+n)^2$.

Here, we use the same re-arrangement as the first proof on this page (the sum of first odd integers is a square). Here's another re-arrangement to see this:
[asy]defaultpen(linewidth(0.7)); unitsize(15); pen heavy = linewidth(2);   /* global configurable variables */ int n2 = 4, n = floor(n2*(n2+1)/2);       // number of colors, number of layers real h = 0.6;                             // scale factor of diagram pair shiftR1 = (n*h+1,0),                 // middle diagram shift offset        shiftR2 = shiftR1 + (n*h+1,0);     // right diagram shift offset  int lvl(int i) { return ceil(((8*i+9)^.5-1)/2); } /* return level of square i */  pen colors(int i) { return rgb(0.5-lvl(i)/5,0.3+lvl(i)/7,1-lvl(i)/6); } /* shading */   /* draw tick line with label, segment between A and B */ void htick(pair A, pair B,pair ticklength = (0.15,0)) { draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); }    /* gradient triangle */ for(int i = 0; i < n; ++i){  for(int j = 0; j < 2*i+1; ++j){   filldraw(shift(shiftR1)*scale(h)*shift((j-i,-i))*unitsquare,colors(i));    /*   if(j % lvl(i) == 0 && j != lvl(i)^2)    draw(shift(shiftR1)*scale(h)*shift((j-i,-i))*((0,0)--(0,1)--(1,1)), heavy);   if(j == 2*i)  // right border    draw(shift(shiftR1)*scale(h)*shift((j-i,-i))*((1,0)--(1,1)--(0,1)), heavy);   */  }  draw(shift(shiftR1)*scale(h)*shift((-i,-i))*((0,0)--(0,1)--(1,1)), heavy);  draw(shift(shiftR1)*scale(h)*shift(( i,-i))*((1,0)--(1,1)--(0,1)), heavy); }   // return kth triangular number (actually, 1+2+...+k)     int tri(int k) { return ((int) (k*(k+1)/2)); }     for(int i = 0; i < n2; ++i) {  draw(shift(shiftR1)*scale(h)*shift((0-tri(i),0-tri(i)))*((0,1)--(2*tri(i)+1,1)), heavy);  if(i % 2 == 0) {   // vertical heavy lines for odd layers   draw(shift(shiftR1)*scale(h)*((-i/2,1-tri(i))--(-i/2,-i-tri(i))), heavy);   draw(shift(shiftR1)*scale(h)*((1+i/2,1-tri(i))--(1+i/2,-i-tri(i))), heavy);     }  else {   // jagged heavy lines for even layers   pair jag1 = (-(i+1)/2,-(i-1)/2-tri(i)), jag2 = (1+(i+1)/2,-(i-1)/2-tri(i));   draw(shift(shiftR1)*scale(h)*(jag1+(0,1+(i-1)/2) -- jag1 -- jag1+(1,0) -- jag1+( 1,-(i+1)/2)), heavy);   draw(shift(shiftR1)*scale(h)*(jag2+(0,1+(i-1)/2) -- jag2 -- jag2-(1,0) -- jag2+(-1,-(i+1)/2)), heavy);       }  }  draw(shift(shiftR1)*scale(h)*shift((-n2*(n2+1)/2,-n2*(n2+1)/2))*((1,1)--(2*n2*(n2+1)/2,1)), heavy);   /* gradient square */ for(int i = 0; i < n; ++i)   for(int j = 0; j < n; ++j)   filldraw(shift(shiftR2)*scale(h)*shift((j,-i))*unitsquare, colors((i>j)?i:j)); for(int i = 0; i < n2; ++i) {  draw(shift(shiftR2)*scale(h)*((0,1-tri(i))--(tri(i),1-tri(i))--(tri(i),1)),heavy);  draw(shift(shiftR2)*scale(h)*shift(tri(i),-i-tri(i))*scale(i+1)*unitsquare,heavy); } /* outside boundary */ draw(shift(shiftR2)*scale(h)*shift((0,1-n))*scale(n)*unitsquare, heavy);    /* n nxn squares  for(int i = 0; i < n2; ++i){  filldraw(scale(h)*shift((-i,-(i+1)*(i+2)/2+1))*xscale(i+1)*yscale(i+1)*unitsquare, colors(floor(i*(i+1)/2)), heavy);   } */ [/asy]

This also suggests the following alternative proof:

[asy]defaultpen(linewidth(0.7)); unitsize(15); pen heavy = linewidth(2);   /* global configurable variables */ int n2 = 4, n = floor(n2*(n2+1)/2);       // number of colors, number of layers real h = 0.6;                             // scale factor of diagram pair shiftR1 = (n*h+1,0),                 // middle diagram shift offset      shiftR2 = shiftR1 + (n*h+1,-1);     // right diagram shift offset  int lvl(int i) { return ceil(((8*i+9)^.5-1)/2); } /* return level of square i */  pen colors(int i) { return rgb(0.5-lvl(i)/5,0.3+lvl(i)/7,1-lvl(i)/6); } /* shading */   /* draw tick line with label, segment between A and B */ void htick(pair A, pair B,pair ticklength = (0.15,0)) { draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); }    /* gradient triangle */  /* for(int i = 0; i < n; ++i){  for(int j = 0; j < 2*i+1; ++j){   filldraw(shift(shiftR1)*scale(h)*shift((j-i,-i))*unitsquare,colors(i));       if(j % lvl(i) == 0 && j != lvl(i)^2)    draw(shift(shiftR1)*scale(h)*shift((j-i,-i))*((0,0)--(0,1)--(1,1)), heavy);   if(j == 2*i)  // right border    draw(shift(shiftR1)*scale(h)*shift((j-i,-i))*((1,0)--(1,1)--(0,1)), heavy);     }  draw(shift(shiftR1)*scale(h)*shift((-i,-i))*((0,0)--(0,1)--(1,1)), heavy);  draw(shift(shiftR1)*scale(h)*shift(( i,-i))*((1,0)--(1,1)--(0,1)), heavy); }   */  // return kth triangular number (actually, 1+2+...+k)     int tri(int k) { return ((int) (k*(k+1)/2)); }     /* for(int i = 0; i < n2; ++i) {  draw(shift(shiftR1)*scale(h)*shift((0-tri(i),0-tri(i)))*((0,1)--(2*tri(i)+1,1)), heavy);  if(i % 2 == 0) {   // vertical heavy lines for odd layers   draw(shift(shiftR1)*scale(h)*((-i/2,1-tri(i))--(-i/2,-i-tri(i))), heavy);   draw(shift(shiftR1)*scale(h)*((1+i/2,1-tri(i))--(1+i/2,-i-tri(i))), heavy);     }  else {   // jagged heavy lines for even layers   pair jag1 = (-(i+1)/2,-(i-1)/2-tri(i)), jag2 = (1+(i+1)/2,-(i-1)/2-tri(i));   draw(shift(shiftR1)*scale(h)*(jag1+(0,1+(i-1)/2) -- jag1 -- jag1+(1,0) -- jag1+( 1,-(i+1)/2)), heavy);   draw(shift(shiftR1)*scale(h)*(jag2+(0,1+(i-1)/2) -- jag2 -- jag2-(1,0) -- jag2+(-1,-(i+1)/2)), heavy);       }  }  draw(shift(shiftR1)*scale(h)*shift((-n2*(n2+1)/2,-n2*(n2+1)/2))*((1,1)--(2*n2*(n2+1)/2,1)), heavy); */   /* gradient square */ for(int i = 0; i < n; ++i)   for(int j = 0; j < n; ++j)   filldraw(shift(shiftR2)*scale(h)*shift((j,-i))*unitsquare, colors((i>j)?i:j)); /* internal heavy  lines */ for(int i = 0; i < n2; ++i) {  draw(shift(shiftR2)*scale(h)*((0,1-tri(i))--(tri(i),1-tri(i))--(tri(i),1)),heavy);  draw(shift(shiftR2)*scale(h)*shift(tri(i),-i-tri(i))*scale(i+1)*unitsquare,heavy); }  // cheating, sorry: draw line for i = 4 draw(shift(shiftR2)*scale(h)*((2,-5)--(2,-9)),heavy); draw(shift(shiftR2)*scale(h)*((6,-1)--(10,-1)),heavy);   /* outside heavy boundary */ draw(shift(shiftR2)*scale(h)*shift((0,1-n))*scale(n)*unitsquare, heavy);  /* construct grid for a m x n rectangle */ void drawGrid(pair pos, int m, int n) {  for(int i = 0; i < m; ++i)   for(int j = 0; j < n; ++j)    draw(scale(h)*shift(pos+(i,j))*unitsquare); }   /* n nxn squares */ for(int i = 0; i < n2; ++i) {  label("$"+(string) (i+1) +"^3$", scale(h)*shift((0,1-tri(i+1)-i))*(0,(1+i)/2), W, fontsize(10));  for(int j = 0; j <= i; ++j) {   if(i % 2 == 0 || i != j) {    filldraw(scale(h)*shift((j*(i+2),1-tri(i+1)-i))*xscale(i+1)*yscale(i+1)*unitsquare, colors(floor(i*(i+1)/2)), heavy);      drawGrid((j*(i+2),1-tri(i+1)-i), i+1, i+1);   }   else {     // half-blocks     filldraw(scale(h)*shift((j*(i+2),1-tri(i+1)-i))*xscale((i+1)/2)*yscale(i+1)*unitsquare, colors(floor(i*(i+1)/2)), heavy);      drawGrid((j*(i+2),1-tri(i+1)-i), (int) ((i+1)/2), i+1);    filldraw(scale(h)*shift((j*(i+2)+(i+1)/2+1/2,1-tri(i+1)-i))*xscale((i+1)/2)*yscale(i+1)*unitsquare, colors(floor(i*(i+1)/2)), heavy);      drawGrid((j*(i+2)+(i+1)/2+1/2,1-tri(i+1)-i), (int) ((i+1)/2), i+1);   }  } }  [/asy]
An animated version of this proof can be found in this gallery.


