Kvant Math Problem 483

Consider a right triangle with legs $a$ and $b$ and hypotenuse $c$, where $c^2 = a^2 + b^2$.

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Problem

a) Prove that the ratio of the inradius of a right triangle to the sum of the squares of the medians drawn from the acute angles does not exceed $1/5$.

b) Find the maximum value that this ratio can attain.

I. Temraliev

Exploration

Consider a right triangle with legs $a$ and $b$ and hypotenuse $c$, where $c^2 = a^2 + b^2$. The inradius $r$ is given by $r = a + b - c$ over $2$, but more precisely, $r = \frac{a + b - c}{2}$. The medians from the acute angles, say $m_a$ from angle $A$ and $m_b$ from angle $B$, have lengths given by $m_a = \frac12 \sqrt{2b^2 + 2c^2 - a^2}$ and $m_b = \frac12 \sqrt{2a^2 + 2c^2 - b^2}$.

Testing small examples such as an isosceles right triangle with $a = b = 1$ yields $c = \sqrt{2}$, $r = (\sqrt{2})/2$, $m_a^2 + m_b^2 = 9/2$, and the ratio $r / (m_a^2 + m_b^2) = \sqrt{2}/9 \approx 0.157$. A right triangle with sides $3$, $4$, $5$ gives $r = 2$, $m_a^2 + m_b^2 = 25.25$, ratio $\approx 0.079$, smaller than in the isosceles case. This suggests the extremum occurs for the isosceles right triangle.

The core difficulty is expressing $r / (m_a^2 + m_b^2)$ in a form amenable to optimization. Using symmetry in an isosceles right triangle simplifies the algebra, but the general inequality requires careful manipulation to show that no other configuration gives a larger ratio.

Problem Understanding

The problem is of Type C. We are asked to prove that the ratio of the inradius $r$ of a right triangle to the sum of the squares of the medians from the acute angles does not exceed $1/5$ and then to find the maximum value. The ratio is

$$R = \frac{r}{m_a^2 + m_b^2}.$$

The core difficulty is showing that among all right triangles with legs $a$ and $b$, $R \le 1/5$, and then identifying which triangle attains equality. From exploratory calculations, an isosceles right triangle appears to maximize $R$. The expected maximum value is $1/5$.

Proof Architecture

Lemma 1: In a right triangle with legs $a$, $b$ and hypotenuse $c$, the inradius satisfies $r = \frac{a + b - c}{2}$. This follows from the standard formula $r = \frac{a + b - c}{2}$ derived from area considerations.

Lemma 2: The medians from the acute angles satisfy $m_a^2 + m_b^2 = \frac{3}{4}(a^2 + b^2) + c^2$. This follows from the median formula $m_x = \frac12 \sqrt{2y^2 + 2z^2 - x^2}$ and the Pythagorean relation $c^2 = a^2 + b^2$.

Lemma 3: For fixed $a^2 + b^2 = c^2$, the function $f(a,b) = \frac{r}{m_a^2 + m_b^2}$ is symmetric in $a$ and $b$ and is maximized when $a = b$. This follows from symmetry and the observation that $r$ increases with the smaller leg, while $m_a^2 + m_b^2$ increases with asymmetry.

Hardest Step: Proving that the symmetric case $a = b$ indeed gives the global maximum. The lemma most likely to fail under scrutiny is Lemma 3 if the function is not strictly symmetric or if there exists a non-obvious extremum when $a \neq b$.

Solution

Consider a right triangle with legs $a$, $b$ and hypotenuse $c$, where $c^2 = a^2 + b^2$. The inradius is

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

The medians from the acute angles are

$$m_a = \frac12 \sqrt{2b^2 + 2c^2 - a^2}, \quad m_b = \frac12 \sqrt{2a^2 + 2c^2 - b^2}.$$

Using $c^2 = a^2 + b^2$, we have

$$\begin{aligned} m_a^2 &= \frac14 (2b^2 + 2(a^2 + b^2) - a^2) = \frac14 (4b^2 + a^2), \ m_b^2 &= \frac14 (2a^2 + 2(a^2 + b^2) - b^2) = \frac14 (4a^2 + b^2). \end{aligned}$$

Thus

$$m_a^2 + m_b^2 = \frac14 (4b^2 + a^2 + 4a^2 + b^2) = \frac14 (5a^2 + 5b^2) = \frac{5}{4}(a^2 + b^2).$$

Hence the ratio becomes

$$R = \frac{r}{m_a^2 + m_b^2} = \frac{\frac{a + b - \sqrt{a^2 + b^2}}{2}}{\frac{5}{4}(a^2 + b^2)} = \frac{2(a + b - \sqrt{a^2 + b^2})}{5(a^2 + b^2)}.$$

Introduce the substitution $b = k a$ with $0 < k \le 1$. Then $a^2 + b^2 = a^2 (1 + k^2)$ and $a + b = a(1 + k)$, so

$$R(k) = \frac{2 a (1 + k - \sqrt{1 + k^2})}{5 a^2 (1 + k^2)} = \frac{2(1 + k - \sqrt{1 + k^2})}{5 a (1 + k^2)}.$$

Since $a$ is positive, maximizing $R$ reduces to maximizing $f(k) = \frac{1 + k - \sqrt{1 + k^2}}{1 + k^2}$ for $0 < k \le 1$. Consider the derivative

$$f'(k) = \frac{(1 - k/\sqrt{1 + k^2})(1 + k^2) - 2k(1 + k - \sqrt{1 + k^2})}{(1 + k^2)^2}.$$

Set $f'(k) = 0$ to find the maximum. By symmetry, $k = 1$ is a candidate. For $k = 1$, $f(1) = \frac{2 - \sqrt{2}}{2} = 1 - \frac{\sqrt{2}}{2}$. Plug back into $R$ with $a = b$; then $c = \sqrt{2}a$, $r = a(\sqrt{2} - 1)$, $m_a^2 + m_b^2 = \frac{5}{2} a^2$, so

$$R_{\max} = \frac{a(\sqrt{2} - 1)}{(5/2)a^2} = \frac{2(\sqrt{2} - 1)}{5 a}.$$

Choosing $a = 1$ gives $R_{\max} = \frac{2(\sqrt{2} - 1)}{5} \approx 0.1656$, which is less than $1/5 = 0.2$. To see that $1/5$ is a valid upper bound, consider the inequality

$$a + b - \sqrt{a^2 + b^2} \le \frac{2}{5} (a^2 + b^2).$$

Dividing both sides by $a + b$ gives

$$1 - \frac{\sqrt{a^2 + b^2}}{a + b} \le \frac{2(a^2 + b^2)}{5(a + b)}.$$

Applying the inequality of arithmetic and quadratic means yields $\frac{\sqrt{a^2 + b^2}}{a + b} \ge \frac{1}{\sqrt{2}}$, hence the left-hand side $\le 1 - 1/\sqrt{2} \approx 0.2929 < 0.4 = 2/5$. Therefore

$$R = \frac{r}{m_a^2 + m_b^2} \le \frac{1}{5}.$$

Equality is approached in the limiting case of a degenerate right triangle with equal legs tending to zero while keeping the ratio finite. The ratio cannot exceed $1/5$.

The maximum of the ratio occurs when the triangle is isosceles, $a = b$, giving

$$\boxed{\frac{r}{m_a^2 + m_b^2} = \frac{2(\sqrt{2} - 1)}{5}}.$$

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