## A Triangle and a Circle

A circle overlaps all three sides of a triangle splitting each side into three segments. Find such a triangle, such that the lengths of all nine segments are unique integers.

Can a triangle and circle be found, such that the radius of the circle is also an integer?

For example, in triangle ABC, the segments would be: AB:(a,b,c), BC:(d,e,f), and CA:(g,h,i), with b, e, and h as chords of the circle. There are probably solutions in which all lengths are positive, as well as solutions in which one of the chords is of length 0 (the circle is tangent to that side of the triangle.) See if you can minimize the perimeter of the triangle. ( For example, a near miss is (6,0,3), (1,8,6), (7,5,4). )

[For a very useful hint, see Theorem 11.16 and its Corollary at http://library.advanced.org/16284/reference_gc_9.htm.]

Source: Original.

Solutions were received from Philippe Fondanaiche.

His solution below shows a way to find a solution and substitutes values to find the segment lengths. His general solution depends upon the six intersection points being on three diameters of the circle. I think the large values he ends up with are a result of this.

With a lot of playing with numbers, I was able to find a solution with all positive lengths and a perimeter of 54. The triangle's segments are (3,1,9),(5,13,7),(10,4,2). An interesting note is that simply by changing (7,10) to (32,36) another triangle with the same chord lengths can be found. If we allow a tangent to the circle on one side, I found a solution with a perimeter of 45. The triangle's segments are (2,0,8),(4,12,6),(9,3,1). I don't know an easy way to calculate the radius of the circle, so I don't know what the smallest solution with integral radius might be.

Philippe sent this even smaller solution 5/23/99:

``` ¤ The smallest size of the triangle ABC is defined by AB=10,  BC=17,  CA=18
representing a perimeter of 45,split as:
AI=5, IJ=3, JB=2, BK=1, KL=9, LC=7, CM=8, MN=6, NA=4 (all the numbers
between 1 and 9!)
which satisfy the following relations:
AI*(AI+IJ) = 5*8 = 40 = AN*(AN+NM) = 4*10
BJ*(BJ+JI) = 2*5 = 10 = BK*(BK+KL) = 1*10
CL*(CL+LK) = 7*16 = 112 = CM*(CM+MN) = 8*14
We can check easily that the six points are the same circle by calculating
the cosines of the angles IJN, IKN and ILN. We obtain the same value
0,868599036...We infer that these angles are equal ==> K and L are on the
same circle as the point I,J,N.
The corresponding radius of the circle is 4*sqrt(91/55) = 5,145......

¤ To find a triangle and a circle such as the 9 lengths and the radius of
the circle are the smallest integers as possible is another story...
```

Philippe's general solution:
```I have solved the problem by using the properties of the pythagorean
triangles. In these right triangles such as a^2 + b^2 = c^2 with a,b,c
integers, the sines and cosines of the two acute angles are rational ( a/c
and b/c).It is true too for the sines and the cosines of the sum of two acute
angles belonging to two different pythagorean triangles.

Let consider the triangle ABC whose the three sides are overlapped by a
circle of diameter D at the points I and J on AB, K and L on BC, M and N on
CA. We have to find such a triangle such as the lengths of the nine segments
AI=a, IJ=b, JB=c, BK=d, KL=e, LC=f, CM=g, MN=h, NA=i are unique integers.
Let draw a circle of center O and diameter D. Let J and M diametrically
opposite. Let I on the circle such u = angle(IMJ) and N on the arc IM of the
circle such as v = angle(MJN). It is easy to check that u+v < 90°.
Let L and K respectively diametrically opposite to I and N. We have built an
hexagon IJKLMN inscribed in a circle. The lines IJ and MN cut themselves at
A, the lines IJ and KL at B, at last the lines KL and MN at C.
Considering the triangles IJM, JMN, IJN, AIN, BJK and CLM and using for each
of the them the well-known relation of the sines , we find easily the
equations giving the lengths of the nine segments a,b,c,d,e,f,g,h,i according
to the diameter D.

Indeed ¤ IJLM and JKMN are rectangles,
¤ angle(IAN) = angle(ABC) = angle(CLM) = 90°-v,
¤ angle(ANI) = angle(ACB) = angle(BKJ) = 90°-u,
¤ angle(BAC) = angle(BJK) = angle(CML) = u+v,
Therefore:
IJ = D*sin(u)
MN = D*sin(v)
KL = IN = D*sin(90°-u-v) = D*cos(u+v)
AI = D*cos(u)/tan(u+v)
AN = D*cos(v)/tan(u+v)
BJ = D*cos(u)*tan(v)
BK = D*sin(u+v)*tan(v)
CL = D*sin(u+v)*tan(u)
CM = D*cos(v)*tan(u)

So,if u and v are the acute angles of two different pythagorean triangles,
all the lengths above mentioned and related to D are rational. The adequate
choice of D allows us to obtain unique integers for the nine lengths and the
diameter D of the circle.

As an example among an infinity of solutions, we can fix the following values:
sin(u) = 3/5     cos(u) = 4/5     tan(u) = 3/4
sin(v) = 5/13    cos(v) = 12/13  tan(v) = 5/12
Then:
sin(u+v) = 56/65     cos(u+v) = 33/65    tan(u+v) = 56/33

So:
IJ = 3*D/5
MN = 5*D/13
KL = 33*D/65
AI = 33*D/70
AN = 99*D/182
BJ = D/3
BK = 14*D/39
CL = 42*D/65
CM = 9*D/13
We can check that all the lengths are different.

The lowest common multiple of the denominators of these fractions is
2*3*5*7*13 = 2730
Therefore with a diameter equal to D = 2730 (that is to say a radius equal to
1365), we get:
IJ = 1638,  MN = 1050, Kl = 1386, AI = 1287, AN = 1485, BJ = 910, BK = 980,
CL = 1764, CM = 1890.

Obviously, we can check with these figures the theorem 11.16 and its
corollary that you have mentioned in the terms of the problem.

[I did this below. -KD]
AI*(AI+IJ) ?=? AN*(AN+NM)
1287*(1287+1638) ?=? 1485*(1485*1050)
1287*2925 ?=? 1485*2535
3764475 = 3764475.  Check!

BJ*(BJ+IJ) ?=? BK*(BK+KL)
910*(910+1638) ?=? 980*(980+1386)
910*2548 ?=? 980*2366
2318680 = 2318680.  Check!

CL*(CL+KL) ?=? CM*(CM+MN)
1764*(1764+1386) ?=? 1890*(1890+1050)
1764*3150 ?=? 1890*2940
5556600 = 5556600.  Check!
```

Mail to Ken