[asy] // To change the value of n shown, edit the line "int n = 5;" to whichever desired value of n. // To edit the size of the diagram, change the line unitsize(15); to the desired size.  unitsize(15); defaultpen(linewidth(0.7));  int n = 5;    // nth pentagonal number real r = 0.2; // dot radius pen p0 = red, p1 = rgb(1,1,0.5), p2 = rgb(0.5,1,0.5), p3 = rgb(0.5,0.5,1);  // dot color  // return the coordinate of the ith point of a regular pentagon with radius s // 0 <= i <= 4, and pentagonalPt(0,s) = (0,0)  pair pentagonalPt(int i, real s) {  return s * (dir(i*72+216+18) - dir(216+18)); }  // draw triangles  filldraw(pentagonalPt(4,1)--pentagonalPt(4,n-1)--pentagonalPt(3,n-1)+(pentagonalPt(4,n-1)-pentagonalPt(3,n-1))/(n-1)--cycle, p1, linewidth(1)); filldraw(pentagonalPt(3,1)--pentagonalPt(3,n-1)--pentagonalPt(2,n-1)+(pentagonalPt(3,n-1)-pentagonalPt(2,n-1))/(n-1)--cycle, p2, linewidth(1)); filldraw(pentagonalPt(2,1)--pentagonalPt(2,n-1)--pentagonalPt(1,n-1)+(pentagonalPt(2,n-1)-pentagonalPt(1,n-1))/(n-1)--cycle, p3, linewidth(1));  draw(pentagonalPt(1, n-1)--(0,0)--pentagonalPt(4, n-1)); for(int i = 1; i < n; ++i) {  for(int k = 2; k <= 4; ++k) {   draw(pentagonalPt(k, i) -- pentagonalPt(k-1, i), linetype("2 2"));   for(int j = 0; j <= i; ++j) {    filldraw(circle(pentagonalPt(k, i) + j*(pentagonalPt(k-1, i)-pentagonalPt(k, i))/i, r), p0);   }  } } filldraw(circle((0,0),r), p0); [/asy]

The $n$th pentagonal number is the sum of $n$ and three times the $n-1$th triangular number.
If $P_n$ denotes the $n$th pentagonal number, then $P_n = 3T_{n-1}+n$.



[asy]defaultpen(linewidth(0.7)); unitsize(15); pen sm = fontsize(10); int n = 5, fib = 1, fib2 = 1, xsum = 1, ysum = 0; real h = 0.15; void fillsq(pair A = (0,0), real s, pen p = invisible, pen l = linewidth(1)){ filldraw(shift(A)*xscale(s)*yscale(s)*unitsquare, p, l); } void htick(pair A, pair B, pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); }  for(int i = 0; i < n; ++i) {  fillsq((0,h*ysum),h*fib2,rgb(0.9,1,0.9));  fillsq((h*xsum,0),h*fib,rgb(1,0.9,0.9));  if(i == n-1){   label("$F_{n}^2$",h*(xsum+fib/2,fib/2),sm);   label("$F_{n-1}^2$",h*(fib2/2,ysum+fib2/2),sm);  }  else if(i == n-2){   label("$F_{n-2}^2$",h*(xsum+fib/2,fib/2),sm);   label("$F_{n-3}^2$",h*(fib2/2,ysum+fib2/2),sm);   }  fib = fib + fib2; fib2 = fib - fib2;  xsum = fib;  ysum = fib2;  fib = fib + fib2; fib2 = fib - fib2; } htick(h*(xsum,0)+(1,0),h*(xsum,ysum)+(1,0)); label("$F_n$",h*(xsum,ysum/2)+(1,0), E, sm); htick(h*(0,ysum)+(0,1),h*(xsum-fib+fib2,ysum)+(0,1),(0,0.15)); label("$F_{n-1}$",h*((xsum-fib+fib2)/2,ysum)+(0,1), N, sm); htick(h*(xsum,ysum)+(0,1),h*(xsum-fib+fib2,ysum)+(0,1),(0,0.15)); label("$F_{n}$",h*((2*xsum-fib+fib2)/2,ysum)+(0,1), N, sm); [/asy]
The identity $F_1^2 + F_2^2 + \cdots + F_n^2 = F_{n} \cdot F_{n+1}$, where $F_i$ is the $i$th Fibonacci number.

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Geometric Series

[asy]defaultpen(linewidth(0.7)); unitsize(15); int n = 10;  /* # of iterations */ real s = 6; /* square size */ pair shiftR = (s+2,0); pen sm = fontsize(10); void fillrect(pair A, pair B = (0,0), pen p = invisible, pen l = linewidth(1)){ filldraw(A--(A.x,B.y)--B--(B.x,A.y)--cycle, p, l); } void htick(pair A, pair B, pair ticklength = (0,0.15)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); }  for(int i = 0; i < 2; ++i) /* left */  fillrect((s/2^(ceil(i/2)),s/2^(floor(i/2)))); for(int i = 0; i < n; ++i) /* right */  fillrect(shiftR,shiftR + (s/2^(ceil(i/2)),s/2^(floor(i/2)))); label("$\frac 12$",(s*3/4,s/2),sm); label("$\cdots$",(s*1/4,s/2),sm);  label("$\frac 12$",shiftR+(s*3/4,s/2),sm); label("$\cdots$",shiftR+(s*1/4,s/2),sm);  label("$\frac 14$",shiftR+(s*1/4,s*3/4),sm); label("$\frac 18$",shiftR+(s*3/8,s/4),sm);  htick((0,-1), (s,-1)); htick(shiftR + (0,-1), shiftR + (s,-1)); label("$1$",(s/2,-1),S,sm); label("$1$",shiftR+(s/2,-1),S,sm); [/asy]

The infinite geometric series $\frac 12 + \frac {1}{2^2} + \frac {1}{2^3} + \cdots = 1$.

[asy] defaultpen(linewidth(0.7)); unitsize(15); int n = 4; real h = 2; pen colors[] = {rgb(0.8,0,0),rgb(0,0.8,0)}; void drawTriGrid(real s){  for(int i = 0; i < 4; ++i){   draw( (-s*3/2,s*(3/2 - i)) -- (s*3/2,s*(3/2 - i)), linetype("2 2"));   draw( (s*(3/2 - i),-s*3/2) -- (s*(3/2 - i),s*3/2), linetype("2 2"));   } } void fillrect(pair A, pair B, pen p){ filldraw(A--(A.x,B.y)--B--(B.x,A.y)--cycle, p, linewidth(1)); }  for(int i = 0; i < n; ++i) {  fillrect( ((-1)^i*-h/3^i*(3/2),-h/3^i*(3/2)) , ((-1)^i*-h/3^i*(1/2),h/3^i*(3/2)) , colors[0]);  fillrect(-((-1)^i*-h/3^i*(3/2),-h/3^i*(3/2)) ,-((-1)^i*-h/3^i*(1/2),h/3^i*(3/2)) , colors[1]);  fillrect( (-h/3^i*(1/2),(-1)^i*h/3^i*(1/2)) , (h/3^i*(1/2),(-1)^i*h/3^i*(3/2)), colors[0]);  fillrect(-(-h/3^i*(1/2),(-1)^i*h/3^i*(1/2)) ,-(h/3^i*(1/2),(-1)^i*h/3^i*(3/2)), colors[1]);  drawTriGrid(h/3^i); } [/asy]

The infinite geometric series $\frac 13 + \frac {1}{3^2} + \frac {1}{3^3} + \cdots = \frac 12$.

