Given an arbitrary triangle ABC, let D be the foot of the perpendicular

from A to BC, let E be the foot of the perpendicular from D to AC, and

let F be a point on the line DE, as illustrated below

Florin Pirvanescu challenged the readers of Mathematics Magazine in

June 1991 to prove that AF is perpendicular to BE if and only if

FE/FD = BD/DC. Several proofs have appeared, based on synthetic,

projective, and vector methods. These proofs are all fairly elaborate,

but there is actually a very simple elementary proof, which is a nice

example of "thinking outside the box".

Let G be the foot of the perpendicular from B to DE, as shown below.

Clearly BDG ~ DAE and DGE ~ DBC, so we have BG/DG = DE/AE and

EG/DG = CB/DB. Also, AEF ~ EGB, giving EG/BG = AE/FE, if and only

if AF is perpendicular to BE. Thus we have

/BG\ /EG\ /DE\ /AE\

( -- )( -- ) = ( -- )( -- )

\DG/ \BG/ \AE/ \FE/

which implies DE/FE = EG/DG = CB/DB and so FED ~ DBC if and only if

AF is perpendicular to BE.

Double Equations from Triangles in Squares

Given the square [1234] in the figure below, is it possible for both

of the inscribed right triangles [125] and [345] to have integer

(or rational) edge lengths?

If we let "a" denote the length of the edge [12] and "b" denote the

intermediate length [25], then obviously the condition for both

triangles to be Pythagorean is that both of the equations

a^2 + b^2 = f^2 a^2 + (a-b)^2 = h^2 (*)

are satisfied, where f and h are integers representing the lengths

of the "hypotenuses" [15] and [45] respectively.

Now suppose that instead of drawing the line from [5] to [4], we

draw a line from [5] to [6] so that the angle (156) is a right angle,

and then we draw the line [6,1]. Thus the original square [1234] is

partitioned into four right triangles. Is it possible for all four of

these to be Pythagorean triangles? Interestingly, the necessary and

sufficient condition for this is the same as for the previous case,

i.e., a (non-trivial) integer solution to the double equation (*).

To show this, let's assign letters to the segment lengths as follows:

[12] = a [25] = b [53] = c

[36] = d [46] = e [15] = f

[16] = g [45] = h [56] = j

By similar triangles we have a/c = b/d = f/j. Also, we can immediately

express c,d,e in terms of a,b as follows

c = a-b

d = bc/a = b(a-b)/a

e = a-d = a - b(a-b)/a

Now, in order for all four of the right triangles partitioning the

square to be rational (which can easily be converted to integers),

we must have in addition to a^2 + b^2 = f^2 the equations

j^2 = c^2 + d^2 g^2 = a^2 + e^2

However, the triangle with edge lengths "cdj" is similar to "abf",

so if the "abf" triangle is rational, then it follows that the "cdj"

triangle is also rational. Specifically we have j^2 = (cf/a)^2.

Hence the only real requirement beyond the "abf" triangle being

rational is that the "aeg" triangle be rational, which is true if

and only if

/ b(a-b) \ 2 a^4 + (a^2 - ab + b^2)^2

a^2 + e^2 = a^2 + ( a - ------ ) = ------------------------

\ a / a^2

is a rational square. Obviously the denominator is a square, so we

need only consider the numerator, which factors as

a^4 + (a^2 - ab + b^2)^2 = [a^2 + b^2][a^2 + (a-b)^2]

The first factor on the right side is already known to be a rational

square, since we have required that "abf" is a rational triangle.

Therefore, the other factor must also be a rational square, and so

we arrive at the same double equation as (*) above.

Of course, the equivalence of the rationality conditions for triangles

[345] and [156] was to be expected, because these two triangles are

obviously similar (noting that f/j=a/c) and have at least one rational

edge assuming that [125] is rational. As a result, we can construct

a new square [1587] and we find that point 7 lies along the line

{364}. Needless to say we have [678] similar to [512], and [679] is

similar to [345] and [516].

