Ken,
I attacked the “triangle areas and perimeters” problem of finding a triangle whose area is M times the
perimeter in a different manner than the solutions you posted, so I thought you might be interested in my
results. I found that a triangle has an area which is M times the perimeter if three integers a, c, and e can be
chosen such that
e
4a  c M 2
. The lengths of the sides of the triangle are then a+c, a+e, and c+e.
ac  4M 2
Also, limits can be placed on the possible values of a and c for a given M, so it is not difficult to generate
all triangles for a given M.
Details:
I first inscribed a circle in a triangle and drew segments from the center of the circle to the vertices of the
triangle. I also drew the three radii (length R) of the circle at the points where they are perpendicular to the
triangle sides (see figure below). The area of AOC is ½ AC  R. Similarly, the area of COB is ½ CB  R
and BOA is ½ AB  R. Adding these three areas gives ½ R ( AC + CB + BA), or ½ R times the perimeter.
Therefore the problem of finding triangles whose areas are M times their perimeters is just that of finding a
triangle with integer-length sides which can be drawn around a circle of radius R = 2M.
C
f
e
a
d
O
A
b
B
c
In the figure above, the circle is divided into three pairs of congruent angles. Each angle has an opposite
segment that makes up part of the perimeter. Since three of the segments/angles are copies of another, we
only have to find the three unique ones. Since each of these angles has a congruent counterpart, the sum of
all three will be 180 degrees. So if we can find three angles that total 180 degrees and each angle’s tangent
multiplied by 2M is an integer, we’re done. AB = a+c. AC = a+e. CB = c+e. Area = 2M(a+c+e).
To get a, c, and e, I draw a half circle of radius 2M. Tangent to
one side I draw a vertical segment a of integer length, and tangent
to the other I draw segment c of integer length. One end point of
each segment is the point of tangency. Then I draw lines g and h
connecting the other endpoints to the center of the circle. These
lines and the diameter of the half circle form three angles. Two
of them, by definition, have tangents that are integers when
multiplied by 2M. The trick is to choose lengths a and c so that
the third tangent will also be an integer when multiplied by 2M.
e
c
h
a
The third segment, e, extends from its point of tangency at the
g
intersection of g and the circle to its intersection with line h. The
slope of line g is a/2M. Line e is perpendicular to this, so its
Z
O 2M
2M
slope is -2M/a. The slope of line h is -c/2M. So if the center of
the circle is at 0,0 then the equation for line h is yh = (-c/2M)xh.
The equation for line e is ye = (-2M/a)(xe – xZ), where xZ is the x
coordinate of Z. The length of g is the same as the length of OZ
because of similar triangles, so xZ = (4M2 + a2)½. The intersection coordinates (xi, yi) of h and e is derived
below:

c
 2M
Xi 
xi  4 M 2  a 2
2M
a
xi 
 4M 2 4M 2  a 2
ac  4M 2
yi 
c
2cM 4M 2  a 2
xi 
2M
ac  4M 2

The coordinates of the point of tangency of e are xt = 2aM/(4M2 + a2)1/2 and yt = 4M2/(4M2 + a2)1/2.
The length of e along the x-axis from its intersection with h to its point of tangency is then
4M 2 4M 2  a 2
4aM 2 a  c 


.
ac  4M 2
4M 2  a 2
ac  4 M 2 4M 2  a 2
2aM


The height of e along the y-axis from its intersection with h to its point of tangency is
2cM 4 M 2  a 2
4M 2
8M 3 a  c 


.
ac  4 M 2
4M 2  a 2
ac  4 M 2 4 M 2  a 2


The length of e is then the square root of the sum of the squares of these two values. This reduces to
4a  c M 2
.
ac  4M 2
For a given M, which values of a and c do we need to try in the above equation? Because we want a, c,
and e to yield integers when summed in pairs, we know a, c, and e must either all be integers, or all be
integers divided by two. If a were l+m/n (j,k,l,m,n are integers; m<n; m,n are relatively prime), c would
have to be k+(n-m)/n to sum to an integer. Also, e would have to be j+(n-m)/n. But then c+e would be
k+j+2(n-m)/n, which can only be an integer if n is 2.
Now if a were j/2 and c were k/2 (j,k are odd integers), e would be 8(j+k)M2/(jk-16M2). For a+e to be an
integer, the denominator of e must also be 2, but since j and k are odd, we know jk-16M2 is also odd.
8(j+k)M2 is an integer, so we know that e cannot have a denominator of 2. Thus a, c, and e must be
integers for the sides of the triangle to be integers.
Let us define a as the smallest of the three segments. It therefore has the smallest angle at the circle’s
center. Since the three angles sum to 180 degrees, and c and e are larger or equal to a, the angle associated
with a can be at most 60 degrees. So a can be at most 2Mtan60, or 3.464M. So we only have to try values
for a between 1 and the floor of 3.464M. ac-4M2 must be positive, so c must be at least (4M2+1)/a. We
can require c to be the second smallest segment. Then c must be less than 4(a+c)M2/(ac-4M2). Solving
for c gives c < (2M/a)(2M+(4M2 + a2)½), which is less than 2M + 8M2/a . These limits can be used to
generate all solutions for a given M.
G. Oliver Stone
November 9, 1999