[asy] defaultpen(linewidth(0.7)); unitsize(15); int n = 10; real h = 6; pen colors[] = {rgb(0.9,0,0),rgb(0,0.9,0),rgb(0,0,0.9)}; pair shiftR = (h+3,0);  void drawEquilaterals(pair A, real s){  filldraw(A--A+s*expi(2*pi/3)--A+(-s,0)--cycle,colors[0]);  filldraw(A--A+s*expi(2*pi/3)--A+s*expi(1*pi/3)--cycle,colors[1]);   filldraw(A--A+s*expi(1*pi/3)--A+(s,0)--cycle,colors[2]); }  for(int i = 0; i < n; ++i)  drawEquilaterals(shiftR + (0,h-h/(2^i) ), (h/(2^(i+1))) *2/3^.5); drawEquilaterals((0,0), h/3^.5); draw((-h/3^.5,0)--(h/3^.5,0)--(0,h)--cycle); label("$\vdots$",(0,3/4*h)); [/asy]

The infinite geometric series $\frac 14 + \frac {1}{4^2} + \frac {1}{4^3} + \cdots = \frac 13$.

[asy] defaultpen(linewidth(1)); unitsize(15); int n = 8;  /* number of layers */ real h = 3; /* square height */  pen colors[] = {rgb(0.8,0,0),rgb(0,0.8,0),rgb(0,0,0.8)};  pair shiftL = (-3*h,0); /* amount to shift second square left by */  void drawSquares(real s, pair A = (0,0)){  filldraw(shift(A)*shift(-2*s,  -s)*xscale(s)*yscale(s)*unitsquare,colors[0]);  filldraw(shift(A)*shift(-2*s,-2*s)*xscale(s)*yscale(s)*unitsquare,colors[1]);  filldraw(shift(A)*shift(-s  ,-2*s)*xscale(s)*yscale(s)*unitsquare,colors[2]); } for(int i = 0; i < n; ++i)  drawSquares(h/2^i); drawSquares(h,shiftL); draw(shift(shiftL+(-2*h,-2*h))*xscale(2*h)*yscale(2*h)*unitsquare); label("$\cdots$",shiftL+(-h/2,-h/2)); [/asy]

Another proof of the identity $\frac 14 + \frac {1}{4^2} + \frac {1}{4^3} + \frac {1}{4^4}+\cdots = \frac 13$.

[asy] unitsize(15); defaultpen(linewidth(1)); pen sm = fontsize(10); real r = 0.7, h = 4.5, n = 10, xsum = 0;  void htick(pair A, pair B, pair ticklength = (0,0.15)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); }  filldraw(xscale(h)*yscale(h)*unitsquare,rgb(0.9,1,0.9)); draw((0,0)--(h/(1-r),0)--(0,h)); for(int i = 0; i < n; ++i){  xsum += r^i;  draw((h*xsum,0)--(h*xsum,h*(1-(1-r)*xsum)));  htick((h*(xsum-r^i),-1),(h*xsum,-1));  if(i < 6)    label("$r^"+(string) i+"$",(h*(xsum-r^i/2),-1),S,sm);  else if(i == 8)    label("$\cdots$",(h*(xsum-r^i/2),-1.2),S,sm); }  /* htick((-1,0),(-1,h),(.15,0)); htick((0,h+1),(h,h+1)); */ htick((h+1,h),(h+1,h*r),(.15,0)); label("$1$",(0,h/2),W,sm); label("$1$",(h/2,h),N,sm); label("$1-r$",(h+1,h*(1+r)/2),E,sm); [/asy]

The infinite geometric series $\sum_{n=0}^{\infty} r^n = \frac{1}{1-r}$.

[asy] unitsize(15); defaultpen(linewidth(1)); pen sm = fontsize(10); real r = 0.55, h = 2.5, n = 7, xsum = 0; pair shiftD = -(0,h*r/(1-r)+2.5); void htick(pair A, pair B, pair ticklength = (0,0.15)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); }  draw((0,h*r/(1-r))--(0,0)--(h*n,0)); for(int i = 1; i < n+1; ++i){  draw((h*i,h*(r/(1-r)-xsum-r^(i)))--(h*i,h*(r/(1-r)-xsum))--(0,h*(r/(1-r)-xsum)));  if(i < 4)    label("$r^"+(string) i+"$", (0,h*(r/(1-r)-xsum-r^(i)/2)), W, sm);  htick((h*i,-1),(h*(i-1),-1));  if(i < n)    label("$1$",(h*(i-1/2),-1),S,sm);  else if(i == n)    label("$\cdots$",(h*(i-1/2),-1.2),S,sm);  xsum += r^i; } draw((0,h*r/(1-r))+shiftD--shiftD--(h*n,0)+shiftD); xsum = 0; for(int i = 1; i < n+1; ++i){  draw(shiftD+(h*i,0)--shiftD+(h*i,h*(r/(1-r)-xsum))--shiftD+(h*(i-1),h*(r/(1-r)-xsum)));  draw(shiftD+(h*i,h*(r/(1-r)-xsum))--shiftD+(0,h*(r/(1-r)-xsum)),linetype("4 4")+linewidth(0.5));  if(i < 4)    label("$r^"+(string) i+"$", shiftD+(h*i,h*(r/(1-r)-xsum-r^(i)/2)), ENE, sm);  htick(shiftD+(h*i,-1),shiftD+(h*(i-1),-1));  if(i < n)    label("$1$",shiftD+(h*(i-1/2),-1),S,sm);  else if(i == n)    label("$\cdots$",shiftD+(h*(i-1/2),-1.2),S,sm);  xsum += r^i; } [/asy]

The arithmetic-geometric series $\sum_{n=1}^{\infty} nr^n = \sum_{n=1}^{\infty} \sum_{i=n}^{\infty} r^i = \sum_{n=1}^{\infty} \frac{r^{n}}{1-r} = \frac{r}{(1-r)^2}$, also known as Gabriel's staircase.[2]

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Geometry

[asy] defaultpen(linewidth(0.7)); unitsize(15); real a = 3.6, b = 4.8, c = (a^2 + b^2)^.5; pair shiftR = (a+b+2,0); pen sm = fontsize(10), heavy = linewidth(1.6); void htick(pair A, pair B, pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } void makeshiftarrow(pair A, real dir, real arrowlength = 0.5){  /* Arrow option resizes */   fill(A--A+arrowlength*expi(dir+pi/8)--A+arrowlength*expi(dir-pi/8)--cycle); }  { /* left side */ filldraw(xscale(a+b)*yscale(a+b)*unitsquare, rgb(1,0.9,0.8));  filldraw((b,0)  --(b,a)--(0,a)  --cycle, rgb(0.9,1,0.9)); filldraw((0,a)  --(a,a)--(a,a+b)--cycle, rgb(0.9,1,0.9)); filldraw((a,a+b)--(a,b)--(a+b,b)--cycle, rgb(0.9,1,0.9)); filldraw((a+b,b)--(b,b)--(b,0)  --cycle, rgb(0.9,1,0.9));  htick((0,-c/10),(b,-c/10),(0,0.15)); htick((-c/10,0),(-c/10,a),(0.15,0)); label("$a$",(-c/10,a/2),W,sm); label("$b$",(b/2,-c/10),S,sm); label("$c$", (a/2,a+b/2),NW,sm); label("$b-a$",(b,(a+b)/2),E,sm); }  { /* right side */ filldraw(shift(shiftR)*xscale(a+b)*yscale(a+b)*unitsquare, rgb(1,0.9,0.8));  filldraw(shift(shiftR)*((0,a)  --(a,a)--(a,a+b)--cycle), rgb(0.9,1,0.9)); filldraw(shift(shiftR)*((a,a+b)--(a,b)--(a+b,b)--cycle), rgb(0.9,1,0.9)); fill(shift(shiftR      )*xscale(a)*yscale(a)*unitsquare,rgb(0.9,0.7,0.7));  fill(shift(shiftR+(a,0))*xscale(b)*yscale(b)*unitsquare,rgb(0.9,0.7,0.7)); filldraw(shift(shiftR)*((a+b,b)--(b,b)--(b,0)  --cycle), rgb(0.7,0.9,0.7)); filldraw(shift(shiftR)*((b,0)  --(b,a)--(0,a)  --cycle), rgb(0.7,0.9,0.7)); draw(shift(shiftR      )*xscale(a)*yscale(a)*unitsquare,heavy);  draw(shift(shiftR+(a,0))*xscale(b)*yscale(b)*unitsquare,heavy); draw(shift(shiftR)*((2*a/3,a+b/3)--(b/3,a/3) ^^ (a+b/3,b+a/3)--(b+2*a/3,b/3))); makeshiftarrow(shiftR+(b/3,a/3),angle((2*a/3,a+b/3)-(b/3,a/3))); makeshiftarrow(shiftR+(b+2*a/3,b/3),angle((a+b/3,b+a/3)-(b+2*a/3,b/3)));   label("$a$",shiftR+(0,a/2),W,sm);  label("$b$",shiftR+(a+b,b/2),E,sm);  label("$c$",shiftR+(a/2,a+b/2),NW,sm);  } [/asy]
The Pythagorean Theorem (first of many proofs): the left diagram shows that $c^2 = 4 \cdot \frac{ab}2 + (b-a)^2 = a^2 + b^2$, and the right diagram shows a second proof by re-arranging the first diagram (the area of the shaded part is equal to $a^2 + b^2$, but it is also the re-arranged version of the oblique square, which has area $c^2$).[3]

