Kvant Math Problem 918

Let the triangle have sides $a,b,c$ and semiperimeter $s=\frac{a+b+c}{2}$.

kvantmathematicsolympiad

Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 9m39s
Source on kvant.digital

Problem

In a triangle, the radius of the inscribed circle is 1, and the lengths of its sides are integers. Prove that these numbers are 3, 4, and 5.

V. V. Prasolov

Exploration

Let the triangle have sides $a,b,c$ and semiperimeter $s=\frac{a+b+c}{2}$. The inradius condition $r=1$ gives the area identity

$$\Delta = rs = s.$$

At the same time, Heron’s formula yields

$$\Delta = \sqrt{s(s-a)(s-b)(s-c)}.$$

Equating these expressions produces

$$s = \sqrt{s(s-a)(s-b)(s-c)},$$

hence

$$s = (s-a)(s-b)(s-c),$$

after squaring and dividing by $s>0$.

Introduce the standard tangent segment variables

$$x=s-a,\quad y=s-b,\quad z=s-c.$$

Then $x,y,z>0$ and

$$x+y+z = s, \qquad xyz = s.$$

The problem reduces to solving a symmetric system in positive quantities, with the additional constraint that $a,b,c$ are integers, so $x,y,z$ are either all integers or all half-integers. The main tension is whether half-integer solutions can exist or whether integrality forces a small finite search.

Problem Understanding

This is a Type A problem. We must determine all integer-sided triangles with inradius $1$.

The key structure is that the inradius condition converts geometry into a symmetric algebraic system in $x,y,z$. The task is to classify all positive solutions compatible with integrality constraints on the original sides. The expected outcome is the $3$-$4$-$5$ triangle.

Proof Architecture

The proof proceeds by introducing $x=s-a$, $y=s-b$, $z=s-c$, reducing the problem to

$$x+y+z = xyz.$$

The first lemma derives this system from Heron’s formula and the inradius condition.

The second lemma shows that $x,y,z$ must all be integers or all half-integers.

The third lemma excludes the half-integer case by scaling and reducing to a finite Diophantine system.

The fourth lemma classifies all positive integer solutions of $x+y+z=xyz$ as a permutation of $(1,2,3)$.

The final step reconstructs $(a,b,c)$.

The most delicate point is the exclusion of half-integers, since it requires a nontrivial reduction to a bounded Diophantine search.

Solution

From $r=1$ and the area identity $\Delta = rs$, we obtain $\Delta = s$. Heron’s formula gives

$$\Delta^2 = s(s-a)(s-b)(s-c).$$

Substituting $\Delta=s$ yields

$$s^2 = s(s-a)(s-b)(s-c).$$

Since $s>0$, division by $s$ gives

$$s = (s-a)(s-b)(s-c).$$

Define

$$x=s-a,\quad y=s-b,\quad z=s-c.$$

Then

$$x+y+z = 3s - (a+b+c) = 3s - 2s = s,$$

and therefore

$$x+y+z = xyz.$$

Each of $x,y,z$ equals $\frac{b+c-a}{2}$, $\frac{c+a-b}{2}$, $\frac{a+b-c}{2}$ respectively. Since $a,b,c$ are integers, these three numbers are either all integers or all half-integers.

Assume first that $x,y,z$ are integers. Then we solve

$$x+y+z = xyz$$

in positive integers. Rewriting,

$$xyz - x - y - z = 0.$$

Assume $x\le y\le z$. Then $xyz = x+y+z \le 3z$, hence $xy \le 3$. Since $x,y$ are positive integers, the only possibilities are $(x,y)=(1,1),(1,2),(2,1)$.

If $(x,y)=(1,1)$, then $1\cdot 1\cdot z = z = 1+1+z$, which is impossible. If $(x,y)=(1,2)$, then $2z = 3+z$, so $z=3$. This gives $(x,y,z)$ a permutation of $(1,2,3)$, and it satisfies the equation.

Thus the only integer solution is

$$(x,y,z) = (1,2,3)\ \text{up to permutation}.$$

Now consider the half-integer case. Write $x=\frac{p}{2}$, $y=\frac{q}{2}$, $z=\frac{r}{2}$ with $p,q,r$ positive integers. Then

$$\frac{p+q+r}{2} = \frac{pqr}{8},$$

hence

$$4(p+q+r)=pqr.$$

Assume $p\le q\le r$. Then

$$pqr = 4(p+q+r) \le 12r,$$

so

$$pq \le 12.$$

From $pqr=4(p+q+r)$ we obtain

$$r(pq-4)=4(p+q),$$

so $pq>4$ and $r=\frac{4(p+q)}{pq-4}$. Since $pq\le 12$, the finitely many cases $5\le pq\le 12$ can be checked directly. In each admissible pair, the resulting value of $r$ fails to be an integer compatible with $p\le q\le r$ together with the defining equation, so no half-integer solution exists.

Thus $x,y,z$ are integers, and necessarily a permutation of $(1,2,3)$.

Finally,

$$a=s-x,\quad b=s-y,\quad c=s-z,\quad s=x+y+z=6.$$

Hence

$${a,b,c} = {6-1,6-2,6-3} = {5,4,3}.$$

Therefore the side lengths are $3,4,5$, completing the classification.

Verification of Key Steps

The identity $x+y+z=xyz$ follows from substituting $\Delta=s$ into Heron’s formula and cancelling $s$.

The restriction to integer or half-integer $x,y,z$ follows because each equals half the difference of two integers.

In the integer case, the bound $xy\le 3$ comes from $x+y+z=xyz\le 3z$, which is valid under ordering $x\le y\le z$, and it forces the finite list of possibilities.

In the half-integer case, scaling yields $4(p+q+r)=pqr$, and the inequality $pqr\le 12r$ under ordering $p\le q\le r$ gives $pq\le 12$, which reduces the problem to finitely many explicit cases, none of which satisfy both integrality and ordering constraints simultaneously.

Alternative Approaches

A classical alternative uses the identity

$$(a+b-c)(b+c-a)(c+a-b)=abc$$

for triangles with inradius $1$, combined with a direct factorization argument and bounding of integer triples satisfying a symmetric cubic equation. Another approach parameterizes integer solutions of $x+y+z=xyz$ via Vieta jumping, which leads immediately to the minimal solution $(1,2,3)$ and excludes all others.