Proofs without words
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]](http://latex.artofproblemsolving.com/1/c/6/1c6d304fd84f1093946f35da3b9ef25c0d48bd3c.png)
The sum of the first odd natural numbers is
.
![[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]](http://latex.artofproblemsolving.com/4/2/3/4234a9f1ce8beb18aab9f27831bf936117909db1.png)
The sum of the first positive integers is
.
![[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]](http://latex.artofproblemsolving.com/b/0/d/b0d29c83d1fc57ecd46538ff58ac9f7da408966b.png)
The sum of the first positive integers is
.[1]
![[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]](http://latex.artofproblemsolving.com/c/5/8/c5855b232fe3f51a7ac4921582cd4f078c678f62.png)
Nichomauss' Theorem: can be written as the sum of
consecutive integers, and consequently that
.
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:
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+1,0); // 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)); /* 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]](http://latex.artofproblemsolving.com/6/f/b/6fbc29c782aefbf321df4ea74a872d77c93fc894.png)
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]](http://latex.artofproblemsolving.com/b/8/c/b8c5626bfc477ab330c923bae0597312aac2b7d2.png)
The th pentagonal number is the sum of
and three times the
th triangular number.
If denotes the
th pentagonal number, then
.
![[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]](http://latex.artofproblemsolving.com/8/5/c/85cd658253f3ee0946038463262b6673d2382545.png)
The identity
![$F_1^2 + F_2^2 + \cdots + F_n^2 = F_{n} \cdot F_{n+1}$](http://latex.artofproblemsolving.com/6/2/4/6246bf460bbd836a5802d62b9c1d42386cd2fde4.png)
![$F_i$](http://latex.artofproblemsolving.com/c/2/e/c2ececdc0bb3ce97a20ed0b4fb7c3b1dd8383e9b.png)
![$i$](http://latex.artofproblemsolving.com/3/4/8/34857b3ba74ce5cd8607f3ebd23e9015908ada71.png)
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]](http://latex.artofproblemsolving.com/5/6/3/563512fa66c538144a78d03360797d66f4af42d9.png)
The infinite geometric series .
![[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]](http://latex.artofproblemsolving.com/c/6/5/c655fc6d2f47f8769aef0bbf6b14eb09d86bffc8.png)
The infinite geometric series .
![[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]](http://latex.artofproblemsolving.com/a/a/4/aa4c7024e5207f42218dcdc03f597170aefeea33.png)
The infinite geometric series .
![[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]](http://latex.artofproblemsolving.com/f/0/b/f0b0d378c16007488639dde8c3db62d0b4b19757.png)
Another proof of the identity .
![[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]](http://latex.artofproblemsolving.com/3/5/c/35ca1caaa3ff0d0eb67f64f8c0c48da6df657b15.png)
The infinite geometric series
![$\sum_{n=0}^{\infty} r^n = \frac{1}{1-r}$](http://latex.artofproblemsolving.com/8/9/7/897d80632398853df6a796efe4b74c75633eac8d.png)
![[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]](http://latex.artofproblemsolving.com/d/9/1/d912b33e1942a66b16f21008945119e3f9cd0917.png)
The arithmetic-geometric series
![$\sum_{n=1}^{\infty} nr^n = \sum_{n=1}^{\infty} \sum_{i=n}^{\infty} r^i = \sum_{n=1}^{\infty} \frac{r^{-n}}{1-r} = \frac{r}{(1-r)^2}$](http://latex.artofproblemsolving.com/a/0/8/a081a904f3a130ee74d736d83aa5dccc5af48aa3.png)
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]](http://latex.artofproblemsolving.com/0/3/e/03efca9d35086370764cd02505bf7e80846486ef.png)
The Pythagorean Theorem (first of many proofs): the left diagram shows that
![$c^2 = 4 \cdot \frac{ab}2 + (b-a)^2 = a^2 + b^2$](http://latex.artofproblemsolving.com/4/6/4/464d2cfe4e10d84b3d566ec24263aacdcb35ece9.png)
![$a^2 + b^2$](http://latex.artofproblemsolving.com/d/7/d/d7d3200f508f2bf0ee872dcc13ec904170623dca.png)
![$c^2$](http://latex.artofproblemsolving.com/2/b/6/2b693026c70424da68ed4c52d3d0d35d9548ee4e.png)
![[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]](http://latex.artofproblemsolving.com/7/3/9/73925776fe9c42e0ee108acf65590198b736e57e.png)
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]](http://latex.artofproblemsolving.com/a/f/1/af1ed80c71b6e50af6c1e6f18f992a9a123a3f98.png)
Another proof of the Pythagorean Theorem; the left-hand diagram suggests the identity
![$c^2 = (a+b)^2 - 4 \cdot \frac{ab}2 = a^2 + b^2$](http://latex.artofproblemsolving.com/3/7/e/37e209892311f0583a33489cd93f60da5024a754.png)
![[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]](http://latex.artofproblemsolving.com/6/f/a/6fabd803733c36b1f39ecdec838548a74087b3b1.png)
COMING: The last (sixth) proof of the Pythagorean Theorem we shall present on this page, this one by dissection.