[asy] defaultpen(linewidth(0.7)); unitsize(15); pen sm = fontsize(10); real a = 3/1.2, b = 4/1.2, c = (a^2 + b^2)^.5, rot1 = acos(a/c); pair shiftR = (a+b+c,0); path top = (0,c)--a*expi(rot1)+(0,c)--(c,c), sq1=rotate(rot1*180/pi)*xscale(a)*yscale(a)*unitsquare, sq2=shift(c,0)*rotate(rot1*180/pi)*xscale(b)*yscale(b)*unitsquare; void htick(pair A, pair B, pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); }  { /* first picture */  filldraw((0,0)--(c,0)--a*expi(rot1)--cycle, rgb(1,0.85,0.7));  fill(sq1, rgb(0.95,1,0.95)); fill(sq2, rgb(0.95,1,0.95)); filldraw(rotate(270)*xscale(c)*yscale(c)*unitsquare, rgb(0.96,1,0.96)); filldraw((0,0)--top--(c,0)--a*expi(rot1)--cycle, rgb(0.5,0.9,0.5)); draw(sq1 ^^ sq2); draw(a*expi(rot1+pi/2)--top ^^ a*expi(rot1)--a*expi(rot1)+(0,c));   label("$a$",a/2*expi(rot1),NW,sm); label("$b$",a/2*expi(rot1)+(c/2,0),NE,sm); label("$c$",(c/2,0),S,sm); }  { /* second picture */ fill(shift(shiftR)*sq1, rgb(0.95,1,0.95)); fill(shift(shiftR)*sq2, rgb(0.95,1,0.95)); filldraw(shift(shiftR)*rotate(270)*xscale(c)*yscale(c)*unitsquare, rgb(0.96,1,0.96)); filldraw(shift(shiftR+(0,-c))*((0,0)--top--(c,0)--a*expi(rot1)--cycle), rgb(0.5,0.9,0.5)); filldraw(shift(shiftR+(0,-c))*((0,0)--(c,0)--a*expi(rot1)--cycle), rgb(1,0.85,0.7));  draw(shift(shiftR)*((0,0)--(c,0) ^^ sq1 ^^ sq2 ^^ a*expi(rot1+pi/2)--top ^^ a*expi(rot1)--a*expi(rot1)+(0,c)));   label("$a$",shiftR+a/2*expi(rot1),NW,sm); label("$b$",shiftR+a/2*expi(rot1)+(c/2,0),NE,sm); label("$c$",shiftR+(c/2,0),S,sm); } [/asy]
Another proof of the Pythagorean Theorem (animated version).

[asy] defaultpen(linewidth(0.7)); unitsize(15); pen sm = fontsize(10); real a = 3/1.2, b = 4/1.2, c = (a^2 + b^2)^.5, rot1 = acos(a/c); pair shiftR = (a+b+c,0); path top = (0,c)--a*expi(rot1)+(0,c)--(c,c), sq1=rotate(rot1*180/pi)*xscale(a)*yscale(a)*unitsquare, sq2=shift(c,0)*rotate(rot1*180/pi)*xscale(b)*yscale(b)*unitsquare, tri = (0,0)--(0,a)--(b,0)--cycle; void htick(pair A, pair B, pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); }  { /* first picture */  filldraw(xscale(a+b)*yscale(a+b)*unitsquare, rgb(1,0.85,0.75)); filldraw(tri, rgb(0.6,0.9,0.6)); filldraw(shift((a+b,0))*rotate(90)*tri, rgb(0.6,0.9,0.6)); filldraw(shift((a+b,a+b))*rotate(180)*tri, rgb(0.6,0.9,0.6)); filldraw(shift((0,a+b))*rotate(270)*tri, rgb(0.6,0.9,0.6)); draw((0,a)--(a+b,b), linetype("2 4")+linewidth(0.7));  label("$a$",(0,a/2),W,sm); label("$b$",(b/2,0),S,sm); label("$c$",(b/2,a/2),NE,sm); }  { /* second picture */ filldraw(shift(shiftR)*xscale(a+b)*yscale(a+b)*unitsquare, rgb(1,0.85,0.75)); filldraw(shift(shiftR+(a,a)  )*rotate(270)*reflect((0,0),(1,1))*tri, rgb(0.6,0.9,0.6));  filldraw(shift(shiftR+(a+b,0))*rotate(90)*reflect((0,0),(1,1))*tri, rgb(0.6,0.9,0.6));  filldraw(shift(shiftR+(a,a)  )*rotate(90)*tri, rgb(0.6,0.9,0.6));  filldraw(shift(shiftR+(0,a+b))*rotate(270)*tri, rgb(0.6,0.9,0.6));  label("$a$",shiftR+(0,a/2),W,sm); label("$a$",shiftR+(a/2,0),S,sm); label("$b$",shiftR+(a+b,a+b/2),E,sm); label("$b$",shiftR+(a+b/2,a+b),N,sm); } [/asy]
Another proof of the Pythagorean Theorem; the left-hand diagram suggests the identity $c^2 = (a+b)^2 - 4 \cdot \frac{ab}2 = a^2 + b^2$, and the right-hand diagram offers another re-arrangement proof.

