Kvant Math Problem 1356

We are given a triangle with sides $a$, $b$, $c$ and the relation

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Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 35m03s
Source on kvant.digital

Problem

Prove that if $abc=4Rrr_c$, where $a$, $b$, $c$ are the sides of a triangle, and $R$, $r$, $r_c$ are the radii of the circumcircle, incircle, and one of the excircles, then the triangle is right-angled. (The excircle touches the side $c$ and the extensions of the other two sides.)

B. Tureshbaev, 11th grade student

Exploration

We are given a triangle with sides $a$, $b$, $c$ and the relation

$abc = 4R r r_c,$

where $R$ is the circumradius, $r$ is the inradius, and $r_c$ is the exradius opposite side $c$. The goal is to prove that this condition implies the triangle is right-angled. Standard formulas express these radii in terms of the triangle’s semiperimeter $s = \frac{a+b+c}{2}$ and area $\Delta$ as

$R = \frac{abc}{4\Delta}, \qquad r = \frac{\Delta}{s}, \qquad r_c = \frac{\Delta}{s-c}.$

Substituting these into the given relation will produce an identity involving only $s$, $\Delta$, and the side lengths. Using Heron's formula

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

will then allow simplification to a relation purely among the sides $a$, $b$, $c$, which can be shown to force $c^2 = a^2 + b^2$.

Problem Understanding

The problem is to show that a specific product of side lengths and radii characterizes right-angled triangles. The natural approach is to eliminate the radii using known formulas and reduce the given condition to an algebraic equation among the sides. The central step is to correctly manipulate the resulting identity and demonstrate that it is equivalent to the Pythagorean theorem.

Proof Architecture

The proof proceeds by substituting the standard formulas for $R$, $r$, and $r_c$ into the given condition, obtaining an equation involving $\Delta$ and $s$. Heron's formula is then applied to replace $\Delta$ with an expression in terms of the sides and semiperimeter. This produces a cubic identity in $s$, which can be systematically simplified to reveal that $c^2 = a^2 + b^2$. The key is careful algebra in the step connecting $(s-a)(s-b)$ and $s(s-c)$ to the Pythagorean condition.

Solution

Let $\Delta$ denote the area of the triangle and $s = \frac{a+b+c}{2}$ the semiperimeter. The standard formulas for the circumradius, inradius, and exradius opposite side $c$ are

$R = \frac{abc}{4\Delta}, \qquad r = \frac{\Delta}{s}, \qquad r_c = \frac{\Delta}{s-c}.$

Substituting these into the given condition yields

$4R r r_c = 4 \cdot \frac{abc}{4\Delta} \cdot \frac{\Delta}{s} \cdot \frac{\Delta}{s-c} = \frac{abc , \Delta}{s(s-c)}.$

Equating this with $abc$ gives

$abc = \frac{abc , \Delta}{s(s-c)} \implies \Delta = s(s-c),$

where division by $abc$ is justified since the triangle is nondegenerate and $abc > 0$.

Applying Heron's formula, we have

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

Equating this to $\Delta = s(s-c)$ gives

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

Expressing each term in terms of the sides $a$, $b$, $c$:

$s = \frac{a+b+c}{2}, \quad s-a = \frac{b+c-a}{2}, \quad s-b = \frac{a+c-b}{2}, \quad s-c = \frac{a+b-c}{2}.$

Substituting these into $(s-a)(s-b) = s(s-c)$ gives

$\frac{b+c-a}{2} \cdot \frac{a+c-b}{2} = \frac{a+b+c}{2} \cdot \frac{a+b-c}{2}.$

Multiplying both sides by 4 produces the equivalent equation

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

Expanding both sides carefully, the left-hand side becomes

$$\begin{aligned} (b+c-a)(a+c-b) &= (b)(a) + (b)(c-b) + (c-a)(a+c-b) \ &= ab + bc - b^2 + ac + c^2 - bc - a^2 - ac + ab \ &= 2ab - a^2 - b^2 + c^2 \ &= c^2 - (a-b)^2 + 2ab. \end{aligned}$$

More systematically, expand $(b+c-a)(a+c-b)$:

$$(b+c-a)(a+c-b) = b a + b c - b^2 + c a + c^2 - c b - a^2 - a c + a b = 2ab - a^2 - b^2 + c^2,$$

which simplifies to

$c^2 - (a-b)^2 + 2ab = a^2 + b^2 + 2ab - (a^2 + b^2 - c^2) = 2ab + c^2 - a^2 - b^2.$

The right-hand side expands as

$(a+b+c)(a+b-c) = (a+b)^2 - c^2 = a^2 + 2ab + b^2 - c^2.$

Equating the two expansions gives

$2ab + c^2 - a^2 - b^2 = a^2 + 2ab + b^2 - c^2 \implies 2c^2 = 2(a^2 + b^2) \implies c^2 = a^2 + b^2.$

Thus the triangle satisfies the Pythagorean theorem. By the converse of the Pythagorean theorem, the angle opposite side $c$ is a right angle, and the triangle is right-angled.

$\boxed{\text{The triangle is right-angled.}}$

Verification of Key Steps

The substitution of standard formulas for $R$, $r$, and $r_c$ is valid. The derivation of $\Delta = s(s-c)$ follows directly from the given condition. Applying Heron's formula to replace $\Delta$ is justified, and the algebra leading from $(s-a)(s-b) = s(s-c)$ to $c^2 = a^2 + b^2$ is carried out systematically with no omitted factors. Each step relies solely on standard triangle identities and nondegeneracy assumptions, making the argument complete.

Alternative Approaches

One may directly express $r$ and $r_c$ in terms of $a$, $b$, $c$ and $R$ without invoking the semiperimeter. Using $r = \frac{2\Delta}{a+b+c}$ and $r_c = \frac{2\Delta}{-a-b+c+2c}$ eventually reduces the condition to a quadratic in $c^2$, which leads to the same conclusion $c^2 = a^2 + b^2$. The approach through semiperimeter and Heron’s formula is more elegant because it reduces the calculation to a single identity and factors naturally.