Kvant Math Problem 841

Consider a right triangle $ABC$ with right angle at $C$.

kvantmathematicsolympiad

Verified: no
Verdicts: SKIP + SKIP
Solve time: 1m17s
Source on kvant.digital

Problem

Prove that the product of the lengths of the segments into which the hypotenuse of a right triangle is divided by the point of tangency of its incircle is equal to the area of the triangle.

Exploration

Consider a right triangle $ABC$ with right angle at $C$. Denote the hypotenuse as $AB$ and let the incircle touch $AB$ at point $D$. Denote the segments $AD = p$ and $DB = q$, so that $AB = p + q$. The claim is that $p \cdot q$ equals the area of triangle $ABC$. Assign side lengths $AC = b$, $BC = a$, $AB = c$, and let $s = \frac{a+b+c}{2}$ be the semiperimeter. From standard formulas for the incenter and tangency points, we know $AD = s - a$ and $DB = s - b$. Then $p \cdot q = (s - a)(s - b)$. The area of a right triangle is $\frac{1}{2}ab$, so the task reduces to showing $(s - a)(s - b) = \frac{1}{2}ab$ in this context. Trying small numeric examples, say $a = 3$, $b = 4$, $c = 5$, yields $s = 6$, $p = s - a = 3$, $q = s - b = 2$, $p \cdot q = 6$, and $\frac{1}{2}ab = 6$. The equality holds. The core point appears to be expressing $p$ and $q$ in terms of $a$, $b$, and $c$ using the semiperimeter and then simplifying.

Problem Understanding

The problem asks to prove that for a right triangle, the product of the segments into which the hypotenuse is divided by the point of tangency of the incircle equals the area of the triangle. This is a Type B problem: a pure proof. The main difficulty is rigorously expressing the segments on the hypotenuse in terms of the triangle’s sides and confirming that their product equals the area without skipping any algebraic justification. The intuitive reason it works is that the tangency points of the incircle satisfy $AD = s - a$ and $DB = s - b$, leading to a symmetric formula that matches the area formula.

Proof Architecture

Lemma 1: In any triangle, the incircle divides each side into segments equal to the semiperimeter minus the opposite side, specifically $AD = s - a$ and $DB = s - b$. This follows from the tangency formulas of the incircle with sides.

Lemma 2: In a right triangle with legs $a$, $b$ and hypotenuse $c$, the semiperimeter is $s = \frac{a + b + c}{2}$, so $s - a = \frac{b + c - a}{2}$ and $s - b = \frac{a + c - b}{2}$. This is simple arithmetic.

Lemma 3: In a right triangle, $c = \sqrt{a^2 + b^2}$, and therefore $(s - a)(s - b) = \frac{1}{2}ab$. The proof requires expanding $(s - a)(s - b)$ algebraically and simplifying carefully to match the area formula $\frac{1}{2}ab$.

The hardest step is Lemma 3 because careless expansion might overlook the right-triangle relation and give a different value.

Solution

Let $ABC$ be a right triangle with right angle at $C$, legs $AC = b$, $BC = a$, and hypotenuse $AB = c$. Denote the incenter by $I$ and let the incircle touch $AB$ at $D$. The semiperimeter is $s = \frac{a + b + c}{2}$. By standard properties of the incircle, $AD = s - a$ and $DB = s - b$. Then

$(s - a)(s - b) = \left(\frac{a + b + c}{2} - a\right)\left(\frac{a + b + c}{2} - b\right) = \left(\frac{b + c - a}{2}\right)\left(\frac{a + c - b}{2}\right).$

Since $c^2 = a^2 + b^2$, we have $c = \sqrt{a^2 + b^2}$, and hence

$b + c - a = b - a + \sqrt{a^2 + b^2}, \quad a + c - b = a - b + \sqrt{a^2 + b^2}.$

Therefore

$(s - a)(s - b) = \frac{(b - a + \sqrt{a^2 + b^2})(a - b + \sqrt{a^2 + b^2})}{4} = \frac{(\sqrt{a^2 + b^2})^2 - (a - b)^2}{4}.$

Expanding the numerator yields

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

so that

$(s - a)(s - b) = \frac{2ab}{4} = \frac{ab}{2}.$

The area of triangle $ABC$ is $\frac{1}{2}ab$, which coincides with $(s - a)(s - b)$. This completes the proof.

∎

Verification of Key Steps

The critical algebra is the expansion $(\sqrt{a^2 + b^2})^2 - (a - b)^2 = a^2 + b^2 - (a^2 - 2ab + b^2) = 2ab$. Testing with numeric examples confirms correctness: for $a = 3$, $b = 4$, $c = 5$, we have $(s - a)(s - b) = 3 \cdot 2 = 6$ and $\frac{1}{2}ab = 6$. The expression $(s - a)(s - b)$ is sensitive to sign errors in $a - b$, so carefully writing it as $(\sqrt{a^2 + b^2} - (a - b))(\sqrt{a^2 + b^2} + (a - b))$ ensures correct factorization.

Alternative Approaches

A geometric approach considers the right triangle as composed of two smaller right triangles sharing the incenter and constructs the segments along the hypotenuse by similarity. One shows that the product of the segments equals the area by the geometric mean property of altitudes and projections. The algebraic approach is preferable because it gives an explicit, general formula without introducing additional constructions or ratios, making it fully rigorous and immediately verifiable for arbitrary right triangles.