![[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]](http://latex.artofproblemsolving.com/c/4/9/c49daf6225720830966a6808cc8337f079bd12fb.png)
The smallest distance necessary to travel between
![$(a,b)$](http://latex.artofproblemsolving.com/8/e/8/8e8da0aef2d19017da7471378386d61620f288f5.png)
![$(c,d)$](http://latex.artofproblemsolving.com/7/9/e/79e31171b9eaf222673b566ed3cf773e0f0b532c.png)
![$b,d > 0$](http://latex.artofproblemsolving.com/8/5/4/8543796ca47c5e9ee7ffc7f4684cfb12d56d55e0.png)
![$\sqrt{(a-c)^2 + (b+d)^2}$](http://latex.artofproblemsolving.com/6/e/d/6eddd2a26bb10fda420bd52754f0cb50471959e6.png)
![[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]](http://latex.artofproblemsolving.com/e/9/2/e92139deedc0ef1697160daa79d632b4ed2e878e.png)
In trapezoid
![$ABCD$](http://latex.artofproblemsolving.com/f/9/e/f9efaf9474c5658c4089523e2aff4e11488f8603.png)
![$\overline{AB} \parallel \overline{CD}$](http://latex.artofproblemsolving.com/a/e/6/ae673cfa079c788363f03f09b685855e459bb074.png)
![$\overline{AC} \perp \overline{BD} \Longleftrightarrow AC^2 + BD^2 = (AB + CD)^2$](http://latex.artofproblemsolving.com/8/6/8/8683269759d813f1b307e7e4cf5672070af72fb3.png)
Miscellaneous
![[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]](http://latex.artofproblemsolving.com/1/7/b/17b7d227148c002ef58e4b0ab8e99024c2d844d1.png)
The Root-Mean Square-Arithmetic Mean-Geometric Mean inequality,
![$\color{red}{ab} \color{black} \le \color{green} \frac{(a+b)^2}{4} \color{black} \le \color{blue} \frac{a^2 + b^2}{2}$](http://latex.artofproblemsolving.com/7/1/2/7125d32816aebd2c625a543e5df33e45bab32bfc.png)
![[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]](http://latex.artofproblemsolving.com/f/b/c/fbc7c9442d05431c8665be2a1052a21c3ec7874e.png)
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]](http://latex.artofproblemsolving.com/1/8/4/184e1a740aad3360df022cc73483e55119ef068c.png)
Fermat's Little Theorem: for
(above
).
References
- ^ MathOverflow
- ^ Wolfram MathWorld
- ^ Attributed to the Chinese text Zhou Bi Suan Jing.
- ^ This is more of a proof without words of the AM-GM inequality
; though the lengths of the segments labeled RMS and HM can easily be verified to have values of
, respectively, it might not be obvious from the diagram. It still serves as a useful graphical demonstration of the inequality.