[asy] import graph; defaultpen(linewidth(0.7)); unitsize(15); pen ds=black, ls = linetype("2 2");  pen fueaev=rgb(1,0.4,0.4), ffffea=rgb(1,1,0.4), evfuea=rgb(0.4,1,0.4), fffuev=rgb(1,0.5,0.3), fzfzfz=rgb(0.8,0.8,0.8), evevfz=rgb(0.4,0.4,1);   fill((0,0)--(1.5,2)--(-0.5,3.5)--(-2,1.5)--cycle,fueaev); fill((1.5,2)--(4.17,0)--(6.17,2.67)--(3.5,4.67)--cycle,fueaev); fill((4.17,0)--(0,0)--(0,-4.17)--(4.17,-4.17)--cycle,fueaev); fill((1.5,2)--(2,2.67)--(2,1.63)--cycle,ffffea); fill((2,2.67)--(2,1.63)--(4.17,0)--(6.17,2.67)--cycle,evfuea); fill((0,3.13)--(0,0)--(1.5,2)--cycle,fffuev); fill((-0.5,3.5)--(0,3.13)--(0,0)--(-2,1.5)--cycle,fueaev); fill((0,0)--(1.5,2)--(4.17,0)--cycle,fzfzfz); fill((2,2.67)--(3.5,4.67)--(6.17,2.67)--cycle,evevfz); fill((0,-4.17)--(1.5,-2.17)--(0,-1.04)--cycle,fffuev); fill((1.5,-2.17)--(0,-4.17)--(4.17,-4.17)--cycle,evevfz); fill((4.17,0)--(3.67,-0.67)--(4.17,-1.04)--cycle,white); fill((3.67,-0.67)--(2.17,-2.67)--(4.17,-4.17)--(4.17,-1.04)--cycle,fueaev); fill((0,0)--(0,-1.04)--(2.17,-2.67)--(4.17,0)--cycle,evfuea); fill((4.17,0)--(3.67,-0.67)--(4.17,-1.04)--cycle, ffffea);  draw((0,0)--(1.5,2)); draw((0,0)--(4.17,0)); draw((1.5,2)--(4.17,0)); draw((0,0)--(1.5,2)); draw((1.5,2)--(-0.5,3.5)); draw((-0.5,3.5)--(-2,1.5)); draw((-2,1.5)--(0,0)); draw((1.5,2)--(4.17,0)); draw((4.17,0)--(6.17,2.67)); draw((6.17,2.67)--(3.5,4.67)); draw((3.5,4.67)--(1.5,2)); draw((4.17,0)--(0,0)); draw((0,0)--(0,-4.17)); draw((0,-4.17)--(4.17,-4.17)); draw((4.17,-4.17)--(4.17,0)); draw((0,0)--(-2,1.5)); draw((-2,1.5)--(-0.5,3.5)); draw((-0.5,3.5)--(1.5,2)); draw((0,0)--(0,-4.17)); draw((0,-4.17)--(4.17,-4.17)); draw((4.17,-4.17)--(4.17,0)); draw((4.17,0)--(6.17,2.67)); draw((6.17,2.67)--(3.5,4.67)); draw((3.5,4.67)--(1.5,2)); draw((0,-4.17)--(0,0)); draw((0,0)--(0,3.13)); draw((1.5,2)--(2,2.67)); draw((2,2.67)--(2,1.63)); draw((2,1.63)--(1.5,2)); draw((2,2.67)--(2,1.63)); draw((2,1.63)--(4.17,0)); draw((4.17,0)--(6.17,2.67)); draw((6.17,2.67)--(2,2.67)); draw((0,3.13)--(0,0)); draw((0,0)--(1.5,2)); draw((1.5,2)--(0,3.13)); draw((-0.5,3.5)--(0,3.13)); draw((0,3.13)--(0,0)); draw((0,0)--(-2,1.5)); draw((-2,1.5)--(-0.5,3.5)); draw((0,0)--(1.5,2)); draw((1.5,2)--(4.17,0)); draw((4.17,0)--(0,0)); draw((2,2.67)--(3.5,4.67)); draw((3.5,4.67)--(6.17,2.67)); draw((6.17,2.67)--(2,2.67)); draw((0,-1.04)--(4.17,-4.17)); draw((0,-4.17)--(1.5,-2.17)); draw((1.5,-2.17)--(0,-1.04)); draw((0,-1.04)--(0,-4.17)); draw((1.5,-2.17)--(0,-4.17)); draw((0,-4.17)--(4.17,-4.17)); draw((4.17,-4.17)--(1.5,-2.17)); draw((4.17,0)--(3.67,-0.67)); draw((3.67,-0.67)--(4.17,-1.04)); draw((4.17,-1.04)--(4.17,0)); draw((3.67,-0.67)--(2.17,-2.67)); draw((2.17,-2.67)--(4.17,-4.17)); draw((4.17,-4.17)--(4.17,-1.04)); draw((4.17,-1.04)--(3.67,-0.67)); draw((0,0)--(0,-1.04)); draw((0,-1.04)--(2.17,-2.67)); draw((2.17,-2.67)--(4.17,0)); draw((4.17,0)--(0,0));   draw((-0.5,3.5)--(1.5,6.17),ls);  draw((1.5,6.17)--(3.5,4.67),ls);  draw((0,3.13)--(0,4.17),ls);  draw((1.5,6.17)--(1.5,2),ls);  draw((2,5.79)--(2,2.67),ls);  draw(rightanglemark((0,0),(1.5,2),(4.17,0))); [/asy]
A dissection proof of the Pythagorean Theorem.[6] (Cut-the-knot)

[asy] defaultpen(linewidth(0.7)); unitsize(15); pen sm = fontsize(10); real a = 3/1.2, b = 4/1.2, c = (a^2 + b^2)^.5, rot1 = acos(a/c);  path top = (0,c)--a*expi(rot1)+(0,c)--(c,c), sq1=rotate(rot1*180/pi)*xscale(a)*yscale(a)*unitsquare, sq2=shift(c,0)*rotate(rot1*180/pi)*xscale(b)*yscale(b)*unitsquare; void htick(pair A, pair B, pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); }  filldraw((0,0)--(c,0)--a*expi(rot1)--cycle, rgb(1,0.85,0.7)); /* draw(rightanglemark((0,0),a*expi(rot1),(c,0))); */ filldraw(sq1, rgb(0.95,1,0.95)); filldraw(sq2, rgb(0.95,1,0.95)); filldraw(rotate(270)*xscale(c)*yscale(c)*unitsquare, rgb(0.96,1,0.96));  label("$a$",a/2*expi(rot1),SE,sm); label("$b$",a/2*expi(rot1)+(c/2,0),SW,sm); label("$c$",(c/2,-c),S,sm); [/asy]
COMING: The last proof of the Pythagorean Theorem we shall present on this page, this one by dissection.

[asy] defaultpen(linewidth(0.7)+fontsize(10)); unitsize(15); real a = 3.6, b = 4.8, c = (a^2 + b^2)^.5; pair shiftR = (a+b+2,0); pen sm = fontsize(10), heavy = linewidth(1); void htick(pair A, pair B, pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } void makeshiftarrow(pair A, real dir, real arrowlength = 0.5){  /* Arrow option resizes */   fill(A--A+arrowlength*expi(dir+pi/8)--A+arrowlength*expi(dir-pi/8)--cycle); }  real s1 = 5, s2 = 7, s3 = 8; // triangle side lengths pen c1 = rgb(0.5,0.5,1), c2 = rgb(0.5,1,0.8), c3 = rgb(0.5,1,0.5); // color pens pair shiftR = (s1+1,0); // distance between two diagrams  pair A=(0,0), B = (s1,0), C = intersectionpoints(Circle(A, s3), Circle(B, s2))[0], I = incenter(A,B,C), D = foot(I,B,C), E = foot(I,A,C), F = foot(I,A,B);  // draw left diagram filldraw(A--I--B--cycle, c1, heavy); filldraw(B--I--C--cycle, c2, heavy); filldraw(A--I--C--cycle, c3, heavy); dot(I); draw(incircle(A,B,C), heavy); draw(I--D, linetype("2 2")); draw(I--E, linetype("2 2")); draw(I--F, linetype("2 2"));  label("$a$",(A+B)/2,S); label("$b$",(B+C)/2,NE); label("$c$",(A+C)/2,NW); label("$r$",(I+F)/2,W); draw(rightanglemark(I,F,A)); draw(rightanglemark(I,D,B)); draw(rightanglemark(I,E,C));  pair reflectC = C + 2*(foot(C,A,I) - C), reflectI = (reflectC.x - I.x, I.y);   // draw right diagram filldraw(shift(shiftR)*(A--I--B--cycle), c1, heavy); filldraw(shift(shiftR+B)*(A--reflectI--reflectC--cycle), c3, heavy); filldraw(shift(shiftR)*(I--B--B+reflectI--cycle), c2, heavy); // dashed perpendiculars; arbitrary coding here draw(shift(shiftR)*(I--F), linetype("2 2")); draw(shift(shiftR)*(B--B+(0,inradius(A,B,C))), linetype("2 2")); draw(shift(shiftR+B)*(reflectI--foot(reflectI,A,B)), linetype("2 2")); draw(shift(shiftR)*rightanglemark(I,F,A)); draw(shift(shiftR)*rightanglemark(B,B+(0,inradius(A,B,C)),I)); draw(shift(shiftR+B)*rightanglemark(reflectI,foot(reflectI,A,B),A)); label("$a$",shiftR+(A+B)/2,S); label("$b$",shiftR+(I+B+reflectI)/2,N); label("$c$",shiftR+B+(A+reflectC)/2,S); label("$r$",shiftR+(I+F)/2,W); [/asy] The area of a triangle is given by $A = \frac{1}{2} \cdot r \cdot (a+b+c) = rs$, where $r$ is the inradius and $s$ is the semiperimeter.[10]
(Comment: we do not need to re-arrange the triangles to a trapezoid to see this, but this re-arrangement works due to alternate interior angles/angle bisector properties of the incenter.)

