Gara di Febbraio

Geometry Difficulty 3.5 AMC 10/12 Find the answer Italy

Problem:

Given a triangle $ABC$ with sides $AB = 13$ , $BC = 14$ and $AC = 15$ , let $H$ be the foot of the altitude relative to side $BC$ , $M$ the midpoint of $BC$ and $N$ the midpoint of $AM$ . What is the length of $HN$ ?

Pick one

Solution

Solution:

The answer is (C). Triangle $AHM$ is right-angled, so we have $\overline{HN} = \overline{AM} / 2$ because $HN$ is the median relative to the hypotenuse. Setting $\overline{BH} = x$ we have $\overline{HC} = \overline{BC} - \overline{BH} = 14 - x$ and, by the Pythagorean theorem, $\overline{AH}^2 = \overline{AB}^2 - \overline{BH}^2 = \overline{AC}^2 - \overline{HC}^2$ , that is $13^2 - x^2 = 15^2 - (14 - x)^2$ , from which we obtain $x = 5$ . We then have
$\overline{AH} = \sqrt{13^2 - 5^2} = 12, \quad \overline{AM} = \sqrt{\overline{AH}^2 + \overline{HM}^2} = \sqrt{12^2 + 2^2} = \sqrt{148} = 2\sqrt{37},$
so $\overline{HN} = \sqrt{37}$ .

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement translated into English from it; metadata (topic, difficulty) added by this project.