But none of this answers the original question, which is whether

such constructions are actually possible, i.e., whether there is

an integer solution of the double equation

x^2 + y^2 = m^2 x^2 + (x-y)^2 = n^2

The question of whether two quadratic forms in two variables has

solutions, and if so, whether it has infinitely many, has been studied

for many years, going back to Diophantus, Bachet, Fermat, Euler, and

so on. Notice that the right-hand equation can also be written in

the form 2x^2 - 2xy + y^2 = n^2. Many different techniques have

been developed to tackle this kind of problem, but it still is not

completely solved for arbitrary pairs of quadratic forms.

To tackle this particular pair of equations, we first note that any

common factor in x and y can be divided out of both equations, so we

can assume that both are primitive Pythagorean triples. From this it

follows that x and y have opposite parity, as do x and x-y, which

implies that x must be even and y must be odd. Consequently we have

coprime integers A,B with opposite parity, and coprime integers C,D

with opposite parity, such that

x = 2AB x = 2CD

y = A^2 - B^2 (x-y) = C^2 - D^2

m = A^2 + B^2 n = C^2 + D^2

This shows that AB = CD, so this product must have four pairwise

coprime factors r,s,R,S (precisely one of which is even) such that

A=rs B=RS C=rR D=sS

Adding the previous expressions for y and x-y gives

x = A^2 - B^2 + C^2 - D^2

= (rs)^2 - (RS)^2 + (rR)^2 - (sS)^2

= (r^2 - S^2)(s^2 + R^2)

Also, since x = 2AB = 2CD = 2rsRS, we have

(r^2 - S^2)(s^2 + R^2) = 2rsRS (1)

Since r,s,R,S are pairwise coprime, we know that both r and S are

coprime to the first factor on the left, and both s and R are

coprime to the second factor. Hence, depending on which of the

two left hand factors is even (recalling that precisely one of

r,s,R,S is even) we have one of two cases:

Case 1: Either r or S is even, and we have

r^2 - S^2 = Rs s^2 + R^2 = 2rS (2a,b)

In this case we can factor the left hand equation to give

(r+S)(r-S) = Rs

and since r,s,R,S are pairwise coprime we have pairwise coprime

integers u,v,U,V such that

uv = r+S UV = r-S uV = R vU = s

Hence we have r = (uv+UV)/2 and S = (uv-UV)/2, and we can insert

these expressions into (2b) to give

(vU)^2 + (uV)^2 = 2(uv+UV)(uv-UV)

Expanding the righthand product and re-arranging, we have

U^2 (v^2 + 2V^2) = u^2 (2v^2 - V^2) (3)

and re-arranging differently gives the alternate form

V^2 (u^2 + 2U^2) = v^2 (2u^2 - U^2) (4)

Since gcd(U,u)=1 we know that u^2 divides v^2 + 2V^2, and so on.

Hence equation (3) can be written as

(v^2 + 2V^2) (2v^2 - V^2)

------------ = ------------ = M

u^2 U^2

for some positive integer M. Likewise equation (4) can be written

in the form

(u^2 + 2U^2) (2u^2 - U^2)

------------ = ------------ = N

v^2 V^2

for some positive integer N. Consequently we have the equivalent

pairs of equations

v^2 + 2V^2 = Mu^2 2v^2 - V^2 = MU^2 (5a,b)

u^2 + 2U^2 = Nv^2 2u^2 - U^2 = NV^2 (6a,b)

Multiplying (5a) by N and making the substitutions for Nv^2 and

NV^2 from equations (6) gives

MNu^2 = Nv^2 + 2(NV^2) = (u^2 + 2U^2) + 2(2u^2 - U^2) = 5u^2

This shows that MN = 5 for positive integers M,N, so either M=1,N=5

or else M=5,N=1. Both of these lead to the same reciprocal pair of

quadratic forms (up to some permutation of the variables)

v^2 + 2V^2 = u^2 2v^2 - V^2 = U^2

Case 2: Either R or s is even, and we have

r^2 - S^2 = 2Rs s^2 + R^2 = rS (7a,b)