[asy]unitsize(15); defaultpen(linewidth(0.7) + fontsize(10)); pen heavy = linewidth(1); real a = 4.5, b = 2.5, c = 2, d = 4; pair A = (a,b), B = (a+c,b+d), C = (c,d), D = IP(B--B+2*(A-B), (0,0)--(a,0)), F = IP(B--B+2*(C-B), (0,0)--(0,d)), G = IP(B--D,(0,d)--(a+c,d)), H = IP(B--F,(a,0)--(a,b+d)), shiftR = (a+c+1,0), shiftR2 = 2*shiftR;  // left diagram filldraw((0,0)--(a,0)--(a,d)--(0,d)--cycle, rgb(0.5, 1, 0.5)); draw((0,0)--A--B--C--cycle);  draw(shift((a,d))*xscale(c)*yscale(b)*unitsquare); draw(A--D ^^ C--F, linetype("2 2"));  draw((0,0)--(a,0)--(a,b)--cycle, heavy); draw(shift(C)*((0,0)--(a,0)--(a,b)--cycle), heavy); label("$a$",(a/2,0),S); label("$d$",(0,d/2),W); label("$b$",B-(0,b/2),E); label("$c$",B-(c/2,0),N); label("$(a,b)$",(a,b),SE,fontsize(8)); label("$(c,d)$",(c,d),NW,fontsize(8));  // middle diagram filldraw(shift(shiftR)*((0,0)--A--(a,d)--G--B--C--(0,d)--cycle), rgb(0.5, 1, 0.5)); filldraw(shift(shiftR)*(G--(a+c,d)--B--cycle), rgb(1,0.5,0.5)); draw(shift(shiftR)*((0,0)--(a,0)--(a,d)--(0,d)--cycle)); draw(shift(shiftR+(a,d))*xscale(c)*yscale(b)*unitsquare); draw(shift(shiftR)*(A--D ^^ C--F), linetype("2 2"));  draw(shift(shiftR)*((0,0)--(0,d)--(c,d)--cycle), heavy); draw(shift(shiftR+A)*((0,0)--(0,d)--(c,d)--cycle), heavy); label("$a$",shiftR+(a/2,0),S); label("$d$",shiftR+(0,d/2),W); label("$b$",shiftR+B-(0,b/2),E); label("$c$",shiftR+B-(c/2,0),N);  // right diagram filldraw(shift(shiftR2)*((0,0)--A--G--(a,d)--H--C--cycle), rgb(0.5, 1, 0.5));  filldraw(shift(shiftR2)*(G--(a,d)--H--B--cycle), rgb(0.3, 0.9, 0.3)); filldraw(shift(shiftR2)*(G--(a+c,d)--B--cycle), rgb(1,0.5,0.5)); filldraw(shift(shiftR2)*(H--(a,b+d)--B--cycle), rgb(1,0.5,0.5)); draw(shift(shiftR2)*((0,0)--(a,0)--(a,d)--(0,d)--cycle)); draw(shift(shiftR2)*(A--D ^^ C--F), linetype("2 2"));  label("$a$",shiftR2+(a/2,0),S); label("$d$",shiftR2+(0,d/2),W); label("$b$",shiftR2+B-(0,b/2),E); label("$c$",shiftR2+B-(c/2,0),N); label("$\overrightarrow{(a,b)}$",shiftR2+(a,b),SE,fontsize(8)); label("$\overrightarrow{(c,d)}$",shiftR2+(c,d),NW,fontsize(8)); [/asy] The area of a parallelogram with adjacent side vectors $\overrightarrow{(a,b)}, \overrightarrow{(c,d)}$ is given by $\overrightarrow{(a,b)} \times \overrightarrow{(c,d)} = ad-bc$.

[asy] defaultpen(linewidth(0.7)); unitsize(15);   real r = 3.5; // radius pair shiftL = (-2.5*r,0); // distance between 2 diagrams  /* returns the vertex of the interior equilateral triangle with one edge shared with the dodecagon */ pair dodecagonPt(int i) {  return r*dir(i*360/12) + rotate(60)*(r*(dir((i+1)*360/12) - dir(i*360/12))); }  /* left diagram */ path dodecagon = shiftL+(r,0)--shiftL+r*dir(30); for(int i = 1; i < 12; ++i)   dodecagon = dodecagon--shiftL+r*dir(i*30); dodecagon = dodecagon--cycle; filldraw(dodecagon, rgb(0.5,1,0.5)); draw(Circle(shiftL, r), linetype("2 2")); dot((0,0)); draw(shiftL--shiftL+(r,0)); label("$R$",shiftL+(r/2,0),S);  /* right diagram */ for(int i = 0; i < 9; ++i) {  filldraw((0,0)--r*dir(i*360/12)--dodecagonPt(i)--cycle, rgb(0,0.8,0));  filldraw((0,0)--r*dir((i+1)*360/12)--dodecagonPt(i)--cycle, rgb(0,0.8,0));  filldraw(r*dir(i*360/12)--r*dir((i+1)*360/12)--dodecagonPt(i)--cycle, rgb(0.8,0.8,0));  if (i % 3 == 1) {   filldraw(r*2^.5*dir(floor(i/3)*90+45)--r*dir(i*360/12)--r*dir((i+1)*360/12)--cycle, rgb(0.8,0.8,0));   filldraw(r*2^.5*dir(floor(i/3)*90+45)--r*dir(i*360/12)--r*dir(floor(i/3)*90)--cycle, rgb(0,0.8,0));   filldraw(r*2^.5*dir(floor(i/3)*90+45)--r*dir((i+1)*360/12)--r*dir(floor(i/3)*90+90)--cycle, rgb(0,0.8,0));  } } for(int i = 9; i < 12; ++i) {  filldraw((0,0)--r*dir(i*360/12)--dodecagonPt(i)--cycle, rgb(0.5,1,0.5), linetype("2 2"));  filldraw((0,0)--r*dir((i+1)*360/12)--dodecagonPt(i)--cycle, rgb(0.5,1,0.5), linetype("2 2"));  filldraw(r*dir(i*360/12)--r*dir((i+1)*360/12)--dodecagonPt(i)--cycle, rgb(1,1,0.5), linetype("2 2")); } [/asy]
The area of a dodecagon is $3R^2$, where $R$ is the circumradius.


[asy] pathpen = linewidth(1); unitsize(15); pen dotted = linetype("2 4"); path xaxis = (-3,0)--(3,0); pair A = (-2,2), B = (1.5,1.5), B3 = (-1.5,0), B2 = (B.x,-B.y), C2 = IP(xaxis, A--B2);   D(xaxis,Arrows(8)); D(D(A)--D(C2)--D(B)); D(D(B2)--C2,dashed+linewidth(0.7));  D(A--D(B3)--B,dotted+linewidth(0.7)); D(B3--B2,dotted); MP("(a,b)",A,W); MP("(c,d)",B,E); MP("(c,-d)",B2,E); [/asy]
The smallest distance necessary to travel between $(a,b)$, the x-axis, and then $(c,d)$ for $b,d > 0$ is given by $\sqrt{(a-c)^2 + (b+d)^2}$.[4]

[asy]defaultpen(linewidth(1)); unitsize(15); pen dotted = linetype("2 4"), sm = fontsize(10); real r = 2; pair A = r*(0,0), B = (r*18/5,A.y), C = r*(16/5,12/5), D = (r*9/5,C.y); pair refl(pair a, pair b = (C+B)/2) { return a+2*(b-a); } void makeshiftarrow(pair a, real dir, real arrowlength = r){  /* Arrow option resizes */   fill(a--a+arrowlength*expi(dir+pi/8)--a+arrowlength*expi(dir-pi/8)--cycle); }  draw(A--B--C--D--cycle); draw(A--C^^B--D); draw(refl(A)--C--B--refl(D)--cycle ^^ C--refl(D), dotted); draw(rightanglemark(A,C,refl(D),15)^^rightanglemark(A,IP(A--C,B--D),B,15), linewidth(0.7)); label("$A$",A,S,sm);label("$B$",B,S,sm);label("$C$",C,N,sm);label("$D$",D,N,sm);label("$A'$",refl(A),N,sm);label("$D'$",refl(D),S,sm);    /* arrow */ draw(arc((B+C)/2,C.y,240,300),linewidth(0.7)); makeshiftarrow((B+C)/2+C.y*expi(pi*300/180),210*pi/180,r/4); [/asy]
In trapezoid $ABCD$ with $\overline{AB} \parallel \overline{CD}$, then $\overline{AC} \perp \overline{BD} \Longleftrightarrow AC^2 + BD^2 = (AB + CD)^2$.

