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Here is a simple question, but the answer is fairly unexpected.

Which of the following triangles has the larger inscribed circle?

A. the triangle with sides 17, 25, and 26

B. the triangle with sides 17, 25, and 28

In fact, they have inscribed circles of the same size! In addition, the radius is an integer.

Such triangle doublets are "miraculous" since:

  1. All of the six sides are positive integers.
  2. One triangle has sides $a$, $b$ and $c$, while the other triangle has sides $a$, $b$ and $d$ ($c$ is not equal to $d$).
  3. They have inscribed circles of the same size.
  4. The radius of each of inscribed circles is a positive integer.

Obviously, if $(a,b,c)$ and $(a,b,d)$ can form such a doublet, then so can $(ka, kb, kc)$ and $(ka, kb,kd)$ for any positive integer $k$. As a result, only "primitive" examples matter.

One of my friend, who is good at programming, offered me some examples:

{{{17,25,26},6},{{17,25,28},6}}

{{{41,50,39},12},{{41,50,73},12}}

{{{61,195,160},21},{{61,195,232},21}}

{{{65,68,57},18},{{65,68,105},18}}

{{{97,169,122},30},{{97,169,228},30}}

{{{97,340,339},42},{{97,340,345},42}}

{{{100,291,289},42},{{100,291,299},42}}

{{{246,365,329},84},{{246,365,455},84}}

{{{255,260,103},42},{{255,260,485},42}}

{{{257,289,168},60},{{257,289,482},60}}

{{{260,629,381},30},{{260,629,879},30}}

{{{305,377,124},42},{{305,377,660},42}}

{{{313,338,155},60},{{313,338,601},60}}

{{{337,625,386},84},{{337,625,888},84}}

{{{353,578,481},120},{{353,578,735},120}}

{{{364,505,545},126},{{364,505,561},126}}

{{{609,740,355},126},{{609,740,1241},126}}

{{{623,884,807},210},{{623,884,1111},210}}

{{{641,841,780},210},{{641,841,1082},210}}

It seems that such "primitive" solutions are infinite, but I can't prove it.

Besides, can we find a formula to generate some primitive solutions?

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    Are you aware of the very recent question on this site: (http://math.stackexchange.com/q/2102886)? – 2017-01-19
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    But it seems that this very recent question concerns only Pythagorean triangles. – 2017-01-19
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    Primitive triangles can be sorted by their area radical. http://demonstrations.wolfram.com/TriangleRadicals/ You're looking at the Heronian triangles. http://www.wm-archive.uni-bayreuth.de/index.php?id=554&L=3 has a long list. – 2017-01-19

1 Answers 1

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The radius equation for the incircle can be given by

$$r=\frac{\Delta}{s}=\frac{2\Delta}{a+b+c}$$

We can solve for c from the above equation by using Heron's formula for $\Delta$, given that $a_1=a_2=a,\ b_1=b_2=b,\ r_1=r_2=r$.

It is no surprise that the equation simplifies to a quadratic equation and in some cases may have two distinct real positive roots which state your case.

Hence we can have $c_1\ne c_2$, and the number of such triangles is infinite as we can form infinite quadratic equations which suits the above case.