[ $[$ Theorem of tangent length and classical inequalities [ Proportional means in a right triangle

Point $M$ lies outside a circle with center $O$. Line $O M$ intersects the circle at points $A$ and $B$, a line passing through point $M$ is tangent to the circle at point $C$, point $H$ is the projection of point $C$ onto $A B$, and the perpendicular to $A B$ erected at point $O$ intersects the circle at point $P$. It is known that $M A=a$ and $M B=b$. Find $M O, M C, M H, M P$ and arrange the found values in ascending order. #
For definiteness, let's assume that $a > b$. Point $O$ is the midpoint of segment $AB$, so

$M O = \frac{M A + M B}{2} = \frac{a + b}{2}$. By the tangent-secant theorem, $M C = \sqrt{M A \cdot M B} = \sqrt{a b}$.

$$
M H = \frac{M C^{2}}{M O} = \frac{a b}{\frac{a + b}{2}} = \frac{2 a b}{a + b}
$$

By the Pythagorean theorem in the right triangle $M O P$, we find that

$$
M P = \sqrt{M O^{2} + O P^{2}} = \sqrt{M O^{2} + O A^{2}} = \sqrt{\left(\frac{a + b}{2}\right)^{2} + \left(\frac{a - b}{2}\right)^{2}} = \sqrt{\frac{a^{2} + b^{2}}{2}}
$$

$M H$ is the leg of the right triangle $M C H$ with hypotenuse $M C$, so $M H < M C$.

$M C$ is the leg of the right triangle $M O C$ with hypotenuse $M O$, so $M C < M O$.

$M O$ is the leg of the right triangle $M O P$ with hypotenuse $M P$, so $M O < M P$.

## Answer

$$
M H = \frac{2 a b}{a + b}, \quad M C = \sqrt{a b}, \quad M O = \frac{a + b}{2}, M P = \sqrt{\frac{a^{2} + b^{2}}{2}} ; M H < M C < M O < M P
$$

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