[asy] // feel free to change these four points! pair A = (0,0), B = (3, -2), C = (5,1), D = (1,3);  // // Rest of code // size(200); defaultpen(linewidth(0.9)); pen lightgreen = rgb(0.6,1,0.6), lightred = rgb(1,0.6,0.6), smdash = linewidth(0.7)+linetype("2 2");  pair E = IP(A--C,B--D), AB = (A+B)/2, BC = (B+C)/2, CD = (C+D)/2, DA = (D+A)/2, ABE = 2*AB-E, BCE = 2*BC-E, CDE = 2*CD-E, DAE = 2*DA-E; path midpts = AB--BC--CD--DA--cycle; filldraw(shift(-E)*scale(2)*midpts,lightgreen); filldraw(midpts,lightred); draw(A--B--C--D--cycle); draw(A--C); draw(B--D); draw((A+ABE)/2--(C+BCE)/2,smdash); draw((B+ABE)/2--(D+DAE)/2,smdash); draw((B+BCE)/2--(D+CDE)/2,smdash); draw((A+DAE)/2--(C+CDE)/2,smdash);  dot(A); dot(B); dot(C); dot(D); label("$A$",A,W);label("$B$",B,S);label("$C$",C,E);label("$D$",D,N); [/asy]

Varignon's theorem: the area of the outer parallelogram is twice the area of the quadrilateral and four times the area of the midpoint parallelogram, so the midpoint parallelogram of a (convex) quadrilateral has area $1/2$ of the quadrilateral.

Proof for Volume of a Cone: http://www.mathematische-basteleien.de/wuerfel16.gif

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Miscellaneous

[asy] import graph; size(220); defaultpen(linewidth(0.7)); Label k; k.p=fontsize(10);  real xmax = pi/2+0.5, xmin = -0.5, ymax = 1.39, ymin = -0.39, lblpt = pi/4 + 0.08;   /* f(x) = sin^2(x) */ real f(real x) { return sin(x) * sin(x); } string pilabel(real x) { if(x > 1) return "$\pi/2$"; else if(x > 0) return "$\pi/4$"; else return "";}  xaxis(xmin,xmax,Ticks(k, pilabel, pi/4),Arrows(6)); yaxis(ymin,ymax,Ticks(k, NoZero),Arrows(6)); filldraw(graph(f,0,pi/2)--(pi/2,0)--(0,0)--cycle,gray(0.7),linewidth(1)); draw((lblpt,f(lblpt))--(lblpt,1),Arrows(6)); draw((lblpt,f(lblpt))--(lblpt,0),Arrows(6)); label("$\cos^2(x)$",(lblpt,f(lblpt)/2+1/2),W,fontsize(10)); label("$\sin^2(x)$",(lblpt,f(lblpt)/2),E,fontsize(10)); draw((0,1)--(pi/2,1),linewidth(1)); [/asy]

$\int_0^{\pi/2} \sin^2 x \, dx = \int_0^{\pi/2} \cos^2 x \, dx = \frac {\pi}{4}$ from $\begin{cases}\sin^2 x + \cos^2 x = 1\\ \sin x = \cos(\pi/2 - x)\end{cases}$. (Source)



[asy] import graph; size(170); defaultpen(linewidth(0.7)); Label k; k.p=fontsize(8);  real xmax = 1.35, xmin = -0.35, ymax = 1.35, ymin = -0.35, lblpt = 1/2 + 0.15, alpha = 2, epsilon = 0.015;   real f(real x) { return x^alpha; }  xaxis(xmin,xmax,Ticks(k, 0.25, NoZero),Arrows(6)); yaxis(ymin,ymax,Ticks(k, 0.25, NoZero),Arrows(6)); filldraw(graph(f,0,1)--(1,0)--(0,0)--cycle,gray(0.7),linewidth(1)); draw(graph(f,1,1.1),linewidth(0.7),EndArrow(4)); draw((lblpt-epsilon,f(lblpt))--(0+epsilon,f(lblpt)),Arrows(4)); draw((lblpt,f(lblpt)-epsilon)--(lblpt,0+epsilon),Arrows(4)); label("$x = y^{1/\alpha}$",(lblpt/2,f(lblpt)),N,fontsize(10)); label("$y = x^{\alpha}$",(lblpt,f(lblpt)/2),E,fontsize(10)); draw((0,1)--(1,1),linewidth(1)); [/asy]

$\alpha > 0 \Longrightarrow \int_0^{1} \left(x^\alpha + x^{1/\alpha}\right) \, dx = 1.$ (Source)



[asy]unitsize(15); defaultpen(linewidth(0.7)); real a=2.5,b=5,s=a+b; pen colors[] = {rgb(0.9,0.2,0.2), rgb(0.2,0.9,0.2), rgb(0.2,0.2,0.9)}; pen sm = fontsize(8); void fillrect(pair A, pair B, pen p = invisible, pen l = linewidth(1)){ filldraw(A--(A.x,B.y)--B--(B.x,A.y)--cycle, p, l); } void htick(pair A, pair B, pair ticklength = (0.2,0)){  draw(A--B);  draw(A-ticklength--A+ticklength);  draw(B-ticklength--B+ticklength); }  fillrect((0,0),(s,s)); fillrect((a,b),(s,s),colors[0]); filldraw((0,a)--(a,a)--(s/2,s/2)--(a,b)--(a,s)--(0,s)--cycle,colors[1],linewidth(1)); filldraw((0,0)--(b,0)--(b,b)--(a,a)--(0,a)--cycle,colors[2],linewidth(1)); draw((0,0)--(a,a),linewidth(1));  draw((s/2,s/2)--(b,a)--(a,a)--(a,b),linewidth(0.7)+linetype("4 2")); htick((s+1,0),(s+1,b)); htick((s+1,b),(s+1,s));   /* in labels, a,b swapped */ label("$a$",(s+1,b/2),E);label("$b$",(s+1,(s+b)/2),E); label("$ab$",(a+s,b+s)/2,sm); label("$\frac{(a+b)^2}{4}$",(a,a+s)/2,sm); label("$\frac{a^2}2$",(s/2,a*2/3),sm); label("$\frac{b^2}2$",(a/4,a*2/3),sm);  [/asy]
The Root-Mean Square-Arithmetic Mean-Geometric Mean inequality, $\color{red}{ab} \color{black} \le \color{green} \frac{(a+b)^2}{4} \color{black} \le \color{blue} \frac{a^2 + b^2}{2}$.

[asy] unitsize(15); defaultpen(linewidth(0.7)); void htick(pair A, pair B,pair ticklength = (0,0.15)){  draw(A--B);  draw(A-ticklength--A+ticklength);  draw(B-ticklength--B+ticklength); } real a=10,b=3,r=(a+b)/2; pen sm = fontsize(8), dark = linewidth(1); pen colors[] = {rgb(0.9,0.2,0.2) + dark,    /* GM */                 rgb(0.2,0.9,0.2) + dark,    /* AM */                 rgb(0.2,0.2,0.9) + dark,    /* QM */                 rgb(0.2,0.9,0.9) + dark };  /* HM */ pair A = (r-b,(r^2-(r-a)^2)^.5),B=foot((A.x,0),(0,0),A);  draw(arc((0,0),r,0,180)--cycle); dot(A); dot((0,r)); dot((A.x,0)); dot((0,0)); draw(B--A,colors[3]);           label("HM",(A+B)/2, E, sm+colors[3]); draw((0,0)--(0,r),colors[1]);   label("AM",(0,r*2/3), NW, sm+colors[1]);  draw((A.x,0)--A,colors[0]);     label("GM",(A.x,A.y/2), SE, sm+colors[0]);  draw((A.x,0)--(0,r),colors[2]); label("RMS",(A.x/5,r*4/5), NE, sm+colors[2]);    draw((-r,0)--A--(r,0), linetype("4 2")); draw((0,0)--B--(A.x,0), linetype("4 2")); draw(rightanglemark((-r,0),A,(r,0))); draw(rightanglemark((0,0),B,(A.x,0)));  htick((-r,-1),(A.x,-1)); htick((A.x,-1),(r,-1)); label("$a$",((-r+A.x)/2,-1),S); label("$b$",((r+A.x)/2,-1),S); [/asy]

The Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality.[5]