In this case the left hand side is even, as is Rs, so we can write

r+S r-S R

--- --- = - s

2 2 2

assuming R is even. These factors are all coprime, so there are

pairwise coprime integers u,v,U,V such that

uv = (r+S)/2 UV = (r-S)/2 uV = R/2 Uv = s

This implies r = uv+UV and S = uv-UV. Inserting these into (7b)

gives

(Uv)^2 + 4(uV)^2 = (uv+UV)(uv-UV)

Expanding and re-arranging gives

U^2 (v^2 + V^2) = u^2 (v^2 - 4V^2)

and

V^2 (4u^2 + U^2) = v^2 (u^2 - U^2)

Proceding in the same way as in Case 1, we find that this leads to

a pair of equations of the form

v^2 + V^2 = u^2 v^2 - 4V^2 = U^2

In this case we see the question is equivalent to asking whether -4

is a "concordant number", defined as an integer N such that x^2 + y^2

and x^2 + Ny^2 can both be squares simultaneously. This is discussed

at length in the note Concordant Forms, where a proof is

given that a large class of positive prime values of N are not

concordant.

A different approach is to cast the problem in the form of an elliptic

curve. Returning to equation (1)

(r^2 - S^2)(s^2 + R^2) = 2rsRS

we see that the first factor on the left can be divided by rS and the

second factor by Rs to give

[r/S - S/r][s/R + R/s] = 2

Thus we have rational numbers X=r/S and Y=R/s such that

/ 1 \ / 1 \

( X - --- )( Y + --- ) = 2

\ X / \ Y /

Multiplying through by XY gives

(X^2 - 1)(Y^2 + 1) - 2XY = 0

If we define the new variable Z such that X = (Z+Y)/(Z-Y), then Z

is rational if X and Y are rational (and of course X is rational if

Z and Y are rational). Notice that if Y=0 then X=1 for ANY value of

Z, and this is a solution of the equation. Substituting for X gives

2Y(2ZY^2 + 2Z - Z^2 + Y^2)

-------------------------- = 0

(Z-Y)^2

Setting aside the trivial solution at Y=0, and assuming Y is not equal

to Z (which corresponds to infinite X) we are left with

2ZY^2 + 2Z - Z^2 + Y^2 = 0

which we can solve for Y^2 to give

Z(Z-2)

Y^2 = ------

(2Z+1)

If we now define the variable W by the bi-rational form Y = W/(2Z+1)

and substitute for Y into the above expression, we get the elliptic

curve

W^2 = Z(Z-2)(2Z+1)

This is nearly identical to the elliptic curve that arises when

proving that there cannot exist four squares in arithmetic progression.

Only the sign in the second factor is different. As discussed in

Weil's historical review of "Number Theory", essentially this same

problem was treated by both Euler and Fermat.

A plot of the real part of the elliptic curve y^2 = x(x-2)(2x+1) is

shown below.

There are obviously at least three rational points on this curve,

given by (x,y) = (0,0), (2,0), and (-1/2,0). Notice that these three

rational points lie along the "horizontal" axis of symmetry of the

curve. Any straight line passing through the closed loop on the

left and striking the open branch of the curve on the right has

three real points of intersection, and obviously if two of those

points are rational then the third must be also. This shows how,

if we are given any two rational points, we can generally construct

a third, simply by drawing a line through the two given points and

locating the remaining intersection point.

For example, suppose the point (x1,y1) on the curve above is a rational

point. In that case we could draw the line through the two points

(-1/2,0) and (x1,y1) as shown, and then we would have a new rational

solution at the point (x2,y2). The equation of the line is

y1 y1

y = -------- x + -------

x1 + 1/2 2x1 + 1

Squaring this and equating it with y^2 = x(x-2)(2x+1) gives a cubic

equation in x whose three roots are the x values of the three points

of intersection between the line and the elliptic curve. We already

know that two of these roots are x=-1/2 and x=x1, so we can easily

determine the third root and the corresponding value of y:

(y1)^2 2x2 + 1

x2 = - ------------- y2 = y1 -------

x1(2x1 + 1)^2 2x1 + 1

Hence if x1 and y2 are both rational, then so are x2 and y2. Once we

have found this new rational point we can draw lines through it and

any previously found rational points to generate still more rational

solutions. Of course, in order to accomplish this we need first to

have one rational point off the axis of symmetry.

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