[asy] unitsize(15); defaultpen(linewidth(0.7)); real r = 0.3, row1 = 3.5, row2 = 0, row3 = -3.5; void necklace(pair k, pen colors[]){  draw(shift(k)*unitcircle);   for(int i = 0; i < colors.length; ++i){   pair p = k+expi(pi/2+2*pi*i/colors.length);   fill(Circle(p,r),colors[i]);   draw(Circle(p,r));  } } void htick(pair A, pair B,pair ticklength = (0.15,0)){  draw(A--B);  draw(A-ticklength--A+ticklength);  draw(B-ticklength--B+ticklength); }   /* draw necklaces */ pen BEADS1[] = {red,red,red},BEADS2[] = {blue,blue,blue},BEADS3[] = {red,red,blue},BEADS4[] = {blue,red,red},BEADS5[] = {red,blue,red},BEADS6[] = {blue,blue,red},BEADS7[] = {red,blue,blue},BEADS8[] = {blue,red,blue};  necklace((-10,(row2+row3)/2),BEADS1);necklace((-7.5,(row2+row3)/2),BEADS2);  necklace((-2.5,row2),BEADS3);necklace((0,row2),BEADS4);necklace((2.5,row2),BEADS5); necklace((-2.5,row3),BEADS6);necklace((0,row3),BEADS7);necklace((2.5,row3),BEADS8);   /* box them and label */ draw((-4,row2-1.3)--(4,row2-1.3)--(4,row2+1.6)--(-4,row2+1.6)--cycle,linewidth(0.9)+linetype("4 2")); draw((-4,row3-1.3)--(4,row3-1.3)--(4,row3+1.6)--(-4,row3+1.6)--cycle,linewidth(0.9)+linetype("4 2")); htick((-4,row2+2),(4,row2+2),(0,0.15)); label("$p$",(0,row2+2),N,fontsize(10));  htick((-11.5,(row2+row3)/2+2),(-6,(row2+row3)/2+2),(0,0.15)); label("$a$",(-17.5/2,(row2+row3)/2+2),N,fontsize(10));  [/asy]

Fermat's Little Theorem: $a^p \equiv a \pmod{p}$ for $\text{gcd}\,(a,p) = 1$ (above $a=2,p=3$).

[asy] defaultpen(linewidth(0.7)); unitsize(30);  real r = 0.2;  pair endPt1 = (-2.5,-1), endPt2 = (2.5,-1); // endpoints of R^1 line real projections[] = {-1.5, -0.75, 1, 2.2}; // x-coordinates of steoreographic projections on y=-1  void makeshiftarrow(pair A, real dir, real arrowlength = r){  /* Arrow option resizes, so draw makeshift arrows */   fill(A--A+arrowlength*expi(dir+pi/8)--A+arrowlength*expi(dir-pi/8)--cycle); }  draw(endPt1 -- endPt2); draw(unitcircle);  for(int i = 0; i < projections.length; ++i) {  draw((0,1) -- (projections[i],-1), linetype("2 2"));  dot((projections[i],-1), Fill(red));  dot(OP((0,1) -- (projections[i],-1),unitcircle), Fill(yellow)); } dot((0,1), Fill(green));  makeshiftarrow(endPt1, 0); makeshiftarrow(endPt2, pi); [/asy][asy] import three; defaultpen(linewidth(0.7)); unitsize(20);   currentprojection = orthographic(0.3,-2,0.6);  pen gridpen = dotted; int gridmin = -2, gridmax = 2, gridlines = 6; pair projections[] = {(1,-1),(1,-2),(2,-2),(2,-1)}; triple IPs[] = new triple[4]; draw((1,-1,-1)--(1,-2,-1)--(2,-2,-1)--(2,-1,-1)--cycle);  for(int i = 0; i < gridlines; ++i) {  draw((gridmin,gridmin+(gridmax-gridmin)*(i+1)/(gridlines+1),-1)--(gridmax,gridmin+(gridmax-gridmin)*(i+1)/(gridlines+1),-1), gridpen);  draw((gridmin+(gridmax-gridmin)*(i+1)/(gridlines+1),gridmin,-1)--(gridmin+(gridmax-gridmin)*(i+1)/(gridlines+1),gridmax,-1), gridpen); }  draw(unitsphere,white); draw(arc((0,0,0), (1,0,0), (-1,0,0), (0,0,1)), linetype("2 2")); draw(arc((0,0,0), (-1,0,0), (1,0,0), (0,0,1))); // draw(circle((0,0,0), 1, (0,1,0)));  // draw projection points for(int i = 0; i < projections.length; ++i) {  real px = projections[i].x/2, py = projections[i].y/2, pxy = 1 + px*px + py*py;  draw((0,0,1)--(2*px, 2*py, -1), linetype("2 2"));  triple IP = (2*px/pxy, 2*py/pxy, (pxy-2)/pxy), OP = (2*px, 2*py, -1);  dot(IP, yellow);  draw(circle(IP, 0.06, (0,1,0)));  IPs[i] = IP;  dot(OP, red);  draw(circle(OP, 0.06, (0,1,0)));  if(i != 0)    draw(IPs[i] -- IPs[i-1], dotted); } draw(IPs[0]--IPs[3], dotted);  dot((0,0,1),green); draw(circle((0,0,1), 0.06, (0,1,0))); [/asy]

There exists a homeomorphism, the stereographic projection, between the punctured hypersphere $S^n \setminus \{(1,\underbrace{0, \ldots, 0}_{n-1\text{ zeroes}})\}$ and $\mathbb{R}^n$ for $n = 1,2$.



Sum of arctangents formula:

[asy]  /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; usepackage("amsmath"); size(13cm);  real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */  pen dotstyle = black; /* point style */  real xmin = -0.4717093412177357, xmax = 7.405441345585962, ymin = -1.1854534297865673, ymax = 7.342957746870971;  /* image dimensions */ pen uququq = rgb(0.25098039215686274,0.25098039215686274,0.25098039215686274); pen aqaqaq = rgb(0.6274509803921569,0.6274509803921569,0.6274509803921569); pen qqwuqq = rgb(0.,0.39215686274509803,0.); pen cqcqcq = rgb(0.7529411764705882,0.7529411764705882,0.7529411764705882);   draw((0.,0.)--(0.,1.)--(1.,1.)--cycle);  draw((1.,1.)--(0.,3.)--(0.,1.)--cycle, uququq);  draw((0.,3.)--(6.,6.)--(1.,1.)--cycle, aqaqaq);  draw(arc((1.,1.),0.3101240427875472,180.,225.)--(1.,1.)--cycle, qqwuqq);  draw(arc((1.,1.),0.3101240427875472,116.56505117707799,180.)--(1.,1.)--cycle, qqwuqq);  draw(arc((1.,1.),0.3101240427875472,45.,116.56505117707799)--(1.,1.)--cycle, qqwuqq);   /* draw grid of horizontal/vertical lines */ pen gridstyle = linewidth(0.7) + cqcqcq; real gridx = 1., gridy = 1.; /* grid intervals */ for(real i = ceil(xmin/gridx)*gridx; i <= floor(xmax/gridx)*gridx; i += gridx)  draw((i,ymin)--(i,ymax), gridstyle); for(real i = ceil(ymin/gridy)*gridy; i <= floor(ymax/gridy)*gridy; i += gridy)  draw((xmin,i)--(xmax,i), gridstyle);  /* end grid */    /* draw figures */ draw((0.,0.)--(0.,1.));  draw((0.,1.)--(1.,1.));  draw((1.,1.)--(0.,0.));  draw((1.,1.)--(0.,3.), uququq);  draw((0.,3.)--(0.,1.), uququq);  draw((0.,1.)--(1.,1.), uququq);  draw((0.,3.)--(6.,6.), aqaqaq);  draw((6.,6.)--(1.,1.), aqaqaq);  draw((1.,1.)--(0.,3.), aqaqaq);  label("$\arctan 1 + \arctan 2 + \arctan 3 = \pi$",(1.544096936901321,0.5925910821953678),SE*labelscalefactor,fontsize(10));   /* dots and labels */ dot((0.,0.),linewidth(3.pt) + dotstyle);  dot((0.,1.),dotstyle);  dot((1.,1.),dotstyle);  dot((0.,3.),dotstyle);  dot((6.,6.),dotstyle);  clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);   /* end of picture */ [/asy]

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References

  1. ^ MathOverflow
  2. ^ Wolfram MathWorld
  3. ^ Attributed to the Chinese text Zhou Bi Suan Jing.
  4. ^ This is more of a proof without words of the AM-GM inequality $\frac{a+b}{2} \ge \sqrt{ab}$; though the lengths of the segments labeled RMS and HM can easily be verified to have values of $\sqrt{\frac{a^2+b^2}{2}}, \frac{2}{\frac 1a + \frac 1b}$, respectively, it might not be obvious from the diagram. It still serves as a useful graphical demonstration of the inequality.